Sheldon <£• Company's Text-l^ooks FRENCH AND GERMAN PEOF. EEETELS' NEW FEENCH SBEIES. The Oral Method with tite French, By Prof. Juax GcSTAVE KEEn"ELS, Author of " Keetels' New Method with the French." lu three parts, ICtao, cloth, [Tlie studer.i is sctccd Vic expense cf a large hook in commencing the study] The Oral :\Icthod of Teaching livis- lang^iages U fnpcricr to ell others In ir.anv respects. li teacht!* the pupil to speak the language he is Icamin;;, end he begins to do to from The first leM- distinctive ideas which constitute the substanceof our Constitution, and which determine the policy of our politics ; and to this end there ought forthwith to be introduced into our schools a simple, comprehensive manual, whereby the needed tuition should be implanted at that early period. JScJimitz's 3Ianual of Ancient History; from the Re- motest Times to the Overthrow of the Western Empire. A. d. 476, with copious Chronological Tables and Index. By Dr. Leoxhakd Schmitz, T. Tl. S. K, Edinburgh. The Elements of Intellectual Philosophy. By Fp.an /? Watland, D.D. 1 vol. 12ino. This clearly-written book, from the pen of a scholar of eminent abil;(y, mn who has had the largest experience in the education of ' .' in nnnil, id unquestionably at the head of text-books in Intellectual PhilooopLy. An Outline of the Necessary Laws of Thought: A Treatise on Pure and Applied Logic. By William Thom- son, D.D., Provost of the Queen's College, Oxford. 1 vol. V^mo. Ooth. This book has been adopted as a regular text-book in Harvard, Yale, Rochester, New York University, &c. Fairchilds' Moral Philosophy ; or, The Science of Obligation, By J. H. FAmcHiLD^, Presi'dent of Oberlin College. 1 vol. 12mo. The aim of this volume is to set forth, more fully than has hitherto been done, the doctrine that virtue, in its elementary form, consists in benevo- lence, and that all forms of virtuous action are modifications of this principle. After ,-resonting this view of obligation, the author takes up the questions of Fiactical Ethics, Government and Personal Rights i*nd Duties, and treat- them in their relation to Benevolence, aiming at a solution 0( the problems of rijjht and wrong upon ♦bis simple principle. :jn,v c' the above sent biJ maiZ, post-paid, on receipt of price. 'OM'^^ k CO \o f 01 CO N Fig. 2. OLNEY'S MATHEMATICAL. SERIES. A TREATISE ON Special or Elementary GEOMETKY. UNIVERSITY EDITION. INCLUDING AN ELEMENTARY, AND ALSO, IN PART III., A HIGHER COURSE, IN PLANE, SOLID, AND SPHERICAL GEOMETRY; AND PLANE AND SPHERICAL TRIGONOMETRY, WITH THE NECESSARY TABLES. BY EDWAED OLISTEY, PROFESSOR OF MATHEMATICS IN THE UNIVERSITY OF MICHIQAN. NEW YORK: Sheldon & Company, No. 8 MURRAY STREET. 1877. Stoddard's lathematical Series. STODDARD'S JUVENILE MENTAL ARITHMETIC - STODDARD'S INTELLECTUAL ARITHMETIC STODDARD'S RUDIMENTS OF ARITHMETIC - STODDARD'S NEW PRACTICAL ARITHMETIC - SHORT AND FULL COURSE FOR GRADED SCHOOLS. STODDARD'S PICTORIAL PRIMARY ARITHMETIC STODDARD'S COMBINATION ARITHMETIC . . - . STODDARD'S COMPLETE ARITHMETIC - . . - Tlie Combination Scliool Arithmetic being Mental and Written Arithmetic in one book, will alone serve for District Schools. For Academies a full high course is obtained by the Complete Arithmetic and InteUectiuil Arithmetic. HIGHER MATHEMATICS, BY EDWAED OL:NrEY, TROPESSOR OF MATHEMATICS IN THE UNIVERSITY OP MICHIGAN. A COMPLETE SCHOOL ALGEBRA, in one vol., 410 pages, Designed for Elementary and higher classes in Schools and Academies. A (.GEOMETRY AND TRIGONOMETRY, in one vol., 8vo, - A GEOMETRY AND TRIGONOMETRY, UNIVERSITY EDITION, in one vol., 8vo, . . - A GENERAL GEOMETRY AND CALCULUS in one vol. - The other books of this Series will be published as rapidly as possible. Entered according to Act of Congress, in the year 1872, by SHELDON & COMPANY. In the OfBce of the Librarian of Congress, at Washington. ^ /'^^ -J I r PEEFACE. '^"^^ This treatise on the Special or Elementary Geometry consists of four parts. Part I. is designed as an introduction. In it the student is made familiar with the geometrical concepts, and with the fundamental definitions and facts of the science. The definitions here given, are given once for all. It is thought that the pupil can obtain his first conception of a geometrical fact, as well, at least, from a correct, scientific statement of it, as from some crude, colloquial form, the language of which he will be obliged to replace by better, after the former shall have become so firmly fixed in his mind, as not to be easily eradicated. No attempt at demonstration is made in this part, although most of the fundamental facts of Elementary Plane Geom- etry are here presented, and amply and familiarly illustrated. This course has been taken in obedience to the canon of the teacher's art, which prescribes " facts before theories." Moreover, such has been the historic order of development of this, and most other sciences; viz., the facts have been known, or conjectured, long before men have been able to give any logical account of them. And does not this indicate what may he the natural order in which the individual mind will receive science ? When the student has become familiar with the things (coAcepts) about which his mind is to be occupied, and knows some of the more important of their properties and relations, he is better prepared to reason upon them. Part II. contains all the essential propositions in Plane, Solid, and Spherical Geometry, which are found in our common text-books, with their demonstrations. The subject of triedrals and the doctrine of the sphere are treated with more than the ordinary fullness. The earlier sections of this part are made short, each treating of a single subject, and the propositions are made to stand out prominently. At the close of each section are Exercises designed to illustrate and apply the principles contained in the section, rather than to extend the pupil's knowledge of geometrical facts. These features, together with the synopses at the close of the sections, practical teachers can- not fail to appreciate. Part III., which is contained only in the University Edition, has 3V • PKEFACE. been written with special reference to the needs of students in the University of Micliigan. Our admirable system of public High- Schools, of which schools there is now one in almost every consid- erable village, promises ere long to become to us something near what the German Gymnasia are to their Universities. In order to promote the legitimate development of these schools, it is necessary that the University resign to them the work of instruction in the elements of the various branches, as fast and as far as they are pre- pared in sufificient numbers to undertake it. It is thought that these schools should now give the instruction in Elementary Geom- etry, which has hitherto been given in our ordinary college course. The first two parts of this volume furnish this amount of instruc- tion, and students are expected to pass examination upon it on their entrance into the University. This amount of preparation enables students to extend their knowledge of Geometry, during the Fresh- man year in the University, considerably beyond what has hitherto been practicable. As a text-book for such students. Part III. has been written. At this stage of his progress, the student is prepared to learn to investigate for himself. Hence he is here furnished with a large collection of well classified theorems and problems, which afford a review of all that has gone before, extend his knowledge of geometrical truth, and give him the needed discipline in original demonstration. To develop the power of independent thought, is the most difficult, while it is the most important part of the teach- er's work. Great pains have therefore been taken, in this part of the work, to render such aid, and only such, as a student ought to require in advancing from the stage in which he has been follow- ing the processes of others, to that of independent reasoning. In the second place, this part contains what is usually styled Applica- tions of Algebra to Geometry, with an extended and carefully selected range of examples in this important subject. A third purpose has been to present in this part an introduction to what is often spoken of as the Modern Geometry, by which is meant the results of modern thought in developing geometrical truth upon the direct method. While, as a system of geometrical reasoning, this Geometry is not philosophically different from that with which the student of Euclid is familiar, and which is properly distinguished as the special or direct method, the character of the facts developed is quite novel. So much so, indeed, that the student who has no knowledge of Geometry but that which our common text-books furnish, knows absolutely nothing of the domain into which most of the brilliant advances of PREFACE. V the present century have been made. He knows not even the terms in which the ideas of such w^riters as Poncelet, Chasles, and Sal- mon, are expressed, and he is quite as much a stranger to the thought. In this part are presented the fundamental ideas concerning Lociy Symmetry, Maxima a)id Minima, Isox>erimeiry, the theory of Trans- versals, Anharmonic Ratio, Polars, Radical Axes, and other modern views concerning the circle. Part IV. is Plane and Spherical Trigonometry, with the requisite Tables. While this Part, as a whole, is much more complete than the treatises in common use in our schools, it is so arranged that a shorter course can be taken by such as desire it. Thus, for a shorter course in Plane Trigonometry, see Note on page 55. In Spherical Trigono- metry, the first three sections, either with or without the Introduc- tion on Projection, will afford a very satisfactory elementary course. A few words as to the manner in which this plan has been executed, may be important. In general, the Definitions are those usually given, with such slight alterations as have been suggested by reflection and experience. There are, however, a few exceptions. Among these is the definition of an Angle. I can but regard the attempt to define an angle as The differ eiice in direction hetiueen tiuo lines, or The amount of divergence, as needlessly vague, abstract, and perplexing to a student, as well as questionable on philosophical grounds. The definition given in the text will be seen to be, at bottom, the old one, the conception being slightly altered to bring it into more close connection with common thought, and also with the idea of an angle as generated by the revolution of a line. As to Parallels, and the definition of similarity, my experience as a teacher is decidedly in favor of retaining the old notions. So also in adopting a definition of a Trigonometrical Function, I am compelled to adhere to the geometrical conception. A ratio is a complex concept, and conse- quently not so easy of application as a simple one. For this reason, among others, I prefer the differential to the differential coefficient, in the calculus, and a line to a ratio, in Trigonometry. Moreover, I have found that students invariably rely upon the geometrical con-^ ception, even when first taught the other ; hence I am not surprised- that all our writers who define a trigonometrical function as a ratio, hasten to tell the pupil what it means, by giving him the geometrical illustrations. Nor are the superior facility which the geometrical conception affords for a full elucidation of the doctrine of the signs of the functions, and its admirable adaptation to fix these laws in the mind, considerations to be lost sight of iu selecting the definition. ▼1 PREFACE. Surely no apology is needed, at the present day, for introducing the idea of motion into Elementary Geometry, notwithstanding the rigorous and disdainful manner with which its entrance was long re- sisted by the old Geometers. And, having admitted this idea, the conception of loci as generated by motion would seem to follow as a logical necessity. In like manner, I take it, the Infinitesimal method must come in. Its directness, simplicity, and necessity in applied mathematics, demand its recognition in the elements. In two or three instances, I have presented the reductio ad ahsicrdum, where the methods are equivalents, and have always in presenting the infinitesimal method woven in the idea of limits, which I conceive to be fundamentally the same as the infinitesimal. Thus we bring the lower and higher mathematics into closer connection. The order of arrangement in Plane Geometry (Chap. L), is thought to be simple, philosophical, and practical. A glance at the table of contents will show what it is. This arrangement secures the very important result, that each section presents some particulai method of 2^roof and holds the student to it, until it is familiar. True, it requires that a larger number of propositions be demonstrated from fundamental truths ; but who will consider this an objection ? To such as consider it the sole province of geometrical demonstra- tion, to convince the mind of the truth of a proposition, not a few theorems in these and ordinary pages must seem quite superfluous. To them. Prop. I., page 121, may afford some merriment. But those who, with myself, consider Geometry as a branch of practical logic, the aim of which is to detect and state the steps which actually lie be- tween premise and conclusion, will see the propriety of such demonstra- tions; and for each individual of the other class, a separate treatise will be needed, since no two minds will intuitively grant exactly the same propositions. To Ex-President Hill, of Harvard, I am indebted for the confir- mation of an opinion which had been previously forming in my mind, that the study of Geometry as a branch of logic, should be preceded by a presentation of its leading facts. The works of Compagxon", Tappan, and our lamented countryman, Chauyexet, have been within reach during the entire work of preparation, and this volume would have been different, in some respects, if any one of these able treatises had not aj^peared before it. In the preparation of Part III. the works of Rouche et Combe- ROUSSE and Mulcahy have been freely used. For the very concise and elegant form in which the principle of Delambre, for the pre- PREFACE. Vil cise calculations of Trigonometrical Functions near their limits, is embodied in Table III., I am indebted to the recent work of Presi- dent Eli T. Tappan, of Kenyon College, Ohio. My long and intimate intercourse with Professor G. B. Merriman", now of the department of Physics in the University, has been a source of great profit to me in the preparation of the entire work. His sound, practical judgment as a teacher of Geometry, and culti- vated taste and skill as a Mathematician, have been ever at my ser- vice, and have done more than I can tell, in giving form to the work, both as respects its matter and its spirit. Edward Olney. University of Michigan, Ann Arbor, January, 1872. COFTEWTS. INTRODUCTION". Paos. SECTION I. Lo(jico-Mathematical Terms Defined and Illustrated 1-3 SECTION II. The Geometrical Concepts Defined and Illustrated 3-11 PART L A FEW OF THE MORE IMPORTANT FACTS OF THE SCIENCE. SECTION I. About Straight Lines 11-19 SECTION 11. About Circles 19-25 SECTION III. About Angles and Parallels 25-30 SECTION IV. About Triangles 31-35 SECTION V. About Equal Figures 35-39 SECTION VI. About Similar Figxhies, especially Triangles 39-44 SECTION VIL About Areas 44-55 SECTION VIII. About Polygons 55-58 X CONTENTS. PART 11. • THE FUNDAMENTAL PROPOSITIONS OF ELEMENTARY GEOMETRY, DEMONSTRATED, ILLUSTRATED, AND APPLIED. CHAPTER I. PLANE GEOMETRY. SECTION I. > Page. Pkrpendicular Straight Lines 60-65 SECTION II. Oblique Straight Lines 65-70 SECTION III. Parallels 70-77 SECTION IV. Relative Positions op Straight Lines and Circumferences 7&-85 SECTION V. Relative Positions of Circidiferences 80-93 SECTION VI. Measurement of Angles 94-103 SECTION vn. Angles of Polygons and the Relation between the Angles and Sides. Of Triangles 104-106 Of Quadi-ilaterals 107-111 Of Polygons of more than Four Sides 112-113 Of Regular Polygons 113-120 SECTION vni. Of Equality. Of Lines and Circles 121-122 Of Angles 122-123 Of Triangles 124-129 Of Quadrilaterals 129-130 Of Polygons of more than Four Sides 130-137 SECTION IX. Op Equivalency and Areas. Equivalency 138-140 Area 140-144 CONTENTS. ti SECTION X. Paor. Of Similakity 144-152 SECTION XL Applications of the Doctrine op Sdhlarity to the Development op Geometrical Properties of Figures. Of the Relations of the Segments of two Lines intersecting each other and intersected by a circumference 153-154 Of the Bisector of an Angle of a Triangle 154-156 Areas of Similar Figures 156-158 Perimeters and the Rectification of the Circumference 158-160 Area of the Cucle 160-163 CHAPTER IL SOLID GEOMETRY. SECTION I. Of Straight Lines and Planes. Paob. Perpendicular and Obhque Lines to a Plane 164-169 Parallel Lines and Planes 169-174 SECTION IL Of Solid Angles. Of Diedrals 176-178 OfTriedrals 178-184 Of Polyedrals 185-186 SECTION IIL Op Prisms and Cylinders 187-199 SECTION IV. Of Pyramids and Cones 199-209 SECTION V. Of the Sphere. Circles of the Sphere 210-211 Distances on the Surface of a Sphere 211-215 Spherical Angles 215-218 Tangent Planes 218-219 Spherical Triangles 219-226 Polar or Supplemental Triangles 226-228 Quadrature of the Surface of the Sphere 229-231 Lunes 231-235 Yoiume of Sphere 235-239 XU CONTENTS. PART IIL AN^ ADVANCED COURSE IN* GEOMETRY. CHAPTER I. EXERCISES IN GEOMETRICAL INVENTION. SECTION I. Page. Theorems m Special or Elementary Geometry 243-267 SECTION II. Problems in Special or Elementary Geometry 267-276 SECTION III. Applications op Algebra to Geometry 276-288 CHAPTER 11. INTRODUCTION TO MODERN GEOMETRY. SECTION I. Of Loci 288-296 SECTION 11. Op Symmetry 296-30 1 SECTION III. Of Maxima and Minima, and Isoperimetry 302-306 SECTION IV. Of Trans\t:rsals 30&-310 SECTION V. Harmonic Proportion, and Harmonic Pencils 310-314 SECTION VI. Anharmonic Ratio 314-318 SECTION VII. Pols and Polar in Respect to a Circle 319-323 SECTION VIII. Radical Axes and Centres op Similitude op Circles 323-329 SPECIAL OR ELEMENTARY GEOMETRY. INTBOD UCTION. SECTION I, LOGICO-MATHEMATICAL TERMS.* JZ, a ^Proposition is a statement of something to be considered or done. III. — Thus, the common statement, " Life is short," is a proposition ; so, also, we make, or state a proposition, when we say, '* Let us seek earnestly after truth."—" The product of the divisor and quotient, plus the remainder, equals the dividend," and the requirement, " To reduce a fraction to its lowest terms," are examples of Arithmetical propositions. 2. Propositions are distinguished as Axioms, Theorems, Lemmas, Corollaries, Postulates, and ProUems. 3. An Axiom is a proposition which states a principle that is so simple, elementary, and evident as to require no proof. Ill,— Thus," A part of a thing is less than the whole of it," " Equimultiples of equals are equal," are examples of axioms. If any one does not admit the Iruth of axioms, when he understands the terms used, we say that his mind is not sound, and that we cannot reason with him. 4. A. Theorem is a proposition which states a real or supposed fact, whose truth or falsity we are to determine by reasoning. III. — '* If the same quantity be added to both numerator and denominator of a proper fraction, the value of the fraction will be increased," is a tJieoi-em. It is a statement the truth or falsity of which we are to determine by a course of reasoniH^. * That i?, terms used in the science in consequence of its logical character. The science of. the Pure Mathematics may be considered as a department of practical logic. 2 LOGICO-MATHEMATICAL TERMS. 5, A T>enionstvation is the course of reasoning by means of which the truth or falsity of a theorem is made to appear. The term is also applied to a logical statement of the reasons for the processes of a rule. A solution tells lioiv a thing is done: a demon- stration tells vjhy it is so done. A demonstration is often called proof. 6, A Lemma is a theorem demonstrated for the pui-pose of using it in the demonstration of another theorem. III. — Thus, in order to demonstrate the rule for finding the greatest common divisor of two or more numbers, it may be best first to prove that " A divisor of two numbers is a divisor of their sum, and also of their difi'erence." This theorem, when proved for such a purpose, is called a Lemma. The term Lemma is not much used, and is not very important, since most theorems, once proved, become in turn auxiliary to the proof of others, and hence might be called lemmas. 7, A Corollary is a subordinate theorem which is suggested, or the truth of which is made evident, in the course of the demon- stration of a more general theorem, or which is a direct inference from a proposition, or a definition. III. — Thus, by the discussion of the ordinaiy process of performing subtrac- tion in Arithmetic, the following Corollary m\g\i\. be suggested: "Subtraction may also be performed by addition, as we can readily obseiTe what number must be added to the subtrahend to produce the minuend." 8, A JPostiilate is a proposition which states that something can be done, and which is so evidently true as to require no process of reasoning to show that it is possible to be done. We may or may not know how to perform the operation. III. — Quantities of the same kind can be added together. 0, A JProblem is a proposition to do some specified thing, and is stated with reference to developing the method of doing it. Ill.^-A problem is often stated as an incomplete sentence, as, " To reduce fractions to a common denominator." — This incomplete statement means that " "We propose to show how to reduce fractions to a common denominator." Again, the problem " To construct a square," means that " We propose to draw a figure which is called a square, and to tell how it is done." 10, A Utile is a formal statement of the method of solving a general problem, and is designed for practical application in solving special examples of the same class. Of course a rule requires a demonstration. THE GEOMETRICAL CONCEPTS. 11. A Solution is the process of performing a problem or an example. It should usually be accompanied by a demonstration of the process. 12, A Scholium is a remark made at the close of a discussion, and designed to call attention to some particular feature or features of it. III. — Thus, after having discussed the subject of multiplication and division In Arithmetic, the remark that " Division is the converse of multiplication," is a scholium. SYNOPSIS. Subject of the section. Lemma. m. Why the Proposition. HI. portant. Varieties of propositions. Corollary. HI. Axiom. ML Postulate. El. One who will not admit the truth Problem. How stated. of axioms. Rule. Theorem. III. Solution. Demonstration. Difference between Scholium. m. a solution and a demonstration. m. SECTION 11. THE GEOxMETRICAL CONCEPTS.* FOUNTS. IS. A JPoint is a place without size, letters. Points are designated by •B III. — If we wish to designate any particular point (place) on the paper, we put a letter by it, and sometimes a dot on it. Thus, in Fig. 3, the ends of the line, which are points, are ^p flesignated as " point A," " point D ;" or, simply, as A and D. The points marked on the line are designated as " point B," " point C," or as B and a B — C. F and E are two points above the line. Fig. * A concept is a thing thought about ;— a thought-object. Thus, in Arithmetic, number is tbe concept ; in Botany, plants ; in Geometry, as will appear in this section, pointP, lines, and solids. These may aleo be said to conetitate the suloect-matter of the science. ELEMENTARY GEOMETRY. LINES. 14, A lAne is the path of a point in motion. Lines are repre- sented upon paper by marks made with a pen or pencil, the point of the pen or pencil representing the moving point. A line is desig- nated by naming the letters written at its extremities, or somewhere upon it. III. — In each case in Fig. 4, conceive a point to start from A and move along ^ the path indicated by the mark to B. The path thus traced is a line. Since a true point has no size, a line has no breadth, though the marks by •which we represent lines have some breadth. The first and third lines in the figure are each designated as "the line AB." The second line is considered as traced hy a point starting from A and coming around to A again, so that B and A coincide. This line may be designated as the line AmnA, or A7?i?iB. In the fourth case, there are three lines represented, which are designated, respectively, as AwiB, A;iB, and Ac3 ; or, the last, as AB. 15. Lines are of Two Kiiuls, Straight and Curved. A straight line is also called a EigJit Line. A curved line is often called simply a Curve. 16, A Straiffht Line* is a Une traced by a point which moves constantly in the same direction. 17. A Curved Line is a line traced by a point which con- stantly changes its direction of motion. Ill's.— Thus in 1, Fig. 4, if the line AB is conceived as traced by a point moving from A to B, it is evident that this point moves in the same direction throughout its course; hence AB is a straight line. If a body, as a stone, be let fall, it moves constantly toward the centre of the earth ; hence its path represents a straight line. If a weight be suspended by a string, the string represents a straight line. Considering the line represented by AiB, Fig. 4, as the path of a point moving from A to B, we see that the direction of motion is constantly changuig. For example, if this were a line traced on a map, we ♦ The word ''line" u*cd aloue Pi^'niSe;? '-straight line. SURFACES. 5 would say, that, starting from A, the point begins to move nearly north, but keeps changing its direction more and more toward the east, until at 3 it moves directly east; and from 3 it continues to change its course and moves more and more toward the south, till at i it is moving directly south. The same general ti-uth is illustrated in 2 and 4, Fig. 4. The path of a ball thrown into the air, in any direction except directly up, represents a curved line. Most of the lines seen in nature are curved, as ^ the edges of leaves, the shore of a river 7/^- or lake, etc. Sometimes a path like that ..^-^-^""^ X. ^ represented in Fig. 5 is called, though im- ^ rv o properly, a Broken Line. It is not a line ^^®- ^• at all ; that is, not one line : it is a series of straight lines. SURFACES. IS. A Surface is the path of a line in motion.* 1.0* Surfaces are of Two Kinds^ Plane and Curved. 20. A I^lane Surface, or simply a Plane, is a surface with which a straight line may be made to coincide in any direction. Such a surface may always be conceived as the path of a straight line in motion. 21. A Curved Surface is a surface in which, if lines are conceived to be drawn in all directions, some or all of them will be curved lines. Ill's. — Let AB, Fig. 6, be supposed to move to the right, so that its extremi- ties A and B move at the same rate and in the same direction, A tracing the line AD, and B, the line BC. The path of the line, the figure ABCD, ,is a surface. This page is a surface, and may be conceived as the path of a line sliding like a ruler from top to bottom of it, or from one side to the other. Such a path will have length and breadth, being in the latter respect unlike a line, which has Fig. 6. only length. As a second illustration, suppose a fine wire bent into the form of the curve AmB, Fig. 7, and its ends A and B stuck into a rod, XY. Now, taking the rod XY in the fingers and rolling it, it is evident that the path of the hne represented by the wire AwB, will be the surfoce of a ball (sphere). ♦ Should it be paid that irrefjular surfaces are not included in this definition, the sufflcient reply is, that such surfaces are not subjects of Geometrical investigation, except approzi* mately, by means of regular surfaces. 6 ELEMENTARY GEOMETRY. nv X A Fig. B Y Again, suppose the rod XY be placed on the surface of this paper so that the wire AmB shall stand straight up from the paper, just as it would be if we could take hold of the curve at m and raise it right up, letting XY lie as it does in the figure. Now slide the rod straight up or down the page, making both ends move at the same rate. The path of AwiB will be like tlie smface of a half-round rod (a semi-cylinder). Thus we see how surfaces plane and cuiTed may be conceived as the paths of lines in motion. ^^ Ex. 1. If the curve A;iB, Fuj. 8, be conceived as revolved about the line XY, the surface of what Y object will its path be like ? X A Fig. S. Ex. 2. If the figure OMNP, Fig. 9, be conceived as revolved about OP, what kind of a path will MN trace? What kind of paths will FN and OM trace? Ans. One path will be like the surface of a joint of stove-pipe, i. e., a cylindrical surface ; and one "will be like a flat wheel, i, e., a circle. Ex. B. If you fasten one end of a cord at a point in the ceiling and hang a ball on the other end, and then make the ball swing around in a circle, what kind of a surface will the string describe ? [Note. — The student is not necessarily expected to give the geometrical name of the surface, but rather to tell in his own way what it is like, so as to make it clear that he conceives the thing itself] Ex. 4. If you were to draw lines in all directions on the surface of the stove-pipe, might any of them be straight ? Could all of them be straight ? What kind of a surface is this, therefore ? Ex. 5. Can you draw a straight line on the surface of a ball ? On the surface of an egg ? What kind of surfaces are these ? Ex. 6. When the carpenter wishes to make the surface of a board perfectly flat, he takes a ruler whose edge is a straight line, and lays this straight edge on the surface in all directions, watching closely ANGLES. 7 to see if ir, always touches. Which of our definitions is he illus- trating by his practice ? Ex. 7. When the miller wishes to make flat the surface of one of the large stones with which wheat is ground into flour, he sometimes takes a ruler with a straight edge, and smearing the edge with paint, applies it in all directions to the surface, and then chips off the stone where the paint is left on it. AVhat principles is he illustrating ? Ex. 8. How can you conceive a straight line to move so that it shall not generate a surface ? ANGLES. 22. A JPlane Angle ^ or simply an A7igle, is the opening be- tween two lines which meet each other. The point in which the lines meet is called the vertex, and the lines are called the sides. An angle is designated by placing a letter at its vertex, and one at each of its sides. In reading, we name the letter at the vertex when there is but one vertex at the point, and the three letters when there are two or more vertices at the same point. In the latter case, the letter at the vertex is put between the other two. III. — In common language an angle is called a corner. The opening between the two lines AB and AC, in wiiich the figure 1 stands, is called the angle A ; or, if we choose, we may call it the angle BAC. At L there are two vertices, so that were we to say the angle L, one would not know whether we meant the angle (cor- ner) in which 4 stands, or that in which 5 stands. To avoid this ambiguity, we say the angle HLR for the former, and RLT for the latter. The angle ZAY is the cor- ner in which 11 stands ; that is, the opening between the two lines AY and AZ. In designating an angle by three lettere, it is im- material which letter stands first so that the one at the vertex is put between the other two. Thus, PQS and SQP are both designa- tions of the angle in which G Fig. 10. 8 ELEMENTARY GEOMETRY. stands. An angle is also frequently designated by putting a letter or figure in it and near the vertex. 23, The Size of an Angle depends upon the rapidity with which its sides separate, and not upon their length. III. — The angles BAG and MON, Fig. 10, are equal, since the sides separate at the same rate, although the sides of the latter are more prolonged than those of the former. The sides DF and DE separate faster than AB and AC, hence the angle EDF is greater than the angle BAG. 24:, Adjacent Angles are angles so situated as to have a com- mon vertex and one common side lying between them. III. — In Fig. 10, angles 4 and 5 are adjacent, since they have the common vertex L, and the common side LR. Angles 9 and 10 are also adjacent, as are also 8 and 9. 25. Angles are distinguished as Right Angles and Oblique Angles, Oblique angles are either Acute or Obtuse. 26, A Itight Angle is an angle included between two straight lines which meet each other in such a manner as to make the adja- cent angles equal. An Acute Angle is an angle which is less than a right angle, i. e., one whose sides separate less rapidly. An Obtuse Angle is an angle which is greater than a right angle, i, e., one whose sides separate more rapidly. B III. — As in common language an angle is called a corner, so a right angle is called a square corner ; an acute, a «/i«rp corner; and an obtuse angle might be called a blunt corner. In Fig. 11, BAG and DAB are right angles. In Fig. 10, 1, 2, 3, 5, 8, 9, and 10 are acute angles, 4 and 6 are obtuse, and 7 is a right angle. A Fig. A SOLID. 27* A Solid is a limited portion of space. It may also be con- ceived as the path of a surface in motion. III. — Suppose you have a block of wood like that represented in Fig. 12, with all its corners (angles) square corners (right angles). Hold it still in your A SOLID. fingers a moment, and fix your mind upon it . Now take the block away and think of the space (place) where it was. This space will be of just the same form as the block of wood, and by a little ef- fort you can think of it just as well as of the wood. This space is an example of what we call a Solid in Geometiy. In fact, the solids of Geometry are not solids ^i^- 1^- at all in the common sense of solids ; they are only just places of certain shapes. Again, hold your ball still a moment in your fingers and then let it drop, and think of the place it filled when you had it in your fingers. It is this place, shaped just like your ball, that we think about, and talk about as a solid, in Geometry. In order to see how a solid may be conceived as the path of a surface, sup- pose you cut out a piece of paper of just the same size as the end of the block represented in Fig. 12. Let ABCD represent this piece of paper. Now, holding the paper in a perpendicular position, as ABCD is represented in the figure, move it along to the right, so that its angles shall trace the lines AC, BH, DE, and CF. When the paper has moved to the position CHFE, its path will be just the same space as the block of wood occupied. This path, or the space through which the surface represented by the piece of paper moved, is the solid. Ex. 1. If a semicircle is conceived as revolved around its diameter, what is the path through which it moves ? See Fig. 7. Ex. 2. If the surface OMNP, Fig. 0, is conceived as revolved around OP, what is the path through which it moves ? Caution. — The student needs to be careful and distinguish between the surface traced by the line MN, and the solid traced by the surface OMNP. Ex. 3. If the surface represented by ABC he con- ceived as revolved about its side CA, what kind of a solid is its path ? [Note.— As has been said before, the student is not necessarily expected to luune these solids, but rather to show, in his own language, that he has the conception.] Ex. 4. As you fill a vessel with water, what is the a solid traced by the surface of the water ? Ans. The same as the space within the yessel. Ex. 5. If a circle is conceived as lying horizontally, and then moved directly up, what will be the solid described, t. e., its path ? Do not confound the surface described with the solid. What de- scribes the surface ? What the solid ? Fig. 13. 10 ELEMENTABY GEOMETBY. EXTEXSIOX AM) FORM. 28. JExtension means a stretching or reaching out. Hence, a Point has no extension. It has only position (place). A Lme stretches or reaches out, but only in length, as it has no width. Hence, a hne is said to have One Dimension, viz., length. A Surface extends not only in length, but also in breadth ; and hence has Tico Dimensions, viz., length and breadth. A Solid has Tliree Di- mensions, viz., length, breadth, and thickness. III.— Suppose we think of a point as capable of sti-etching out (extending) in one direction. It would become a line. Now suppose tlie line to stretch out (extend) in another direction— to widen. It would become a surface. Finaliy, suppose the surface capable of thickening, that is, extending in another direc- tion. It would become a sohd. 29* The Limits (extremities) of aline are points. The Limits (boundaries) of a surface are lines. The Limits (boundaries) of a solid are surfaces. 30, 3Iaf/nitucle (size) is the result of extension. Lines, sur- faces, and solids are the geometrical magnitudes. A point is not a magnitude, since it has no size. The magnitude of a line is its length ; of a surface, its area ; of a solid, its volume. 31, Fiffure or Form (shape) is the result of position of points. The form of a line (as straight or curved) depends upon the relative position of the points in the line. The form of a surface (as plane or curved) depends upon the relative position of the points in it The form of a solid depends upon the relative position of the points in its surface. Lines, surfaces, and solids are the geometrical figures.* III. — In Fig. 14, it is easy to conceive the form of the lines by knowing the position of points in the lines. By taking a quantity of common pins of different lengths, sticking them upright in a board, and conceiv- 1 / \ ing the heads to represent points in a surface, V__^ ( I we can readily see how the position of the points in a surface determine its form. Fig. 14. Ex. 1. Suppose a line to begin to con- * Line?, surfaces, and solids are called mao^nitadea when reference is had to their extent^ and fig:ure:^ w hc-n reference is had to their /or//i. SYNOPSIS. 11 tract in length, and continue the operation till it can contract no longer, what does it become ? That is, what is the minor limit of a line? Ex. 2. If a surface contracts in one dimension, as width, till it reaches its limit, what does it become ? If it contracts to its limit in both dimensions, what does it become ? Ex. 3. If a solid contracts to its limit in one dimension, what does it pass into ? If in two dimensions ? If in three dimensions ? Ex. 4. What kind of a surface is that, every point in which is equally distant from a given point ? S2. Geometry treats of magnitude and form as the result of extension and position. The Geometrical Concepts are points, lines, surfaces (including plane and spherical angles), and solids (including solid angles). The Object of the science is the measurement and comparison of these concepts. Plane Geometry treats of figures all of whose parts are confined to one plane. Solid Geometry, called also Geometry of Space, and Geometry of Three Dimensions, treats of figures whose parts lie in different planes. The division of Part II. into two chapters is founded upon this distinction. In the Higher or General Geometiy these divisions are marked by the terms " Of Loci in a Plane,'' and " Of Loci in Spaced o JO < A SYNOPSIS. What. — How designated. — 111. Point \ Dimensions of. Limit of Line. — Surface. — Solid. Line r "What. How designated. Dimensions of. Limit of Surface, ( Straight.— What.— 7^?. Kinds \ Curved.— What.— I/^. ( Broken (?). r What. Dimensions of. Limit of Solid. T^;„ria J Plane.— What— 72?. J^mas ^ Curved.— What.— 7?Z. iWhat. — Size depends on what Adjacent. ( Right.— What.— //?. Kinds \ ni.i:^„^ j Acute.-What— //?. I Oblique J^ obtuse.-What.-/??. Solid What— 7ZZ.— Examples. Macmitude.- What— Result of what Surface r rr f p S Macmitude.- What— Result ot what I > ^'^^^^ ^^ \ Figure or form.— AVhat.— Result of what \ Concepts. — Wh t Object— What Geometry... Concepts.-What PART L A FEW OF THE MORE IMPORTANT FACTS OF THE SCIENCE. Ai ^B p, -HQ p, -H li SECTION L ABOUT STRAIGHT LINES. SS* JProh, — To measure a straight line with the dividers and scale. Solution.— Let AB, Fig. 15, be the line to be measured. Take the dividers, Fig. 2 (frontispiece), and placing the sharp point A firmly upon the end A of the line AB, open the dividers till the other point B (the pencil point) just reaches the other end of the line B. Then letting the dividers re- )K main open just this amount, ^^^- ^^- place the point A on the lower end of the left hand scale, as at o, Fig. 1, and notice where the point B reaches. In this case it reaches 3 spaces beyond the figure 1. Now, as this scale is iiXiihes and tentJis of inches,* the Ime AB is 1.3 inches long. Ex. 1. What is the length of CD ? A7is. .15 of a foot Ex. 2. What is the length of EF ? Ans. .75 of an inch. Ex. 3. What is the length of CH ? Ans. H inches. Ex. 4. What is the length of IK ? Ans. .18 of a foot. Ex. 5. Draw a line 3 inches long. Ex. 6. Draw a line 2.15 inches long. Ex. 7. Draw a line 1.25 inches long. Ex. 8. Draw a line .85 of an inch long. * The next, scale to the right is divided into lOths and lOOths of a foot. Thus frouxp to IC 1 tentl* of a foot. ai.d the pmaller divisions are hundredths. ABOUT STRAIGHT LINES. 13 [Note. — Suppose a fine elastic cord were attached by each of its ends to the points A and B of the dividers ; when they were opened so as to reach from C to D, Fig. 15, the cord would represent the line CD, Now applying the di- viders to the scale is the same as laying this cord on the scale. Without the cord, we can imagine the distance between the points of the dividers to be a line of the same length as CD.] Ex. 9. Find in the same way as above the length and width of this page. Also the distance from one corner (angle) to the opposite one (the diagonal). 34:, IProb, — To find tlie sum of tivo lines. Solution. — To find the sum of AB and CD, 1* first draw the indefinite line Ejx. With the dividers I obtain the length of AB, by placing one point on A and extending the other to B, A' 'B This length I now lay ofi" on the indefinite line E^, by putting one point of the dividers at E and E C>- F G with the other marking the point Fig. ig. F. EF is thus made equal to AB. In the same manner taking the length of CD with the dividers, I lay it off from F on the line F.r. Thus I obtain EG=EF + FC=AB + CD. Hence, the sum of AB and CD is EG. [Note.— The student may measure EG by (55) and find the sum of AB and CD in inches or feet; but it is most important that he be able to look upon EG as the sum itself.] Ex. 1. Find the sum of AB and EF, Fig. 15. Ex. 2. Find the sum of EF, CD, and GH, Fig. 15. Ex. 3. Make a line twice as long as CD, Fig. 16. Three times as long. ' 35. JProb, — To find the difference of two lines. Solution.— To find the difference of AB and CD, I take the length of the less line AB with-the dividers ; and placing one point of the dividers at one extremity of CD, as C, make Ce = AB. Then is eD c« • «D the difference of AB and CD, since eD = Fi«. 17. CD - C(3 = CD - AB. Ex. 1. Find the difference of IK and EF, Fig. 15. Ex. 2. Find the difference of GH and CD, Fig. 15. * Theee elementary solutions are sometimes put in the singular, as the more simple style. 14 ELEMENTARY GEOMETEY. Ex. 3. Find how much longer IK, Fig. 15, is than the sum of EF, Fig. 15, and CD, Fig. 16. Ex. 4. Find tlie difference of the sum of AB and CH, and the sum of CD and EF, Fig. 15. 36. Prob. — To compare the lengths of two lines ; that is, to find their ratio (approximately*). Solution. — To compare the lengths of AB and CD, I lay off AB, the shorter, upon CD, as Ca. (If AB could be C" — ^ — y- — y^D applied two or more times to CD, I should apply it as many times as ^

/S the remainder of CD, viz., aD, to AB, ^^^- ^^- as many times as AB will contain it, which is once with the remainder 5B. This remainder I now appl}" to ianiefer of a Circle is a line passing through the centre and terminating in the circumference. 20 ELEMENTARY GEOMETRY. Fig. 27. Ii.L. — A circle may be conceived as the path of a line, like OB, Fig. 27, one end of which, O, remains at the same point, while the other end, B, moves around it in the plane (say of the paper). OB is the Badius, and the path described by the point B is the Circumference. AB is a diameter. In Fig. 28, the curved line ABCDA (going clear around) is the Qir- cumference, is the Centre, and the space within the circumference is the Circle. Kiaj part of a circum- ference as AB, or any of the curved lines BB, Fig. 27, is an arc. So also AM and EF, Fig. 29, are arcs. EF is an arc drawn from O' as a centre, with the radius O'B. 52. A Chord is a straight line joining any two points in a circumference, but not passing through the centre, as BC or ad. Fig. 28. The portion of the circle included between the chord and its arc, as kmD, is a Segment. 53, A Tangent to a circle is a straight line which touches the circumference, but Fig. 29. does uot iutcrsect it, how far soever the line be produced. 54:, A Secant is a straight line which intersects the circumfer- ence in two points. Ex. 1. Suppose DC, Fig. 11, to represent a small wooden rod, and BA a wire stuck into it at right angles. Now if you take the end C of the rod in your fingers and place the end D on the table so that the rod shall stand upright, and then revolve the rod once around like a shaft, what will the wire describe ? What the end B ? What any point in BA ? If you only revolve the rod a little way, what will the point B describe ? What does BA represent ? Ex. 2. If you take a string, OP, and hold one end at a particular point, O, on your slate or blackboard, while with the other hand you hold the other end, p, of the string upon the end of a pencil or crayon, and then move the end P around 0, making a mark as it goes, what will the mark made represent when the pencil or crayon has gone clear ^^x Pro. so. ABOUT CIECLES. 21 around ? What will the string represent ? What is the surface passed over by the string ? Ex. 3. If you take the dividers, Fig. 1, and open them (say 2 inches), and then place the sharp point. A, firmly on the paper while you turn them around, making the pencil point, B, mark the paper as it goes, what kind of a line will be described ? What is the line joining the points of the dividers ?* What line describes the cir- cle ? If the dividers only turn a little way, what is the line described ? Ex. 4. If a boy skating on the ice makes a curve which bends everywhere just alike, Avhat kind of a path will he make ? Does the boy describe a circle ? How might you conceive the circle in- closed by his path, as described ? Is a circle described by a point or by a line ? [Note. — The word "circle" is used in common language as equivalent to " ciicumference." It is also thus used in General Geometry. But, however the words may be used, the pupil should be taught to mark the distinction between the plane surface inclosed and the bounding line.] Ex. 5. In how many points can a straight line intersect a circum- ference? In how many points can one circumference intersect another ? Ex. 6. There is a piece of ground in the form of a circle, the radius of which is 100 rods, by which run two roads; one road runs within 80 rods of the centre, and the other within 100 rods. How do the roads lie with reference to the ground ? Ex. 7. When you unwind a thread by drawing it off a spool in the ordinary way, what geometrical line does the unwound thread represent ? "Ex. 8. In a circle whose diameter is 50 feet, there are drawn two chords, one is 20 feet long, and the other 30 feet. Which is nearer the centre ? Ex. 9. There are two circles whose radii are respectively 12 and 18 feet. The distance from the centre of one to the centre of the other is 25 feet. Do the circumferences intersect ? Would they in- tersect if the centres were 3 feet apart ? How would they lie in ref- erence to each other in the latter case ? How if their centres were 30 feet apart ? How if they were 35 feet apart ? * The imagination may be aided by supposing a tine elastic cord stretched between the points of the dividers and carried by them. 22 ELEMENTARY GEOMETRY. Ex. 10. What kind of a line is represented by water flying from a swiftly-revolving grindstone ? Ex. 11. If you draw two chords in the same circle, one of which is twice as long as the other, will the arc cut off by the longer chord be twice as long as the arc cut off by the shorter ? Will it be more than twice as long, or less ? 55, TJieorem* — The cliord of a sixth part of the circumference of a circle is just equal to the radius of the same circle. III. — If I draw a circle, and then, being careful not to open or close the di- viders, place the shaip point on tlie circumference at some point, as A, and mark the cu'cumferencc at another point, as B, with the pencil point, and then move the sharp point to B and mark again, as C, 1 find that when I have measm'ed oflF six such chords, each equal to the radius, I return exactly to A, the point of starting. Moreover, if I draw the chords AB, BC, etc., I have a regular figure with six equal sides. A figure ^^^- 31- with six sides is called a hexagon. This hexagon is called regular, because its sides are equal each to each, and its angles are also mutually equal. Again, if I unite the alternate angles of the regular hexagon, as FB, BD, and DF, I have a regular triangle, called an equilateral triangle. 56, Inscribed Figures are figures drawn in a circle, and having the vertices of all their angles in the circumference, as the hexagon and triangle in the last illustration. When the figure is without, and all its sides touch but do not cut the circumference, it is circumscribed about the circle. Ex. 1. Draw a regular hexagon whose side is two inches. Ex. 2. Inscribe an equilateral triangle in a circle whose radius is one inch. 57, JProh, — To find the centre of a circle ichen the circumference is draivn (or, as we usually say, known). Solution. — The circumference of my circle is drawn, but the centre is not ABOUT CIRCLES. 23 marked. So I want to find the centre. I draw any two chords, as AB and CD (the nearer they are at right angles to each other the better for accu- racy). I then bisect each chord with a perpen- dicular, as AB with the perpendicular MN, and CD with RS {39). The intersection of these two perpendiculars, as 0, is the centre of the circle. [The pupil must do everything with his pencil, ruler, and dividers, just as he says. He must not be of those who " say and do not." He must do the Fig. 32. things told, *' over and over," till he can do them neatly and easily.] S8, JProb, — To pass a circumference through three given pomts. Solution.— I wish to pass a circumference through the three given points A, B, and C. [The pupil should first designate three points by dots on his paper, slate, or board, and then proceed according to the solution.] In order to do this, I join A and B with a line, and also B and C. I now bisect these lines with the perpendiculars MN and RS, as in the last problem. The intersection of these per- pendiculars, 0, is the centre of the required circle. Now setting the sharp point of the dividers upon and ^ opening them till the pencil point just reaches A (B or "N^ C will answer as well), I draw the circumference with Fig. 33. as its centre and the radius OA, and find that it passes through the three given points A, B, and C. Ex. 1. To pass a circumference tlirougli the three vertices of a triangle, i, e., to circumscribe a circumference about a triangle, as this operation is technically called. SuG.— This is just like the last, A, B, and C being the vertices of the triangle. The four figures in the margin represent the successive steps in the solution. First draw the given triangle. Then take the first step in the solution, then the second, etc. Ex. 2. Given the centre of a circle and a point in the circumference, to draw the circle. SuG. — Make a dot on the board to indicate the centre, and an- other dot to indicate the point in the circumference to be found. This is what is given. You are Fi?. 34. 24 ELEMENTAKY GEOMETRY. then to draw the circumference, which shall pass through the latter point, and have the former for its centre. Ex. 3. Draw an arc of a circle, and rub out the mark, if you make any, at the centre, so that you cannot see where the centre is. Then find the centre, and complete the circumference according to these problems. SuG. — Mark three points in the given are, and tnen the example is just like the last [Do not fail to do it, " over and over," till you can do it quickly and neatly. These exercises require much care in order to get good figures.] S9» Tlieoreni, — Tlie circumference of a circle is ahciit 3.1416 times its diameter. The Greek letter n (called p) is used to repre- sent this niimher ; and hence the circumference is said to be n times the diameter, III. — The pupil can illustrate this fact by taking any wheel which is a true circle, and measuring the diameter with a narrow band of paper (something that will not stretch), and then wrapping this measure about the circumference. He will find that it takes a little more than three diameters to go around. Of course he cannot tell exactly how much more. In fact, nobody knows exactly. But the number given above is near enough for most purposes. For many pur- poses 3f is sufllciently accurate. By drawing a circle veiy carefully, say 1 inch in diameter, as in the margin, and dividing the diameter into lOths inches, a nice pair of dividers can be opened one 10th inch and made to step around the circumference. If it is all done with nicety, it wUl be found to be a little over 31 steps around, when it is 10 across. Ex. 1. The distance across a wagon-wheel Fig. a3. (the diameter) is 4 feet, how long a bar of iron will it take to make the tire ? Ex. 2. Suppose the crown of your hat is a circular cylinder 7 inches in diameter, how much ribbon Avill it take for a band, allow- ing \ of a yard for the knot ? Ex. 3. How many times will the driving-wheel of an engine, which is 6 feet in diameter, revolve in going from Detroit to Chicago, a distance of 288 miles, allowing nothing for slipping ? Ex. 4. A boy's hoop revolved 200 times in going around a city- square, a distance of 140 rods. What was the diameter of his boon ? ABOUT ANGLES. 25 Ex. 5. What is the radius of a circle whose semi-circumference is ;r ? In -a circle whose radius is 1, what part of the circumference 7t 7t does -X represent ? What part — ? What part does %7t represent ? SECTION IIL ABOUT ANGLES. 60. I^roh, — To sJioiv koto angles are generated and measured. III. — An angle is generated by a line revolving about one of its extremities. Thus, suppose OB to have started from coincidence with OA, and, remaining fixed, the line to have revolved to the position OB, the angle BOA would have been generated. When the revolving line has passed one-quarter the way around, as to DO, it has generated a right angle ; when one-half way around, as to FO, two right angles ; when entirely around, four right angles. ^ Now, if any circle be described from as a cen- tre, the arc included by the sides of any angle having its vertex at 0, is (he same part of a quarter of this circumference as the angle is of a right angle. Hence the angle is said to be measured by the arc included by its sides. Thus, the angle COA is measured by tlie arc ac, i. e., it is the same part of a right angle that arc ac is of arc ad. (See Trigonometry, 3-10.) Fig. 36. 61. Tlieof^em, — The relative lengths of arcs described luith the same radius can he found in a manner altogether similar to that give?i in {36) for comparing straight lines. III. — If I wish tp compare the two arcs db and cd described with tlie sama rfidii, I take the dividers, and placing the sharp point on d (one end of the shorter arc), open them till the other point is at c. I then measure this distance off on ab as many times as I can, — in this case 2 times, with a remain- der /&. This remainder, fb, I measure off in the same way upon f?c, and find it goes once with a remainder gc. This remainder, gc, I apply to the arc/6, and find it goes once with a remainder ?ib. This last remainder I find is contained in the last preceding, gc, 2 times. Then, count- ing up the parts, I find that dc is made up of 5 parts each equal to ?ib, and ab of 13 such parts. Therefore, nb is 2} times as long as dc. [The angle is therefore 2f times the angle C] FiQ. 37. ELEMENTARY GEOMETRY. Ex. 1. Draw an acute angle and also an obtuse angle, and then compare them as above. Ex. 2. Draw a small acute angle and a large acute one, and then compare them as above. Ex. 3. Draw a small acute angle, and then draw another angle 3 times as large. Ex. 4. Draw an acute angle, and also a right angle, and com- pare them as above. SuG.— Article {39) shows how to draw a right angle. Ex. 5. Draw any angle, and then draw another equal to it. Ex. 6. Show that the angles or, b, and c are respectively ^, |, and .6 of a right angle.* Fig. m. Fig. 38. Ex. 7. Show that angles a and b. Fig. 39, are respectively 1| and 1^ times a right angle. Ex. 8. Draw a regular inscribed hexagon, as in Fig. 31, and then comparing any one of its angles with a right angle, find that it is 1^ times a right angle. Ex. 9. Draw an equilateral triangle, as in Fig. 31, and find that any angle of it is I of a right angle. Ex. 10. Show that a right angle is ^-- measured by ^^ of a circumfei-ence. Solution.— If CD is perpendicular to AB, the four angles fomied are equal, and each is a right angle. But, as all of them taken together are measured by the whole circumference, one of them is measured by i of the circumference. * Of course, absolnte accuracy is not to be expected in such eolations. ABOUT ANGLES. 27 62, An Inscribed Angle is an angle whose vertex is in the circumference of a circle, and whose sides are chords, as A, Fig, 41. 63. Theorem.. — An inscribed angle is measured hy one-half the arc included letween its sides. III. — The meaning of this is that an inscribed angle like A, which includes any particular arc, as cd, is only half as large as an angle would be at the centre, as cOd, whose sides included the same arc, cd, or an equal arc. Thus, in this case, drawing the arc ah from A as a centre, with the same radius, O^Z, as cd is drawn with, I find that ah which measures A is ^ of c^ which measures cOd. Fig. 42. Ex. 1. Which of the angles a, I, c, d, e is the largest ? What is a measured by? What^? Whatc? WhattZ? Whate? i^/^. 42. Ex. 2. Which is the greatest angle, a, d, or c, Fig. 43 ? By what is a measured ? By what Z* ? By what c ? What is the measure of a right angle ? [See Example 10 in the preceding set] Fig. 43. Fig. 44. Ex. 3. Suppose I take a square card like CEDF, with a hole in one corner as at c, and sticking two pins firmly in my paper, as at A and B,, place the corner of the card between them, as in Fig. 44, and then, keeping the sides of the card snug against the pins, put a 28 ELEMENTARY GEOMETRY. pencil through the hole C and move it around to A and then back to B ; what kind of a line will the pencil trace ? Will it make any difference whether c is a right angle or not ? If any difference, what? Ex. 4. By what part of a circumference is an angle of a regular inscribed hexagon measured ? See {So), and Fig. 31. How many right angles is the angle of the hexagon equal to ? ^Miat is the sum of the six angles equal to ? Ans. to last, 8 right angles. Ex. 5. Show, from the way in which an equilateral triangle is constructed in Fig. 31, that one of its angles is measured by J- of a circumference, and hence is f of a right angle. 64:, Hieorein. — When two lines intersect, they form either four right angles, or two equal acute and two equal obtuse angles. III. — [The pupil can illustrate this for himself by drawing lines and noticing what angles are equal.] Ex. 1. Having a carpenters square, an instrument represented by MON, I wish to test the angle O and ascer- ^ tain whether it is, as it should be, a right angle. I draw an indefinite right line AB, -] and placing the angle at some point c on Q ^ this line with ON extending to the right on ^:|:'^ CB, I draw a line along OM. Turning the |i square over so that ON shall lie on CA, I I draw another line along OM. Three cases 1 may occur. — 1st. Suppose the first line ^ ^ ^ drawn along OM is CF, and the second CE ; what kind of an angle is ? 2d. Suppose the first line drawn is CE and the second CF ; what kind of an angle is ? 3d. Suppose the first and second lines drawn along OM coin- cide and are CD ; what kind of an angle is O ? Ex. 2. Show that the sum of all the angles formed by drawing lines on one side of a given line, and to the same point in the line, is two right angles. S5, I^rob, — To bisect a given angle. Solution. — I wish to divide the angle AOB into two equal parts, t. e., to ABOUT ANGLES. 29 [•adius, as 0«, I Fig. 46. bisect it. With 0, the vertex, as a centre, and any convenient strike an arc, as ba, cutting the sides of the angle. Then from a and h as centres, "with the same radius in each case, I strilie two arcs intersecting as at P. Drawing a line through P and 0, it bisects the angle; i. e., the angle POA = angle BOP. [Let the pupil try this by cutting out the angle AOB, and then folding the paper along the line P, or cut- ting it through in the line OP, and then putting one angle on the other, and thus see if they do not fit.] Ex. 1. Draw an angle equal to \ of a right angle. SuG. — First draw a right angle and then bisect it Ex. 2. Draw an angle equal to ^ of a right angle. SuG. — Draw a circle. Inscribe an equilateral triangle. [Do it neatly, by rule, as in {55).] Then bisect any angle of this triangle. This will be ^ of a right angle, since the whole angle is |. See Ex. 9 {61). Ex. 3. How does it appear that the angle EOF, Fig. 31, is J of a right angle ? GO. Parallel Straight Lines are such as, lying in the same plane, will not meet how far soever they are produced either way. III. — The sides of this page are parallel lines, ■ as are also the top and bottom. The lines in Fig. 47 are parallel. 67. Proh.—To drato a line through ^'^- ^^• a given point and parallel to a given line. r Solution. — I wish to draw a line through the point O and parallel to the line AB. [The pupil should first draw some line, as AB, and mark some point, as O.] Ic— i ^ D take O as a centre, and with a radius * greater \ \ than the shortest distance to AB, as Oa, draw an ^ indefinite arc aP. Then with a as a centre, and tlie same radius, I draw an arc from O to the line AB at h. Taking the distance Oh (the chord) in the dividers, I put the sharp point on a and strike a small arc intersecting this indefinite arc, as at P. Fi- nally, drawing a line through O and P, it is the parallel sought. * This means " put the sharp point of the dividers on O and open them till the distance be- tween the points (the radiu?) is more than the distance from to AB-" ■B Fig. 48. 30 ELEMENTARY GEOMETEY. 68, Theorem, — Tioo parallel lines are everyiuhere the same dis- tance apart. III. — Let AB and CD be two parallel lines. I will examine them at the two points O and P. To find how for apart the — B lines are at these points I draw the perpen- diculars OM and PN. [The pupil should not f^ D guess at these, but actually draw them as in- FiG. 49. structed in (^4).] Measuring these, I find them equal. We can understand that this proposition must be true, since the lines could not approach each other for awhile and then separate more and more without being crooked ; or, if they kept on approaching each other, they would meet after awhile, and so not be parallel. M 69, Hieorem, — Parallel lines mahe no angle luith each other. III. — Let AB be a straight line, and suppose CD another sti-aight line passing through the point O. Now let CD turn around, first into the position D'C, then into D"C", etc., all the time passing through O. It is evident that the angle which this line makes with the line AB is all the time growing less, i. e., a' < a, and a" < a'. It is also evi- dent that this angle will become when the lines become parallel ; for it becomes less and less all the time, but is always somethuig so long as the lines are not parallel. 70, Hieorem, — Parallel lines have the same direction with each other. III.— Thus, in Fig. 47, the parallel lines all extend to the right and left, i. e., in the same direction. Ex. 1. How shall the farmer tell whether the opposite sides of his farm are parallel ? Ex. 2. If we wish to cross over from one parallel road to another, is it of any use to travel farther in the hope that the distance across will be less ? Ex. 3. If a straight line intersects two parallel lines, how many- angles are formed? How many angles of the same size? May they all be of the same size ? "When ? When will they not be all of the same size ? ABOUT TRIANGLES. 31 SECTION IV. ABOUT TRIANGLES. 71. A Plane Triangle, or simply A Triangle, is a plane figure bounded by three straight lines. 7^. With respect to their sides, triangles are distinguished as Scalene, Isosceles, and Equilateral. A scalene triangle has no two sides equal. An isosceles triangle lias two sides equal. An equi- lateral triangle has all its sides equal. 73, With respect to their angles, triangles are distinguished as acitte angled, right angled, and obtuse angled. An acute angled triangle has three acute angles. A right angled triangle has one right angle, and the side opposite the right angle is called the hypotenuse. An obtuse angled triangle has one obtuse angle. Ex. Fig. 51 affords illustrations of all the different kinds of triangles. Let the pupil point them out until he is perfectly familiar with the terms. He should also practise drawing the different kinds of triangles, for the purpose of familiarizing the names applied to the different kinds. Fig. 5L 74. Theorem, — The sum of the angles of a triangle is tioo right angles. B ; show how an attempt ^/f to apply one to the other fails. Solution. — Since angle D = angle A * I ap- ply the vertex D to the vertex A, and make DE take the direction AB. AsDE^rAB, E will fall on B, and the sides DE and AB will coincide. Again, since D = A, the side DF will take the direction AC when the planes of the triangles coincide ; and the extremity F will fall in AC, or in AC produced (really in AC produced, in this case). Finally, since E > B, EF will fall to the right of BC, and the application fails. yig. G8. Ex. 2. Construct two trapeziums with their respective sides equal, as AC = HE, AB = HG, BD =: OF, and CD = EF, but with their angles unequal ; and show how an attempt to apply one to the other fails. Ex. 3. If the sides of two trapeziums, as in the last figure, are equal, and two of the angles including a side in one are respectively equal to the corresponding angles in the other, as A = H, and B = C, can one be applied to the other ? If so, give the details of the process. fig. c9. SECTION VL ABOUT SIMILAR FIGURES, ESPECIALLY TRIANGLES. 87, Similar Figures are such as are shaped alike — i. e., have the same form. A more scientific definition is. Similar Figures are such as have their angles respectively equal, and their homologous (correspond- ing) sides proportional. * Be careful to distinguish betweea the vertex, which is a point, and the angle, which is the opening between the lines. ^ ELEMENTARY GEOMETRY. Fig. 70. 88. Honioloffous^ or Corresponding Sides of similar figures, are those which are included between equal angles in the respective figures. In Similar Triaxgles, the Homologous Sides are those OPPOSITE THE equal AJh^GLES. III. — The triangles ABC and DEF are similar, for they are of the same shape. But it is easy to see that ABC is not similar to IHK or MON. The pupil should notice that A = D, C r= F, and B = E. Also, side e is IJ times 5, side /is 1^ times c, and side rf is 1^ times a; so that f : c : : e :b, and / : c : : d : a, and d \a : : e : b. Now there are no -such relations existing between the parts of ABC and IHK. The angles B and K are nearly equal, but A is much larger than H, and C is smaller than I. So these triangles are not mutually equiangular, i. e., each angle in one has not an equal angle in the other. Again, as to their sides, IH is a little less than AC, but H K is greater than AB. These two triangles are, there- fore, not similar. In the similar triangles ABC and DEF, 6 is homologous with e, since they are opposite the equal angles B and E. For a like reason a is homologous with dy and c with f. It may also be observed, that the shortest sides in two similar triangles are homologous with each other ; the longest sides are also homolo- gous with each other, and the sides intermediate in length are homologous with each other. Ex. 1. Can a scalene triangle be similar to an isosceles triangle ? Can an obtuse angled triangle be similar to a right angled triangle ? Ex. ^. Are all squares similar figures ? SuG. — First, are the angles equal? Second, is any one side of one square to some side of another square as a second side of the fii*st is to a second side of the second, etc. ? Ex. 3. A farmer has two fields, each of which has 4 sides and 4 right angles. The first field is 20 rods by 50, and the other 40 by 80. Are they similar ? SuG.— Are they mutually equiangular? Then are the lengths in the same ratio as the widths? If they are not similar, how long would the second have to be in order to make them similar? Draw two such ligures, and see if they look alike in shape- ABOUT SIMILAR FIGURES, ESPECIALLY TRIANGLES. 41 89. I*roh, — To find a fourth proportional to three given lines. Solution. — I have the three given lines A, B, and C, and wish to find a fourth line ^ such that Q A shall be to B as C is to X]ie fourth Um, i. d., A : B : : C : fourth line. To do this, I draw two indefinite lines OX and OY, from a common point 0. On one of these, as OX, I lay off" Oa = A, and Oc = B. Then on the other 1 make Ob = 0, and draw ab. Finally, drawing a parallel to ab through the point c (67), I have Od as the line sought. Thus, calling OfZ, D, the proportion is Fig. 71. Oa : Oc A : B Ob Od.OT D. N.B. — The order in which the lines are taken, and the loay of drawing the lines ab and cd, are essential. The following directions will insure correctness : Lay off the FIRST a?id second on the same line, as on OX ; and the third on the OTHER LINE, as on OY. Then join the extremities of the first and third, and draw tlie parallel through the extremity of the second. Fig. Ex. 1. Show that if the order of the proportionals in Fig. 71 is B : A : : C : fourth line, the fourth proportional is e, Fig, 71. Ex. 2. Show that a fourth propor- tional to A, B, and C is D. Also, that a fourth proportional to C, A, and B is.E. Show that, if the order he A : C : : B : fourth line, D is still the fourth proportional. Show that B : C : : A : 2C, nearly. Ex. 3. Solve the proportion 3 : 8 : : 5 : a;, and find x geometrically. SuG.— Using the scale of lOOths of a foot, the figure is that in the margin. OD is the foiu'th proportional, or x = OD, which is found bi/ measurement to be 10 j, as it is by arithmetic. Fio. 73. 42 ELEMENTAEY GEOMETRY. 00, I^rob, — To draio a triangle similar to a given tria?igle, and having a given side. Solution. — 1st Method. — I have a triangle ACB, and want to make another similar to it, but having the side homologous to BC equal to a. I draw an indefinite line, and on it take EF, equal to a. Then at F I make an angle equal to C, and make the side indefinite. Now I find a fourth proportional to BC, EF, and AC. Having found this, as in the last article, I lay it ofi" from F, as FD. Drawing DE, I have DEF, the triangle required. I can readily satisfy myself that DEF is simi- lar to ABC, for besides the foct that it looks as if it were of the same shape, by measuring the other two angles, I find that E := B, and D = A. More- over, I know that BC, EF, AC, and DF are pro- portional, because I made them so. And, by finding a fourth proportional to BC, EF, and AB, I find it exactly equal to DE. In like manner constructing a fourth proportional to AC, DF, and A B, I find it to be DE. So that the two triangles are mutually equiangular, and have their homolo- gous sides (those opposite the equal angles) pro- portional. Hence, the triangles are similar. 2d Method. — But an easier way to construct DEF is to make the angle F = C as before, and then make E = B, and produce the sides till they meet in D. The triangles will then be similar, and the proportionality of the sides can be tested. Ex. 1. Given a triangle whose sides are 7, 11, and 15, to construct a similar triangle having the side corresponding to the one which is 11 in the given triangle, 8. Ex. 2. Construct two triangles with equal angles, and then com- pare the sides, and see whether you can make two triangles whose angles shall be respectively equal, and their sides not be propor- tional. SuG. — Having the triangle ABC, make DEF equiangular with it, and then compare the homologous sides. In the figure D is made equal to C, and F to A; whence E = B. DE and BC are homologous sides, because opposite the equal angles F and A. DF is homolo- FiG. 75. ABOUT SIMILAR FIGURES, ESPEOTATJiY TRIANGLES. 43 gons with AC, because it is opposite angle E, which equals B. For a simi- lar reason. EF is homologous with AB. Now, taking two sides of ABC, as BC and AB, and a side of DEF homologous with one of them, as DE, and finding a fourth proportional Oc, it will be found exactly equal to EF; so that BC : DE : : AB : EF (= Oc). Ex. 3. Make two triangles, two of whose angles shall be, one f and the other J of a right angle ; but make the side included between these angles twice as great in the second triangle as in the first. What will be the ratio of the side opposite the angle f in the first triangle to the homologous side in the second ? What the relation of the sides opposite the angles ^ ? Ex. 4. If you make one triangle whose sides are 5, 8, and 3 ; and a second wiiose sides are 15, 24, and 9, will they be mutually equi- angular? Which angles are the equal ones ? W^hich are the homol- ogous sides? Ex. 5. There are three pairs of similar triangles in Fig. 76. Can you point them out? Also point out their homologous parts. Are all the triangles which you can make out from the figure similar to each other ? I'iG. 76. Fig. 77. Ex. 6. Wishing to know the height EC of a house, I set up a stake DB 5 feet long; and putting my eye close to the ground, I moved back from the stake to A, so that the top of the stake and the top of the house were just in range (in a line). Then by meas- uring I found AB = 10 feet, and AC = 80 feet. What was the height of the house ? Ex. 7. If you take three sticks of different lengths and put them together by joining their ends two and two, so as to represent a triangle ; can you, by putting together the same sticks in a different order, make a triangle of different form from the first ? Will the angles opposite the same sticks always be the same ? Ex. 8. If you take more than three sticks (say 4), and make of them the boundary of a figure, by putting their ends togetlier two and two, can you put them together so as to make another figure of different form ? Can you make figures having different angles ? Ex. 9. If you take three sticks, A 3 inches long, B 5 inches, 44 ELEMENTARY GEOMETRY. and c 6 inches ; and also three other sticks, D inches long, E 15 inches, and F 18 inches ;* can you place them together so as to make dissimilar triangles ? Will the corresponding angles of the two tri- angles be equal however you may arrange the sticks ? If the sides of two triangles are proportional, will their angles be equal and the triangles similar? Ex. 10. If you take four sticks, A 3 inches long, B 5 inches, C 6 inches, and K 4 inches ; and also four other sticks, D 9 inches long, E 15 inches, F 18 inches, and L 1^ inches ;* can you place them to- gether so as to make four-sided figures which shall be dissimilar {i. e., not of the same shape) ? Will the corresponding angles of the two figures be necessarily equal? If the sides of a four-sided figure are proportional, does it follow that the corresponding angles are equal, and the figures similar ? Ex. 11. Why do the braces in the frame of a building stiffen it? Is a four-sided figure stiff? i. e., are its angles incapable of change while its sides remain of the same length ? Can the angles of a triangle be changed while the sides remain un- changed ? Fig. 78. SECTION VIL ABOUT AREAS. 91. A Quadrilateral is a plane surface inclosed by four right lines. 92. There are three Classes of quadrilaterals, viz., Trapeziums, Trapezoids, and Parallelograms. 93. A Trapezium is a quadrilateral which has no two of its sides parallel to each other. 94. A Trapezoid is a quadrilateral which has but two of its sides parallel to each other. * Notice that the sides are proportional i. e., in the same ratio taken two and two. ABOUT AREAS. 45 95o A Parallelogram is a quadrilateral which has its oppo- site sides parallel. ^6. A Rectangle is a parallelogram whose angles are right angles. 97* A Square is an equilateral rectangle.* 98, A Rhombus is a parallelogram whose angles are not right angles, and ail of whose sides are equal. 99. A Rhomboid is a parallelogram whose angles are not right angles, and two of whose sides are greater than the other two. III. — The figures in the margin are all quadrilat- erals. A is a trapezium. (Why?) B is a trapezoid. (Why?) C, D, E, and F are / parallelograms. (Why ?) D and E are rectangles, although D is the form usually referred to by the teiTa rectangle. So C is the form usually referred to when a parallelogram is spoken of, without saying what kind of a parallel- ogram. C is also a rhom- boid. (Why?) E is a square. (Why?) F is a rhombus. (Why ?) This page is a rectangle; so also are the common panes of glass. Fig. 79. 100. A Diagonal is a line joining two angles of a figure, not adjacent. III. — In common language, a diagonal is a line running "from comer to corner." Ex. 1. To construct a square, having given a side; or, in other words, to construct a square on a given line. *The pupil Phould be able to give this and all similar definitions at length. Thus, A Square is a surface inclosed by four equal right lines making right angles with each other. 46 ELEICENTARY GEOMETRY. Y N — 1st Method.— Let A be the given side. Draw the indefinite line OX, and lay off OM = A. At M erect a perpendicular MY, as taught in (44). On this take M N = A. From N and as centimes, with a radius equal to A, describe arcs iutei-sect- ing, as at P. Draw NP and PO. 2d Metliod. — Let Q be the given side. Con- struct equal angles at the extremities of Q, and produce tlie sides till they meet, and one of them till it will meet another side of the square proposed. With S as a centre, and ST or SR as radius, describe a semicircle. Draw RV, and it forms a right angle at R. The construction can now be finished as before. Ex. 2. Construct a rhombus whose side is 2 inches, and one of whose acute angles is I of a right angle. ^^^- ^^- Ex. 3. Construct a rectangle whose ad- jacent sides are 3 and 5.* Ex. 4. Construct a rhomboid whose adjacent sides are 3 and 7, and their included angle J a right angle. Ex. 5. How many diagonals has a triangle? How many has a quadrilateral ? How many has a figure with five sides (a pentagon) ? Of six? Of eight? 101, TJie Area of a surface is the number of timos it contains some other surface taken as a unit of measure ; or it is the ratio of one surface to another assumed as a standard of measure. 102* The Uiiit of Area usually assumed is a square, a side of which is some linear unit: thus, a square inch, a square foot, a square yard, a square mile, etc. By these terms is meant a square 1 inch on a side, one foot on a side, one yard on a side, etc. The acre is an exception to the general rule of assuming the square on some linear unit as the unit of area, there being no linear unit in use whose length is the side of a square acre. III. — The area of a board is the number of squares 1 foot on a side which it would take to cover it. The area of a floor may be spoken of in square yards, and is the same as the number of square yards of carpeting it would take to cover it. * TaVe any convenient unit, ap \ inch, 1 inch. ABOUT AREAS. 47 103, The Altitude of a parallelogram is the distance between its opposite sides ; of a trapezoid, it is the distance between its parallel sides; of a triangle, it is the distance from any vertex to the side opposite or to that side produced. ±04t, TJie Bases of a parallelogram or of a trapezoid are the sides between which the altitude is conceived as taken ; of a triangle, it is the side to which the altitude is perpendicular. III.— The dotted hnes in B, C, D, and F, Fig. 79, represent When the altitude is the distance between two parallels, the figure bases. The altitude of a parallelogram may be reckoned between either pair of parallel sides, but it is most common to conceive it as the distance between the two longer sides. The altitude of a rectangle is the same as either side to which it is parallel. A triangle may have three altitudes, and any side of a triangle may be conceived as its base. In Fig. 81, AB is conceived as the base in each case, and CD the altitude. altitudes, has two Ex. What side of a triangle must you conceive as the base, in order that the altitude shall fall upon it, and not upon its pro- longation ? a case ? From what angle will the altitude be reckoned in such 105. TJieorem, — 77ie area of a rectangle is the product of its two adjacent sides; or, luhat is the same thing, the product of its altiticde and base. III. — Let ABCD represent a rectangle, of which AB is 8 units long, and AC 5. Now, let us conceive a square a constructed on one of these units. Using this surface as the unit of area, it is evident that in the rectangle cABd there will be 8 such. Hence, the area of this rectangle is 8 (square units). Now, drawing parallels to the base through the several points of division of the altitude, it is evident that the whole rectangle ABCD is made up of as many rectangles like cAB^ as there are units in the altitude — in this case 5, Hence the whole area is 5 times the area of cABfZ, i.e., 5 times 8 (square units) = 40 (square units). N.B. — The pupil should be careful to observe that the language '^product of ba^e into cUtitude^^ is only a convenient form of abbreviated expression. It is ^ A/ ^ ' /f. 5 6 7 ffQ Fig. 82. 48 ELEMENTARY GEOMETRY. just as absurd to talk about multiplying a line bj^ a line, as to talk about multi- plying dollars by dollars. Thus 8 inches in length can be taken 5 times, and makes 40 inches in length. But what does 8 inches in length, multiplied by 5 inches in length mean ? Or what is 8 dollars taken 5 dollars times ? The multi- plier must always be an abstract number, and the product be like the multipli- cand, from the *\'erv nature of multiplication. With this the explanation given above agrees. When we say that the area of ABCD = 8 x 5, we mean 5 times 8 squai-e units, which equals 40 square units. 100, TJieoreni. — TJie area of aiiy parallelogram is the same as the area of a rectangle having the same base and altitude as the parah lelogram, and hence is the product of its base and cdtitude. III. — This truth is easily illustrated by cutting out a parallelogram, as ABCD. Then, cutting off the triangle DEC, being careful to make DE perpendicular to BC, and placing DC upon AB so as to bring the triangle DEC into the position AFB, the two parts will just make up the rectangle AFED. Hence we see that the area of ABCD is the same as the area of AFED, which latter is a rectangle having the same base AD, and the same altitude ED, iis the given parallelogram. Fig. S3. 107* Tlieorenu — The area of a triangle is half the j^roduct of its base and altitude. III.— To illustrate this truth, cut out two triangles A and B just alike. By placing them together, a /P\^ /^\" " 7 parallelogram can be / i ^\ / j ^\^ A .-' formed whose base and / I A ^\ / I ^ ^\^ , ' altitude are the same as ^ ' -^ ^ '■ the base and altitude of ^''G. &4. tiie triangle. The area of the parallelogram is the product of its base and altitude. Hence the area of one of the ti'iangles is one-half the product of its base and altitude. In fact, by cutting one of the triangles, as A, into two triangles, its parts can be put with B so as to make a rectangle having the same base and altitude as the ti'iangles. [The pupil should do it.] 108, llieoreni,—The area of a trapezoid is ilie product of its altitude into the line joiimig the middle points of its inclined sides, III.— To illustrate this truth, cut out any trapezoid, as ABCD, and through ABOUT AREAS. 49 the middle of the inclined sides, as a and J, cut off the triangles kam and B6/i, being careful to cut in lines' am and hn perpendicular to the base. These can be applied as indicated in the figure, so as to fill out the rectangle omnp. Hence we see that the area of the trapezoid is A nv TV B Fig. 85. just equal to the product of its altitude into the line joining the middle points of its inclined sides, as ah. Ex. 1. How many square yards of plastering in the walls of a room 20 feet by 30, and 15 feet high, including the ceiling ? Ans. 233f Ex. 2. A salesman is selling a piece of velvet which is worth 88 per yard. The velvet is cut " on the bias," as the technical phrase is, i. e., obliquely, instead of square across. The piece he is selling is measured along the selvedge in the usual way half a yard. He is disposed to charge the customer somewhat more than 84. Is he right ? The customer claims that he is getting but half a yard of velvet, and so ought to pay but 84. Is he right ? Ans. Both are right, — the salesman in his demand, and the customer in his statement. How is it ? Ex. 3. There are two parallel roads one mile apart. A has a farm which extends along one of the roads half a mile, and the lines run perpendicularly from one road to the other. B has a farm l}ing be- tween the same roads, and half a mile front on each road, but run- ning obliquely across. Which has the larger farm ? C D E F Ex. 4. Of the four triangles ACS, ADB, AEB, and AFB, Fig. 86, which has the greatest area, CF being parallel to AB ? fig. sg. Ex. 5. Which is the largest triangle which can be inscribed in a semicircle, having the diameter for its base ? Ex. 6. Can you vary the area of a triangle while the sides remain of the same length? Can you vary the area of a quadrilateral while the sides remain of the same length ? Ex. 7. If you have two lines each 5 inches long, and two each 3 inches long ; into what kind of a parallelogram must you form them in order to have its area the greatest ? 4 Fig. 87 50 ELEMENTARY GEOMETRY. Ex. 8. Rough boards are usually narrower at one end than at the other, for which reason the lumberman measures their width in the middle. What is the number of square feet in the following : 12 boards 16 feet long, 10 inches wide (in the middle) ; 15 boards 11 feet long, 9 inches wide " " ; 8 boards 10 feet long, 13 inches wide " " ? What principle is involved in such measurement ? Ex. 9. What is the area of a triangle whose altitude is 6 feet, and base 10 feet ? Are these elements sufficient to fix the form of the triangle ? Ex. 10. If a line be drawn from any angle of a triangle to the middle of the opposite side, what is the relation of the areas of the two partial triangles ? Why ? THE PYTHAGOREAN PROPOSITION.' 109* Hieorem, — Tlie square described on ilie liyiMenuse of a right angled triangle is equivalent to the sum of tlie tico squares described on the other two sides. III. — The meaning of this proposition may be illustrated thus : Let ABC be a right angled triangle, right angled at C, and the sides AC and CB be 4 and 3 respectively. Then measuring AB, it will be found to be 5, and we observe that 4"^ + S'' = 5*. This is also seen from the figure, in which the square on AC contains 4"= 16 square units, and that on CB 3^= 9 ; while that on AB contains 5' = 25, i. e., as many as on both the other sides. We cannot so readily illus- trate the truth of the proposition when the ratio of the sides is any other, than that of 3, 4, and 5, but it is equally true in all cases, as will be proved in the next part of this book. Ex. 1. Can you make a right angled triangle whose sides shall be 5, 8, and 10 ? SuG.— As 10 is the longest side, it will have to be the hypotenuse. Now 5^ + 8* = 25 + 64 = 89. But 10''^ = 100. Hence, 10 is too long for the hypote- nuse of a right angled triangle whose other sides are 5 and 8. Ex. 2. Can you make a right angled triangle whose sides shall be 9, 12, and 15? THE PYTHAGOREAN PROPOSITION. 51 Ex. 3. A carpenter has framed the four sills of a building to- gether, and placed them on the foundation. He then wishes to adjust them so that the angles shall be right angles. He places one end of his ten foot pole ah at a, 6 feet from c ; and, holding it in position, orders his attendants to move the sill AB to the right. How far will the end h of the pole be from c when the angle B is a right angle ? Ex. 4. A gate is to be 10 feet long and 4 feet high. How long must the brace be to go in as a diagonal and hold the gate in the form of a rectangle ? Ex. 5. The angles of a room are all right angles, and its dimen- sions are 20 feet by 30 on the floor, and 15 feet high. AVhat is the length of the longest diagonal extending from one corner on the floor to the opposite corner in the ceiling ? Ans, A little more than 39 feet. Ex. 6. The numbers 3, 4, and 5 are much used by artizans as parts of a right angled triangle. Will any equi-multiples of them answer the same purpose, as twice them, i. e., 6, 8, and 10 ; or three times them, as 9, 12, and 15, etc. ? Ex. 7. In an obtuse angled triangle, is the square of the side oppo- site the obtuse angle greater or less than the sum of the squares of the other two sides ? How is it with the square of the side opposite an acute angle ? SuG.— In the right angled tinangle ABC, AC = CB^ + AB'^ In the obtuse angled triangle C'B is equal to CB in the right angled triangle. But AC"* is greater than AC^ hence AC'' > BC'^ + AB^ By 'a similar inspection the other case may be determined. Fig. 90. no* JProh, — To find a mean proportional hetween two lines. Solution. — I wish to find a mean propor- tional between the lines M and N, ^. e., a line aj, such that M : a; : : a; : N, whence a;' = M X N, and x= -v/M X N. I draw a line AB equal to the sum of M and N, making DB = M, and AD = N. I draw a 52 ELElfENTARY GEOMETRY. semicircumference on AB, and at D erect CD perpendicular to AB. CD is a;, the mean proportional required. Ex. 1. To construct a square which shall be equal in area to a given rectangle. SuG. — Draw any rectangle. Then find a mean proportional between its adjacent sides as described above. A square constructed on this line will be equal in area to the rectangle ; since, if .r is the side of the square, and M and N are the adjacent sides of the rectangle, x"^ = M x N. But x^ is the area of the square, and M x N is the area of the rectangle. Ex. 2. To find the square root of 15 by means of the ruler and compasses. SuG.— Since 15 = 3 x 5, if DB = 3 and AD = 5, Fig. 91, x (CD) = V 3 x 5 = V 15. Therefore, making a figure having DB and AD of these lengths, CD can be measured, and thus the square root of 15 obtained, approximately, in numbers. N. B. — In such a case CD represents exactly tJie required root, although we may not be able to express the value exactly in numbers. In this case geometry does exactly what arithmetic can only do approximately. Ex. 3. Draw a line which shall represent, exactly, the square root of 5. SuG.— Make DB = 1, and AD = 5. Ex. 4. Draw a rectangle whose adjacent sides are 2 and 3, and then draw a square of the same area. 111. Theoi^em,— Tlie areas of similar triangles are to each other as the squares of their homologous sides. III.— The meaning of this is, that if ABC and DEF K^^ are similar, and any side of ABC is 2 times as great as the 1 \. homologous side of DEF (as is the case in the figure, CB \ \. being = 2FE, CA to 2FD and AB to 2DE) the area of r;; '^\. ^^^ i'^ 4 times the area of DEF. In fact, in a simple \ V. \ \^^^ case like this, we can di\nde ABC into four triangles \ '^-..j N^ exactly equal to DEF, as is done by the dotted lines. A B Ex. 1. A and B have triangular pieces of land, K which are similar to each other, and similarly 1 N. situated. But A's front is to B's as 5 to 3 ; bow I X much more land has A than B ? ° p ^92 Ans. 2| times as much. AREAS OF SIMILAR TRIANGLES. 53 Ex. 2. In order that one triangle may be similar to and 4 times as great as another, how must any side of the first compare with the ho- mologous side of the second ? Ex. 3. In order that the areas of two similar triangles may be to each other as 4 to 9, what must be the ratio of their homologous . , „ ^ Fig. 93. Sides ? 112. Theore^n.—The homologous sides of similar triangles arc to each other as the square roots of their areas. This theorem is involved in the theorem that the areas of similar triangles are to each other as the squares of their homologous sides. It is illustrated in the preceding examples. Ex. Construct a triangle with one of its sides 2 in length. Then construct a similar triangle 1| times as large. What must be the length of the side of the second triangle which is homologous with the side 2 of the first. Solution.— Let CAB be the given triangle, whose side AB is 3. Since the second is to be li times as great as the first, the ratio of the areas is 2 : 3. Hence, V'^ ■ V^ :: AB (or 2) : x, the side of the re- quired triangle ho- mologous with side 2 of the given tri- angle. Construct the square roots of 2 and 3, as ab and ac in the figure, and then find a fourth proportional to ab, ac, and AB. This is found to be ^^^ ^^ ay. Taking DE zr ay, construct on it a triangle DEF similar to ABC, and it will be U times as large. 54 ELEMENTABY GEOMETRY. H Fig. 95. THE AREA OF A CIRCXE. 113* TJieorem, — The area of a circle whose radius is r, is ;rr'» i.e., 3.1416 times the square of its radius. III. — If we take a circle whose radius is r and circumscribe about it a square ABCD, we observe that the area of this square is 4?"'. Hence we see that the area of a cu-cle is less than 4 times the square of its radius. Again, drawing two diametei"s EF and CH at right angles to each other, ^ and joining their exti'emities, we have the inscribed square GEHF. The area of this square is equal to the area of the two triangles CEF and EHF. But area CEF = iCO x EF = ^r x 2r = ^^ ; and in like manner EHF = ?•='. Hence area GEHF = 2;-='. TVe thus see that the area of a chcle is more than two times the square of the radius. The area of a circle is therefore somewhere between two and four times the square of its radius. Just how many times 7^ the area is, we do not propose to find in this place, but only say that it has been found to be 3.1416 times r'^. TVe must also remark that this is not exact ; but it is near enough for practical purposes. In fact, nobody knows exactly how many times the square of the radius the area of a circle is. Ex. 1. If Yoii cut from a square the largest possible circle, show that you cut away a little less thau ^ of the square, or more exactly .2146 of it. Ex. 2. What is the area in acres of a circle whose diameter is 3 miles? ^W5. 4523.904. Ex. 3. A horse is so tied to a tree that he can graze on every side of it to a distance of 100 feet. What is the area in acres over which he can graze ? A7is. A little less than J of an acre. Ex. 4. What is the area of a circle whose radius is 1 ? [Remember this result.] Ex. 5. Wliat is the area of a circle whose radius is 2 ? 3 ? 4 ? How do these areas compare with the area of a circle whose radius isl? 114. TJieorem, — 77ie areas of circles are to each other as the squares of their radii. III. — This is readily seen from the last theorem. Thus the area of a circle whose radius is 5 is 2o7r ; and of one whose radius is 6, the area is 36;r. Now, OF POLYGONS. 55 the ratio of these areas 25;r : SGtt is the same as 25 : 36, i. e., as the squares of the radii of the two circles. Ex. 1. In the figure the radius of the outer circle is twice that of the inner. How do their ^ — ] — -^ areas compare ? How do the 4 parts into which X ; X the larger circle is divided compare with each / 3 /^ ^\ 2 \ other? \ \ ^ ) \ Ex. 2. The radii of 2 circles are 3 and 5 re- V' ^^ — 'J spectively ; what is the relation of their areas ? ^--^.^^-^^ Ans. 9 : 25 ; or one is 2 } times as large as the ^^^ ^ other. Ex. 3. I have a circle whose radius is 5, and wish to make another whose area is twice as great ; what must be its radius ? Ans. "v/50, or 7.071 nearly. Ex. 4. Can we compare the areas of circles by means of the squares of their diameters as well as by means of the squares of their radii ? How much greater is the square of the diameter of any circle than the square of the radius? Ex. 5. Two 5-inch stovepipes run together into one 7-inch pipe. Is the caj^acity of the one pipe equal to that of the two ? Ex. 6. Two men bought grindstones of equal thickness. The stones cost $4 and $9 respectively. One was 2 feet in diameter and the other 3. What was the difference in the rates paid ? SECTION VIIL OF POLYGONS. 115. A Polygon is a portion of a plane bounded by straight lines. The word polygon means many-angled ; so that with strict propriety we might limit the definition to plane figures with five or more sides. This limita- tion in the use of the word is frequently made. 116 * A polygon of three sides is a triangle ; of four, a quadrilat- eral; of five, a pentagon ; of six, a hexagon ; of seven, a heptagon ; of eight, an octagon; of nine, a 7ionagon ; of ten, a decagon; of twelve, a dodecagon. 56 ELEMENTAEY GEOMETRY. 11'^. A Regular Polygon is a polygon whose sides are equal each to each, and whose angles are equal each to each. US, Hie VeriTYieter of a polygon is the distance around it, or the sum of the bounding lines. 119, Theorem, — Any polygon may he divided hy diagonals df'aiun from any angle, into as many triangles as the polygoji lias sides, less tivo sides. Iix. — In the figure the polygon has 7 sides. By drawing the diagonals from C to the other angles, we divide the polygon into 5 (7—2) tiiangles. Fig. 97. 120, Theore^n,—The sum of the an- gles of any polygon is twice as many right angles as the ptolygon has angles {or sides), lessfonr right angles. III. •Draw a polygon, as ABCDEFC, and the arcs a, b, c, d, e,f, g, measuring its angles. With the same radius draw a circle. Beginning at some point, as O, lay off OA = a, AB = i, EC = c, CD = d, DE = e, EF =/, and FG = g. It is found in this case that the sum of these meas- ures is two circumferences and a half. Now, one circumference is the measure of 4 right angles. Hence, 2^ circumfer- ences measure 2^ x 4 r= 10 right angles. Thus it appears that the sum of all the angles of the polygon is 10 right angles. This agrees with the theorem ; for, .by that, the sum should be 2 right angles x 7 — 4 right angles, which is 10 right angles. ^121, Troh. lar polygon. To draw a regu- Fig. 98. as the pol5'gon has sides. of the polygon. Solution. — Draw a circle, and divide the circumference into as many equal arcs The chords of these arcs will constitute the perimeter OF POLYGONS. 67 The practical difficulty lies in dividing the circumference as required. The circumference can be divided into 6 equal arcs by (J5). Drawing radii to these points of division, and bisecting the included angle, a division into 13 equal parts is effected. These can be again bisected, and the division into 24 equal parts effected, etc. Again, the circumference can be divided into 4 equal parts by drawing two diameters at right angles to each other (see Fig. 95). These arcs can be bisected as indicated above, and the division into 8 equal parts effected. Bisecting the latter arcs, we have 16 equal parts, etc. There is also a way to divide the circumference into 10 equal parts, but it is too difficult to be given here. For all regular polygons except those of 3, 6, 12, 24, etc., and 4, 8, 16, etc., sides, the pupil, at this stage of his progress, is expected to eflfect the division by trial. EXERCISES. . 1. By drawing diagonals from any one angle, into how many tri- angles can a pentagon* be divided ? Show it with a figure. Into how many an octagon ? A dodecagon ? A nonagon ? A hexagon ? 2. What is the sum of the angles of a hexagon ? Determine the number mentally, and then measure the angles geometrically, as in the sohition of (120), observing that the latter result verifies the former. In like manner determine the sum of the angles of a pen- tagon. Of an octagon. Of a decagon. Of a nonagon. Of a tri- angle. Of a quadrilateral. 3. If the angles of a hexagon are equal each to each — that is, if the hexagon is equiangular — what is the value of any one angle ? Ans. IJ right angles. [Note. — A regular polygon is equiangular.] 4. What is the value of any angle of a regular octagon? Of a regular pentagon ? Of a regular dodecagon ? Anstuer to the last, If right angles. 5. Construct a regular dodecagon. 6. Construct a regular heptagon. Sdg's.— Observing that as the chord for the hexagon is the radius, and hence the chord for the heptagon is a little less, we can readily find by ^rm^ just how wide to open the dividers so that they shall step around the circumference at 7 steps. This is not a veiy scientific Avay of constructing a figure, it is true, but it is the only way we can get the chord in this case. f7o. 99. * Polygons are not to be assumed regular unless they are so designated. 58 ELEMENTARY GEOMETRY. 7. Construct a regular octagon. SuG.-See the general solution {121). Fia. 101. 8. Construct a regular nonagon. Solution.— First get a quarter of the cir- cumference by marking the points where two diameters at right angles to each other would cut the circumference. AX is an arc of 90°. Then from A take AY = 60° by using radius as a chord. YX is therefore an arc of 30°. Divide this into three equal parts by trial. Measure YB equal to two- thirds of YX, and AB and BC are arcs of 40°, and the chords AB and BC are chords of the regular nonagon. 9. To draw a five-point star. Solution.— Draw a circle, and dividing the circumference into five equal parts, join the alternate points of division, as in the figure. 10. To circumscribe a square about a circle {56)- Also an equi- lateral triangle, and a regular hexagon. SYNOPSIS OF PLANE FIGURES. 59 SYNOPSIS OF I^LANE FIGURES. ^ What? O < Oh 03 o o ^^ o 0^ Q w Q P 1^ O pq What? Triakgles. Sides. Perimeter. What ? Altitude. ' Scalene. Isosceles. Equilateral. Diagonal. Base. ^3 ^ r Acute. I t Right. Quadrilat- erals. Pentagon. Hexagon. Heptagon. Octagon. Nonagon, etc. What? Trapezium. Trapezoid. Parallelo- J gram. ' f Rhombus. Rhomboid. Rectan- i With unequal \ sides. T \ sides, t g^l^^^- 1 Square. Regular. What? Circle. What? Circumference. Centre. Radius, Diameter. r Ellipse. CoKic Sections.* \ Parabola. ( Hyperbola. Higher Plane Curves.* * These are inserted simply to give completeness. Of course, the student ia not expected to know more than their names. PART II. THE FUIN^DAMENTAL PEOPOSITIONS OF ELEMENT- AEY GEOMETRY, DEMONSTRATED, ILLUS- TRATED, AND APPLIED. CHAPTER I. FLAXE GEOMETRY. SECTION L OF PERPENDICULAR STRAIGHT LINES. PROPOSITION L 122. OTJieoreni. — At any point in a straight line, one perpen- dicular can he erected to the line, and only one, luhich shall lie on the same side of the line. Dem. — Let AB* represent any line, and P be any point therein ; then, on the same side of AB there can be one and only one perpendic- ular erected at P. For from P draw any ob- lique line, as PC, forming with AB tlie two angles CPB and CPA. Now, while the ex- Q tremity P, of PC, remains at P, conceive the line PC to revolve so as to increase the less of Fig. 102. the two angles, as CPB, and decrease the greater, as CPA. It is evident that for a certain position of CP, as C'P, these c /c / / / -.. A r * In class recitation the pnpil shoiiM jjo to the blackboard, after having had hip proposition assigned hira, and first draw the figure required for the demonstration. This ehould be done neatly, accurately, with dispatch, and ivifhout any aids. The figure being complete, he stands at the board, pointer in hand, enunciates the proposition, and then gives the demon- Btration as it is in the text, pointing to the several parts of the figure as they are referred to. OF PERPENDICULAR LINES. 61 angles will become eqnPCA. Now, while the extremity P of this line remains fixed, conceive the line to revolve so as to make the greater angle PCB decrease, and the less angle PC A increase. At some position of the revolving line, as PD, the two angles which it makes with the line AB will become equal. When these adjacent angles are equal, the line, as PD, is per- pendicular to AB {20 f 4:3). Moreover, there is only one position of the line in which these angles are equal ; hence, only one perpendicular can ])e drawn from a given point to a given line. q. e. d. D E Fig. 105. C B Ps Fig. 106. PROPOSITION IT. 128, Theorein.—From a point with- out a straight line, a perpendicular is the shortest distance to the line. Dem. — Let AB be any straight line, P any point without it, PD a perpendicular, and PC any oblique line ; then is PD < PC. Produce PD, making DP' = PD, and draw P'C. Then let the portion of the plane of the lines above AB be revolved upon AB as an axis, until it coincides with the portion below AB. Since PP' and AB intersect at right angles, PD will fall in DP' {126) ; and, since PD = DP', P will fall at P', and PC = P'C, since they coincide when applied. Finally, PP' being a straight line, is shorter than OF PERPENDICULAR LINES. 63 PCP', which is a broken line, since a straight line is the shortest distance be- tween two points. Hence PD, the half of PP', is less than PC, the half of the broken lin^ PCP'. q. e. d. PROPOSITION V. 129. Uieorem. — If a perpendicular he erected at the middle point of a straight line, 1st. Any pfoint in the perpendicular is equally distant from the extremities of the line. 2(1. Any point without the perpendicular is nearer the extremity of the line on its own side of the perpendicular. Dem. — 1st. Let PD be a perpendicular to AB at its middle point D. Then, O being any point in the perpendicular, OA = OB. For, revolve the figure OBD upon OD as an axis until it falls in the plane on the other side of PD. Since ODB and ODA are right angles, DB will fall in DA {126) ; and, since DB := DA, B will fall at A. Hence, OA and OB coincide, and OA = OB. 2d. O' being any point without the perpendicular on the same side as B, 0'B<0'A. For, drawing O'A and O'B, let O be the point at which O'A cuts the perpendicular. Draw OB. Now O'B < BO + 00', since O'B is a straight and O'OB is a broken line. But, as OA=OB, we may substitute it in the inequality, and have O'B <0A + 00', which sum = O'A. 130. Cor. — // each of two points in one line is equally distant from the extremities of another line, the former line is perpendicular to the latter at its middle point. 'Dem.— Every point equally distant from the extremities of a straight line lies in a perpendicular to that line at its middle point, by the proposition. But, two points determine the position of a straight line. Hence, two points, each equally distant from the extremities of a straight line, determine the position of the perpendicular at the middle point of the line. A EXERCISES. 1. JProh. — To erect a perpendicular to a given line at a given voint in the line. 64: ELEMENTARY PLANE GEOMETRY. A C Fig. 108. Solution. — [The process is given in {44), and should be repeated here ex- actly as given there, with the reasons for the solution, as follows.] A is one point in the line OA, which is equally distant from B and C, by construction, and is another. Hence OA is perpendiculai* to BC at A, by [130). 2. I* rob. — To bisect a given line. Fig. 109. Solution. — [For the process see {39). The student should first do it as he did then. The reason why this process bisects AB is as follows.] Since m is one point equally distant from the ex- tremities A and B, and n another, there are two points in mn each equally distant from the ex- tremities of AB. Hence mn \s, perpendicular to AB at its middle point 0, by {130). [The reason for the process in Fig. 20 is the same. Let the student give this method, and show how the cor- ollary {130) applies.] XT Fig. 110. Cx 3. I^roh^ — From a point witliout a given line, to let fall a perpendicular upon the line. Solution. — [Repeat the process as in (45), and give the reason for it as follows.] O is one point equally distant from B and C, and D is another. Hence a line drawn from to D is perpendicular to BC by {130). 4. Wishing to erect a line perpendicular to AB at its centre, I take a cord or chain somewhat P' ft^ P Fig. 111. longer than AB, and, fastening its ends at A and B, take hold of the middle of the cord or chain and carry it as far from AB as I can, first on one side and then on the other, sticking pins at the most remote points, as at P and P'. These points determine the perpendicular sought. What is the principle involved? 5. Two boys are skating together on the ice, and both start from OF OBLIQUE LINES. 65 the same point at the same time, one skating directly to the shore and the other obliquely. They both reach the shore at the same time. Which skates the faster ? What principle is involved ? 6. Several persons start at different times from the same point in a straight road that runs along a wood, and each travels directly away from the road. Will they come out at the same, or at different points on the opposite side of the wood ? What principle is involved ? What is the geometrical language for the colloquial phrase " Directly away from the road" ? 7. If I go from A to B, Fig, 111, by first passing over AP, will I gain or lose in distance by going on a little farther in the direction of AP before I turn and go straight to B ? What principle is in- volved ? Would I gain or lose by stopping short of P on the line AP? Why? SECTION II. OF OBLIQUE STRAIGHT LINES. PROPOSITION I. 131, Theore^n, — When an oblique line meets another straight line forming tzuo adjacent angles, the sum of these angles is two right angles. Dem. — Let the oblique line CD meet the straight , line AB forming the two adjacent angles CDB and CDA ; then CDB + CDA equals two right angles. For suppose CD to revolve toward the position of the perpendicular CD ; the angle CDB will increase at the same rate that CDA diminishes; hence their „ sum will remain constant {i. e., the same). But, when CD becomes perpendicular, the sum of the adjacent angles formed with AB is two right angles by definitions {26, 43). Therefore CDB + CDA = two right angles. Q. e. d. 132. Cor. — The sum of all the consecutive angles formed hy any number of lines meeting a given line on the same side and at a given point is ttvo right angles. 5 66 ELEMENTARY PLANE GEOMETRY. D Fig. 113. Dem.— Thus ADC" + C^DC'" + C'^DC" + C"'DC" + CDC + CDC + CDB = ADC + CDB, which sum is two right angles by the proposition. Or, in general tenns, the angles thus formed can always be united into two groups, constituting respectively the two adjacent angles formed by one line meeting another. 133, Def. — Two angles whose sum is two right angles, are called StippUmenial Angles. Hence, the Supplement of an angle is what remains after subtracting it from two right angles. PROPOSITION n. 1.34:. Tlieorem, — WJien any tivo straight lines intersect, the opposite or vertical angles are equal to each other, and the sum of thv four angles formed is four right angles. Dem.— Let AB and CE intersect at D ; then CDA = the opposite angle BDE, ADE = the opposite or vertical angle CDB, and ADC + CDB + BDE + EDA rl four right angles. For, since CD meets AB, ADC is the supplement of CDB {131, 138). Also, since BD meets CE, BDE is the supplement of CDB. Hence ADC — BDE. In a sim- ilar manner ADE can be proved equal to CDB. [The student should give the proof.] Again, since ADC + CDB = two right angles, and BDE + EDA = two right angles, by adding the cor- responding members together, we have ADC + CDB 4- BDE + EDA = four right angles. Fig. 114. 135. Cor.— The sion of all the consecutive angles formed hy any number of lines meeting at a common point is four right angles. Dem.— The truth of this corollary is rendered apparent by drawing a line through the common vertex, and observing that the sum of all the angles on each side thereof is two right angles; whence the sum of all the angles on both sides, which is the same as the sum of all the consecutive angles foi-med by the Une, is four right angles. [Let the student put letters on the figme,and demonstrate by means of it] Fig. 115. OF OBLIQUE LINES. 67 PROPOSITION III. 136. Theorem, — If two supplemental angles are so situated as to le adjacent to each other, the two sides not common tuillfaU in the same straight line, Dem. — Let the sum of the two angles BOA and CO'D be two right angles. Prolong CO', forming the angle DO'E. Then is DO'E supplemental to CO'D {131, 133) f and hence equal to BOA, which is supplemental to CO'D by hypothesis. Now, if AOB be placed adjacent to CO'D, the vertex being at 0', and the side OA falling in O'D, OB will fall in O'E, since BOA = DO'E. Hence, when the angles are so situated, OB becomes the prolonga- tion of CO'. Q. E. D. PROPOSITION IV. 137. TJieorem. — If from a point withotit a straight line a per- pendicular he draivn, oblique lines from the same point cutting the line at equal distances from the foot of the perpendicular are equal to each other ; the angles which they form iDxth the perpendicular are equal to each other ; and the angles lohich they form with the line are equal to each other, Dem. — Let AB be any straight line, P any point without it, PD a perpen- dicular, and PC and PE oblique lines cutting AB at C and E, so that DC=DE ; then PC = PE, angle CPD = angle DPE, and angle PCD = angle PED. Revolve the figure PDE upon PD as an axis, until it falls in the plane on the other sic^e of PD. Since AB is perpehtlicular to PD, DB will fall in DA; and, since DE = DC, E will fall at C. Now, as P remains stationary, the triangles PDE and PDC coincide. Hence, PC = PE, angle CPD = angle DPE, and Angle PCD = angle PED. q. e. d. Fig. 117. Query. -How does the equality of PE and PC follow from {129) ? PROPOSITION V. 138. Tlieorem.— If from a point without a line a perpendicu- Jar he draion to the line, and also from the same point two ohlique 68 ELEMENTARY PLANE GEOMETRY. lines making equal angles with the perpendicular, the oblique lines are equal to each other, cut the line at equal distances from the foot DC. Fig. 119. Then is PF > PC. Produce PD, making DP' = PD, and draw FT and P'C, producing the latter until it meets PF in H. Revolve the figure FPD upon AB as an axis, until it falls in the plane on the opposite side of AB. Since PP' is perpendicular to AB, PD will fall in DP' ; and, since PD = DP', P will fall at P'. Then P'C = PC and P'F = PF. Now the broken line PCP' < than the broken line PHP', since the straight line PC < the broken line PHC. For a like reason the broken line PHP' arc CwD, chord AB > chord CD. But both of these conclusions are contrary to the hypothesis. Hence, as arc AmB can neither be equal to not greater than arc CtiD, it must be less. PROPOSITION VI. 167* TJieorem, — In the same or in equal circles, of tivo unequal cliords, tlie less is at the greater distance from the centre. Dem. — Let CE < AB, then is the perpendicular OD, which measures the distance of CE from the centre, greater than OD' which measures the distance of AB from the centre. From A lay off AE' =: CE, and draw the perpen- dicular OD". Then OD" = OD, since equal chords are equally distant from the centre. As arc AE' < arc AB, AB cuts OD" in some point as H. NoV OH > OD' since the former is oblique and the latter per- pendicular to AB. Also OD" > OH. Much more then is OD" > OD'. Therefore OD (which equals OD") > OD'. Q. E. D. Fig. 136. 168, Cor.— Conversely, Of tiuo chords which are unequalhj distant from the centre, that which is at the greater distance is the Dem. — Thus, if CE is at a greater distance from the centre than AB, CE < AB. For, if CE were equal to AB, it would be equally distant from the centre. And if CE were greater than AB, it would be at a less distance from the centre. Hence, as CE cannot be at an equal distance from the centre with AB, nor at a less distance, it miist be at a greater. PROPOSITION vn. 169. Theorem.^A straight line can intersect a circumference in only two points. Dem.— The distances from the centre to the intersections, being radii, are equal. Hence, as there can be only two equal straight linos drawn from ai point to a straight line, there can be only two intersections. Q. e. d. 82 ELEMENTARY PLANE GEOMETRY. PROPOSITION vin. 170, Tlieorem, — A straight line which intersects a circumfer- e7ice in one j^oint intersects it also in a secojid 2)oint. Dem. — Let LM intersect the circumference at A ; then does it intersect at some other point, as B. For, since LM intersects the circumference, it passes within it, and has points nearer to than A. The radius OA is, therefore, an oblique line. Now two equal oblique lines can be drawn from to the straight line LM. But all points in the plane at the distance OA from 0, are in the circumference. Hence there is a second point, as B, common to LM and the circumference, q. e. d. Fig. ]3T. 171* Cor. — Any line which is oUique to a radius at its extremity, is a seca7it line. PROPOSITION IX. 172, Theorefii, — A straight line tvhich is perpendicular to a radius at its extremity is tangent to the circumference, Dem. — The line touches the circumference because the extremity of the radius is in the circumference. Moreover, it does not intersect the circum- ference, since, if it did, it would have points nearer the centre than the extremity of the radius; but these it cannot have, as the perpendicular is the shortest distance from a point to a line. Hence, as a line which is perpendicular to a radius at its extremity touches the circumference but does not intersect it, it is a tangent {53). Q. e. d. 173. Cor. — ConTersely, A tangent to a circumference is perpen- dicular to a radius at the poinl of contact. For, as a tangent to a circumference does not pass within, the point of contact is the nearest point to the centre, and hence i* the foot of a perpendicular from the centre. STRAIGHT LINES AND CIRCUMFERENCES. 83 PROPOSITION X. 174, Theorem, — Two parallel secants intercept equal arcs. Dem.— Let the parallels LM and RS intersect the circumference AECF ; tlien Hre the intercepted arcs AB and DC equal. Draw the diameter EF perpendicular to one of the parallels, as LM, whence it will be per- pendicular to the other {144). Draw the radii OB and OD. Revolve the portion of the figure on the right of EF, upon EF until it falls on the plane on the left of EF. Then, since RS and LM are perpendicular to EF, IS will fall in IR, and HM in HL. Moreover, as there cannot be two equal oblique lines on the same side of a perpendicular, and from the same point {140), OD and OB must coincide, and D fall at B. In like manner C falls at A, and CD coincides with AB. Therefore CD := AB. q. e. d. (V PROPOSITION XI. 17S, Theorein* — If a secant le parallel to a tangent^ tlie arcs intercepted hetioeen the intersections and the point of tangency are equal, Dem. — Let the secant LM be parallel to p the tangent RS ; then is CP - EP. For, draw the radius OP to the point of tangency ; it will be perpendicular to the tangent {173), and also to the parallel LM {144). But a radius which is perpen- dicular to a chord, as OP to CE, bisects the subtended arc {159), hence CP — EP. In like manner, if VU is parallel to LM, CB = EB. Q. E. D. ^ ^ ^ ^ Fig. 139. 176* Cor. — Tioo parallel tangents inchide eq^ial arcs hetween the points of tangency ; and these arcs are semi-circumferences» R /-^ ~"~\ s L 0/ D \e m 64 ELEMENTARY PLANE GEOMETRY. EXERCISES. 1. Draw a circle ana divide it into two equal parts. What proposi- tion is involved ? 2. Given a point in a circumference, to find where a semi-circum- ference reckoned from this point terminates. What proposition is involved ? 3. JProb, — To bisect a given arc. OL B "^ ^'--.^ A Solution. — Let AB be an arc which wo. wish to "\ / I bisect.* Draw its cliord AB, and erect 00' bisecting the chord, by {130). Now, as 00' is perpendicular to the chord at its middle point, it bisects the arc by {162), since there can be but one perpendicular at the middle point of the chord. The arc AB is, therefore, Fig. 140. bisected at 0, i. e., AC = CB. 4. Proh, — To bisect a given ajigh. SuG. — The method of solving this is given in Part I. The student should do it as there directed, and then point out the principle upon which the method depends. 5. In a circle whose radius is 11 there are drawn two chords, one at 6 from the centre, and one at 4. Which chord is the greater ? By what proposition ? 6. In a certain circle there are two chords, each 15 inches in length. What are their relative distances from the centre ? Quote the principle. 7. There is a circular plat of ground whose diameter is 20 rods. A straight path in passing runs within ^ rods of the centre. What is the position of the path with reference to the plat ? What is the position of a straiglit path whose nearest hy^^ point is 10 rods from the centre ? One .'-''' ^^ X whose nearest point is 11 rods from the / y^ yp centre? y^\ X 8. Pass a line through a given point, A. \ y^ and parallel to a given line, by the prin- y^P ciples contained in (174) and {165). Fig. 141. * Thi? solntion and many others are given, not so mnch that it is feared that the student will not be able to «olve the problem?, as to afford models for describing the process. In this case an arc should be drawn first, and all trace of the centre obliterated. Then proceed aa directed. STRAIGHT LINES AND CIRCUMFERENCES. 85 9. Prob, — To draw a tangent to a circle at a given point in the circumference. Solution. — Let P be the point at which a tan- gent is to be drawn. Draw the radius OP to the given point of tangency, and produce it any con- venient distance beyond the circle. Erect a per- pendicular to this line at P, as MT ; then is MT a tangent to the circle {172). Fig. 141*. 10. I?r6h. — To find the centre of a circle whose circumference is hnoiun, or of any arc of it, SuG. — The pi'ocess is given in Pakt I, Do the work as there directed, and then show upon what proposition in this section it is founded. w . O CO < o o ^ m ^ S ^ %^ 03 o Oh > H W SYNOPSIS. Diameters. Prop. I. How divide circles and circumferences. Chords. Secants. Tangents. Parallels. Exercises. r Prop. II. Prop. III. Prop. IV. Prop. V. Prop. VI. Bisects angle. Converse of Prop. r Cor. 1. Radius perp. J Cor. 2. to chord. I Cor. 3. " *' " [ Cor. 4. Dist. from centre. Distance of equal chords from centre. Equal arcs, and converse. Unequal arcs. Unequal chords. Dis- ^ ^ Converse, tance from centre. '^ Prop. VII. Intersect in only two points. Prop. VIII. If a lin^e intersect m one ^ ^^^ ^ine oblique ^rlnanoTh^r'^J to radius at extr. Prop. IX. Line perpendicular to ) ^ Converse, radius at extremity. ) Prop. X. Prop. XI. Secant par. to tangent. Parallel secants intercept equal arcs. Cor. Two parallel tangents. r Proh. To bisect an arc. ' Proh. To bisect an angle. Prob. To draw a tangent at a point in circumference. Proh. To find centre of circumference or arc. 86 ELEMENTABY PLANE GEOMETRY. SECTION V, OP THE RELATIVE POSITIONS OF CIRCUMFERENCES. PROPOSITION L 177, Theorem, — All the circumferences wTiich may le passed tlirougli three ])oints not in the same straight line coi^icide, and are one and the same. Dem. — Let A, B, and C be three points not in the same straight line ; then all the circumferences which can be passed through them will coincide. For join the points, two and two, by straight lines, as AB and BO. Bisect these lines with perpendiculars, as DF and EH. Since ..■ p."""--., AB and BC are not in the same straight line, DF and / \ Q EH will meet when sufficiently produced, at one and • /■-. only one point, as O, because they are straight lines I -f / j Now, every point in FD is equally distant from A and ^^ / 7^/ B, and eveiy pomt in HE is equally distant from B and V"---:.._ / / c {129). Hence is equally distant from the three X^ ,.--'' B points A, B, and C; and, if a circumference be drawn '" with as a centre, and a radius AG, it will pass through the three points. Moreover, every circumference pass- ing through these points must have O for its centre, since the centre must be in FD (otherwise it would be unequally distant from A and B), and also in HE {129). But these lines intereect only in O. Also, every circumference with as its centre, and passing through A, must have AG for its radius. Hence, as all circles having the same centre and the same radius coincide, all those passing through three poiats. A, B, and C, coincide. Q. e. d. 178, Cor. 1. — Tlirough any three points not in the same straight line a circumference can le passed. 179, Cor. 2. — T7iree poi7its not in the same straight line determine n circumference as to position and extent ; i. e., in all respects. ISO, Cor. 3. — Tiuo circumferences can intersect in 07ily tioo points. For, if they have three points common, they coincide, and form one and the same circumference. RELATIVE POSITIONS OF CIRCUMFERENCES. 87 PROPOSITION II. 181, TlieQvem. — Two circumferen- ces ivMch intersect in one point, intersect also in a second point. Dem. — Let M intersect N at P. As M passes from without to witliio the circle N, it has points both without and within. Now, for M to re- turn into itself from any point within N, as Y, to any point without, as X, it must intersect N ; but it cannot intersect in P, for a circumference dDcs not intersect itself Hence, it intersects in I second point, as P'. q. e. d. Fig. 143. PROPOSITION in. 182. TJieorem, — If a straight line he drawn through the cen- tires of two circles, of the intersections of either circumference luith tlat line, the one on the side toward the centre of the other circle is tie nearest jmnt in this circumference to that centre, and the one on tie opposite side is the farthest point from that centre. Dem. — Let M and N, or M' and N', be two circumferences whose centres are C and O'. Draw an indefinite line through these centres. Let A and H be the intersections of M or M' with this line, of which A is on the side of M or M' toward the centre O', and H is on the opposite side. Then is A the nearest point in M or M' to 0', and H the farthest point from 0'. Mrst, To show that A is nearer 0' than any other point in the circumference. A will lie between and 0', in 0', or beyond 0'. When A lies between O and 0', as in M, let P be any other point in M, and draw OP and O'P. Now 00' being a straight line, is less than OPO', a broken line. Subtracting OA from the former, and its equal OP from the latter, we have AO' < PO'. When A falls at 0' the truth is self-evident. When A lies beyond O', as in M', let P be any other point in M', and draw OP and O'P. Now O'P + 00' > OP (= OA). Subtracting 00' from both, we have O'P > OA — 00' (= O'A). Hence, in any case, A is the nearest point in M or M' to 0'. Second, To show that H is the farthest point in M or M' from 0'. In either Fig. 144. 88 ELEMENTARY PLANE GEOMETRY. figure, let P' be any other point in the circumference than H, and draw OP and O'P'. >'ow, PO + 00' > P'O'. But P'O = HO. Hence HO + 00' (= HO') > P'O' PROPOSITION IT. 183. Theorem. — When tlie distance between the centres of two circles is greater than the sum of their radii, the circumferences are wholly exterior the one to the other. Dem. — Let M and N be the circumferences of two circles whose centres ar« O and O'. Let 00' be greater than the sum of the radii. Then are M and N wholly exterior the one to the other. / For A, the intersection of M witi 00', is between O and 0', since OA <" 00', Now, by hj-pothesis, 00' > OA + BO'. Subtracting OA from both, \ge have AO* > BO'. Hence, as the neareJt point in M is farther from O' than the circumference of the latter circle, M lies wholly exterior to N. q. e. d. Fig. 145. 184, Cor. — Conversely, W7ie7i two circumferences are exterior th one to the other, the distance between their centres is greater than tliB sum of their radii. Dem. — For, join the centres 00' with a straight line. Now tlie point A where this line cuts the circumference M is the nearest point in this circumfer- ence to the centre 0'. But, by hypotliesis, this (and every other point in cir- cumference O) is without circle O'. Hence, AO' > BO'. To each add OA, and OA + AO' (or 00') > OA + BO'. PROPOSITION T. 185. Tlieoreni, — When the distance hetiueen the centres of two circles is equal to the sum of their radii, the circumferences are tan- gent to each other externally. Dem. — Let M and N be two circumferences, and 00', the distance between their centres, be equal to 00 + O'C, the sum of their radii; then are the cir- cumferences tangent to each otlier externally. RELATIVE POSITIONS OF CIRCUMFERENCES. 89 The point A, where M cuts the line join- ing the centres, is between and 0', since OA < 00' by hypothesis. Moreover, A is the nearest point in M to the centre 0'. Again, as 00' = 00 + O'C, subtracting OA from the first member, and its equal 00 from the other, we have O'A = O'C; that is, A is in the circumference N. Hence, as A lies in N, and all other points in M are more distant from 0' than the length of the radius O'C, M is entirely without N, except the point A, and the circles are other externally, q. e. d. to each ISO, Cor. 1. — Conversely, Wlien two circumferences are tangent to each other externally, the distance betiveen their centres is equal to the sum of their radii. Dem. — M being tangent to N externally, the point in M nearest the centre O' must be in N, while all other points in M are exterior to N. Now, the point in M nearest to 0' is A on the line joining their centres {182). A is therefore the point of tangency, and 00' = OA + O'A. 187 » Cor. 2. — When two circumferences are ta7igent to each other externally, the point of tangency is in the line joining their centres. PROPOSITION VI. L- 188, Uieorem, — WJie^i the distance hetiueen the centres of two circles is less than the sum and greater than the difference of their radii, the tivo circumferences intersect. Dem. — Let M and N be the circumferences of two circles whose centres are O and 0'. Let the radius of M be equal to or greater than ^ — --\ N the radius of N, Now, if 00' <0A + O'B, and > OA— O'B, M and N intersect. For, when 00' > OA, 00' < OA + O'B gives 00' — OA (= AO') < O'B ; and when Fig. 147. 00' < OA, 00' > OA - O'B gives O'B > OA - 00' (= AO')- Hence the nearest point in M to 0' lies within N. Again, to the first member of 00' > OA — O'B add HO, and to the second its equal OA, and we have 00' -I- HO (=: HO') > 20A - O'B. Now, since O'B ^ OA,* by hypothesis, the difference 20A - O'B ^ O'B. Hence, HO' > O'B, and H lies without N. As, ♦ Read " O'B is equal to or less tliau 0A-' 90 ELEMENTARY PLANE GEOMETRY. therefore, M has oue pomt at least within N and one without, M and N inter- sect. Q. E. D. 189, CoE. — Conversely, WJien tiuo circumferences intersect, the distance heticeen their centres is less than the sum and greater than the difference of their radii. Dem. — Let the radius of N be equal to or less than the radius of M. As the circumferences intersect the farthest point H' of N from must be farther from O than the length of the radius of M, i. e., must lie without that circle. So we have by hypothesis H'O > OA. Subtracting H'O' from the first menil)er and its equal BO' from the second, we have H'O — O'H' (= 00') > OA - BO'; that is, the distance between the centres is greater than the difi'erence of the radii. Again, as the nearest point in M to O' must lie within N, we have AO' < BO', and adding OA to both members, OA + AO' (= 00') < OA + BO'; that is, the distance between the centres is less than the sura of the radii. PROPOSITION TIL 190, Theorem. — When the distance hetween the centres of two unequal circles is equal to the difference of their radii, the less cir- cumference is tangent to the other internally. Dem.— Let M and N be the circumferences of two circles whose centres and O' are so situated that 00' = 00 - O'C ; then are the circles tangent to each other internally. For, let N be the circumference of the less circle, so that 00 > O'C Let HH' be a diameter of M. Bj hypothesis 00' = 00 — O'C. Now, subtracting each member of this eq tj,lity from OH', we have OH' — 00' (= O'H') = O'C. Whence it appears that H', the point in N at the greatest distance from O, is in M ; and, con- FiG. 148. sequently, that every other point in N is within M. Hence, N is tangent to M internally. Q. e. d. 191. Cor. 1. — Conversely, Wlien a less circumference is tangeiit to a greater internally, the distance hetween their centres equals the difference of their radii. Dem. — The less circumference N being tangent to the greater M, internally, the point in N at tlie greatest distance from the centre of M, must be in M, while all other points of N lie within M. Now H' in the line passing through the centres is the point of N at the greatest distance from O. Hence we ob- serve that 00' - OH' - O'H', i. e., the difi'erence between the radii. RELATIVE POSITIONS OF CIRCUMFERENCES. 91 192. Cor. 2. — When one circumference is ta?ige?it to another in- iernally^'the i^oint of iangency is in the line passi?ig through their centres. 193, ScH. — If the radii are equal the two circumferences coincide. PROPOSITIOX Tin, 194. Tlieore'in, — When the distance ietiueen the centres of two unequal circles is less thari the difference of their radii, the less cir- cumference is wholly within the greater. Dem. — Let N be a less circumference than M, and 00', the distance between their centres, be less than OA — O'H', the difference of their radii ; then is N wholly within M. For, to each member of 00' < OA — O'H' add O'H', and wehaveOO' + 0'H' i-i E-t ints. •< Cor. 2. ( Cor. 3. A circf. can be passed. A circf. determined by. Intersections of two circf 's. Prop. II. Two circumferences which intersect in one point. Prop. III. Points in one circumference nearest to and farthest from tlie centre of another. > ' Prop. IV. Greater than sum of radii. ■{ Cor. Converse. Prop. V. Equal to sum of radii. \ ^^'' l ^^P^.^^f ; ^ { Cor. 2. Pomt of tangency. Prop. VI. Less than sum and greater than ( n r^ difference of radii. J Cbr. Converse. Prop. VII. Equal to diff. I Cor. of radii. < Cor. (Sch. Cor. 1. Converse. 2. Point of tangency. Radii equal. Prop. VIII. Less than diff. of radii. Cor. Converse. Sch. Concentric, Coincident Prop. IX. Perpendicular to common chord. Prop. X. Common tangent to two circles tangent to each \ q^^ rj,^ ^^i Exercises -i -^'^^^' '^^ ^^^^ circumference through three points. I Prob. To circumscribe a triangle with a circumference. 94. ELEMENTABY PLANE GEOMETRY. SECTION VL OF THE MEASUREMENT OF ANGLES. 200» Angles are said to be measured by arcs, according to the principles developed in the three following propositions. PROPOSITION I. 201., Theorem, — In the same or in equal circles, equal arcs subtend equal angles at the centre. -In the equal circles let arc AB = arc DC; then will the angles O and 0', called angles at the centre, be equal. For, placing N upon M so that C shall fall on 0, and CD on OA, since the circles are equal, D will fall on A ; and since, by hypothesis, arc DC = arc AB, C will fall on B. Hence, O'C will coincide with OB, and angle O' = angle O, because they coincide when applied, q. e. d. 202, Cor. 1. — Conversely, In the same or in equal circles, eqtial angles at the centre intercept equal arcs. Dem. — If, by hypothesis, angle O' = angle O, in the equal circles M and N, arc DC = arc AB. For, placing circle N upon M, so that O' shall foil on O, and O'D on its equal OA, D falls on A, and, since angle O' = angle 0. O'C takes the direction OB, and, being equal to it, C falls on B. Hence, DC and AB co- incide and are equal. 203, Cor. 2. — A right angle at the centre intercepts a quarter of a circumference, and is said to he r)ieasured hy it. Hence, a semi- circumference is the measure of tioo right angles, and a whole circum,- ference of four. MEASUREMENT OF ANGLES. 95 r PROPOSITION n. Fig. 153. 204:. Theorein, — In the same or in equal circles, arcs wJiich are in the ratio of tiuo luhole 7iumhers subtend angles at the centre ivhich have the same ratio, whe?ice the angles are to each other as the arcs which subtend them. Dem. — In tlie equal circles M and N, let the arcs EF and IH, which subtend the angles and O' at the centre, be in the ratio of 5 to 8 ; then are the angles O and 0' in the ratio of 5 to 8, and we have angle : angle O' :: arc EF : arc I H . For, divide EF into 5 equal parts, as E«, ab, etc., then IH can be divided into 8 such parts, le, ef, etc. Draw the radii Oa, Ob, Oc, etc., and O'e, O'/, O'g, etc. ; and, since these partial arcs are equal, the partial angles which they subtend are equal, by the preceding proposition. Now, O is composed of 5 of these angles, and 0' of 8; whence angle : angle O' : : 5 : 8. But, a?-c EF : arc IH :: 5 : 8. Hence, the two ratios being equal, we have angle 0' : angle : : arc IH : arc EF. As the same method could be pursued in case the arcs were to each other as any other two whole numbers, the argument is general. 206, Cor. — Conversely, In the same or in equal circles, angles at the centre ivhich are in the ratio of tivo tvhole numbers are to each other as their intercepted arcs. Dem.— Thus, let angle O'^be to angle O in the ratio of 8 to 5. Conceive O' divided into 8 equal partial angles, then will be divisible into 5 such partial angles. Now, the partial angles being equal, their intercepted arcs are equal, by the preceding proposition, Cor. 1. Whence, arc IH : arc EF But, angle O' : angle Hence, arc IH : arc EF And the same method could be pursued with angles having the ratio of any other whole numbers. : 8 : 5. : 8 : 5, by hypothesis. : angle 0' *: angle 0. 96 ELEMENTARY PLANE GEOMETRY. PROPOsmox m. 206. TJieoreni, — In the same circle or in equal circles, tvw hi- conimensurahle arcs are ^ to each other as the angles which they suUend at the centre. Fig. 1S4. Dem. — In the equal cir- cles M and N, let EF and IH be incommeusurable arcs. Now there is some arc to which EF beai-s the same ratio as angle to angle O'. If that arc is not IH let it be IL, an arc less than IH, so that angle : angle (y : : arc EF : arc IL * Conceive EF divided into equal parts, each of which is less than LH,f the as^ sumed difference between IH and IL. Then conceive one of these equal parts to be applied to I H as a measure, beginning at I. Since the measure is less than LH, at least one point of division must fall between L and H. Suppose K to be such a point. Draw O'K. Now, the arcs EF and IK are commensurable, and by the last proposition angle : angle lO'K : : arc EF : arc IK. But^we assumed that ■ angle O : angle lO'H : : arc EF : arc IL. In these proportions the antecedents being alike, the consequents should be pro- portional, so that angle lO'K should be to angle lO'H : : arc IK : arc IL. But this proportion is false, since angle lO'K < angle lO'H, whereas arc IK > arc IL. ' In a manner altogether similar (the student should supply it) we can show that angle is not to angle O' : : arc EF : any arc greater than IH. Hence, as the fourth term of the proportion cannot be less or greater than IH, it must be IH itself; and angle : angle O' : : arc EF : arc IH, q. e. d. 207, Cor. — Conversely, In the same or in equal circles, two incom- mensurable angles at the centre are to each other as the arcs which they intercept. * This ig a false hypothesis, and the object of the argument following is to show its falsity, + This can be done by snpposing EF bisected, then the halves bisected, then the fonrths bisected, and this process of bisection continued until the parts are each less than LH. MEASUREMENT OF ANGLES. 97 Dem. — In the equal circles M and N, O and O' being incommensurable angles at the centre, are to each other as the arcs EF and IH, If not, let us sup- pose arc EF : arc IH : : angle : angle lO'L, an angle less than 0'. Divide into equal partial angles, each less than LO'H, the assumed differ- ence between lO'H and lO'L. Also conceive this angle to be applied as a measure to lO'H, beginning at O'l. At least one line of division will fall be- tween O'L and O'H. Let O'K be such a line. Now, as and lO'K are com- mensurable, we have by {205), arc EF : arc IK : : a7igle : angle I O'K. But by supposition arc EF : arc IH : : angle : angle I O'L. Therefore, since the antecedents are the same, arc IK should be to arc IH : : angle lO'K : angle lO'L. But this is false, since arc IK < arc IH, whereas angle lO'K > angle lO'L. Whence we learn that the fourth term of the proportion cannot be less than angle lO'H. In a similar manner it can be shown (let the student do it) that it cannot be greater. Hence 4t must be lO'H itself; and arc EF : arc IH : : angle O : angle lO'H. 208. ScH. — Out of the truths developed in the three preceding proposi- tions grows the method of representing angles by degrees, minutes, and seconds, as given in Trigonometry (Part IV., 3-6). It will be observed, that in all cases, if arcs be struck 20it7i the same radius, from the vertices of angles as centres, the angles bear the same ratio to each other as the arcs intercepted by their sides. Hence the arc is said to measure the angle. Though this language is convenient, it is riot quite natural; for we naturally measure a quantity by another of like kind. Thus, distance (length) we measure by distance, as when we say a line is 10 inches long. The line is length ; and its measure, an inch, is length also. So, likewise, we say the area of a field is 4 acres : the quantity measured is a surface ; and the measure, an acre, is a surface also. Yet, not- withstanding the artificiality of the method of measuring angles by arcs,, instead of directly by angles, it is not only convenient but universally used ,- and the student must know just what is meant by it. For example, a circumference is conceived as divided into 360 equal arcs, caU'ed degrees. Hence, as a right angle at the centre is subtended by one-fourth of the circumference, it is called an angle of 90 degrees. 180 degrees is the measure of two right angles, 45 de- grees, of half a right angle, etc. Thus we get a i)erfectly definite idea of the; 7 98 ELEMENTARY PLANE GEOMETRY. magnitude of an angle from the statement of the number of degrees which measure it ; and, for brevity, the angle is spoken of as an angle of the same number of degrees as the intercepted arc. 209^ An Inscribed Angle is an angle whose vertex is in a circumference, and whose sides are chords, or a chord and diameter, of that circumference. PROPOSITION IV. 210. TJieorem. — An inscribed angle is measured by half the arc intercepted betwee^i its sides. Dem. — Mr St, when one side is a diameter. Let APB be an inscribed angle, and PB a diameter; then is APB measured by one-half of arc AB. For, through the centre O, draw the diameter DC parallel to the chord PA; then COB = POD {134), whence arc CB = arc PD {202), also COB = APB {152); and arc PD = arc AC {174), whence PD = CB = iAB. Now COB is measured by CB {208) \ hence APB is measured by Fig. 155. CB-^AB. Q. E. D. Second, when both sides are chords and the centre of ih£ circle lies between them. Let APB be such an angle. Draw the diameter PC. Now, by the preceding part, APC is measured by-^AC, and CPB by iCB. Hence APC + CPB, or APB, is measured by ^AC + iCB, or iAB. Q. E. D. ^ Third, when both sides are cliords and tlie centre lie.i without the angle. Let APB be such an angle. Draw the diameter PC. Now APC is measured by iAC, and BPC by I BC. Hence APC — BPC, or APB, is measured by iAC-^BC, or ^AB. q. e. d. Fig. 157. MEASUREMENT OF ANGLES. 99 211. Cor. — In the same or equal circles all angles inscribed in the same segment , intercept equal arcs, and are consequently equal. If the segment is less than a semicircle, the angles are obtuse ; if a semicircle, right ; if greater than a semicircle, acute, III.— In each separate figure the angles P are equal, for they are each measured by half the same arc. ... In 0, each angle P is acute, being measured by \m, which is less than a quarter of a circumference. ... In 0', each angle P is a right angle, being measured by \w! ^ which is a quadrant (quarter of a circumference). ... In 0", each angle P is obtuse, being measured by ^m", which is greater than a quadrant. SCH. — The converse. of this proposition is usually taken for granted ; i. e., that if the several angles P, P, etc., are equal and subtended by the same chord, their vertices lie in the circumference. This is readily proved rigorously after the next two propositions. Thus, if vertex P were without, the angle would be measured by ^AB — |^ the other intercepted arc ; and if within, by ^AB + 1 the other intercepted arc. PROPOSITION V. 212. Theorem,— Any angle formed by two chords intersecting in a circle is measured by one-half the sum of the arcs intercepted between its sides and the sides of its vertical, or opposite, angle. Dem.— Let the chords AB and CD intersect at P; then is APD, or its equal CPB, measured by •i(AD +CB); and APC, or its equal BPD, is measured by ^(AC + BD). For, through C draw CE parallel to BA ; whence ECD = APD {152\ and CB = EA {174). But ECD is measured by i ED {210\ which equals ^ (AD + EA) = i(AD + CB). That APC, or its equal BPD, is measured by i(AC + BD), appears from the fact that the sum of the four angles about P being equal to four right angles, is measured by a whole circumference {208). But APD + CPB is measured by AD + CB; whence APC + BPD, or 2 APC, is measured by the whole circumference minus (AD + CB) ; that is, by AC -fBD. Then is APC measured by ^ (AC + BD). Fio. 159. 100 ELEMENTARY PLANE GEOMETRY. 213, Sen. — The case of the angle included between two chords passes into that of the inscribed angle in the preceding proposition, by conceiving AB to move parallel to its present position until P arrives at C and BA coincides with CE. The angle APD is all the time measured by half the sum of the in- tercepted arcs; but, when P has reached C, CB becomes 0, and APD becomes an inscribed angle measured by half its intercepted arc. In a similar manner we may pass to the case of an angle at the centre, by supposing P to move toward the centre. All the time APD is measured by i(AD + CB); but, when P reaches the centre, AD = CB, and i (AD + CB) = \ (2AD) = AD ; i. 5). Solution.— Through P to draw a parallel to AB. Fiom P as a centre, with any radius greater than the distance from P to AB, describe an arc cut- . ting AB, asac. From a as a centre, ij\ 9<-= r^f N with the same radius, strike an arc \ through P, intersecting AB, as Ph. Take the chord Pb and apply it ^5^ — ^ from a on the arc ac, as aO. These chords being equal, the arcs Pb and Fig. 163. 102 ELEMENTAKY PLANE GEOMETRY. Fitt. 164. rtO are equal {165). Again, angle Pdb = anglo OPa, since they are measured by equal arcs struck with the same radius (201). These alternate angles being equal, MN is parallel to AB {149). 2. In Fig. 164 there are 4 pairs of equal angles. Which are they, and why ? Show also that COB = ABD + CDB, by (210), and (212). Show also that DOB = ABC + DAB. 3. I^rob, — From a point without a circle to dratv a tangent to the circle. SoLUTiox. — Let be the given circle, and P the given point. Join P with the centre O, and upon PO as a diameter describe a circle. Let T and T' be the intersections of the two cir- cumferences. Now, if lines be drawn from P through T and T', they will be tangent to the cir' cle O. For OTP and OT'P, being inscribed in semi-circles, are right angles {211). Hence, PM is perpendicular to radius OT at its extremity T, and is therefore a tangent {172). In like manner PT' is shown to be a tangent, and we see that from a point without a cir- cle two tangents can he drawn to ifie circle. Fig. 165. 4. I^rob, — On a given Ji?ie, to consti'uct a segment ichich shall contain a given angle. SoLUTiox. — Let AB be the given line, and O the given angle. At one ex- tremity of the given line, as B, construct an angle ABC equal to the given angle O, which shall lie on the oppo- site of AB from that on which the required segment is to lie. Erect a perpendicular to the line CB at B, and also a peq^endicular bisecting AB. Let FB and FE be MEASUREMENT OF ANGLES. 103 these perpendiculars, intersecting at F. From F as a centre, with a radius equal to FB, describe a circle. Then is AHB the segment required. For, CB being perpendicular to radius FB at its extremity, is tangent to the circle, and angle ABC (= angle 0) is measured by ^ of arc AmB {210). Now, any angle - inscribed in the segment km'm"B^ as AHB, has ^ AwB for its measure, and is, consequently, equal to 0. Another Solution. — On the side of AB on w^hich the segment is to lie, draw any line through either extremity of AB, making an acute angle with AB. Let CB be such a line. At any point in CB, as C, draw a line CE, making angle ECB = the given angle 0, Fig. 166. Through A pass a parallel to CE (see Ex. 1), as AD. Pass a circumference through A, D, and B. Any angle inscribed in segment AtwB is equal to O. [Let the student give all the reasons, and make the construction. The requisite marks for the construction are made in the figure. Why is it said, make CBA an acute angle? When would a right angle answer? When an obtuse angle?] Fic. 167. SYNOPSIS. CO o < o H 1^ How angles are measured. r Prop. L Equal arcs subtend equal ( g^^' \ S['°Iure'of 1, 2, and angles at the centre. ] 4 right angles. O Prop. II. Commensurable arcs in the same ratio as their subtended angles. Cor. Converse. Prop. III. Incommensurable arcs. Cor. Converse. iSch. Method of measuring angles. Inscribed angle, what ? Prop. IV. Inscribed angle, how measured. | ^''^'- "^'^g^emSh^le.^' ^'^^ ^ Prop. V. Angle between two chords. ^Scli. Compared with preceding. Prop. VI. Angle between two secants. [ Sch. Compared with Prop. IV. Prop. VII. Angle between tangent and chord. Prop. VIII. Angle between two tangents. \ Sch. Compared with Prop. VI. f Prob. To draw a parallel through a given point. ■p^..^nTci.« 1 P^'o^- To draw a tangent to a circle from a point without j!.XERCisES. < p^.^j rj.^ coustruct a segment on a given line which shall i. contain a given angle. 104 ELEMENTARY PLANE GEOMETRY. SECTION VII. OF THE ANGLES OF POLYGONS, AND THE RELATION BETWEEN THE ANGLES AND SIDES. OF TRIANGLES. PROPOSITION I. %19, Theorem, — The sum of the three angles of a triangle is tiuo right ajigles. Dem. — Conceive a circumference passed through the vertices of the triangle, as aitc, through the ver- tices of the triangle ABC {58). The angle A is measured by \ arc a, B by i h, and C by i c. Hence, A + B + C is measured by i (a + 6 + c), or a semi-circumference, and is equal to two right angles {2031 Q- E. D. Fig. 168. 220. CoR. 1. — A triangle can have only m^e right angle, or one oUuse angle. Why ? 221, Cor. 2. — Two angles of a triangle, or their su7n, being given, the third may he found hy subtracting this sum from two right angles, i. e., either angle is the siqyplement of the other two. 222, Cor. 3. — The sum of the two acute aiigles of a right-angled tri- angle is equal to one right angle; i.e., they are comjjlements of each other. 223, Cor. 4. — If the angles of a triangle are equal each to each, any one is one-third of two right angles, Oi tico-thirds of one right angle. PROPOSITION n. 224, TJieoreni, — The sides of a triangle sustain the same GENERAL relation to each other as their opposite angles ; that is, the greatest side is 02)posite the greatest angle, the second greatest side opposite the second greatest angle, and the least side opposite the least angle. OF THE ANGLES OF TRIANGLES. 105 Fio Dem.— In the triangle ABC let C > B > A be the order of the values of the angles; then AB > AC > BC is the order of the values of the sides. For, circumscribe the circumference dbc. The angle C be-ing greater than B, the arc c, the half of which measures C, is greater than the arc 5, the half of which measures B. Now, the greater arc has the greater chord {166). Hence, AB > AC. In like manner, if B > A, arc b > arc a, and AC > BC. If either angle, as C, is obtuse, AB is greater than AC or BC, because it lies nearer the centre {167)- 225, Cor. 1.— Conversely, The order of the magnitudes of the sides being AB > AC> BC, the order of the magnitudes of the angles is C> B > A. [Let the student give the demonstration in form.] 220, Cor. 2. — An equiangular tria^igle is also equilateral ; and, conversely, an equilateral triangle is equiangular. Dem.— If A = B = C, arc a = arc h — arc c, and, consequently, chord BC = chord AC = chord AB. Conversely, if the chords are equal, the arcs are, and hence the angles subtended by these arcs. 227, Cor. 3. — In an isosceles triangle the angles opposite the equal sides are equal; and, conversely, if tioo angles of a triangle are equal, the sides opposite are equal, and the triangle is isosceles. Dem. — If AB = BC, arc a = arc c; and hence, angle A, measured by i a, = angle C, measured by i c. Conversely, if A = C.arca — arc c ; and hence chord BC = chord AB. 228, ScH.— It should be observed that the proposition gives only the general relation between the angles and sides of a triangle. It is not meant that the sides are in the same ratio as their opposite angles : this is not true. Thus in Fig. 172 angle c is twice as great as angle a ; but side c is not twice as great as side a, although it is greater. Trigonometry discovers the exact relation which exists between the sides and angles. Fro. 172. 106 ELEMENTARY PIA^'E GEOMEITIY. PROPOSITION m. 229, Tlieoreni, — If from any j^oint witJiui a triangle lines be draiun to the extremities of any sidcy ths included angle is greater than the angle of the triangle opposite this side. Dem. — Let OA and OB be two lines drawn from any point within the triangle ABC, to the extremi- ties of the side AB ; then angle AOB > ACB. For, circumscribe a circle about the triangle. Xow, ,y ACB is measured by \ AwB, but AOB is measured by '^"'n^' i(A7iB + EwiD). Therefore, AOB > ACB. q. e. d. Fig. 173. 230, An JExterior Angle of a polygou is an angle formed by any side with its adjacent side produced, as CBD, Fig. 174. PROPOSITION IT. 231, TJieorem, — ^n exterior angle of a triangle is equal to the sum of the tivo interior non-adjacent angUs. Dem.— Let ABC be any triangle, and CBD an ex- terior angle; then CBD = A + C. For CBD is the supplement of CBA by {131), and CBA is the supplement of A + C by {221). Hence, CBD rr A + C. Q. E. D. 232. Cor.— Either angle of a triangU not adjacent to a specified exterior angle, is equal to the difference of this exterior angle and the other non-adjacent angle. Thus, since CBD = A + 0, by transposition, CBD — A = C, and CBD — C = A, OF THE ANGLES OF QUADRILATERALS. 107 OF QUADRILATERALS. PROPOSITION V. 233, Theorem, — The sum of the angles of a quadrilateral is four right angles. Dem. — Let ABCD be any quadrilateral ; then DAB + ABC + BCD + CDA = four right angles. For, draw either diagonal, as AC, di- viding the quadrilateral into two triangles. Then, as the sum of the angles of the two triangles is the same as the sum of the an- gles of the quadrilateral, and the sum of the angles of the triangles is twice two right angles {219) ; the sum of the angles of the quadrilateral is four right angles. Q. e. d. PROPOSITION TI. 234, Theorem. — The oi^posite angles of any quadrilateral which can he inscribed in a circle are supj^lementary. Dem. — Let ABCD be any inscribed quadrilateral; then A -f C = two right angles^ and D -f B = two right angles. For, A is measured by | arc BCD, and C is meas- ured by i arc DAB (210). Hence, A + C is meas- ured by one-half a circumference, and is, therefore, equal to two right angles {203). In like manner D is measured by i arc ABC, and B by i arc ADC. Consequently, D + B is measured by one-half a cir-' cumference, and is, therefore, equal to two right angles. PROPOSITION m, 23S, Hieorem, — The opposite angles of a parallelogram are equal, and the adjacent angles are supplementary. Dem.— ABCD, Fig. 177, being any parallelogram, A = C, B r= D, and B -h C, C -f- D, D 4- A, and A + B, each = two rigid angles. 108 ELEMENTARY PLANE GEOMETRY. For, produce any side, as AB, fonn- ^ ing the two exterior angles EAD and y/^ CBF. Since CB and DA are parallel, / and FE cuts them, the corresponding an- ^ P gles, CBF and DAB are equal {152). '^ FiQ. 177. Again, since EF and DC are parallel, and CB cuts them, the alternate interior* cngles CBF and C are equal {152). Hence, as DAB and C are each equal to CBF, they are equal to each other. Li a similar manner D can be proved equal to CBA. [Let the student give the proof.] That the angles B and C of the parallelogram are supplemental is evident from {150), which proves that the sum of two interior angles on the same side of a secant cutting two parallels is two right angles. For a like reason A -i- D = ttco right angles^ etc. 236. CoK. 1. — Tlie ttvo angles of a trape- / \ zoid adjacent to either one of the two sides / \ not parallel are supplemental. Fig. its. [Let the student show why.] 237. Cor. 2. — If one angle of a parallelogram is right, the others are also, and the figure is a rectangle. PROPOSITION Tin. 238, TJieoreni, — Conversely to the last, If the adjacent angles of a quadrilateral are supplementary, or the opposite angles equal, the figure is a parallelogram. Dek.— If A + D = two right angles, AB and DC are parallel by {147). For a like reason, if D + C = two right angles, DA and CB are parallel. Again, if A = C and D = B, by adding we have A + D = C+B. ButA + D + C + B =r four right angles {233). Hence, A + D = two right angles, and ^'''- ^'^- AB and CD are parallel. So, also, A + B can be shown to be equal to two right angles ; and, consequently, AD and CB are parallel. * Interior with reference to the parallels (146). OF THE SIDES OF QUADRILATERALS. 109 PROPOSITION EX. , 239* Theorem. — If two op])osite sides of a quadrilateral ar^ equal and parallel, the figure is a 2:>arallelogram. Dem.— In (a) let DC be equal and parallel to AB ; then is ABCD a parallelogram. For, drawing the diagonal AC, it makes the angles ACD and CAB equal, since they are altern- ate intenor angles (15^). Con- ceive the quadrilateral divided in this diagonal into two tri- angles, as in (b). Reverse the triangle ACB and place it as in (c). Draw DB. Since angle DCA = angle CAB, and DC = BA, if CBA be revolved upon AC, AB will take the direction CD, B will fall in D, and CBA will coin- cide with ADC. Hence, angle ACB = angle DAC. But in (a) these are alternate interior an- gles made by AC with AD and BC Q. E. D. Therefore, AD is parallel to BC (149). PROPOSITION X. 240, Tlieoveni, — If the opjjosite sides of a quadrilateral are equal, the figure is a parallelogr'am. Dem.— In (a) let AB = DC, D C and AD = BC ; then is ABCD a parallelogram. For, divide the quadrilateral in the diagonal AC, and revers- ing the triangle ABC, place it as in (c), and draw DB. Since AB ^ CD, and CB = AD, DB is perpendicular to CA {130). Now, revolving ABC upon CA, it will coincide with ADC. Hence, angle DCA = angle CAB, and AB is parallel to DC. Also, angle DAC = angle BCA, and AD is parallel to BC. There- fore, ABCD is a parallelogram. Q. E. D. Fio. 181. 110 ELEMENTARY PLANE GEOMETRY. PROPOSITION XI. 241. TJieorem, — Conversely to the last, Tlie opposite sides of a parallelogram are equal. D C Dem.— Let ABCD be a paral- lelo. /(^) / -__/ I-\ B A^ q;.- («») OF THE DIAGONALS OF QUADRILATERALS. Ill PROPOSITION XHL 245, TJieorem. — The diagojials of a rJiomhus bisect each other at right angles. Dem. — Let AC and DB be the diagonals of the rhombus ABCD ; then are they at right angles to each other, and bisected at Q. For, since AB = AD, and DC = CB, AC has two of its points equally distant from D and B, and is, therefore, perpendicular to DB, at its middle point {130). In like manner D and B are each equidistant from A and C, whence Q is the middle point of AC. .-"'Q- A Fig. 184. -/C 246. Cor. — The diagonals of a rhomhus lisect its angles. For, revolve ABC upon AC as an axis, and it will coincide with ADC. Hence angles A and C are bisected.' In like manner revolve DAB upon DB, and it will coincide with DCB. Hence D and B are bisected. 247. Theorem.- gle are equal. PROPOSITION XIV. Tlie diagonals of a rectan- Dem.— Let AC and DB be the diagonals of the rectan- gle ABCD ; then AC = DB. For, upon AC as a diameter describe a circle. Since D and B are right angles, they are inscribed in semicir- cles {211), and DB is a diameter. Therefore, AC = DB. Q. E. D. Fig. 185. 248. CoR. — Conversely, If the diagonals of a parallelogram arc equal, the figure is a rectangle. Dem.— Since the diagonals of a parallelogram bisect each other, if they are equal, a circumference described from their intersection as a centre, with a radius equal to half of a diagonal, will pass through the vertices of the parallel- ogram. Hence the diagonals will be diameters, and the angles will be inscribed in semicircles, and consequently will be right angles. 112 ELEMENTARY PLANE GEOMETRY. OF POLYGONS OF MORE THAX FOUR SIDES. 240, A Salient Anr/le of a polygon is one whose sides, when produced, can only extend wifJiout the polygon. 250, A JRe'entrant Anr/le of a polygon is one whose sides, when pro- duced, can extend with 171 the polygon. III.— In the polygon ABCDEFC, all the an- gles are salient except D, which is re-entrant. 251. A Convex JPolyf/ofi is a polygon which has only salient angles. A polygon is always supposed to be con- vex, unless the contrary is stated. A Concave or Re-entrant JPolygon is a polygon with at least one re-entrant angle. 0x0 PROPOSITION XT. 253, TJieoreni, — The sum of the interior angles of a polygon is equal to tiuice as many right angles as the polygon has sides, less four right angles. F Dem. — Let n be the number of sides of any polygon ; then the sum of its angles is n times two rigTit angles — 4 right angles. For, from any point O, will) in, draw lines to the vertices of the angles. As many triangles will thus be formed as the polygon has sides, that is, n. The sum of the angles of these triangles ia n times two Hght angles {219). But this exceeds the sum of the angles of the polygon by the sum of the angles at the common vertex 0, that is, by 4 right angles. Hence the sum of the angles of the polygon is n times two right angles — 4 right angles, q. e. d. 25^, ScH. 1. — The sum of the angles of a pentagon is 5 times tico right an- gles — 4 right angles^ or 6 right angles. The sum of the angles of a hexagon is 8 right angles; of a heptagon, 10 ; of an octagon, 12, etc. Fig. 187. OF THE ANGLES OF REGULAR POLYGONS. 113 255, ScH. 2.— This proposition is equally applicable to triangles and to quadrilaterals. Tlius the sum of the angles of a triangle is 3 times two right angles — 4 nght angles (or 6 — 4) = 2 right angles. So also the sum of the angles of a quadrilateral is 4 times tico Tight angles — 4 right angles, or 4 right angles. 256, ScH. 3. — To find the value of an angle of an equiangular polygon, that is, one whose angles are equal each to each, divide the sum of all the angles by the number of angles. PROPOSITION XYI. m:.^ 257. Theore'in, — If the sides of a polygon he produced so as to form one exterior angle {and only one) at each vertex, the snm of these exterior angles is four rigid angles. Dem. — Let n be the number of sides of ahy polygon. At each of the n angles, there is an interior and an exterior angle, whose sum, as A + a, is two right angles. Hence the sum of all the exterior and interior angles is n times two right angles. Now, from this sum subtracting the sum of the exterior angles, the remainder is the sum of the interior angles. But, by tlie preceding proposition, 4 nglit angles subtracted from n times tico right angles, leaves the sum of the interior angles. Therefore the sum of the exterior angles is 4 right angles, q. e. d. Fig. 18S. OF REGULAR POLYGONS* PROPOSITION xrn. 258. Tlieorem. — The angles of an inscribed equilateral polygon are equal ; and the polygon is regular, Dem.— Let ABCDEF be an inscribed polygon, with AB = BC = CD,etc.: then is angle A = B = C = D, etc., and the polygon is regular. For, from the centre of the circle draw OF, OA, and OB, and also the per- pendiculars Oa and 06. Revolve OFA upon OA as an axis, until it falls in the 8 114 ELEMENTARY PLANE GEOAIETRY. plane of OAB. D Since the chords FA and AB are equal, the arc FA = arc AB, and F falls at B. Hence the triangles OFA and OAB coincide. The angle A of the ])oiygon is therefore bisected by OA ; that is, OAF = OAB. In the same manner OBA can be shown equal to OBC. More- over, since O A and OB are equal oblique lines drawn from a point in the perpendicular, angle OAB = angle OBA. Hence, as the half of A equals the half of B, A = B. In like manner, B can be shown equal to C, C to D, D to E, etc. Therefore the polygon is equiangular, as well as equilateral, and consequently regular {117). Q. e. d. PROPOSITION XTin. 2o0» TJieovem. — T7ie sides of an inscribed eqxdanguJar polygon are equal wlien their number is odd ; and the polygon is regular. Dem. — Let ABCDEFG be an inscribed equiangular polygon of an odd number of sides; then is side AB = BC = CD, etc., and the polygon is regular. For, from the centime of the circle draw the radii OA, OB, etc., to tlie vertices of the polygon, and Oa, Ob, etc., perpendicular to the sides. Re- volve the quadrilateral GGA«, upon Oa as an axis until it falls in the plane of OCBa. Since 0^ is perpendicular to the chord AB, ^a = aB, and A will f;\ll at B. Also, as the angle A of the poly- gon = B, AG will fall in BC. Now G falls at the same time in the arc BCD {138) and in BC, and hence falls at their intersection C. Therefore AG = BC. In like manner revolving OBCc upon Oc as an axis, BC is found equal to ED. So also we can show that ED = FG ; then that FG = AB ; then that AB = DC ; and finally, that DC = EF. Hence we have GA = BC = ED = FG = AB = DC = EF ; and as the polygon is equiangular by hypothesis, it is regular {117)- Q- e. d. 260, ScH.— It is easy to see that the above argu- ment would fail in the case of a polygon of an even number of sides, because, in going around the second time the same sides would coincide as in going around the first time. Moreover, we can readily inscribe an equiangular polygon of an even number of sides which shall not be regular. Fie. 191. Fig. 190. OF THE SIDES AND ANGLES OF REGULAR POLYGONS. 115 PROPOSITION XIX. < 2S1* Theovein, — TJie sides of a circiunscrihed equiangular polygon are equal ; and the polygon is regular. Dem. — Let ABCDEF be a circumscribed polygon, with angle A = B = C, etc. ; then is AB = BC = CD, etc., and the polygon is regular. For, from the centre of the circle, draw OA, OB, etc., to the vertices of the polygon, and Oa, 06, etc., to the points of tangency. The latter will be per- pendicular to the sides by {173). Now reverfe the triangle AaO, and apply it to A60, placing Oa in its equal 06 ; aA will take the direction 6A. Then will OA of the triangle AaO, fall in OA of the triangle A&O, since there cannot be two equal oblique lines on the same side of Oh {140). Hence angle bAO = angle aAO, and bA = ak. In the same way it can be Fig. 192. shown that OB, OF, etc., bisect the other angles, and that bS = Be, etc. Whence, as the polygon is equiangular, these halves are equal, that is, OAa = OFa, etc. Then, as OA and OF make equal angles with AF, they cut off equal distances from a, and Aa — aF. So, likewise, we can show that Ab = 6B, and that each side is bisected at the point of tangency. Therefore, as the halves of the sides are equal, the polygon is equilateral, as well as equiangular, and consequently regular {117). Q. e. d.'- PROPOSITION XX. 262, Hieorem, — The angles of a circumscribed equilateral polygon are equal when their number is odd ; and the polygon is regular. Dem. — Let ABODE be a circumscribed polygon with AB = BC = CD, etc. ; then is angle A = 3 = C = D, etc., and the polygon is regular. In the same manner as in the preceding demon- stration, we may show that OA, OB, etc., bisect the angles of the polygon. [The student should go through the process.] Then revolve the tri- angle AOE upon AO as an axis till it falls in the plane of AOB ; and as angle OAE = angle OAB, and AE =: AB, the triangles will coincide. Hence angle OEA, the half of angle E of the polygon. Fig. in.3. equals angle OBA the half of B, and E = B. In like manner revolving AOB upon OB, we can show that A = C. So also we find B = D, and D = A. Therefore the polygon is equiangular as well as equilateral, and consequently regular, q. e. d. 116 ELEMENTAEY PLANE GEOMETRY. 263, ScH.— That the above style of argument fails in the case of a polygon of an eveji number of sides, may be observed by attempt- ing to apply it. Thus, from Fig. 192, we would have A = C, B = D, C = E, D =r F, E = A, and F = B. From these we have A = C =: E, and B = D = F. But the process will not give any one of the first three Fig. 194. angles equal to any one of the second set. That is, it Joes not follow that two adjacent angles are equal in case the number of sides is even. We can readily construct a circumscribed equilateral polygon which shall not be equiangular. PROPOSITION XXI. 264, TJieorem, — A circumference may he circumscribed about any regular 'polygon, Dem. — Let ABCDEFbe a regular polygon. Bisect A F with a perpendicular Oa. Any point in this perpendicular is equidistant from A and F. Bisect AB, adjacent to AF, with a perpendicular, as Oh. Any point in this perpendic- ular is equidistant from A and B. Hence the inter- section of these perpendiculars, O, is equidistant from A, F, and B, and a circumference described from as a centre, with a radius OA, will pass through F and B. Now revolve the quadrilateral FOM upon Oh as an axis until it falls in the plane of COJB, hk will fall in its equal 6B ; and since angle A = angle B, and side AF = side BC, F will fall at C. Thus it appears that the circumference described from 0, and passing through F, A, and B, also passes through C. In a similar manner it can be shown that the same cu-cumfcrence passes through all the vertices, and hence is circumscribed. Q. e. d. 265, Cor. 1. — A circumference may be inscribed in any regular polygon. Dem.— For, having circumscribed one about it, the equal sides become equal chords, aud hence are equally distant from the centre. If, therefore, a circle be drawn from as a centre, with Oa as a radius, it will touch every side of the polygon at its middle point. 266, Cor. 2. — Tlie ceiitres of the inscribed and circumscribed circles coincide. 267, Tlie Centre of a regular polygon is the common centre of its inscribed and circumscribed circles. OF THE SIDES OF POLYGONS. 117 268. An Angle at the Centre of a regular polygon is the angle included by two lines drawn from the centre to the extremities of a side, as FOA, AOB. 269, Co^. 3. — The angles at the centre of a regular polygon are equal each to each; and any one is equal to four right angles divided by the number of sides of the polygon. 270* The Apothem of a regular polygon is the distance from the centre to any side, and is the radius of the inscribed circle. PROPOSITION XXII. 271. Theorem. — The side of a regular inscribed hexagon is equal to the radius. Dem. — Let ABCDEF be a regular inscribed hexagon ; then is any side, as BC, equal to OB, the radius. In the triangle BOC the angle O is measured by the arc BC, or i of a circumfejence, and hence is \ of 4 right angles, or | of a right angle. Angle ABC is measured by ^ arc CDEFA, or f of a circumfer- ence. Hence angle OBC, which is ^ of ABC, is measured by ^ of f , or | of a circumference, and is, consequently, equal to BOC. So also OCB, the half of DCB, is measured by ^ of a circumference. Hence OCB is equiangular, and consequently equilateral {226), and BC = OB. q. e. d. 272. A Broken Line is said to be Convex when no one of its parts will, when produced, enter the space included between it and a line joining its extremities. PROPOSITION xxin. 273. Theore'kn. — A Convex broken line is less than any broken line luhich envelops it and has the same extremities. Dem. — Let khcdB be a broken line enveloped by the broken line ACDEFB, and having the same extremities A and B ; then is KbcdS < ACDEFB. For, produce the parts of ^hcdB till they meet the enveloping line, as A6 to e, be to /, and cd to g. Now, since a straight line is the shortest path between two points, Ae < ACe, bf < beDEf, Fig. 197. 118 ELEMENTARY PLANE GEOMETRY. ^9 < rf^9, and dB < dgS. Hence, if a point starts from A to move to B, AeDEFB will be a shorter path than ACDEFB, A6/FB shorter than A^DEFB, KhcgB shorter than A5/FB, and AJcrfB shorter than khcgB. Therefore, kbcdB is shorter than ACDEFB. Q. E. D. 274, Cor. 1. — The sum of any two sides of a triangle is greater than the third side. This is the same as the axiom that the shortest distance between two points is a straight line. 275, Cor. 2. — The difference hetween any tivo sides of a tria^igle is less than the third side. Dem. — Let rt, 6, and c be the sides. By Corollary 1st, a + & > c. Therefore, transposing, a> c — b. 276, Cor. 3. — If from any point ivithin a tria^igle liiies he draton to the extremities of any side, the sum of these lines is less than the sum of the other two sides of the triangle. Fio. 198. EXERCISES. 1. Giyen two angles of a triangle, to find the third. Bug's. — The student should draw two angles on the blackboard, as a and b, and then proceed to find the thhd. The figure will suggest the method. The third angle is c. The solution is effected also by constructing the two given angles at the extremities of any line, and producing the sides till they meet. 2. Two angles of a triangle are re- spectively f and 1^ of a right angle. What is the third angle ? 3. The angles of a triangle are respectively |, J, and f of a right angle. Which is the greatest side ? Which the least ? Can you tell the ratio of the sides ? 4. What is the value of one of the equal angles of an isosceles tri- angle whose third angle is ^ of a right angle ? 5. Two consecutive angles of a quadrilateral are respectively f and I of a right angle, and the other two angles are mutually equal to OF THE SIDES AND ANGLES OF POLYGONS. 119 each other. What is the form of the quadrilateral ? What the value of each of the two latter angles ? 6. One of the angles of a parallelogram is f of a right angle. What are the values of the other angles ? 7. The two opposite angles of a quadrilateral are respectively -I and f of a right angle. Can a circumference be circumscribed ? If so, do it. 8. Two of the opposite sides of a quadrilateral are parallel, and each is 15 in length. What is the figure ? Do these facts determine the angles ? 9. Two of the opposite sides of a quadrilateral are 12 each, and the other'two 7 each. What do these facts determine with reference to the form of the figure ? 10. What is the value of an angle of a regular dodecagon ? 11. What is the sum of the angles of a nonagon ? What is the value of one angle of a regular nonagon ? Of one exterior angle ? 12. What is the^ regular polygon, one of whose angles is l|f right angles ? /^/^^J^-^ 13. What is the regular polygon, one of whose exterior angles is f of a right angle ? 14. Can you cover a plane surface with equilateral triangles with- out overlapping them or leaving vacant spaces? With quadrilat- erals? Of what form? With pentagons? Why? With hexagons? Why ? What insect puts the latter fact to practical use ? Can you cover a plane surface thus with regular polygons of more than 6 sides ? Why ? 15. Is an equilateral hexagon circumscribed about a circle neces- sarily regular ? A heptagon ? An octagon ? A nonagon ? 16. Is an equiangular circumscribed quadrilateral necessarily reg- ular ? A pentagon ? A hexagon ? A heptagon ? 17. Is an equilateral inscribed pentagon necessarily regular ? An octagon? How is it if they are equiangular; are they necessarily equilateral and regular ? 120 ELEMENTAEY PLANE GEOMETRY. SYNOPSIS. CO o c o 'O m Q »2 c 9 5zi Q -< C» cc '"^ (=J ^ t^ <: c >^; HZ •< S p; o S < Prop. L Sum of angles. f Cor. 1. I a>r. 2. 1 Cor. 3. I (7(>r. 4. Only one right or obtuse. Two angles given. Acute angles if right angled One angle if equiangular. Cor Cor. Prop. II. Sides and opp. angles. ■{ Cor. I ScJi. I Converse. 2. Equiangular, equilat- eral, and converse. 3. Isosceles, equiangular, and converse. These only general rela- tions. Prop. Ill, Angle within a triangle. Def. Exterior angle. I Prop. IV. Exterior angle.— Cplemental. Dem.— Let ABC and DEF be two angles, having BC and ED parallel, and extending in the same direction from the vertices, and AB and EF parallel, and extending in opposite directions from the vertices ; then are ABC and DEF supplements of each other. For, produce the two sides not parallel, if necessary, till they meet. Now, BHD is the supplement of BHE by {131), BHE = the al- ternate interior angle DEF, and BHD = the corresponding angle ABC. Therefore, ABC is the supplement of DEF. q. e. d. [This demonstration is adapted to the upper cut; let the student adapt it to the lower.] PROPOSITION T. 283, Theorem, — Iftioo angleshave their sides respectively perpendicular to each other, the angles are either equal or supplementary. Dem, — Let BA be perpendicular to EF or to E'F', and BC to ED ; then is ABC = DEF. For, through B draw BO and BN, respectively parallel to ED and EF; then by the preceding jTropositions NBO = DEF, and is the supple- ment of F'E'D. But NBA =: OBC, since both are right angles. Take away OBA from each, and we have NBO = ABC; and as NBO is the supplement of F'E'D, ABC is also the supple- ment of F'E'D. Q. e. d. 124 ELEltENTABY PLANE GEOMETEY. OF TRIMGLES. PROPOSITION VI. 284. Theorem. — Tiuo tricmgles ivMcli have hoo sides and the included angle of one equal to two sides and the included angle of the other, each to each, are equal, Dem.— Let ABC and DEF be two triangles, having AC = DF, AB = DE, and angle A = angle D ; then are the triangles equal. For, place the triangle ABC in the position (6), the side AB in its equal DE, and the angle A adjacent to its equal angle D. Then re- volving ABC upon DB, until it falls in the plane on the opposite side of DB, since angle A = angle D,AC will take the direction DF ; and as AC = DF, C will fall at F. Hence BC will fall in EF, and the triangles will coincide. Therefore the two triangles are equal, q. e. d. We may also make the application of ABC to DEF directly, as in {85). The method here given is used for the purpose of uniformity in this and the follow- ing. We may observe that in this, as in the other cases, DB is perpendicular to PC, and bisects it at 0. This fact might easily be shown, and the demonstra- tion be based upon it 2S5. ScH. — This proposition signifies that the two triangles are equal in all iespects, 1. e., that the two remaining sides are equal, as CB = FE; that angle C = angle F, angle B = angle E, and that the areas ai'e equal Fig. 204. PROPOSITION TIL 286. Theorem. — Tiuo triangles ivhich have two angles and the included side of the one equal to two angles a7id the included side of the other, each to each, are equal. EQUALITY OF TRIANGLES. 125 Pig. 205. Dem.— Let ABC and DEF be two tiiuDgles, having angle A = angle D, angle B = angle E, and side AB = side DE; then are the triangles equal. For, place ABC in the position (b), the side AB in its equal DE, the angle A adjacent to its equal angle D> and B adjacent to its equal angle E. Then revolving ABC upon DB till it falls in the plane on the same side as DFE, since angle A = angle D, AC will take the direction DF, and C will fall somewhere in DF or DF produced. Also, since angle B = angle E, BC will take the direction EF, and C will fall somewhere in EF, or EF produced. Hence, as C falls at the same time in DF and EF, it falls at their intersection F. Therefore the two triangles coincide, and are consequently equal, q. e. d. 287, Cor. — // one triangle has a side, its opposite angle, and one adjacent angle, equal to the corresponding parts in another triangle, each to each, the triangles are equal. For the third angle in each is the supplement of the sum of the given angles, and they are consequently equal. Whence the case is included in the pro- position. 288* ScH. — A triangle may have a side and one adjacent angle equal to a side and an adja- cent angle in another, and the second adjacent angle of the first equal to the angle opposite the equal side in the second, and the triangles not be equal. Thus, in the figure, AB := C'^', A = A', and B = B' ; but the triangles are evidently not equal. [Such triangles are, however, similar, as will be shown hereafter.] ® PROPOSITION rm. 289. Theore^n, — Two triangles ivhich have two sides and an angle opposite one of these sides, in the one, equal to the corresponding 126 ELEMENTARY PLANE GEOMETRY. parts ill the other, are equal, if of these tiuo sides the one opposite the giveii a?igle is equal to or greater than the one adjacent. Dem.— In the triangles ABC and DEF, let AC = DF, CB = FE, A = D, and CB {= FE) = AC {= DF); then are the triangles equal. For, apply AC to its equal C F DF, the point A falling at D and C at F. Since A = D, AB will take the direc- tion DE. Let fall the per- pendicular FH upon DE, or DE produced. Now, CB being ^ DF, cannot fall between it and the perpen- dicular, but must fall in FD or beyond both. As there can be but one line on the same side of the perpen- dicular equal to CB, and as FE = CB, CB must fall in FE. Hence, the two triangles coincide, and are consequently equal. Q. e. d. 290. ScH. l.-If A and D are acute and CB {= FE) = AC {= DF), the tri- angles are isosceles. If A and D are right or obtuse, CB (=FE) must be greater than AC (= DF), in order that there may be a triangle, since the right or obtuse angle is the greatest angle in a triangle, and the greatest side is opposite the greatest angle. This impossibility appears also from the demonstration above. 291. ScH. 2.— If A and D are a€ui£, and the side opposite A, i. e., CB, is less than AC, it must be equal to or greater than the pei-pendicular CI (= FH) in order to have a triangle. Then, applying AC to DF, and ob- serving that AB takes the direction DE, and that EF, which = CB, being intermediate in length between DF and FH, may lie on either side of FH, we see that ABC may or may not coincide with DEF. "Whether it does or not will depend upon whether angle C = angle F, or whether AB = DE. This is the ajibiguous case in the solution of triangles, and should receive special attention. EQUALITY OF TRIANGLES. 127 PROPOSITION IX. 292,17ieore^n. — Tiuo triangles wliich have the three sides of the one equal to the three sides of the other, each to each, are equal. Dem.— Let ABC and DEF be two triangles, in whicli AB = DE, AC = DF, and BC = EF; then are tlie triangles equal. For, place the triangle ABC in the position (&)^ and the side AB in its equal DE, so that the other equal sides shall be adjacent, as AC ad- jacent to DF, and BC to EF. Draw FC. Now, since DC = DF, and EC = EF, DB is perpendicular to FC at its middle point {130). Hence, p^^ 209 revolving ABC upon DB, it will coincide with DEF when brought into the plane of the latter Therefore the two triangles are equal, q. e. d. 293, Cor. — In tioo equal triangles, the equal angles lie opposite the equal sides. 294, Sen. — If the triangles compared, as in the three preceding propositions, have an obtuse angle, and the two sides first brought together are sides about the obtuse angle, the figure will take the form in the mar- gin ; but the demonstration will be the same. When the three sides are the given equal parts, the form of figure given in the demonstration above can always be secured by bringing together the two greatest sides. Fio. 210. PROPOSITION X. 29S, Theorem, — If tioo triangles have tiuo sides of the one respectively equal to tiuo sides of the other, and the included angles unequal, the third sides are unequal, and the greater third side belongs to the triangle having the greater i^icluded angle. VzS ELEMENTARY PLANE GEOMETRY. Dem.— Let ACB and DEF be two tri- angles having AC = DF, CB = FE, and C > F; then is AB > DE. For, placing the side DF in its equal AC, since angle F < angle C, FE will fall within the angle ACB, as in CE. Then let the triangle ACE = the triangle DFE. Bi- sect ECB with CH, and draw HE. The triangles HCB and HCE have two sides and the included angle of the one, respec- tively equal to the corresponding parts of the other, whence HE = HB. Now AH -i- HE > AE; but AH + HE = AH + HB =r AB. Therefore, AB > AE. q. e. d. 296. Cor. — Conversely, If two -P^ 211 ^^^^^ ^f ^'^^ triangle are respectively equal to tivo sides of another^ and the third sides iinequah the aJigle opposite this third side is the greater in the triangle ivhich has the greater third side. Dem.— If AC = DF, CB = FE, and AB > DE, angle C > angle F. For, if C = F, the triangles would be equal, and AB = DE {284); and, if C were less than F, AB would be less than DE, by the proposition. But both these conclu- sions are contrary to the hypothesis. Hence, as C cannot be equal to F, nor less than F, it must be greater. PROPOSmON XL 297. Hieovein, — Tioo right angled triangles tohich have the hypotemise and one side of tlw one equal to the hypotenuse and one side of the other, each to eacJi, are equal. Dem.— In the two triangles ABC and DEF, right angled at B and E, let AC = DF, and BC = EF; then are the triangles equal. For, place BC in its equal Q PC ^^> so that tlie right angles shall be adjacent, the angles A and D lying on opposite sides of EF. as in {h\ Since E and B are right angles, DA is a straight line. Now, since equal oblique lines, as Fig. 212. ^^ ^^fl CA, cut off equal distances from the foot of the perpendicular {14:1\ DE = BA; and revolving CAB upon FB, the two EQUALITY OF QUADKILATERALS. 129 triangles will coincide when CAB falls in the plane on the side D. Therefore, the triangles are equal, q. e. d. PROPOSITION xn. 298. TJieorem, — Two right angled triangles liaving the hypo- tenuse and one acute angle of the one equal to the hypotenuse and an acute angle of the other, are equal. Dem.— One acute angle in each being equal, the other acute angles are equal, since they are complements of the same angle {222). The case is, then, that of two angles (the acute angles in each), and their included side (the hy- potenuse), and falls under {286). PROPOSITION xni. 299, Tlieorem, — Two right angled triangles having a side and a corresponding acute angle in each equal, are equal. This also falls under {286). Let the student show why. OF QUADRILATERALS. PROPOSITION xiy. 300, Tlieorem, — T2V0 quadrilaterals are equal when the follow- ing parts are equal, each to each, in both quadrilaterals, and similarly 1. The triangles into which either diagonal divides the quadrilaterals. 2. The four sides and either diagonal. 3. The four sides and one angle. 4. Three sides and the two included angles. 5^. Three angles and any tioo sides, if the other two sides are non- parallel. Dem. — 1. This case is demonstrated by applying one quadrilateral to the other. 2. This case is reduced to the former by {292). 3. Drawing the diagonal opposite the known angle, this case is reduced to the first by {284), and {2i)2). 4. This is demonstrated by applying one quadrilateral to the other ; or, draw- ing a diagonal, it may be reduced to case first by {284). 5. In this case the quadrilaterals are mutually equiangular by {25o). If, then, the two Bides are adjacent, by drawing the diagonal joining their extremi- 9 130 ELEMENTARY PLAN"E GEOMETRY. ties the case is brought under the first by (284), and (280). If, however, the y\ ^ two known sides are non-adjacent, by pro- ^y \ / \ ducing the unknown sides two triangles are formed in each figure, which are mu- tually equal by (280), and the case comes under the axiom concerning equals sub- ^ 2lT ' ' ^ tracted from equals. [Let the pupil draw the figures and give the demonstrations in full form. Trapeziums should be used ; although it should be seen in each case — except the fifth— that the truth applies to any other form of quadrilateral.] PROPOSITION XT. 301, Theorem. — Two 2)arallelograms having tivo sides and the incUided angle of the one equal to tioo sides and the included angle of the other, each to each, are equal. Dem.— Let AC and EG be two parallelograms, with AD =: EH, AB = EF, and A = E ; then are they equal. For, applying the angle E to A, since EH ^ AD, H will fall at D ; and since EF = AB, F will fall at B. Now, through D but one Hue can be drawn parallel to AB ; hence HG will fall in DC, and C will be found in DC, or in DC produced. In like manner, since but one parallel to AD can be drawn through B, YiQ^ 214. ^^ must fall in BC, and G be found in BC, or in BC produced. Therefore, as G falls at the same time in DC and BC, it falls at C, and the parallelograms coincide. 302, Cor. — Two rectangles of the same base and altitude are equal. / 7 / 7' A B hi r' / / / 7G / / / L F OF POLYGONS. PROPOSITION XVI. S03. Theorem. — Two polygons of the same number of sides, having all the parts of the one except three angles, known to he respec- tively equal to the corresponding parts of the other ^ are equal. Dem. — If the two polygons AE* and A'E', have all the parts of the one equal to the corresponding parts of the other, each to each, except three angles ; then are the pol5''gons equal. * It is often more convenient to read a polygon by two letters, instead of all those at the vertices. EQUALITY OF POLYGO>'S. 131 Fig. 215. 1st. When the three un- known angles are consecu- tive, as G, F, E, and C, F', E'. Draw CE, and G'E'. Apply polygon A'E' to AE, beginning with g' in its equal ^r; A' = A, and a' = a; hence, B' falls at B: B' = B, and b' = b, hence, C falls at C ; etc. Thus, we may show that the perime- ters coincide till we reach E' and E. Then will G'E' = GE, and the triangles GFE and G'F'E', having their corresponding sides respectively equal, are them- selves equal, and the polygons coincide throughout. 2d. When two of the unknown angles are consecutive, and the third not con- secutive with these, as G, E, D, and G', E', D'. From the angle which is not consecutive with the other two, draw diagonals to the other angles, as GE, GD, and G'E', G'D'. Now, G'A'B'C'D' can be applied to GABCD, and G'F'E' to GFE, in the ordinary way. Hence, the triangles G'E'D' and GED are mutually equi- lateral, and consequently equal. Therefore the polygons are equal. 3d. When no two of the three unknown angles are consecutive, as G, B, D, and G', B', D'. Join the unknown angles by diagonals, as GB, GD, BD, and G'B', G'D', B'D'. Now, polygon G'F'E'D' can be applied to GFED, D'C'B' to DCB, and G'A'B' to GAB in the ordinary way. Hence, the triangles G'D'B' and GDB are mutually equilateral, and consequently equal. Therefore the polygons are equal.* S04, Cor. — Two quadrUaterals having their corresponding sides eqitaly and an angle in one equal to the corresponding angle in the other, are eqtial. PROPOSITION xyn. SOo, Theorem, — Tivo polygons of the same number of sides, having all the parts of the one except tico angles and the i^icluded si^e, hnown to he respectively equal to the corresponding parts of the other, are equal. Dem.— Let the unknown parts be C, c, D, and C, c' , D'. From any other two of the mutually equal angles, as G and G', draw the diagonals GO, GD, GX', G'D', * Notice that in each case the unknown angles are to form the vertices of triangles, which the argument shows to be equilateral, and therefore equal. In Case 1st, we have to draw only one line in order to give the triangles, as two sides are sides of the polygon ; in Case 2d, we have to draw two sides , and in Case 3d, three sides, for analogous reasons. 133 ELEMENTARY PLANE GEOMETET. to the unknown angles. Then the polygon G'F'E'D' can be applied in the ordinary way to GFED, /' being placed in /, etc So also G'A'B'C can be applied to GABC, begin- ning with g' in its equal g. Hence, angle F'G'D' = FGD, A'G'C'=AGC; and, adding, F'G'D' + A'G'C = FGD + AGO. Subtracting these equals from G' = G, we have C'G'D' = CGD. Whence the triangles C'G'D' and CGD have two sides and their included angle equal in each, and are equal; therefore the polygons are equal in all their parts. SOG, ScH.— When the unknown angles are both separated from the unknown side, the polygons may or may not be equal — the case is am- biguous. Thus, if C and E are the unknown angles and AH the un- known side, the polygons ABCDEFG, and A'B'C'D'EFG fulfill the condi- tions, but are not equal. By draw- ing CE, CA, and EH, the case is re- |-1 duced to that of two quadrilaterals having all the parts equal, each to each, except two angles and their non adjacent side ; in which case the quadrilaterals are not necessarily equal. So, also, when one of the unknown angles is adjacent to the unknown side and the other separated, the polygons may or may not be equal. Thus, let the unknown parts be D, c, C, and D', c' , C. From the separated angle draw the diagonals to the extremities of the unknown side, as GC, CD (or GDi), and C'C, CD'. In the usual way G'A'B'C' can be applied to GABC, and G'F'E'D' to EQUALITY OF POLYGONS. 133 CFED. Whence C'C = GC, CD' = CD, and angle C'C'D' = CCD. Thus the case is reduced to that of two triangles having two sides and an angle oppo- site one of them mutually equal, and is, therefore, ambiguous. The polygon (a) may have the part corresponding to C'F'E'D' situated as CFED, or as CFiEiDi, In the former case the polygons are equal, in the latter not. 307* Cor. — Two quadrilaterals having three sides and the corre- sponding angles included by these sides equal, are equal. This falls under the 1st case. 308* ScH. — If the three unknown or excepted parts are all sides, the poly- gons are not necessarily equal, as will appear by an inspection of the figure. The Fig. 219. unmarked sides being the excepted ones, the polygons may be those included by the continuous lines, or those included in part by the broken lines, all the parts being equal in each two, except the three unknown ones. PROPOSITION xvni. 309, Theorem, — Two iMijgons of the same numler of sides, having tiuo adjacent sides and the diagonals drawn from the included angle, in the one, respectively equal io the corresponding parts in the other, and their corresponding included angles equal, are equal figures. Dem.— The demonstration is based upon {284). Let the student draw the figures, and make the applications. PROPOSITION XIX. 310. Theorem. — Two polygons of the same number of sides, having all the parts {sides and angles) of the one respectively equal to the corresponding parts of the other, exce2^t two parts, are equal, un- less the exapled partA arc parallel sides. 134 ELEMENTARY PLANE GEOMETRY. Dem. — The demonstration can be supplied by the pnpil, as it i3 similar to the several preceding. The cases will be, 1st, When two angles are excepted, (a) they being consecutive, {b) they not being consecutive ; — 2d, An angle and a side, (a) consecutive, {b) not consecutive ; — 3d, Two sides, (a) consecutive, {b) not consecutive. EXERCISES. 1. JProb. — Having two sides and their included angle given, to construct a triaiigle. Sug's. — The student should draw two lines on the blackboard, and a detached angle, as the given parts. Then, making an angle equal to the given angle {200), he should lay off the given sides from the vertex on the sides of the angle, and join their extremities. The triangle thus formed is the one required, for any other triangle formed with these two sides and this angle will be just like this by {284). 2. I^i*oh, — Having two angles and their included side give?i, to C07istruct a triangle. 3. JProb, — Having the three sides of a triangle given, to construct the triayigle. Solution. — Let a, b, and c, be the given sides. Draw an indefinite line CX, and on it take CB — a. From C as a centre with 5 as a radius, describe an arc as near as can be discerned where the angle A will fall. From B, with a radius c, describe an arc intersecting the former. Then is ABC the triangle required, since any other triangle having the same sides would be equal to ABC {292). 4. ^roh. — To inscrihe a circle in a given triangle. Solution. — For the method of doing it see Part I. {70). To prove the method correct, we observe that the triangles ODB and QBE have OB common, and are mutually equiangular ; hence they are equal, and CD = OE. In like manner triangle OEC = OFC, and OE = OF. [Tiiangle OFA = ODA ; but we do not need the fact in the demonstration.] Since OD = OE = OF, the circumference stnick from O as a centre with a radius OD, passes through E and F. More- over, since each side of the U'iangle is per- pendicular to a radius at its extremity, it is tangent to the circle {172) ; and the circle is inscribed. Fig. 2iJ0. OP EQUALITY. 135 5. JProb, — Having tivo sides and an angle opposite one of them given, to construct the triangle. Solution. — 1st. When the given angle is right or obtuse, the side opposite must be greater than the side adjacent, as the greatest side is opposite the greatest angle {224^), and the greatest angle in such a triangle is Xu.^ right or obtuse angle. In this case let m and o be the given sides, and the angle oppo- site o. Draw an indefinite line O'X, construct 0' equal to 0, and take O'N' equal to m. . From N' as a centre, with a radius equal to 0, describe an arc cut- ting O'X, as at M'. Draw N'M'. Then is N'M'O' the triangle required, since all triangles having their corresponding parts equal to w', o\ and 0' are equal. 2d. When the given angle is acute, as A, there will be no solution if the given side, a, opposite A, is less than the perpendicular ; one solution if a = p, or if a > than both j) and 5, and two solutions if a > ^, and less than h. This will appear from the construction, which is the same as in Case 1st. 6. If a perpendicular be let fall from the right angle C of the triangle ACB upon the hypotennse, as CD, show from {222) that the three triangles in the figure are mutually equiangular. 7. Given the sides of a triangle, as 15, 8, and 5, to construct the triangle. 8. Given two sides of a triangle a = 20, Z» = 8, and the angle B opposite the side h equal | of a right angle,* to construct the triangle. 9. Same as in the 8th, except h = 12. Same, except that i = 25. 10.' Construct a triangle with angle A = f of a right angle, angle B = 4- of a right angle, and side a opposite angle A, 15. 11. Construct a right angled triangle whose hypotenuse is 16, and ♦ To construct this angle, bisect an angle of an equilateral triangle. 136 ELEMENTARY PLANE GEOMETRY. one of the otlier sides 7. The same with one acute angle f of a right angle, and a side about the right angle 12. Will there be any difference in the shape of the triangles if one is constructed with the given angle adjacent to the given side, and the other with it oppo- site ? Will there be any difference in the size 9 12. Construct a right angled triangle having its hypotenuse 20, and one acute angle ^ of a right angle. 13. Construct a quadrilateral three of whose sides are 20, 12, and 15, and the angle included between 20 and the unknown side | of a right angle, and that between 15 and the unknown side -} a right angle. f of a right angle, and h = 20. From D as a centre, with a radius 12, strike the arc on. At any point on side a, make an angle B' = ^ a right angle. Take B'7n = 15, and draw Cm parallel to AB'. Fi*om the intersection C draw CB parallel to mS'. Draw CD. Then is ABCD the quadrilateral required. Queries. — If d + cis less than the perpendicular from D upon AB, then what ? If equal to the perpendicular, then what? Is it necessary to consider angle B in answeriug the two pre- ceding queries ? 14. Construct a parallelogram whose two adjacent sides are' 6 and 8, and whose included angle equals 1^ right angles. 15. Construct a heptagon whose sides in order are a == 4:, b = 6, c = 5, d = 6,e = 6,f = d, (/ = 4:; and the angle included between a and b, 1^ right angles; between b and c. If ; c and d, IJ; d and Sug's.— See Fig. 187. Proceed in ordei, laying off the parts as given, from A to F. Draw AF. From F as a centre, with a radius/ = 3, strike an arc, and also from A, with a radius g = 4. The intersection of these arcs will determine C. Queries.— ^\\a.i is tlie limit of the sum of the possible values of the given angles ? "What the limit of the sum of tne sides included between the unknown anglwi ? OF EQUALITY. 137 SYNOPSIS. f WUat? How shown? Prop. I. Of straight lines. Prop. II. Of circles. it i Prop. III. Sides parallel. Direction same or opposite. Prop. IV. " *' " one same, other opposite. Prop. V. " perpendicular. Prop. VI. Two sides and included angle. ■{ Sch. All parts equal. Prop. VIL Two angles and ( ^'='^- ^^'l^' ^^^ fcljacent and one oppo- mcluded side, j . ; -p ' f^^"" '''^'''''^' { Sell. Exception. Prop. VIII. Two sides and angle j Sc7i. 1. When isosceles. opposite one. ( Sch. 2. Wlien ambiguous. T>r,^T. TV Ti ,.^^ .• i^o i (^^^'- Equal angles opposite equal sides. Prop. IX. Three sides, j ^^;, ^^^^ ^^ ^g^^^^^ i^j^g^^ ^.^^^^ ^^ ^.^ -Prop. X. Two sides equal, included angles un- ) ^^^ Converse Prop. XI. Hypotenuse and one side. Prop. XII. Hypotenuse and one acute angle. <* I Prop. XIII. Side and one acute angle. Prop. XIV. Three sides and non-included angles equal. Prop. XV. Two parallelograms having two ( Co?: Rectangles of sides and the included angles equal. same base and altitude. Prop. XVI. Three angles excepted. ■{ Cor. Quadrilaterals. V1TTT m 1 J ( ScJi. 1. The ambiguous case. ! Prop. XVIL Two angles and one \ ^^,, Quadrilaterals. i side excepted. ^ ^^^^ ^ ^liree sides excepted. Prop. XVIII. Two sides and included diagonals. Prop. XIX. Any two parts excepted. Exercises. Prob. In a triangle, given two sides and included angle. Prob. " " " angles " side. Prob. " " " sides and angle opposite one. Prob. " " " three sides. Prob. To inscribe a circle in a triaui^le. 138 ELEMENTARY PLANE GEOMETRY. 3 SECTION IX. OF EQUIVALENCY AND AREA. Sll. Equivalent Fif/iires are such as are equal in magni- tude. PROPOSITION I. 312. TJieorenu — Parallelograms having equal bases and equal aUitudes are equivalent. Dem. — Let ABCD and EFCH be two parallelograms having equal bases, BC and FC, and equal altitudes; then are they equivalent. A E' D H' E H *^^^' P^^^^ FG in its equal 7 \ 7 J 7 BC ; and, since the altitudes \ / / / are equal, the upper base EH \ / / / will fall in AD or AO pro- / \/ / / duced, as E'H'. Now, the ^ C F G two triangles AE'B and DH'C Fig. 223.* are equal, because the three sides of the one are respectively equal to the three sides of the other. Thus AB = DC, being opposite sides of the same parallelogram. For a like reason, E'B = H'C. Also, E'H' = BC = AD. From AH' taking E'H', AE' remains, and taking AD, DH' remains. Therefore AE' = DH'. These triangles being equal, the quadrilateral ABCH' - the triangle AE'B = ABCH' - DH'C. But ABCH' - AE'B = E'BCH' = EFCH ; and AB^CH' - DH'C = ABCD. Hence, ABCD = EFCH. Q. E. D. 313, Cor. — Any parallelogram is equivalent to a rectangle having the same base and altitude. PROPOSITION n. 314, TJieorem, — A triangle is equivalent to one-half of any parallelogram having an equal base and an equal altitude with the triangle. OF EQUIVALENCY. 139 DKM.~Let ABC be a triangle. Through C draw CD parallel to AB ; and through A draw AD parallel to BC. Then is ABCD a parallelogram, of which ABC is one- half {243). Now, as any other parallelogram having an equal base and altitude with ABCD is equivalent to ABCD {312), ^BO is equiva- lent to one-half of any parallelogram having an equal base and altitude with ABC. q. E. D. 315, Cor. 1. — A triangle is equivalent to one-half of a rcctamjle having an equal base and an equal altitude with the triaixgle. 316. Cor. 2. — Triangles of equal lases and equal altitudes are equivalent, for they are halves of equivalent parallelograms. Fig. -ZiA. PROPOSITION III. 317, Theorein, — The square described on a line is equivalent to four times the square described on half the line, nine times the square described on one-third the line, sixteen times the square on one-fourth the line, etc. Dem. — Let AB be any line. Upon it describe the square ABCD. Bisect AB, as at«c?, and AD, as at a. Draw dc parallel to AD, and ah parallel to AB. Now, the four quadrilaterals thus formed are parallelograms by construction, ^ r n hence their opposite sides and angles ' are equal ; and as A, B, C, and D are right angles, and ka — kd = dB — bB = etc., the four figures 1, 2, 3, 4, are equal squares. Hence kdoa = i ABCD, In like manner it can be shown that the nine figures into which ^^^- ^^• the square on A'B' is divided by draw- ing through the points of trisection of the sides, lines parallel to the other sides, ^re equal squares. Hence AV, the square on ^ of A'B', is i of the square A'B'C'D'. The same process of reasoning can be extended at pleasure, show- ing that the square on i a line is -iV the square of the whole, etc. n o c 4 \ 3 <7 d\ / i 2 7i A d. ti a 8 \ 9 G . 5 \4r or 7n\ 4 '. 2 \ 3 l/" y" PROPOSITION IV. 31S. Tlieorem. — A trapezoid is equivalent to t%vo triangles having for their bases the upper and lovjer bases of the trapezoid, and for their common altitude the altitude of the traijezoid. 140 ELEMENTAKY PLANE GEOMETRY. Dem.— By constructing any trapezoid, and drawing cither diagonal, the student can show the truth of this theorem. 319. Proh, PROPOSITION Y. To reduce any i}olygon to an equivalent triangle. Solution.— Let ABCDEFbca polygon which it is proposed to reduce to an equivalent triangle. Produce any side, as BC, indefinitely. Draw the diagonal EC and DH parallel to it, ^ Draw EH. Now, consider the triangle CDE as cut off from the polygon and replaced by CHE. The magnitude of the polygon will not be changed, since CDE and CHE have the same base CE, and the same altitude, as their vertices lie in DH parallel to EC. From the polygon thus reduced we cut the triangle FHE, and replace it by its equivalent FHI, by drawing the diagonal FH, and the parallel El. In like manner, by drawing FB and the parallel AG, we can replace FBA by its equivalent FGB. Hence, CFI is equivalent to ABCDEF. It is evident that a similar process would reduce a polygon of any number of sides to an equiva- lent triangle. AREA. PROPOSITION TI. 320, Theorem, — The area of a rectangle is equal to the product of its base and altitude. Dem. — Let ABCD be a rectangle, then is its area equal to the base AB multi- plied by the altitude AC. If the sides AB and AC are commensurable, take some unit of length, as E, which is contained a whole number of times in each, as five times in AC, and eight times in AB, and apply it to the lines, dividing them respectively into five and eight equal parts. A f 2 s ^ 3 e 7 e^ Frcm the several points of division draw lines through Fig. 2-27. ^^^^ rectangle perpendicular to its sides. The rect- angle will be divided into small parallelogram^, which are all equal squares, as the angles arc all right angles, and the- sides all O E D & i : : i i i ; • c — i — i—l— J— h'.- .{/'—.— ^ ^\ \ \ \ \ \ \ OF AREA. 141 equal to each otlier. Each square is a unit of surface, and the area of the rect- angle is expressed by the number of these squares, which is evidently equal to the number in the row on AB, multiplied by tlie number of such rows, or the number of linear units in AB multiplied by the number in AC'i If the two sides of the rectangle are not commensurable, take some very small unit of length which will divide one of the sides, as AC, and divide the rectangle into squares as before ; the number of these squares will be the measure of the rectangle, except a small part along one side, not covered by the squares. By taking a still smaller unit, the part left unmeasured by the squares will be still less, and by diminishing the unit of length E, we can make the part unmeasured as small as we choose. It may, therefore, be made infinitely small by regarding the unit of measure as infinitesimal, and consequently is to be neglected.* Plence, in any case, the area of a rectangle is equal to the pro- duct of its base into its altitude, q. e. d. 321, Cor. 1. — The area of a square is equal to the secoiid poicer of one of its sides, as in this case the base and altitude are equal. 822. Cor. 2. — The area of any parallelogram i.^ equal to the pro- duct of its base into its altitude; for any parallelogram is equivalent to a rectangle of the same base and altitude {'J 13). 823, Cor. 3. — The area of a tria7igle is equal to one-half the pro- duct of its base and altitude; for a triangle is one-half of a parallelo- gram of the same base and altitude {314). 324, Cor. 4. — Parallelograms or triangles^ of equal bases are to each other as their altitudes ; of equal altitudes, as their bases ; and in general they are to each other as the products of their bases by their altitudes. PROPOSITION VII. 825 • Theorem, — The area of a traptezoid xs equal to the product of its altitude into one-half the sum of its parallel sides, or, what is the same thing, tlie product of its altitude and a line joining the middle points of its inclined sides. * This principle may be thus stated: An infinitesimal is a quantity conceived, and to be treated, as less than any assignable quantity ; hence, as added to or subtracted from finite quantities, it has no value. Thus, suppose — = a, w, n, and a bein;? finite quantities. Let c represent an infinitesimal; then — — -, or — ^, or "^^n.", i^ to be considered as still equal to n n ± c n r c a, for to consider it to differ from a by any amount we might name, would be to assign som4 value to c. t By this is meant the areas of the figures. 142 ELEMENTARY PLANE GEOMETRY. 71 B Dexi.— In the trapezoid ABCD draw either diagonal, as AC. It is thus divided into two triangles, whose areas are to- gether equal to one- half the product of their common altitude (the altitude of the trapezoid), into their bases DC and AB,or this altitude into i (AB + DC). Secondly, if ab be drawn bisecting AD and CB, then is «J = ^ (AB + CD). For, through a and h draw the perpendiculars om and pn^ meeting DC produced when necessary. Now, the triangles aoD and iKam are equal, since Aa = aD, angle o = m, both being right, and angle oaD = Aam being opposite. Whence Am = oD. In like manner we may show that Cp = nB. Hence, ab — \{pp + mn) = i(AB + DC); and area ABCD, which equals altitude into i(AB + DC), = aliitucU into ab. q. E. d. Fig. 228. PROPOSITION vm. 326. Theorem, — Tlie area of a regiilar polygon is equal to one- half the product of its apothem into its perimeter. Dem. Let ABCDEFG be a regular polygon whose apothem is Oa\ then is its area equal to i Oa (AB + BC + CD + DE + EF + PC + GA). Drawing the inscribed circle, the radii Oa, Ob, etc., to the points of tangency, and the radii of the circumscribed circle OA, OB, etc. {264, 205), the polygon is divided into as many equal triangles as it has sides. Now, the apothem (or radius of the inscribed circle) is the common altitude of these tri- angles, and their bases make up the perimeter of the polygon. Hence, the area = |Oa(AB + BC + CD 4- DE + EF -f- PC + CA). Q. E. D. B 3 C Fig. 229. 327, Cor. — The area of any polygon in which a circle can le inscribed is equal to one-half the product of the radius of the in- scribed circle into the perimeter. The student should draw a figure and observe the fact. It is especially worthy of note in the case of a triangle. See Fig. 60. PROPOSITION IX. 328. Theorem. — Tlie area of a circle is equal to one-half the product of its radius into its circumference. OF AKEA. 143 Dem. — Let Oa be the radius of a circle. Circum- scribe any regular polygon. Now the area of this polygon is one-half the product of its apothem and perimeter. Conceive the number of sides of the polygon; indefinitely increased, the polygon still continuing to be circumscribed. The apothem con- tinues to be the radius of the circle, and the perim- eter approaches the circumference. When, there- fore, the number of sides of the polygon becomes in- finite, it is to be considered as coinciding with the cir- cle, and its perimeter with the circumference. Hence the area of the circle is equal to one-half the pro- duct of its radius into its circumference, q. e. d. 329, Def. — A Sector is a part of a circle included between two radii and their intercepted arc. Similar Sectors are sectors in differ- ent circles, which have equal angles at the centre. 330, Cor. 1. — The area of a sector is eqital to one-half the product of the radius into the arc of the sector. 331, Cor. 2. — The area of a sector is to the area of the circle r?.5 the arc of the sector is to the circumference^ or as the angle of the sector is to four right angles. EXERCISES. 1. What is the area in acres of a triangle whose base is To rods and altitude 110 rods? 2. What is the area of a right angled triangle whose sides about the right angle are 126 feet and 72 feet? 3. If 2 lines be drawn from the vertex of a triangle to the base, dividing the base into parts which are to each other as 2, 3, and 5, how is the triangle divided ? How does a line drawn from an angle to the middle of the opposite side divide a triangle ? 4. Review the exercises on pages 49 and 50, giving the reasons, in each case. lU ELEMENTARY PLANE GEOMETRY. SYNOPSIS. pa o M f Definition. I Prop. I. Of parallelograms. •{ Cor. Paral. and rectangle. ■r. rr r\P * ' i i Coj'. 1. Triangle and rectangle. I Prop. II. Of triangles. ^ ^^^ ^ Of equal bases and equaUltitudea. Prop. III. Square on i, ^, i a line, etc. Prop. IV. Trapezoid. Prop. V. To reduce a polygon to a triangle. f Cor. 1. Of square. Cor. 2. Any parallelogram. Prop. VI. Of rectangle, i Cor. 3. Of triangle. I Coi: 4. Relation of parallelograms and [ of triangles. Prop. VII. Of trapezoid. Prop. VIII. Of regular polygons. Cor. Of any circumscribed polygon. Prop. IX. Of a circle. Def. Of sector. Cor. 1. Area of sector. Cor. 2. Relation of sector to circle. Exercises. SECTION X, OF SIMILARITY. 55^. The primary notion of similarity is UJceness of form. Two figures are said to be similar which hav-e the same shape, although they may differ in magnitude.* A more scientific definition is as follows : 333, Shnilar Ftr/ures are such as hare their angles respec- tively equal, and their homologous sides proportional. 334. Homoloffous Sides of similar figures are those which are included between equal angles in the respective figures. * The student phould be careful, at the outset, to mark the fact that nirniarify involves Itco things, equalitt op angles and proportionalitt op sides. It will appear that, in the case of triangles, if one of these facts exists, the other does also ; but this is not so in other polygons, as is illustrated in Part I. OF SIMILARITY. U5 In Similar Triangles, the Homologous Sides are those opposite the equal angles. PROPOSITION I. 33S. Theorem, — Triangles luliich are mutually equiangular .are similar. Dem.— Let ABC and DEF be two mu- tually equiangular triangles, in which A=D, B=E, and C = F; then are the sides opposite these equal angles propor- tional, and the triangles possess both requisites of similar figures ; ^. e., they are mutually equiangular and have their homologous sides proportional, and are consequently similar. To prove that the sides opposite the equal angles are proportional, place the triangle DEF upon ABC, so that F shall coincide with its equal C, CE'=FE, and CD'=FD. Draw AE', and D'B. Since angle CE'D'=CBA, D'E' is parallel to AB, and as the triangles D'E'A and D'E'B have a common base D'E' and the same altitude, their vertices lying in a line parallel to their base, they are equivalent {316), Now, the triangles CD'E' and D'E'A, having a common alti- tude, are to each other as their bases (324). Hence, Fig. 231. For like reason CD'E' : D'E'A : : CD' : D'A. CD'E' : D'E'B : : CE' : E'B. Then, since D'E'A and D'E'B are equivalent, the two proportions have a com- mon ratio, and we may write CD' : D'A : : CE' : E'B. By composition CD' : CD' + D'A : : CE' : CE' + E'B, or CD' : CA : : CE' : CB, or FD : CA : : FE : CB. In a similar manner, by applying angle E to B, we can show that I^E : CB : : ED : BA. Therefore, FD : CA : : FE : CB : : ED : BA. Q. E. D. S3G, CoR. 1. — If two triangles have tiuo angles of the one respec- tively equal to two angles of the other, the third angles heing equal (221), the triangles are similar. 337, CoR. 2. — A line drawn through a triangle parallel to any side divides the other sides proportionally. Thus D'E' being parallel to AB, it is shown in the proposition that CD': D'A ::CE':E'B. 10 146 ELEMENTAEY PLANE GEOMETRY. 338, CoE. 3. — If any two lines cut a series of paraUels, they are divided proportionally, O, /O' Dem. — If the two 'secant lines are parallel, as OA and O'B', the intercepted parts are equal, i. e.^ ac = hd, ee = dL eg = fh, etc. {24=2). Hence, ac : bd :: ce : df :: eg : fh. Secondly, if the secant lines are not parallel, let them meet in some point, as O. Then, by the propo sition, we have Oa : ac :: Ob : bd (1), and also Oc : ce :: Od : df (2). Taking the first by composition, it becomes Oa + ac : ac : : Ob + bd : bd, or Oc : ac :: Od : bd (3). Now, as the antecedents in (2) and (3) are the same, we have ac : bd :: ce : df, ot ac : ce :: bd : df. In like manner, we may show that ce : df : : eg : fh, or ce : eg :: df : fh. PROPOSITION n. 339, T7ieorein,—Con\eYse\y, If tiuo triangles have their cor- responding sides p)roportional, they are similar. Dem.— In the triangles ABC and DFE, let FD : CA : : FE : CB : : DE : AB ; then are the triangles similar. As one of the chai*acteristics of simi- larity, viz., proportionality of sides, exists by hypc-thesis, we have only to prove the other, i. e., that the triangles are mu- tually equiangular. Make CD' equal toFD, and draw D'E' parallel to AB. By the preceding proposition CD' (= FD) : CA : : D'E' : AB. But, by hypotliesis, FD : CA :: DE : AB. Whence, D'E' = DE. In like manner CE' : CB : : CD' (=FD) : CA. But, by hypothesis, FE : CB : : FD : CA. Whence CE' = FE ; and the tiian- gle CD'E' is equal to the triangle FDE {292). Now, CD'E' and CAB are mutu- ally equiangular, since D'E' is parallel to AB (152)^ and C is common. Hence. Fig. 2.33. OF SIMILARITY. 147 the triangles ABC and DEF are mutually equiangular, and consequently similar. Q. E. D. 34:0, Sen. — As we now know that if two triangles are mutually equiangular, they are similar; or, if they have their sides proportional, they are similar, it will be sufficient hereafter, in any given case, to prove either one of. these facts, in order to establish the similarity of two triangles. For, either fact being proved, the other follows as a consequence. See Section VI., Part I., for familiar illustra- tions of this most important subject. * — PROPOSITION III. 34:1, Theorem, — Tioo triangles which have the sides of the one respectively 2^(tralld or perpendicular to the sides of the other, are similar. Dem.— Let ABC and A'B'C be two triangles whose sides are respectively parallel or perpendicular to each other, then are the triangles similar. For, any angle in one triangle is either equal or supplementary to the angle in the other which is included between the sides which are parallel or perpendicular to its own sides. Thus A either equals A', or A + A' = 2 right angles {281, 282, 283). Now, if the corresponding angles are all supplemen- taiy, that is, if A + A' = 2 R.A., B + B' :=2R.A.,andC + C = 2 R.A., the sum of the angles of the two triangles is 6 right angles, which is impossible. Again, if one angle in one triangle equals the corresponding angle in the other, as A = A', and the other angles are supple- mentary, the sum is 4 right angles plus twice the equal angle, which is impossible. Hence, two of the angles of one ^triangle must be equal respectively to two angles of the other ; and, if two are equal, the third angle in one is equal to the third in the other {221). Hence, the tiiangles are mutually equiangular, and therefore similar {335). Q. e. d. PROPOSITION IV. "^42, OrJieorem. — Two triangles, which have an angle in each equal, and the sides about the equal a7igles proportional, are similat. 148 ELEMENTAEY PLANE GEOMETRY. Fig. 235. Dem.— In the triangles ABC and DEF let C = F, and AC : DF :: CB : FE; then are the triangles similar. For, place F on its equal C, and let D fall at D'. Draw D'E' parallel to AB. Then AC : D'C (= DF) :: BC : CE' (337). But by hypothesis AC : DF : : BC : FE. .'. CE' = FE, and the triangles D'CE' and DFE are equal (284). Therefore, D'CE' being equiangular with ACB, is similai to it (835) ; and as DFE is equal to D'CE', DFE is similar to ACB. q. e. d. PROPOSITION y. 343. Theorem, — In any right angled triangle, if a line he drawn from the right angle perpendicular to the hypotenuse, it divides the triangle into two triangles, which are similar to the given triangle, and consequently similar to each other. Dem.— Let ACB be a triangle right-angled at C, and CD a perpendicular upon the hypotenuse AB ; then are ACD and CDB similar to ACB, and consequently to each other. For, the triangles ACD and ACB have the angle A common, and a right angle in each; hence they are mutually equiangular, and consequently similar(556). For a like reason CDB and ACB are similar. Finally, as ACD and CDB are both similar to ACB, they are Q. E. D. A D B Fig. 236. similar to each other. 344, Cor. 1. — Either side about the right angle is a mean propor- tional letween the wJiole hypotenuse a7id the adjacent segment. Dem. — This is a direct consequence of the similarity of the partial triangles with the whole trinngle. Thus, comparing the homologous sides of ACD and ACB, we have AD : AC : : AC : AB ;* and from CDB and ACB, we have DB : CB :: CB : AB. 34o. Cor. 2. — The perpendicular is a mean proportional between the segments of the hypotenuse. Dem. — This is a consequence of the similarity of ACD and CDB. Thus, AD : CD : : CD : DB. * Notice that AD of the triangle ACD is opposite angle ACD, and AC, i^8 consequent, is of the triangle ACB, and opposite the angle B, which equals angle ACD- The student must be sure that he knows in what order to take the sides, and ivhy. OF SIMILARITY. 149 Queries.— -To -which triangle does the fii-st CD belong ? To which the second ? Why is CD made the consequent of AD ? Why, in the second ratio, are CD and DB to be compared ? 346, Cor. 3. — The square described on the hypotenuse of a right angled triangle is equivalent to the sum of the squares described on the other tiuo sides. Dem.— From Ow. 1, AC* =r AB x AD and also C B'' = ABxDB . Therefore, adding, AC' + CB' =AB (AD + DB) = AB". 547. Cor. 4. — If a perpe^idicular be let fall from any point in a circumference upon a diameter, this perpendic- ular is a mean ptroportional between the seg- ments of the diameter. Dem.— Thus, AD : CD : : CD : DB, or CD' =:AD x DB. For, drawing AC and CB, ACB is a right angle, and the case falls under Cor. 2. The chords AC and CB are mean proportionals between the whole diameter and their adjacent segments by Cor. 1. 34:8, ScH.— This proposition, with its corollaries, is perhaps the most fruit- ful in direct practical results of any in Geometiy. Coi'. 3 will be recognized as a demonstration of the Pj-thagorean proposition {109), Part I. There are many other demonstrations of exceeding beauty, some of which will be given in Part III. The one here given is the simplest, and shows best the way in which this truth grows out of the more general fact of similarity. PROPOSITION VI. w 340, Tlieorein, — Regular p)olygons of the same number of sides are similar figures. Dem. — Let P and P' be two regular polygons of the same number of sides,* a, 5, c, d, etc., being the sides of the former, and a', b\ c', d\ etc., the sides of the latter. Now, by the definition of regular polygons, the sides «, b, c, d, etc., are equal each to each, and also a\ b\ c\ d\ etc. Hence, we have a :a' : :b :b' : '. e: c' : : d: d', etc. Again, the angles are equal, since n being the number of sides of each polygon, each angle is 71 X 2 right angles — 4 right angles ^^-^/^n n Hence the pol5''gons are mutually equiangular, and have their sides proportional ; that is, they are similar, q. e. d. * The etudent may conetnict two regular hexagons, if thought desirable. 150 ELEMENTARY PLANE GEOMETRY. 350. Cor. 1. — Tlie corresponding diagonals of regular polygons of the same number of sides are in the same ratio as the sides of the polygons. Let the student draw a figure and demonstrate the fact. 351. Cor. 2. — The radii of the inscribed, and also of the circuni- P^^_^^g^ scribed circles, of regular polygons of the same nxiniber of sides, are in the same ratio as the sides of the polygons. Dem. — Since the angles F and/ are equal, and bisected by Fd the right angled triangles OSF, 0^/ are equiangular, and hence similar. There- fore FS : /5 : : SO : «0 or FO : /O. Whence, doubling both terms of the first couplet, Fig. 2^. FA : /a : : SO : sO or FO : /O. PROPOSITION ^TL ■Circles are similar figures. Dem. — Let Oa and OA be the radii of any two cii'cles. Place the circles so that they shall be concentric, as in the figure. Inscribe the regu- lar hexagons, as abcdef, ABCDEF. Conceive the arcs AB, BC, etc., of the outer circumference, bi- sected, and the regular dodecagon inscribed, and also tiie corresponding regular dodecagon in the inner circumference. These are similar figures hj {34:9). Now, as the process of bisecting the arcs of the exterior circumference can be con- ceived as indefinitely repeated, and the coiTespouding regular polygons as in- scribed in each circle, the circles may be considered as regular polygons of the same number of sides, and hence similar, q. e. d. 353. Cor. — Arcs of similar sectors are to each other as the radii of their circles; i. e., arc /e : arc FE : : 0/: OF. ScH. — The circle is said to be the Limit of the inscribed polygon, and the circumference the limit of the perimeter. By this is meant that as the number of the sides of the inscribed polygon is increased it approaches nearer and nearer to equality with the circle. The apothem approaches equality with the radius, and hence has the radius for its limit. It is an axiom of great importance in mathematics that, Whatever can he sfiown to he true of a magni- iude as it approaches its limit indefinitely, is true of that limit. OF SIMILARITY. 151 ABC B A EXERCISES. 1. l^Toh, — To divide a given line into ixirts which shall le proportional to several given lines. ' Solution. — Let it be required to divide OP into parts proportional to tlie lines A, B, C, and D. Draw ON making any convenient angle with OP, and on it lay off A, B, C, and D, in succession, terminating at M. Join M with the extremity P, and draw parallels to MP through the other points of division. Then by reason of the parallels we shall have A:B:C:D::«:6:c:d, {338). 2. I^rob. — To find a fourth proportional to three given lines. For the solution see {89). Repeat the process, and give the reasons. 3. I^rob, — To find a third propor- tional to tiuo given lines. Solution. — This may be solved as the two preceding. Thus, take any two lines, as A and B, for the given lines. We are to find a third line X, such that A : B : -. B . x. The figure will suggest the details. The following is a solution based on {347). Draw an indefinite line AM. Take AD = A, and erect BD = B. Join AB, and bisect it by the perpendicular ON. Then with O as a cen- tre, and OA as a radius, describe a'semi-circum- _ ference. This will pass through B. (Why?) M Also AD : BD :: BD : CD (= x). (Why?) 4. Draw any straight line on the blackboard, and divide it into 5 equal parts, upon the principle used in the preceding solutions. 5. Review the exercises under (89, 90), and give the reasons. 6. IProh, — To find a mean proportioiial betiveen two given lines. For the solution see {110). Repeat the process, and give q the reasons for the method. • 7. DE being parallel to BO, prove that the triangles DOE and BOG are similar, and hence that OD : 00 :: OE : OB. Are the following proportions true? OD : 00 :: OE : OB, OD : DE : : 00 : BC, OD : OE :: 00 : OB, OB : BC : : OE : DE. O D Fig. Si4\ 152 ELEMENTARY PLANE GEOMETRY. 8, Show that if ABCDEF is a regular polygon, kbcdef is also regular, Ic, cd, etc., being parallel to BC, CD, etc. Show that any two similar polygons may be placed in similar relative positions, and hence show that the corresponding diagonals are in the same ratio as the homologous sides. Fig. iM3. homologous with latter ? 9. The sides of one triangle are 7, 9, and 11. The side of a second similar triangle, side 9, is 4J. What are the other sides of the 10. The diameter of a circle is 20. What is the perpendicular distance to the circumference from a point in the diameter 15 from one extremity ? What are the distances from the point where this perpendicular meets the circumference to the extremities of the diameter? SYNOPSIS. O Primary notion of similarity. Definition of similarity. Homogeneity of sides. In general. 5 r S3 fa 5 S § Prop. L Mutually equiangular In triangles. {Cor. 1. Two angles equal. Cor. 3. A parallel to a side. Cor. 3. Lines cutting parallels. ■I Sell. Either of two facts sufficient Prop. II. Sides proportional. Prop. III. Sides parallel or perpendicular. Prop. IV. An angle equal in each, and sides proportional. ' Cor. 1. Side about right angle. Prop. V. Perpendicular from right angle upon hypotenuse. Prop. VI. Regular polygons similar. Prop. VII. Circles similar. Cor. 2. Perpendicular, Cor. 3. Square on hypotenuse. Cor. 4. Perpendicular on diameter. Sch. Importance of this Prop, and Cor's. Cor. 1. Corresponding diagonals. Cor. 2. Radii of inscribed and cir- cumscribed circles. Sch. Circle limit of polygon. Exercises. Prob. To divide a line into proportional parts. Prob. To find a fourth proportional. Prob. To find a third proportional. Prob. To find u mean proportiouiil. APPLICATIONS OF THE DOCTRINE OF SIMILARITY. 153 SECTION XL APPLICATIONS OF THE DOCTRINE OF SIMILARITY TO THE DEVELOPMENT OF GEOMETRICAL PROPERTIES OF FIGURES. 354:, The doctrine of similarity, as presented in the preceding section, is the chief reliance for the development of the geometrical properties of figures. This section will be devoted to the investiga- tion of a few of the more elementary properties of plane figures, which we are able to discover by means of this doctrine. OF THE RELATIONS OF THE SEGMENTS OF TWO LINES INTERSECT- ING EACH OTHER, AND INTERSECTED BY A CIRCUMFERENCE. PROPOSITION I. 355. Tlieovenu — If tioo cliords intersect each other in a circle, their segments are reciprocally j)v02)ortional ; whence the product of the segments of one chord equals the 2)roduct of the segments of the other. Dem.— Let the chords AC and BD inter8ect at ; then is AG : BO : : DO : CO, whence AO x CO == BO x DO. For, draw AD and BC. The two triangles AOD and BOC ^^'Jt~^\ are equiangular, and hence similar ; since the angles at O /^ /"\ J^ are vertical, and consequently equal {134), and D = C, / / ^ because both are measured by i arc AB {210). (A = B / / /^^ because both are measured by i arc DC ; but it is only I /y^ ^ necessary to show that two angles are equal in order to }\ show that the triangles are equiangular, and hence simi- \ lar.) Now, comparing the homologous sides (those oppo- ^ ^-^B site the equal angles), we have AO : BO : : DO : CO ; Fig. 244. whence, AO x CO = BO x DO. q. e. d. Queries.— Why is AO compared with BO? Why DO with CO? Would AO : CO : : BO : DO be true? Would AO : DO : : BO : CO? What is the force of the word "reciprocally," as used in the proposition? 154 ELEMENTARY PLANE GEOMETRY. PROPOSITION n. 356. TJieoreni, — If from a point without a circle, two seca7its he draiun terminating in the concave arc, the lohole secants are recip- rocally proportional to their external segments; ivhence the product of one secant into its external segment equals the product of the other Q^ into its external segment. Dem. — OA and OB being secants, OA : OB : : OC : OD, and consequentlj' OA x OD = OB x OC. For, drawing AC and DB, the two triangles AGO and BDO have angle common, and A = B, since both are measured bj' \ DC ; hence the triangles are similar, and we have OA : OB : : OC : OD, and consequently OA x OD=;OB x OC. q. e. d. Same queries as under the preceding demon stra- FiG. 245. tion. PROPOSITION in. ^ 357. TJieorem. — If from a point without a circle a tangent he dravm, and a secant terminating in the concave arc, the tangent is a mean proportio?ial hetween the whole secant and its external seg- ment; whence the square of the tangent equals the product of the secant into its external seg- ment. Dem.— OA being a tangent and OB a secant, OB : OA : : OA : OC, whence OA^ = OB x OC. For, drawing AB and AC, the triangles OAB and ACO have angle O common, and OAC = B, since each is measured by i arc AC; hence the triangles are similar, and OB : OA : : OA : OC, whence OV = OB x OC. q. e. d. Fig. 246. OF THE BISECTOR OF AN ANGLE OF A TRIANGLE. PROPOSITION IV. 358. Tlieorein. — A line whicli bisects any angle of a triangle divides the opposite side into segments proportional to the adjacent sides. BISECTOR OF AN ANGLE OP A TRIANGLE. 155 Dem.— Let CD bisect the angle ACB; then AD : DB ; : AC : CB. For, draw BE parallel to CD, and produce it till it meets AC produced, as at E. Now, by- reason of the parallels CD and EB, angle ACD = AEB, and angle DCB = CBE {152). Whence, as ACD = DCB by hypothesis, E = CBE, and CE = CB {227)- Also, since CD is parallel to EB, AD : DB : : AC : CE {337). Substituting for CE its equal CB, we have AD : DB : : AC E. D. PROPOSITION Y. S50» Tlieorenn, — If a line le draivn from any vertex of a triangle hisecting the exterior a?igle and intersecting the ojjposite side jyroduced, the distances from the other vertices to this intersection are lyrojportional to the adjacent sides. Dem.— Through the vertex C let CD be drawn, bisecting the exterior angle FCB, and intersecting AB produced in D ; then AD : BD : : AC : CB. For, draw BE parallel to CD. By rea- son of these parallels, angle FCB=: CEB, andBQD=CBE(i5;?). Hence CEB = CBE, and CB = CE. Also, by reason of the parallels, AD : BD :: AG : CE, or its equal CB (555). Q. E. d. PROPOSITION yi. 360, TJieorem. — If a line be draiun hisecting any angle of a triangle and intersecting the opposite side, the rectangle of the sides ahoiit the bisected angle equals the rectangle of the segments of the third side,pflf^s the square of the bisector. Dem.— Let CD bisect the angle ACB ; then AC x CB =^ AD X DB + CD". For, circumscribe the circle about the triangle, pro- duce the bisector till it meets the circumference at E, and draw EB. The triangles ADC and CBE are simi- lar, since angle ACD = EQB, by hypothesis, and A = E, because each is measured by i arc CB. Therefore, AC : CE : : CD : CB, whence AC x CB = CE x CD = (DE + CD) CD = DE x CD + CD*. For DE x CD, substituting its equivalent AD x DB {355), we have AC X CB = AD X DB +. CD^ Q. e. d. -\B E. Fig. a49. 156 ELEMENTARY PLANE GEOMETRY. PROPOSITION yn. 361, Theorem, — The bisectors of the angles of a triangle all pass through the same point, which is the centre of the inscribed circle. Dem. — Draw two lines bisecting two of the angles, and from their inter- section draw a line to the other angle. Then show that the latter angle is bisected. By (Ex. 4, page 134) this point is shown to be the centre of the in- scribed circle. [The student should fill out the demonstration.] AEEAS OF SIMILAR FIGURES. PROPOSITION Yin. 362, Theore'in, — Tlie areas of similar triangles are to each other as the squares described on their homologous ^ides. Dem. — Let ABC and DEF be any two similar triangles ; then is area ABC : area DEF : : CB' : FE" : : AC ' : DF' : • AB" : DE' For, place the largest angle of the ti'iangle DEF, as D, on its equal angle A, of the triangle ABC*; let E fall at E', F at F', and draw ET'; then is triangle AE'F' = DEF {284), and E'F' is parallel to BC. Let fall a pei-pendicular from A to CB. Then Al is the altitude of AE'F', and AH of ABC. Now, by similar triangles we liave CB : F'E' : : AH : Al. But iAH : iAI : : AH : Al ; and, multiplying corresponding terms, ^ AH x CB : ^Ai x F'E' : : AH*: AlV Whence, since •^AH X CB = area ABC, and lAI x F'E' = area AE'F'= area DEF, and AH : Al : : CB : FE : : AC : DF : : AB : DE, or AH* :"aI' : : CB' : FE' : : A'c' :~DF" Q. E. D. PROPOSITION IX. 363, Hieoreni, — The areas of similar polygons are to each other as the squares of the homologous sides of the polygons. * The only object in taking the largest angle is to make the perpendicular fall wUhin the triangle. Any two equal angles may be applied, and the demonstration is essentially the same. ^ : Al: :CB FE ::AC: DF:: AB:DE or ah' : Al' ::CB' :FE' AB':: DE% we have area ABC : area DEF :: CB ^ FE^: AC' : DF' : : AB': DE'. AREAS OF SIMILAR FIGURES. Dem.— Let ahcdef and ABCDEF be two similar polygons. Desig- nate the former by p, and the lat- ter by P. Then p: P ::ab' :: AB^* or as the squares of any other two homologous sides. For, from the equal angles a and A drawing the diagonals, the corresponding partial angles into which a and A are divided are yig. 2o1. equal. [Let the student show why hj 34:2^] Now take A&'=«6, and draw I'c', making angle A6'c' = h. Then h'c' = be, and Ac' = ac, since the triangles abc and A&'c' have two angles and the included side of one equal to two angles and the included side of the other. In like manner draw c'd' making angle b'c'd'= bed, e'd'= ed, and AcZ'= ad. So, also, making angle e'd'e'= cde, and angle d'e'f = def, die' — de, e'f — ef, and /'A =fa. Hence the polygon Nb'c'd'e'f'=p, and its sides are respectively parallel to the corresponding sides of P. Now, let m, n, 0, and s represent the triangles in which they stand, and M, N, 0, and S the corresponding triangles of P, as AFE, etc. Triangles m and M being similar, and also n and N, we have m : M : Cf^^ : AE", and ti : N : : a7'' : AE*. Whence w : M : : ?t : N. In like manner we can show that 7i : N : : o : 0, and that o : : : 8 :S. Whence m : M : : n -M : - o : : : s : S. By composition, (m + n + o + s) (or j9) : (M + N + f S) (or P) : : « : S, But « : S : ~M'^ {ov~ab^) :'AB\ Therefore p : P : :'ab^ : AB\ or as the squares of any two homologous sides. Q. e. d. 364, CoR. 1. — Similar polygons'^ are to each other as the squares of their c07'responding diagonals. In the demonstration we have « : S : : Ac* (or ac) : AC^ Whence p : P : : ac" : AC*. The same may be shown of any other corresponding diagonals. SOS, Cor. 2. — Regular polygons* of the same numler of sides are to each other as the squares of their homologous sides ; since they are similar figures (349). 366, Cor. 3. — Regular polygons* of the same number of sides are to each other as the squares of their apothems. For their apothems are to each other as their sides. Hence the squares of their apothems are to each other as the squares of their sides. 367, Cor. 4. — Circles are to each other as the squares of their radii {3d2), and as the squares of their diameters. * This is a common elliptical form for " The areas of, etc." 158 ELEMENTARY PLANE GEOMETRY. OF PERIMETERS AND THE RECTIFICATION OF THE CIRCUxMFERENCE. 368, Tlie Rectification of a curve is the process of finding its length. The term rectijication signifies making straight, and is applied as above, under the conception that the process consists in finding a straight line equal in length to the cuitc. PROPOSITION X. 369* Hieorem, — The perimeters of similar polygons are to each other as their homologous sides, and as their corresponding diagonals. Dem.— Let a, J, c, d, etc., and A, B, C, D, etc., be the homologous sides of two similar polygons whose perimeters are p and P ; then p : P : : a : A : : b : B : : c : C, etc. ; and r and R being corresponding diagonals, p : P : : r : B. Since the polygons are similar, a : A : . b : B : : c : C : : rf : D, etc. By composition, {a + b + c + d+ etc.) (or ^) : (A + B + C + D + etc.) (or P) : : a : A, or as any other homologous sides. Also, as the homologous sides are to each other as the corresponding diagonals {350), p : P : : r : M. Q. e. d. 370, Cor. 1. — The perimeters of regular polygons of the same numder of sides are to each other as the apothems of the polygons. For the apothems are to each other as the sides of the polygons {351). 371, Cor. 2. — Tlie circumferences of circles are to each other as their radii, and as their diameters ; since they may be considered as regular polygons of the same number of sides (352), PROPOSITION XI. 372, Theorem, — The circumference of a circle tvhose radius is 1, is 27r. the numerical value of n being approximately 3.1416. Dem. — We will approximate the circumference of a ch-cle whose radius is 1, by obtaining, 1st, the perim- eter of the regular inscribed hexagon ; 2d, the perim- eter of the regular inscribed dodecagon; 3d, the perimeter of the regular inscribed polygon of 24 sides ; then of 48, etc. In order to do this, let us find the relation between the chord of an arc and the chord of \ the arc in a Fig. 252. circle whosc radius is 1. Call the chord of an arc as RECTIFICATION OF CIRCUMFERENCE. 159 AB, (7, and the cliorcl of half the arc, as CB, c. Now, BDO is right angled at D, whence "BO' ="60' + DO'' (346), or DO = ^/bo'-BD". But in the present case BO = 1 ; hence DO = VT^W\ Taking DO from CO, we have CD = 1 - -/T^fC*- From the right angled triangle BDC we have CB (or c) = -/bd" + W, or substituting i G for BD, and 1 - VT^^JC^ for CD, this re- duces to c=v'2^~P5z^^ or, [2-(4-0'2)^]^ By the use of this formula, we make the following computations : No. sides. Form of Compntetion. Length of Side. Perimeter. 6. ^ee{271) 1.00000000 6.00000000 12. e=^^Zvi^= V2- Vd, or (2-3')^ . = .51763809 6.21165708 24. c= ■J2-[4-(2-3^)]^p = [2-(2 + 3^)']^ . = .26105238 6.26525722 = j 2- [2 + (2 + 8^)']^ } ^ = .13080626 6.27870041 96. c = (2- -j 2 + [2 + (2 + 8')']^ \^y . . . = .06543817 6.28206396 192. c = [2-(2+-J2 + [2 + (2 + 3^y]'p)*]' . . =.03272346 6.28290510 384. c= ]2-[2 + (24-]2 + [2 + (2-f3'y]'p)']4' = -01636228 6.28311544 768. c = (2--{2 + [2 + (2+|2 + [2 + (2 + 3^)]4^")']4')'= -00818121 6.28316941 It now appears that the first four decimal figures do not change as the num- ber of sides is increased, but will remain the same how far soever we proceed. We may therefore consider 6.28317, as approximately the circumference of a circle whose radius is 1, i. e., 27r = 6.28317, nearly ; and re = 3.1416, nearly. 373. Sen. — The symbol 7t is much used in mathematics, and signifies, primarily, tJie semi-circumference of a circle w/iose radius is 1. ^tc \s therefore a symbol for a quadrant, 90°, or a right angle. :| ;r is equivalent to 45°, etc., the radius being always supposed 1, unless statement is made to the contrary. The numerical value of 7t has been sought in a great variety of ways, all of which agree in the conclusion that it cannot be exactly expressed in decimal numbers, but is approximately as given in the proposition. From the time of Archimedes (287 B.C.) to the present, much ingenious labor has beeu bestowed upon this problem. The most expeditious and elegant methods of approxi- mation are furnished by the Calculus. The following is the value of tc extended 10 fifteen places of decimals : 3.141592653589793. PROPOSITION XII. 374. TJieoretn, — The circumference of any circle is %7tr, r being the radius. Dem.— The circumferences of circles being to each other as their radii, and 160 ELEMENTARY PLANE GEOMETRY. 27t being the circumference of a circle whose radius is 1, we have 27t : circf. : : 1 : r, whence circf. — 27tr. 375. Cor. — The circumference of a circle is ttD, D being the diameter, since 2;rr = ;r2r = nD. ABEA OF THE CIRCLE- PROPOSITION xni. 37 6 » TJieorem, — The area of a circle ivhose radius is l,is n, Dem.— The area of a circle is | r x circumference {328). When r = 1, dr^ cumference = 2it {372) ; hence area of circle icTiose radius isl ■= \x2Tt = Tt. q. e. d. PROPOSITION xiy. 577. Theorem, — Tlie area of any circle is nr^, r leing the radius. "' - Dem. — The areas of circles being to each other as the squares of their radii, and 7t being the area of a circle whose radius is 1, we have TT : aj'ea of any circle : : 1" : ?*", whence area of any circle = nr^. q. e. d. 378, Cor. — Tlie area of any sector is such a f)art of the area of the circle as the angle of the sector is of four right angles. 370. ScH. — As the value of tt cannot be exactly expressed in numbers, it follows that the area cannot. Finding the area of a circle has long been known as the problem of Squaring the Circle, i. e., finding a square equal in area to a circle of given radius. Doubtless many hare-brained visionaries or igno- ramuses will still continue the chase after the phantom, although it has long ago been demonstrated that the diameter of a circle and its circumference are incommensurable. It is, however, an easy matter to conceive a square of the same area as any given circle. Thus, let there be a rectangle whose base is equal to the circumference of the circle, and whose altitude is half tlie radius; its area is exactly equal to the area of the circle. Now, let there be a square whose side is a mean proportional between the altitude and base of this rect- angle ; the area of the square is exactly equal to the area of the circle. APPLICATIONS OF THE DOCTRINE OF SIMILARITY. ICl EXERCISES. 1. Show that if a chord of a circle is conceived to revolve, varying ^ _in length as it revolves, so as to keep its extremities in the circum- ference while it constantly passes through a fixed point, the rect- angle of its segments remains constant. 2. The two segments of a chord intersected hy another chord are 6 and 4, and one segment of the other chord is 3. What is the other segment of the latter chord ? 3. Show how Prop's L, IL, and III. may be considered as differ- ent cases of one and the same proposition. Sxjg's.— By stating Propositions I. and II. tlius, The distances from the inter- section ofthelines to their intersections with the circumference, what follows? In Fig. 245, if the secant AO becomes a tangent, what does OD become? 4. In a triangle whose sides are 48, 36, and 50, where do the bisec- tors of the angles intersect the sides ? 5. In the last example find the lengths of the bisectors. 6. Keview the examples under {III9 112, 113, 114), and give the reasons. 7. In a circle whose radius is 20, what is the length of the arc of a sector whose angle is 30° ? What is the area of this sector ? 8. If a circle whose radius is 24 is divided into 5 equal parts by concentric circumferences, what are the diameters of the several cir- cles ? If the radius is r, and number of parts 71 ? 9. I*rob,—To divide a line in extreme and mean ratio ; that is, so that the whole line shall he to the greater segment, as the greater segment is to the less. Solution.— Let it be proposed to divide the line AB in extreme and mean ratio. At one extremity of the line, as B, erect a perpendicular equal to half the line, that is, make BO = ^ AB. With as a centre, describe a circumference passing through B. Draw AO, and take AC equal to AD. Then is AB divided in extreme and mean ratio at C, so that AB : AC : : AC : CB. To prove it, produce AO to E. Now, AE : AB : : AB : AD {357\ or by inver- sion, AB . AE : : AD : AB. By division, AB : AE - AB (= AE - DE) (or AD) (= AC) : : AD (= AC) : AB - AD (= AB - AC) (or CB). That is, AB : AC : : AC : CB. 11 16:^ ELEMENTAP.Y PLANE GEOMETRY. 10. I^rob* — To inscribe a regular decagon in a circle, and hence a, regular pentagon, and regular j^olygons of 20, 40, 80, etc., sides. Solution. {a) X .:^ Fig. 254. -Divide the radius iu extreme and mean ratio, as at (a). Then is the greater segment ac the chord of a regular inscribed decagon, as ABCD,ctc. To prove this, draw A and OB, and taking OfA = ac = AB, a side of the polygon, draw BM. Now, OA : OM : : OM : MA by con- struction. As OM = AB^ we have OA : AB ; : AB : MA. Hence, con- sidering the antecedents as belong- ing to the triangle OAB, and the consequents to the triangle BAM we observe that the two sides about the angle A, which is common to botli tri- angles, are proportional, hence the triangles are similar (342). Therefore, ABM is isosceles, since OAB is, and angle BMA = A = OBA, and MB = BA = OM This makes 0MB also isosceles, and the angle O = OBM. Again the exterior angle BMA — + OBM = 20; hence A, which equals BMA = 20. Hence also OBA, which equals A, = 20. Wherefore is i the sum of the angles of the triangle OAB, or i of 2 right angles, = tV of 4 right angles. The arc AB is, therefore the measure of -^ of 4 right angles, and is consequently -A- of the circumference. To consti'uct the pentagon, join the alternate angles of the decagon. To construct the regular polygon of 20 sides, bisect the ai-cs subtended by the sides of the decagon, etc. 11. The projection of one line upon another in the same plane is the distance between the feet of two perpendiculars let fall from the extremities of the former upon the latter. Show that this pro- jection IS equal to the square root of the difference between the square of the line and the square of the difference of the perpen- diculars. 12. In the triangle ABCi? being a perpendicular upon BA, prove that 7)1 + n {= c) '. a + 1) '.: a — b'.m — n. State the fact as a proposition. Give the neces- sary modification when the perpendicular falls without the triangle. SuG. a' — 7n* = &* — n*, whence a' — &' = m' — n", etc. 13. The three sides of a triangle being 4, 5, and 6, find the seg- ments of the last side, made by a perpendicular from the opposite angle. Ans. 3.75, and 2.25. SYN-OPSIS. 163 14. Same as above when the sides are 10, 4, and 7, and the perpen- dicular is let fall from the angle included by the sides 10 and 4. Draw the figure. Why is one of the segments negative ? 15. What is the area of a regular octagon inscribed in a circle whose radius is 1? What is its perimeter? What if the radius is 10? 16. What is the side of an equilateral triangle inscribed in a circle whose radius is 1 ? SYNOPSIS, W H O CO » « H I— ( sg « O 02 ., W ^ H »-( « 23 W H ft, O O O 5 K H O Importance of this doctrine. 1 S a ^ 2 s z H H W w o § Prop. I. Of chords. Prop. II. Of secants. Prop. III. Of secants and tangents. o U H J W fc o S « to <=> ^ w cc J s; < o !3 w Prop. IV. How divide sides. Prop. V. Of exterior angles. Prop. VI. Length of in relation to other parts. Prop. VII. All intersect at one point. l' Prop. VIII. Of triangles. Prop. IX. Of polygons. < Definition of rectification. Prop. X. Perimeters of similar poly- gons. Prop. XI. Rectification of circum- ) c^i ference whose radius >• isl. ) Prop. XII. Circumference of any circle = 27rr. Cor. 1. As squares of diagonals. C'(9?'. 2. Regular polygons. Cor. 3. As squares of apothems. Cor. 4:. Of circles. Cor. 1. Regular polygons. Cor. 2. Circumferences as radii. Signification and importance of tc. ^Cor. Also icD. Prop. XIII. Whose radius is 1 Prop. XTV. Of any circle, j Exercises. Cor. Of sector. Sch. Squaring the circle. Prob. To divide a line in extreme and mean ratio. Prob. To inscribe a regular decagon, etc. 164 ELEMENTARY SOLID GEOMETRY. OHAPTEE II. SOLID GEOMETRY.* SECTION I OF STRAIGHT LINES AND PLANES. PERPENDICULAR AND OBLIQUE LINES. 880* Solid Geometry is that department of geometry in which the forms (or figures) treated are not limited to a single plane. 381, A JPlane (or a Fla7ie Surface) is a surface such that a straight line joining two points in it lies wholly in the surface. III.— The surface of the blackboard is designed to be a plaDC. To ascertain ■whether it is truly so, take a ruler with a straight edge, and apply this edge in all directions upon it. If it always coincides, i. e., touches throughout its whole length, the sui-face is a plane. Is the surface of the stove-pipe a plane ? "Will a straight line coincide with it in any direction? Will it in every direction ? PROPOSITION I. 382. Theorem, — Tliree points not in the same straight line determine the position of a i^lane, Dem.— Let A, B, and C be three points not in the same straight line ; then one plane can be passed through them, and only one ; i. e., they determine the position of a plane. * In some respects, perhaps, "Geometry of Space" is preferable to this term; but, as neither is free from objections, and as this has the advantage of simpUcity and long use, the author prefers to retain it. OF STRAIGHT LINES AND PLANES. 165 For, pass a straight line through any two of these points, as A and B. Now, conceive any plane con- 'C taining these two points; then will the line passing through them lie wholly in the plane (381). Con- ceive this plane to revolve on the line as an axis until the point C falls in the plane. Thus we have one plane passed through the three points. That there can be only one is evident, since when C tails in the plane, if it be revolved either way, C will not be in it. ^^^- 255. The same may be shown by first passing a plane through B and C, or A and C. There is, therefore, only one position of the plane in which it will contain the third point, q. e. d. 383. Cor. 1. — Through one line, or two points, an infinite nuni' her of planes can he passed, 384:. Cor. 2. — Two intersecting lines determine the position of a plane. Dem. — For, the point of intersection may be taken as one of the three points requisite to determine the position of a plane, and any other two points, one in each of the lines, as the other two requisite points. Now, the plane passing through these points contains the lines, for it contains two points in each line. 385. Cor. S.^Ttvo parallel lines determine the position of a plane. Dem.— For, pass a plane through one of the parallels, and conceive it revolved until it contains some point of the second parallel ; then as the plane cannot be revolved either way from this position without leaving this point without it, it is the only plane containing the first parallel and this point in the second. But parallels lie in the same plane {66), whence the plane of the par- allels must contain the first line, and the specified point in the second. There- fore, the plane containing the first line and a point in the second is the plane of the parallels, and is fixed in position. 386. Cor. 4. — The intersection of tioo pla^ies is a straight line. For two planes cannot have even three points, not in i7ie same straight line^ common, much less an indefinite number, which would be required if we con- ceived the intersection (that is, the common points) to be any other than a straight line. 387. A I^erpenclicular to a JPlane is a line which is perpendicular to all lines of the plane passing through its foot. Conversely, the plane is perpendicular to the line. 166 ELEMENTARY SOLID GEOMETBT. PROPOSITION n. 388. Tlieorem. — A line which is perpendicular to two lines of a plane, at their intersection, is perpendicular to the plane, Dem.— Let PD be perpendicular to AB and CF at D ; then is it perpendicular to MN, the plane of the lines AB and CF. Let OQ be any other line of the plane MN, passing through D. Draw FB intersecting the three linea AB, CF, and OQ. Produce PD to P', making P'D = PD, and draw PF, PE, PB, P'F, P'E, and P'B. Then Ls PF = P'F, and PB = P'B, since FD and BD are per- pendicular to PP', and PD = P'D {284:). Hence, the triangles PFB and P'FB are equal {292)\ and, if PFB be revolved upon FB till P falls at P', PE will fall in P'E. Therefore OQ has E equally distant from P and p', and as D is also equidistant from the same points, OQ is pei-pendicular to PD at D {130). Now, as OQ is any line, PD is perpendicular to any line of the Fig. 226. plane passing through its foot, and consequently per- pendicular to the plane {387). Q- e. d. 389. Cor. — If one of two perpendiculars revolves alout the other as an axis, its path is a plane perpe?idicular to the axis. Thus, if AB revolves about PP'as an axis, it describes the plane MN. P f. M iK / K 'J>^ 1 p^ tri P' f PROPOsmox nii 300, Theorem, — At any point in a plane one perpendicular can he erected to the plane, and only one. Dem. — Let it be required to show that one perpen- p ^ dicular, and only one, can be erected to the plane MN at D. Through D draw two lines of the plane, as AB and CE, at right angles to each other. CE being perpendicular to AB, let a line be conceived as starting from the position ED to revolve about AB as an axis. It will remain perpendicular to AB {389). Conceive it to have passed to P'D. Now, as it con- tinues to revolve, P'DC diminishes continuously, and at the same rate as P'DE grows greater; hence, in one position of the revolving line, and in only one, as PD, PDE will equal PDC, and PD will be perpendicular to CE. Therefore, PD is peri:)endicular to two lines of the plane, at their intersection, and is the only line that can be thus perpendicular, whence it is perpendicular to the plane {388), and is the only perpendicular, q. e. d. M r /> PE'. For, take ED = E'D ; then PE = PE', Fia. 258. by the preceding part of the demonstration. But PF > PE by {139). Hence, PF > PE'. q. e. d. 392, The Inclination of a line to a plane is measured by the angle which the line makes with a line of the plane passing through the point in which the line pierces the plane and the foot of a perpendicular to the plane from any point in the line. Thus PFD is the inclination of PF to the plane MN. 393, Cor. 1. — The angles luhich oUiquc lines draion from a com- mon point in a perpendicular to a plane, and ptiercing the plane at equal distances from the foot of the perpoidiciilar, make with the perpendicular, are equal ; and the inclinations of such lines to the plane are equal. Thus the equality of the triangles, as shown in the demonstration, shows that EPD = E'PD = E"PD = E"'PD, and PED = PE'D = PE"D = PE"'D. 394, Cor. 2. — Conversely, If the angles which oilique lines draion from a point in a perjjendicular to a plane, make toith the perpendicular, are equal, the lines are equal, and p>ierce the plane at equal distances from the foot of the perpendicular. DEM.-Thus, in the figure, let DPE' = DPE"; then PE' = PE" and DE' = DE". For, revolve DE'P about PD ; DE' will continue in the plane MN, and when angle DPE' coincides with its equal DPE", PE' coincides with PE", and DE' with DE '. 168 ELEMENTABY PLANE GEOMETRY. 39o, Cor. 3. — Also, conversely, Equal ollique lines from the same point in the perpendicular, pierce the plane at equal distances from the foot of thep)erpendicular, Dem.— Let PE' = PE" ; then is DE' = DE". For, let PDE' revolve npon PD until ED foils in E"D ; then, if DE' were less thaa DE", PE' would be less than PE" ; and, if DE' were greater than DE", PE' would be greater than PE". But both of these conclusions are contrary to the hypothesis. Hence, as DE' can neither be less nor greater than DE ", it must equal it. This corollary follows also from {297). 396, Cor. 4. — The perpendicular is the shortest line that can le drawn to a plane from a point without, and measures the distance of the p)oint from the p lane. PROPOSlllOX T. 397 • TJieoreui, — From a point ivithout a plane one perpendic- ular can he drawn to the plane, and only one, Dem. — Let it be required to show that one perpen- dicular can be drawn from P to the plane MN, and only one. Take AB, any line of the plane, and con- ceive PD' perpendicular to it. Through D' draw EF, a line of the plane, perpendicular to AB, Now, if PDE = PD'F, they are both right angles, and PD' is perpendicular to two lines of the plane passing through its foot, and hence perpendicular to the plane (388). If, however, PD'E does not equal PD'F, in the first in- stance, but PDF < PDE, conceive the line AB to move along the plane, continuing parallel to its primitive position, so as to cause D' to move towards F, thus diminishing PDE and increasing PD'F. At the same time observe that, if necessary in order to keep PD A = PD'B*, EF can move along the plane parallel to its first position. Now, as PD F increases, passing through all successive values, and PD'E dimin- ishes in the same way, there will be some position of PD', as PD, in which PDF = PDE, and as by hypothesis PDA' remains = PDB', PD becomes perpen- dicular to two lines passing through its foot, and hence perpendicular to the plane. That there can be only one perpendicular is evident, since, if there were two, as PD' and PD, there would be two right augles in the triangle PD D. * According to the conception here used it would not be necessary. PARALLEL LINES AND PLANES. • 169 S98, Cor. — Tlirough a given point in a line one plane can be passed perpendicular to the line, and only one. Dem. — Let D be the point in the line PD. Pass two lines through D, as EF, and A''B', each perpendicular to PD ; the plane of these lines is perpendicular to PD. Moreover, the required plane must contain both these lines, for if it passed through D and did not contain DF, there would be one line of the plane, at least, which would pass through D and not be perpendicular to PD, which is impossible. Hence, there can be no other plane than the plane of the two per- pendiculars EF and A'B' which shall be perpendicular to PD, through D. PROPOSITION TI. 399. Theorem, — If from the foot of a perpendicular to a plane a line he drwwn at right angles to any line of the pilcLne, and this intersection he joined with any point in the •perpendicular, the last line is perpendicular to the line of the plane* Dem. — From the foot of the perpendicular PD let DE be drawn perpendicular to AB, any line of the plane MN, and E joined with 0, any point of the per- pendicular ; then is OE perpendicular to AB. Take EF = EC, and draw CD, FD, CO, and FO. Now, CD = DF (?)* whence CO = FO (?), and OE has O equally distant from F and C, and also E. There- Fig. 2G0. fore, OE is perpendicular to AB (?). q. e. d. 4:00. Cor. — The line DE measures the shortest distance between PD a7id AB. For, if from any other point in AB, as C, a line be drawn to D, it is longer than DE(?) ; and if drawn from C to a, any other point in PD than D, Ca is longer than CD (?), and consequently longer than DE (?). PARALLEL LINES AND PLANES. 401, A Line is Parallel to a plane when the two will not meet, how far soever they be prodnced. The plane is also said to be parallel to the line. * Hereafter the reason will be often left out, and the mark (?) will be used to indicate that the student is to supply it. 170 ELEMENTARY SOLID GEOMETRY. ;^ @ PROPOSITION Til. ^ 402. Hieore^n. — One of tim parallel lines is parallel to every plane containing the other. Fig. 261. Dem. — AB being parallel to CD is parallel to 1 any plane MN containing CD. / Since AB and CD are in the same plane (?)^ and as the intersection of their plane with MN is [sj CD (?), if AB meets the plane MN, it must meet it in CD, or CD produced. But this is impossi- ble (?). Whence AB is parallel to MN. q. e. d. 403. Cor. l. Tlirough any given line a plane may be passed parallel to any other given line not in the plane of tlie first. For, through any point of the line through which the plane is to pass, con- ceive a line parallel to the second given line. The i)lane of the two intersecting lines is parallel to the second given line (?). 404, Cor. 2. — Tlirough any point in space a plane may Repassed parallel to any two lines in space. For, through the given point, conceive two lines parallel to the given lines ; then is the plane of these intersecting lines parallel to the two given lines (?). PROPOSITION Tm* 405 • Theoreifn. — If 07ie of two parallels is perpendicular to a plane, the other is perpendicular also. Dem.— Let AB be parallel to CD, and perpendicular to the plane MN ; then is CD perpendicular to MN. For drawing BD in the plane MN, it is perpendicular to AB (?), and consequently to CD (?). Through D draw EF in the plane and perpendicular to BD, and join D with any point in AB, as A; then is EF perpendicular to AD (?). Now, EF being perpendicular to two lines, AD / and BD of the plane ABDC, is perpendicular to the plane, / " y^ I and hence to any line of the plane passing through D, ^ ~ — 4vl as CD. Therefore CD is perpendicular to BD and EF, ■ Fig 262. and consequently to the plane MN (?). q. e. d. 40G. Cor. 1. — Two lines which are perpendicular to the same plane are parallel. Thus, AB and CD being perpendicular to the plane MN, if AB is not parallel to CD, draw a line through B which shall be. By the proposition this Ime is perpendicular to MN, and hence must coincide with AB {390). M PARALLEL LINES AND PLANES. 171 4:07* Cor. 2. — Two lines parallel to a third not in their oiun plane are parallel to each other. Dem. — If AB and CD are parallel to EF, they are par- allel to each other. Let MN be a plane passing through EF at F', and to which EF is perpendicular; then areAB and CD respectively perpendicular to MN (?), and hence parallel (?). Q. e. d. ;: — TN B ^ Fig. 263. 408, Parallel Planes are such as do not meet when indefi- nitely produced. PROPOSITION IX. 409, Theorem. — Two planes perpendicular to the same line are parallel to each other, Dem.— For, if they could meet in some point, as 0, conceive two lines drawn from O, one in each plane, to the points where the perpendicular pierces the planes. We should then have two lines from the same point, perpendicular to the same line(?), which is impossible. Hence, as the planes cannot meet, they are parallel. Q. e. d. PROPOSITION X. 4^0. Theorem* — If a plane intersect two parallel planes^ the lines of intersection are parallel, Dem. — Let AD intersect the parallel planes MN and PQ in AB and CD ; then is AB parallel to CD. For, if AB and CD could meet, the planes MN and PQ would meet, as every point in AB is in MN, and every point in CD in PQ. Hence, AB and CD lie in the same plane, and do not meet how far soever they be pro- duced ; they are therefore parallel. Q. e, d. 411, Cor. — Parallel lines intercepted hetioeen parallel planes are equal. Thus AC "BD if they are parallel. For, the intersections AB and CD, of the plane of these parallels, are parallel (?), and the figure A B DC is a paral- lelogram ; whence AC = BD (?). Fig. 264. PROPOSITION XI. 412, Theorem, — A line which is perpendicular to one of two par aUcl planes, is perpendicular to the other also. 172 ELEMENTARY SOLID GEOMETRY. M. 'N B Dem. — Let AB be perpendicular to MN ; then is it perpendicular to PQ, paral- lel to MN. For, pass two planes through AB, and let £-. Y CA, DB, and EA, FB, be their intersections with the planes MN and PQ. Now CA and EA are perpendicu- lar to AB (?) ; hence DB and FB being parallel to CA and EA (?) are pei-pendicular to AB (?). Therefore AB is perpendicular to two lines of the plane PQ, and consequently to the plane (?). q. e. d. 4:13. Cor. 1. — Tlirough any 'point out of a ' ^ ° "" Q plane, one plane can he passed parallel to the Fig. 265. given plane, and only one. Dem. — To pass a plane through B parallel to MN, draw the perpendicular BA from B upon MN. Draw any two Imes in the plane MN, through A, as CA and EA. Through B draw DB and FB parallel to CA and EA; then is PQ, the plane of these lines, perpendicular to AB, and hence parallel to MN. As the plane parallel to MN must contain FB and DB, and as but one plane can be passed through these lines, there can be only one plane through B parallel to MN. 4:14, Tfie JDistance hetween two parallel planes at any point is measured iy the perpendicular, 415 • Cor. 2. — Parallel planes are every- where equally distant from each other. De3I. — Let A and B be any two points in the plane MN, and AC and BD tlie perpendiculars from these points, let fall on the parallel plane PQ ; then are they perpendicular to MN by the proposition, and since the figure ABCD is a parallelogram (?) [a rectangle, also (?)], Fig. 266. AC = BD. PROPOSITION xn. 416. TJieorem. — Two angles lying in different planes, hut hav- ing their sides parallel and extending in the same direction, are equal, and their planes are parallel Dem.— Let A and A' lie in the different planes MN and PQ, and have AB parallel to A'B', and AC to A'C ; then A = A', and M N and PQ are parallel. For, take AD = A'D', and AE = A'E', and draw AA', DD', EE', ED, and E'D'. Now, AD being equal and par- allel to A'D', AA' = DD' (?). For like reason AA' = EE', therefore EE' — DD'. Again, since EE' and DD' are respectively parallel to AA', they are parallel to each other (?), whence EDD'E' is a parallelogram (?), and ED = E'D'. Hence the triangles ADE and A'D'E' are Fig. 267. OF STRAIGHT LINES AND PLANES. 173 mutually equilateral, and A, opposite ED, is equal to A', opposite E'D' equal to ED. Again, the plane of the angle BAC, MN, is parallel to PQ, the plane of B'A'C. For, let a plane be passed through AC and revolved until it is parallel to PQ. It must cut DD', which is parallel to AA', and EE', so that DD' shall equal AA' and EE' (?); hence it must pass through D. 4zl7, Cor. 1. — Ifttoo intersecting planes lecut ly parallel planes, the angles formed ly the i^itersections are equal, Th«s, AB' and AE' being cut by the parallel planes MN and PQ, AD is parallel to A'D' (?), and lies in the same dh'ection, and AE to AE'. Hence BAC= B'A'C (?). 418, Cor. 2. — If the corresponding exiremities of three eqical parallel lines not in the saine plane, be joined, the triangles formed are equal, and their planes parallel. Thus, if AA' = DD' = EE', the sides of the triangle AED are equal to the sides of A'E'D', since the figures AD', DE', and EA' are parallelograms (?), and the corollary comes under the proposition (?). PROPOSITION XIII. 410, Theorem, — The corresponding seg- ments of lines cut by parallel planes are propor- tional. Dem.— LetAB.CD and EF be cut by the parallel planes MN, PQ, RS, and TU ; then Aa : Cd : : a6 : «/ : : 5B : /D, and Aa : Ei : : ah :ik :: bB : kF, and Ce : Ei : : ef : ik : : /D :kF. For, join the extremities A and D, and E and D, and conceive the intersections of the plane of AB and AD with the parallel planes to be BD, bd, and ac. These lines are parallel (?), and Aa : fiiC :: ab : cd :: bB :dD (?). For a similar reason, Ce : Ac :: ef : cd :: fD : dD (?). Whence, the consequents of the proportions being the same, the antecedents give Aa : Ce : : ab : ef :: bB : /D. In like manner we can show that Ce : Ei :: ef : ik : :/D : kF. [Let the student give the details.] From these proportions we have Aa : Ei ;: ab : ik : : bB : kF (?). Q. E. D. lt>/. pf ;/ '^ I ^ 1 1 1^ R ! I Q f-^ UL T / ' M ' '- 1 Fig. 268. EXERCISES. 1. Designate any three points in the room, as one corner of the desk, a point on the stove, and some point in the ceiling, and show how you can conceive the plane of these points. ft 174: ELEMENTARY SOLID GEOMETRY. 2. Show the position of two lines which will not meet, and yet are not parallel. 3. Conceiye two lines, one line in the ceiling and one in the floor, which shall not be parallel to each other. What is the shortest dis- tance between these lines ? 4. The ceiling of my room is 10 feet above the floor. I have a 12 foot pole, by the aid of which I wish to determine a point in the floor directly under a certain point in the ceiling. How can I do it ? SuG.— Consult Prop. IV. 5. Upon what principle in this section is it that a stool with three legs always stands firm on a level floor, when one with four may not? 6. By the use of two carpenter's squares you can determine a per- pendicular to a plane. How is it done ? 7. If yon wish to test the perpendicularity of a stud to a level floor, on how many sides of it is it necessary to measure the angle which it makes with the floor ? By applying the right angle of the carpen- ter's square on any two sides of the stud, to test the angle which ^t makes with the floor, can you determine whether it is perpendicular or not ? 8. We see in straight lines. If a line* be placed between our eye and a surface, it covers a certain space on the surface ; this figure or space is said to be the projection of the line on that surface. Upon what principles in this section is it that the projections of straight lines are straight ? Why is it that the projections of parallels which are parallel to the plane upon which we see them projected, are ■parallel, while parallel lines which are inclined to this plane are pro- jected in oblique lines ? 9. If a line is drawn at an inclination of 23° to a plane, what is the greatest angle which any line of the plane, drawn through the point where the inclined line pierces the plane, makes with the line ? Can you conceive a line of the plane which makes an angle of 50° with the inclined line? Of 80° ? Of 15° ? Of 170° ? Sereafter^the student should make the sxjnopses. * The term line is here ased in its colloquial sense, and refers to a material representation, as a cord, the edge of a board, etc. OF SOLID ANGLES. 175 SECTION II. OF SOLID ANGLES. 4:20 • A SolictJLngle^ is the opening between :two or more^ plane s, each of which intersects alFlhe others. The line s of in tei'.-^ section are called .£^£65, and the planes, or the portion of the planes between the edges, where there are more than two, are called Faces.^ 421. A Dieclral Angle, or simply a Diedral, is the opening between hvo intersecting planes. 422. A Polyedral Angle, called also simply a Polyedral, is the opening between three or more planes which intersect so as to have one common point, and only one. In the case of three inter- secting planes the angle is called a Triedral. The point common to all the planes is called the Vertex. The plane angles enclosing a polyedral are the Facial angles. 423. A Diedral {Angle) is measured by the plane angle included by lines drawn in its faces from any point in the edge, and perpen- dicular thereto. A diedral angle is called right, acute, or obtuse, according as its measure is right, acute, or obtuse. Of course the magnitude of a solid angle is independent of the distances to which the edges may chance to be produced. Ill's.— The opening between the two planes CABF and DABE is a Diedral (angle), AB is the Edge, and CABF and DABE are the Faces. Let MO lie in the plane AF, perpendicular to the edge; and NO in AE, and also perpendicular to the edge ; then the plane angle MON is the measure of the diedral. A diedral may be re ad by t he letters on the edge, -when there 176 ELEMENTARY SOLID GEOMETEY. would be no ambiguity, or otherwise by these letters and one in each face"; thus, tht^ diedral in (a) may be designated as AB, or as C-AB-D. In (6) we have a Ti-iedral (angleX, The edges are SA, SB, and SC; HolQ faces ASB, BSC, and ASC: the facial angles are ASB, ASC, and BSC ; and S is the vertex. Such an angle, and any polyedral (angle), may be read by naming the angle at the vertex, when there would be no ambiguity, or otherwise by naming the letter at the vertex, and then one in each edge ; thus S-ABCDE designates the polyedral (o). The openmg between the planes is the angle, in each case. OF DIEBBALS. 4:24:» A Diedral may be considered as generated by the revolution of a plane about a line of the plane, and hence we may see the pro- priety of measuring it by the angle included by two lines in its faces perpendicular to its edge, as stated in the preceding article. III. — Let A3 be a line of the plane GB. Conceive gB perpendicular to AB. Now, let the plane revolve upon AB as an axis, whence gB describes a circle (?) ; and at any posi* tion of the revolving plane, as/BAF, since fBg measures the amount of revolution, it may be taken as the measure of the diedral f-Bk-g. When^-B has made i of a revolution, the plane will have made i of a revolution, and the diedral will be right. 425, Coit. — Opposite diedrals are equal. Thus, if C-AB-D is measured by MON, c-AB-d is measured by the equal angle nOm. PROPOSITION 1. 4:26, TJieorem. — Any line in one face of a rigid diedral, per- pendicular to its edge, is perpendicular to the other face, Dem.— In the face CB of the right diedral C-AB-D, let MO be perpendicular to the edge AB ; then is it perpendicular to the face DB. For, draw ON in the face DB, and perpendicu- lar to AB. Now, since the diedral is right, and MON meas- ures its angle, MON is a right angle; whence MO is perpen- dicular to two lines of the plane DB, and consequently per- Fio. 272. pendicular to the plane, q. e. d. ' OF DIEDRALS. 177 427 • Cor. — Conversely, If one plane contain a line which is perpendicular to another plane, the diedral is right. Thus, if MO is perpendicular to the plane DB, C-AB^D is a right diedral. For MO. is perpendicular to every line of DB passing tlirough its foot (?); and hence is perpendicular to ON, drawn at right angles to AB. Whence C-AB-D is a right diedral, for it is measured by a right plane angle. M PROPOSITION II. 428. Theo7*ein, — If tivo planes are perpendicular to a third, their intersection is peipendicular to the tidrd p>lcine. Dem. — If CD and EF are perpendicular to the plane MN, then is AB perpendicular to MN. For, EF being perpendicular to MN, D-FG-E is a right diedral, and a line in EF and perpendicular to FC at B is perpendicular to MN ; also a line in the plane CD, and perpendicular to DH at B, is perpendicular to MN (?). Hence, as there can be one and only one perpendicular to MN at B, and as this perpendicular is in both planes, CD and EF, it Q. E. D. N Fig. 273. is their intersection. PROPOSITION III. 420. TJieorem, — If from any point perpendiculars he drawn to the faces of a diedral angle, their included angle will be the supple- ment of the angle lohich measures the diedral, or equal to it. Dem. — Let BD and AD be any two planes including the ^ diedral A-SD-B, then will two lines drawn from any point, perpendicular to these planes, include an angle which is the supplement of the measure of the diedral, or equal to it. If the point from which the lines are drawn is not in the edge SD, we may conceive two lines drawn through any point, as S, in this edge, which shall be parallel to the two proposed, and hence include an equal angle, and have their plane parallel to the plane of the proposed angle {416). Let the latter lines be SO and SP. We are to show that OSP is supplemental to the measure of A-SD-B. A plane passed through S, perpendicular to the edge SD, will contain the lines SO and SP {388) ; and its intersec- tions with the faces, as SB and SA, will form an angle (ASB) which is the measure of the diedral {423). Now, PSA = a right angle (?), and OSB = a right angle (?). Hence, PSO and ASB are either equal or supplemental {283). Q. E. D. 178 ELEMENTARY SOLID GEOMETRY. 430. Triedrals ^^^ — '--._^^,^^ B ■!'■ Fig. 275. OF TRIEDRALS. are Rectangular, Biredangular, or Trirectan- gnlar, according as they have one, two, or three, right diedral angles. Ill's. — The corner of a cube is a Trirectangular triedral, as S-ADC. Conceive the upper portion ol" the cube removed by the plane ASEF; then the angle at S, i. e., S-AEC is a Biredangular triedi-al, A-SC-E and A-SE-C being right diedrals. 431, An Isosceles Triedral is one that has two of its facial angles equal. An Bquilateral Triedral is one that has all three of its facial angles equal. 432, Two Sf/minetrical Triedrals are suoh as have the facial angles of the one equal to the facial angles of the other, each to each; but in which the equal facial angles are not similarly situated, and hence the triedrals are not necessarily capable of superposition. Ill's. — Let the edges of the triedral S-ABC, be pro- duced beyond the vertex, forming a second triedral S-abc ; then are the two triedrals symmetrical, i. e., the faces are equal plane angles, but disposed in a different order. Thus, ASB = aSb, ASC = aSc, and BSC = bSc ; but the triedrals cannot be made to coincide. To show this fact, conceive the upper triedral detached, and the face aSc placed in its equal face ASC, Sa in SA, and Sc in SC. Now, the edge Sb, instead of falling in SB will fall on the left of the plane ASC. Sj'mmetrical solids are of frequent occurrence: the two hands form an illustration ; for, though the parts may be exactly alike, the hands cannot be placed so that their like parts will be similarly situated; in short, the left glove will not fit the right hand. 433, Two triedrals are SiiJU^lementary when the edges of one are perpendicular to the faces of the other. (See 438a,) Fig. '2\\i. PROPOSITION IV. 434, Tlieoreni, — The sum of any tvjo facial a^igles of a trie' dral is greater than the third. OF TRIEDRALS. 179 Dem. — This proposition needs demonstration only in case of the sum of the two smaller facial angles as com- pared with the greatest (?). Let ASB and BSC each be less than ASC ; then is ASB + BSC > ASC. For, make the an- gle ASb'' = ASB, and Sb' = S6, and pass a plane through b and &', cutting SA and SC in a and c. The two triangles aSb and aSb' are equal (?), whence ab' = ab. Now, ab + be > ac (?), and subtracting ab from the first member, and its equal (d>' from the second, we have be > b'c. Whence the two tri- angles bSc and b'Sc have two sides equal, but the third side be > than the third side b'e, and consequently angle bSc > b'Sc. Adding ASB to the former, and its equal AS5' to the latter, we have ASB + BSC > ASC. q. e. d. Fig. a77. 43 S, Cor. — The difference hettoeen any tioo facial angles of a triedral is less than the third facial angle (?). PROPOSITION V. 436, Theorem, — The sum of the facial angles of a triedral mag he anything hePween and four right angles. Dem.— Let ASB, BSC, and ASC be the facial angles enclosing a triedral ; then, as each must have some value, ,.'q the sum is greater than 0, and we have only to show that ASB -f- ASC -1- BSC < 4 right angles. Produce either edge, as AS, to D. Now, in the triedral S-BCD, BSC < BSD + CSD. To each member of this inequality add ASB -I- ASC, and we have ASB + ASC + BSC < ASB + ASC + BSD + CSD. But, ASB 4- BSD = 3 right angles (?), and ASC + CSD - 2 right angles; whence ASB + ASC -i- BSD ■{■ CSD = 4 right angles; and consequently, ASB + ASC -[- BSC < 4 right angles, q. e. d. PROPOSITION TI. 437 m Theorem, — Two iriedrals having the facial angles of the one equal to the facial angles of the other, each to each., and similarly arranged, are equal. 180 ELEMENTARY SOLID GEOMETRY. Dem.— In the triedrals S and «, let ASB = asb, BSC = bsc, and CSA = cm ; and let these ftlcial angles occur in the same order, then is S-ABC = s-abc. Take C, any point in SC, and make «c=SC, and draw AC and BC perpendicular to SC, and ac and he perpendicular to sc ; then ACB measuues tlie diedral A-SC-B, and ach the die- dral a-scb. Now the triangles SCB and sch are mutually equiangular (?) and have SC = SC ; hence SB = ab^ and CB = cb. For a like reason AC = ac, and SA = sa. Hence AB — ab (?), and the triangles ACB and acb are equal (?). Now, since angle ACB, measuring the diedral A-SC-B, equals angle cicb, measuring the diedral a-8C-b, these diedrals are equal, and can be applied to each other. Applying these diedrals, since angle ASC = asc, and BSC = bsc, the edges AS and as coincide, as also do BS and bs, whence the triedrals coincide throughout, and are conse- quently equal, q. e. d. Fio. 279. y^"^ PROPOSITION yn. 438. Theorem, — Of tivo s^ipplementary triedrals, the facial angles of the one are the supplements of the diedrals of the other, Dem.— Let S-ABC be any triedral ; if a second triedral be foi-med with its edges perpendicular to the faces of S-ABC, one to each face respect- ively, then are the facial angles of the one, supplements of the die- drals of the other. If the vertex of the second trie- dral is not at the vertex of the firet, we may conceive a triedral formed by drawing three lines tlirough the vertex S, as SE, SD, and SF, paral- lel to the edges of the given trie- dral ; then will these edges be per- pendicular to the same planes as the edges to which they are paral- lel {405), and hence the angle thus formed (S-EDF) will be a triedral supplemental to S-ABC {433\ and the facial angles of the two having their edges parallel will be equal {416), and consequently the triedrals equal {437)- Now, SE being perpendicular to ASB, and SF to ASC, angle ESF is supplemental to the diedral B-SA-C {429.) In like manner, S3 Lcing perpendicular to BSC/DSE is supplemental to A-SB-C, and DSF to A-SC-B. Thus we have shown that the facial angles of S-EDF are the supplements Fig. 280. OF TEIEDRALS. 181 of the diedrals of S-ABC. We are now to show that the facial angles of S-ABC are supplements of the diedrals of S-EDF ; i. s., that ASB is the sup- plement of D-SE-F, BSC of E-SD-F, and ASC of D-SF-E. Since SE is by hy- pothesis perpendicular to ASB, it is perpendicular to AS (387) ; and since SF is perpendicular to ASC, it is perpendicular to AS {387). Hence AS is per- pendicular to the face FSE (?). In like manner we may show that SB is per- pendicular to DSE, and SC to DSF ; whence it follows from the preceding part of the demonstration, or directly from {429), that angle ASB is the supple- ment of D-SE-F, BSC of E-SD-F, and ASC of D-SF-E. 438a. SCH.— If any edge of S-EDF, as OS, is produced beyond S, another triedral is formed which has its edges perpendicular to the faces of S-ABC. Thus in all 4 triedrals can be formed with their edges perpendicular to the faces of S-ABC ; but the proposition holds only for S-EDF. PROPOSITION nil. 439, TJieorem, — The stem of the diedrals of a triedral mai/ be anything leiween two and six right angles. Dem.— Each diedral being the supplement of a plane angle of the supple- mentary triedral, the sum of the three diedrals is 3 times 2 riglit angles, or 6 right angles — the sum of tJie angles of the supplementary triedral But this latter sum may be anything between and 4 right angles (?). Hence the sum of the diedrals may be anything between 3 and 6 right angles, q. e. d. PROPOSITION IX. 440. Theorem, — Aii isosceles triedral and its sym- metrical triedral are eqiial. Dem. — Let S-ABC be an isosceles triedral with the facial angle ASB = BSC ; then is it equal to its symmetrical triedral S-abc. For, revolve S-abc about S until Sb falls in SB, and bring the plane Sba into the plane SBC ; then, since the diedrals CSB-A and a-Sb-c are opposite, they are equal {425)* and the plane Sbc will Fall in SBA. Moreover, Sa will fall in SC, since angle BSC = ASB (by hypothesis) = bSa (vertical to ASB). In like manner Sc will fall in SA, and the triedrals will coincide, and are therefore equal. Q. E. D. 441. ScH.— If angle ASB is not equal to BSC, it is easy to see that the ap * Should the papil have difficulty in perceiving this, let him notice that CSB and cSb art parts of one and the same plane ; and ASB and aSb are parts of another. Now 6B is tbe inter, eection of these planes, and the diedrals mentioned are on opposite sides of this line of inten eection. 182 ELEMENTARY SOLID GEOMETRY. plication will fail, notwithstanding the diedrals are equal, and the triedraLs symmetrical. 442. Cor. 1. — The diedrals opposite the equal facial angles of an isosceles triedral are equal. The diedral b-Sa-c = B-SAC being opposite, and b-Sa-c — B-SC-A as shown in the demonstration ; hence B-SA-C = B-SC-A. 443, Cor. 2. — Conversely, If two diedrals of a triedral are equal, the triedral is isosceles. Dem.— If B-SA-C = B-SC-A, SABC is isosceles. For, revolving S-ahc as be- fore till the facial angle aSc falls in its equal (?) ASC, since the diedral B-SC-A = B-SA-C (by hypothesis) and the latter equals its opposite h-Sa-c, the plane hSa will fall in the plane BSC ; and, for like reasons, the plane hSc will fall in BSA. Now, as these planes coincide, their intersections Sh and SB coincide, and the triedrals are equal ; and the facial angle BSC = hSa. But &Sa = ASB (?); hence ASB = BSC ; i. e., the triedral S-ABC is isosceles. 444, TJieoreni, d PROPOSITION X. —Two symmetrical triedrals are equivalent. Dem.— S-ABC is equivalent to its symmetrical triedral S-dbc. For, let dD be so drawn that the angles DSA, DSC, and DSB shall be equal, and consequently dSa = dSc = dSb, and the latter respectively equal to the former. Then the isosceles • symmeti"ical triedral S-DCB = S-dcb, S-DCA = S-dea, and S-ADB = S-adb. Whence the polyedral S-ABCD = S-abcd. Now, from the former subtracting S-ADB, and from the latter S-adb, there remains S-ABC = S-abc. Q. e. d. 445, ScH.— If dD fell within the given triedrals, they would be made up of the three equal isosceles triedrals, and hence equivalent. Fig. 282. PROPOSITION XI. 446. TJieorem, — Tiuo triedrals which have two facial angles and the included diedral equal, each to each, are equal, or symmetri- cal and equivalent. Dem.— If the equal faces are on the same sides of the diedral in the two tiiedrals, the one figure can be applied to the other ; and if they are on different OF TBIEDRALS. 183 sides, the edges of one triedral may be produced, forming the symmetrical tri- edral, to which the other given triedral may be appUed. [Let the student con- struct figures, and go through with the application.] PROPOSITION xn. 447* TJieorem, — Two triedrals which have two diedrals and the included facial angle equal, are equal, or symmetrical and equiva- lent. Dem. — [Same as in the preceding. Let the student draw figures like those for the preceding, and go through with the details of the application.] 44S, Cor. — It loill he observed that in equal or in symmetrical triedrals, the equal facial angles are opposite the equal diedrals. PROPOSITION xni. 449. TJieorem, — Two triedrals which have tivo facial angles of the one equal to tzvo facial angles of the other, each to each, and the included diedrals unequal, have the third facial angles unequal, and the greater facial a7igle belongs to the triedral having the greater in- cluded diedral. Dem. — Let ASC = asc, and ASB = ash, while the diedral C-SA-B > c-sa-h\ then CSB S s > csh. Jf\ A. For, make the diedral C-SA-c? =: c-sa-b: and /rA / \\ taking AS(9 = asby bisect the diedral o-SA-B with / |i i \ / \ \ the plane ISA. Draw o\ and oC, and conceive / jl \ \ / \ \ the planes oS\ and oSC. Now, the triedral y\-v[i't;^ o/,-------^-->^ S-AoC = s-abc, since they have two facial angles / ^:f^-\o ^ ' i\ and the included diedral equal {446). For a i i like reason S-AI(? = S-AIB, and the facial angle „ „^„ Fio 2S3 oS\ = ISB. Again, in the triedral S-bC, oSI + ISC > oSC {434), and substituting ISB for oS\, we have ISB + ISC (or BSC) > oSC, or its equal bsc. q. e. d. 450. Cor. — Conversely, If the tico facial angles are equal, each to each, in the ttvo triedrals, and the third facial angles imequal, the diedral opposite the greater facial angle is the greater. That is, if ASB = ash, and ASC = asc, while BSC > bsc, the diedral B-AS-C > b-as-c. For, if B-AS-C = b-as-c, BSC = bsc {44C>\ and if B-AS-C < b-as-c, BSC < bsc, by the proposition. Therefore^ as B-AS-C cannot be equal to nor less than b-as-c, it must be greater. 184 ELEMEXTABY SOLED GEOMETRY. PROPOSITION xiy. 451. Tlieorem. — Two triedrals which have the three facial angles of the one equal to the three facial angles of the other ^ each to each, are equal, or symmetrical and equivalent. Dem.— Let A, B, and C represent the facial angles of one, and a, 5, and c the coirespondmg facial angles of the other. If A = rt, B = &, and C = c, the triedrals are equal. For A being equal to a, and B to h, if, of their included die- drals, SM were greater thau.«?«, C would be greater thanc; andifdiedral SM were less than diedral sm, C would be less than c, by the last corollary. Hence, as diedral SM can neither be greater nor less than diedral sm, it must be equal to it. For like reasons, diedral SH = diedral sn, and diedral SO = diedral so. Therefore, the triedrals are equal, or syinmetrical, according to the arrangement of the faces. Thus, if SN and m are both considered as lying on the same side of the planes MSO and mso, the triedrals are equal ; but, if one lies on one side and the other on the opposite side of those planes (SN in front, and sn behind, for example), the diedrals are symmetrical, and hence equivalent Fig. 2S4. PROPOSITION XT. 4:52 1 Theorem. — Tioo triedrals which have the three diedrals of the one equal to the three diedrals of the other, each to each, are equ ABC, SOB + SOD > BCD, etc. {4=34). Therefore the sum of the angles at S is less than the sum of the angles at 0, i. e., less than 4 right angles. Q. e. d. Fio. 285. EXERCISES. 1. I have an iron block whose corners are all square (edges i-ight diedrals, and the vertices tri rectangular, or right, trie- 'drals). If I bend a wire square around one of its edges, as cS'd, at what angle do I bend the wire ? If I bend a wire ob- liquely around the edge, as asby at what angle can I bend it? If I bend it ob- liquely, as eS'f at what angle can I bend it? \^; i S' y FiQ. -.ifeO. 186 ELEMENTARY SOLID GEOMETRY. 2. Fig. 286 represents the appearance of a rectangular parallelepi- ped, as seen from a certain position. Now, all the angles of such a solid are right angles : why is it that they nearly all appear oblique ? Can you see a right parallelopiped from such a position that all the angles seen shall appear as right angles? 3. The diedral angles of crystals are measured with great care, in order to determine the substance of which the crystals consist. How must the measure be taken ? If we measure obliquely around the edge, shall we get the true value of the angle ? 4. Cut out any triedral from a block of wood (or a potato), and stick three pins into it, as near the vertex as you can, one iu each face, and perpendicular to that face. What figure do the three pins form ? What relation does the angle included between any two ad- jacent pins sustain to one of the diedrals of the block ? Which ones are they that sustain this relation ? 5. Can three planes intersect each other and yet not form a trie- dral angle ? In hoAv many ways ? Can they all three have a common point, and yet not form a triedral ? 6. From a piece of pasteboard cut two figures of the same size, like ABCDS and alcds. Then drawing SB and SC so as to make 1 the largest angle and 3 the smallest, cut the pasteboard almost through in these lines, so that it will readily bend in them. Now fold the edges AS and DS together, and a triedral will be formed. From the piece leads form a triedral in like manner, only let the lines sc and 8a be drawn so as to make the angles 1, 2, and 3 of the same size as before, while they occur in the order Pig. 267. given in heads. Now, see if you can slip one triedral into the other, so that they will fit. What is the diffi- culty ? 7. In the last case, if 1 equals I of a right angle, 2 = ^ of a right angle, and 3 = f of a right angle, can you form the triedral ? Why ? If you keep increasing the size of 1, 2, and 3, until the sum becomes equal to 4 right angles, will it always be possible to form a triedral ? How is it when the sum equals 4 right angles ? OF PRISMS AND CYLINDERS. 187 SECTION III, OF PRISMS AND CYLINDERS. 4:57* A JPrisni is a solid, two of whose faces are equal, parallel polygons, while the other faces are parallelograms. The equal par- allel polygons are the Bases, and the parallelograms make up the Lateral or Convex Surface. Prisms are triangular, quadrangular, pentagonal, etc., according to the number of sides of the polygon forming a base. 4a8. A Hight I*rism is a prism whose lateral edges are per- pendicular to its bases. An Oblique prism is a prism whose late- ral edges are oblique to its bases. 459, A ^Regular Prism is a right prism whose bases are regular polygons ; whence its faces are equal rectangles. 460, The Altitude of a prism is the perpendicular distance between its bases : the altitude of a right prism is equal to any one of its lateral edges. 461, A Truncated JPrism is a portion of a prism cut off by a plane not parallel to its base. A section of a prism made by a plane perpendicular to its lateral edges is called a Right Section Ill's. — In the figure, (a) and (6) are both prisms : {a) is oblique and (6) right, PO represents the altitude of (a); and any edge of (ft), as 5B, is its altitude. ABCDEF, and ahcdef, are lower and upper bases, respectively. Either por- tion of ip) cut ojff by an oblique plane, as a'h'c'd'e\ is a truncated prism. 462, A I^arallelopiped is a prism whose bases are parallel- ograms : its faces, inclusive of the bases, are consequently all parallel- FiG. 28 188 ELEMENTARY SOLID GEOMETRY. ograms. If its faces are all rectangular, it is a rectangular parallel- epiped. 463, A Cube is a rectangular parallelepiped whose faces are all equal squares. PROPOSITION I. 40 4* Theorem, — Parallel plane sections of any prism are equal polygons. Dem.— Let ABCDE and cibcde be parallel sections of the prism MN ; then are they equal polygons. For, the intersections with the lateral faces, as ah and AB, etc., are parallel, since they are intersections of parallel planes by a third plane {410). Moreover, these intersections are equal, that is, ab = AB, be = BC, cd = CD, etc., since they are parallels included between parallels (242). Again, the corresponding angles of these polygons are equal, that is, a = A, 6 = B, c = C, etc., since their sides are parallel and lie in the same direction {416). Therefore the polygons ABCDE, and abcde, are mutually equilateral and equiangular; that is, they are equal. <^. e. d. 46S, Cor. — Any plane section of apristn, parallel to its base, is equal to the base ; and all right sections are equal. Fig. 289. PROPOSITION IL 466» Theorem, — If three faces including any triedral of one prism are equal respectively to three faces including a triedral of the other, and similarly placed, the prisms are equal, Dem. — In the prisms AfZ, and k'd', let ABCDE equal A'B'C'D'E', fKBba = A'B'b'a', and BCcb = B'C'c'b'; then are the prisms equal. For, since the facial angles of the triedrals B and B' are equal, the trie- drals are equal {451), and being ap- plied they will coincide. Now, con- ceiving A'rf' as applied to Arf, with B' in B, since the bases are equal poly- gons, they will coincide throughout ; and the faces aB and a'B' will also coincide. Whence, as a'b' falls in a&, OF PRISMS. 189 and Vc' in he, the upper bases, which are equal because equal to the equal lower bases, will coincide. Therefore the remaining edges will have two points com- mon in each, and will consequently coincide. 467* Cor. 1. — Two right prisms having equal bases and equal altitudes are equal. If the faces are not similarly arranged, one prism can be inverted. 4:68, Cor. 2. — The above proposition applies also to truncated prisms. PROPOSITION m. 469, Tlieorem* — Any oblique prism is equivalent to a right prism, whose bases are right sections of the oblique prism, and whose edge is equal to the edge of the oblique prism. Dem. — Let LB be an oblique prism, of which abcde and fghil are right sections, and gh — OB-^ then is lb equiva- lent to LB. For the truncated prisms 10 and eB have the faces including any triedral, as C and B, equal and simi- larly placed (?), whence these prisms are equal {466). Now, from the whole figure, take away prism 10, and there remains the oblique prism LB ; also, from the whole take away the prism eB, and there remains the right prism Jb. Therefore, the right prism lb is equivalent to the oblique prism LB. q. e. d. PROPOSITION IT. ■The opposite faces of a parallelopiped are 470, Theorem. equal and parallel. Dem. — Let. Ac be a parallelopiped, AC and ac being its equal bases (462) ; then are its opposite faces equal ' and parallel. Since the bases are parallelograms, AB is equal and parallel to DC ; and, since the faces are parallelograms, aA is equal and parallel to dD. Hence angle «AB = e?DC, and their planes are parallel, since their sides are parallel and extend in the same directions. Therefore aB and dO are equal {301) and parallel parallelograms. In like manner it may be shown that aD is equal and parallel to bO. 190 ELEMENTARY SOLID GEOMETRY. PROPOSITION Y. 4:71* Theorem^ — Tlie diagonals of a parallelopiped bisect each other, Dem. — Pass a plane through two opposite edges, as &B and dD. Since the bases are parallel, hd and BD will be parallel {410), and bBDd will be a paral- lelogram. Hence, bD and dB are bisected at o (?). For a like reason, passing a plane through dc and AB, we may show that dB and cA bisect each other, and hence that cA passes through the common centre of dB and bD. So also aC is bisected by 6D, as appears from passing a plane through ab and DC. Hence, all the diagonals are bisected at o. q,. e. d. Fig. 293. 4 72, Cor. — Tlie diagonals of a rectangular parallelopiped are equal. PROPOSITION TL 47 3 • Theorem, — A parallelopiped is divided into ttvo equiva- lent triangular prisms hy a plane passing through its diagonally opposite edges. Dem.— Let H-ABCD be a parallelopiped, divided through its diagonally opposite edges FA and HC; then are the triangulai* prisms H-ABC, and L-ADC equivalent. For this parallelopiped is equivalent to a right parallelopiped having a right section N}cd for its base, and AF for its edge {469), i. e., H-ABCD is equiva- lent to h-fiibcd. For the same reason the oblique triangular prism H-ABC is equivalent to the right triangular prism h-Abc; and L-ADC is equivalent to l-Mc. But Ji-Mc is equal to h-Mc, as they are right prisms with equal bases {467) and a com- mon altitude. Hence, H-ABC is equivalent to L-ADC, they are equivalent to two equal prisms, q. e. d. '^ Fig. 294. PROPOSITION yn. 474, Theoretn, — Any parallelopiped is equivalent to a redan' gular parallelopiped having an eq\dvalent base and the same altitude. OF PRISMS. 191 Dem. — Let H-ABCD be any parallelopiped with all its faces oblique. 1st. By making the right section rtclHe, and completing the paral- lelepiped adHebcCf, we have an equivalent right parallelopiped (409). 2d. Through the edge ef of this right parallelopiped make the right section ea'b'f and complete the parallelopiped ea'b'/Hd'c'C, and we have a rectangular paral- lelopiped equivalent to the one previously formed {469), and hence equivalent to the given one. Now, the base of this rectangular parallelopiped, i. e., a'b'c'd', is equal to abed (?), which in turn is equivalent to ABCD (?). Moreover, the altitude of the rectangular parallelopiped is the same as that of the given one, since their bases lie in the same parallel planes Ac' and EG. Therefore, the parallelopiped H-ABCD is equivalent to the rectangular parallelopiped H-a'b'c'd', which has an equivalent base and the same altitude, q. e. d. A A Jl / t\ Ll\ V / k //i'u^_ // 1/ ■»' B 295. "i/b a Fig. •6 PROPOSITION vm. \D 47S, Theorem, — The area of the lateral surface of a right prism is equal to the product of its altitude into the perimeter of its base. I)em. — The lateral faces are all rectangles, having for their common alti- tude the altitude of the prism {460). Whence the area of any face is the product of the altitude into the side of the base which forms its base ; and the sum of the areas of the faces is the common altitude into the sum of the bases of / / / / / the faces, that is, into the perimeter of the base of / / / •' •' -^ Ihe prism, q. e. d. / / .g . . 476. A Cylindrical Surface is a curved surface traced by a straight line moving so as to remain constantly parallel to its first position, while any point in it traces some curve. The moving line is called the Gener- atrix, and the curve traced by a point of the line is the Directrix. Fig. 296. III. — Suppose a line to start from the position AB. and move towards N in 192 ELEMENTARY SOLID GEOMETRY. such a manner as to remain all the time parallel to its first position AB, whUe A traces the curve A123456 M. The surface thus traced is a Gyliil drical Surface; AB is the Geiierattix, and the curve ANM the Directrix. 477. A Circular Cylinder, called also a Cylinder of Revolu- tion, is a solid generated by the revolution of a rectangle around one of its sides as an axis. B" III.— Let COAB be a rectangle, and conceive it re- volved about CO as an axis, talviDg successively the positions COA'B', COA"B", etc. ; the solid generated is a Circular Cylinder, or a cylinder of revolution. Tlie re- volving side AB is the generatrix of the surface, and the circumference OA (or CB) is the directrix. This is tlie only cylinder treated in Elementary Geometry, and is usually meant when the word Cylinder is used without specifying the kind of cylinder. 478, Tlie Axis of the cylinder is the fixed side of the rectangle. The side of the rectangle opposite the axis generates the Convex Surface ; while the other sides of the rectangle, as OA and CB, generate the Bases, which in the cylinder of revolution are circles. Any line of the surface corresponding to some position of the generatrix is called an Element of the surface. 479. A Right Cylinder is one whose elements are perpen- dicular to its base. In such a cylinder any element is equal to the ftxis. A Cylinder of Revolution {477) is right. 480, A prism is said to be inscribed in a cylinder, when the bases of the prism are inscribed in the bases of the cylinder, and the edges of the prism coincide with elements of the cylinder. PROPOSITION IX. 481, Theorem. — The area of the convex surface of a cylinde? of revolution is equal to the product of its axis into the circumfer' ence of its base, i. e., 2;rRH, H bei7ig the axis and E the radius of the base. OP PRISMS AND CYLINDERS. 193 Dem.— Let a right prism, with any regular polygon for its base, be inscribed in the cylinder, as k-abcdef, in the cylinder whose axis is HO. The area of the lateral surface of the prism is HO (= hb) into the perimeter of its base, i.e., HO k {ah + hc + cd + de + ef + fa). Now, bisect the arcs ah, he, etc., and inscribe a regular polygon of twice the number of sides of the preceding, and on this polygon as a base construct the right inscribed prism with double the number of faces that the first had. The area of the lateral surface of this prism is HO x the perimeter of its base. In like manner conceive the operation of inscribing right prisms with regular polygonal bases continually repeated ; it will ahoays be true that the area of the lateral surface is equal to HO x t?ie perimeter of the base. But the circum- ference of the base of the cylinder is the limit toward which the perimeters of the inscribed polygons forming the bases of the prisms constantly approach, and the convex surface of the cylinder is the limit of the lateral surface of the inscribed prism. Therefore, the area of the convex sur- face of the cylinder is HO into the circumference of the base. Finally, if R i» the radius of the base, 2;rR is its circumference. This multiplied by H th» altitude, i. e., H x 27rR, or 27rRH, is the area of the convex surface of the cylinder. Fig. 298. PROPOSITION X. 4:82, Tlieoj^ein. — The volume of a rectangtdar parallelopiped is equal to the product of the three edges of one of its triedrals. 1^ Dem. — Let H-CBFE be a rectangular paral- lelopiped. 1st. Suppose the edges commen- surable, and let BC be 5 units in length, BA 4, and BF 7. Now conceive a cube, as d-fbBg whose edge is one of these linear units. This cube may be used as the unit of volume. Con- ceive the parallelopiped 0-caBb, whose length is 7, and whose edges ca and cb are 1 (the linear unit of measure assumed). This paral- lelopiped will contain as many of the units of volume as there are linear units in BF: we suppose 7. Again, conceive the paral- lelopiped whose base is EGBF and altitude PE, one of the linear units. This parallelopiped will contain as many of the former as there nrc linear units in BC: we suppose 5. Hence this last volume is 5 x 7 = 35. Finally, there will 13 FiH. 29!). 194 ELEMENTARY SOLID GEOMETRY. be as many times this number of imits of volume in the whole parallelepiped as AB contains linear units, or 4 x 35 = 140. Hence, when the edges are commensurable, the volume is the product of the three edges including a triedral. 2nd. Wlien the edges are not commensurable, we reach the same conclusion by takini; successively a smaller and smaller linear unit. Thus, for a first approximation take some aliquot part of one edge, as i^g of FB. Now, by hypo- thesis this is not contained an exact number of times in BC, nor in BA. But conceive it as applied to BC as many times as it can be ; the remainder will be less than -,^ FB. In like manner conceive it applied to AB. The volume of the parallelopiped included by these edges will be measured by the product of the edges. Now conceive the linear unit smaller. The unmeasured portion will be less. Thus, b}^ supposing the linear unit to diminish indefinitely, we see that it will cdways remain true tliat the measure is the product of the three edges forming a triedral. 483, Cor. 1. — The volume of a cuhe is the third power of its edge. 4S4, ScH.— This fact gives rise to the term cuhe, as used in arithmetic and algebra, for " third power." 485. Cor. 2. — The volume of a rectangular parallelopiped is equal to tlie product of its altitude into the area of its base, the linear unit being the same for the measure of all the edges. 486. Cor. 3. — T7ie volume of a7ii/ parallelopiped is equal to the product of its altitude and the area of its base. For any parallelopiped is equivalent to a rectangular parallelopiped having an equivalent base and the same altitude {474). 487. Cor. 4. — Parallelepipeds of the same or equivalent bases are to each other as their altitudes, and those of the same altitudes are to each other as their bases. And, in general, jmrallelopipeds are to each other as the products of their bases and altitudes. PROPOSITION XI. 488. Tlieorem. — The volume of any prism is equal to the pro ■duct of its altitude into its base. OF PRISMS AND CYLINDERS. 195 Dem. — 1st. Let E-ABD be a triaDgular prism. Com- plete the parallelepiped E-ABCD. Then is E-ABD = i E-ABCD {473). But the vokime of E-ABCD is equal to its altitude into its base ; hence the volume of E-ABO is equal to its altitude into ^ABCD, or ABD. 2d. Any prism may be divided into partial, tri- angular prisms, by passing planes through one edge and all the other non-adjacent edges, as in the figure. Let H be the altitude of the whole prism, then is it also the common altitude of the partial prisms. Now, the volume of each triangular prism is H into its base ; hence, the sum of the volumes is H into the sum of the bases, i.e., H into the base of the whole prism. 489. Cor. 1. — T/ie volume of a right prism is equal to the product of its edge into its base. 490. Cor. 2. — Prisms of the same altitude are to each other as their bases ; and prisms of the same or equivalent bases are to each other as their altitudes; and, i?i general, prisms are to each other as the products of their bases and altitudes. PROPOSITION xn. 491, Theorem.— Tlie volume of a cylinder of revolution is equal to the product of its base and altitude, i. e., ;rR'H, H being the altitude and K the radius of the base. Dem. — Inscribe any regular right j)i-ism in the cylinder, as in {4:81). The volume of this prism is equal to the product of its base and altitude ; and this continues to be the fact as the number of sides of the polygon forming the bftse is successively doubled, and the prism approaches equality with the cylinder. Hence, as the volume of the prism is always equal to the product of its base and alti- tude, and as the altitude of the prism remains equal to the altitude of the cylinder, this fact is true when the number of the sides of the base of the prism is infinitely multiplied ; whence the volume of the cylinder is equal to the product of its base and altitude. Now, K being the radius of the base, the area of the base is ^rR^ (?) : hence, the volume of the cylinder is equal to ;rIl^H. 196 ELEMENTAKY SOLID GEOMETRY. 492, Cor. — Tlie volume of any cylinder is equal to the product of its base into its altitude. This can be demonstrated in a manner altogether analogous to the case given in the proposition. 493. Similar Solids are such as have their corresponding solid angles equal and their homologous edges proportional. 494. Similar Cylinders of reyolution are such as have their altitudes in the same ratio as the radii of their bases. 495. Homologous Edges of similar solids are such as are included between equal plane angles in corresponding faces. Ill's.— The idea of similarity in the case of solids is the same as in the case of plane figures, viz., that of liksness of form. Thus, one would not think such a cylinder as one joint of stovepipe, similar to another composed of d hundred joints of the same pipe. One would be long and very slim in propor. tion to its length, while the other would not be thought of as slim. But, if w« have two cylinders the radii of whose bases are 2 and 4, and whose lengths ar« respectively 6 and 12, we readily recognize them as of the same shape: they are similar. PROPOSITION xin. 496. TJieorem^. — The lateral surfaces of similar right prisma are to each other as the squares of their edges (or altitudes) and as the squares of any tiuo homologous sides of their bases, i. e., as tht squares of any two homologous lines. Dem.— Let A,B, C, D, and E, be the sides of the base of one right prism •whose edge (equal to its altitude) is H, and a, b, c, d, and e, the homologous sides of a similar piism whose edge is h. Letting A + B +C + D+ E = P^ and a + b + c + d+e = p,we have P:p :: A: a ::B:b ::C:c, etc. (?). But by h3rp«tiiesis, H :h:: A : a : : B : b, etc. Hence, A P : p : : H . h (?). Now, H:h::R:h {?). Whence, P y. H : p x h : : ET" :h} (?). And as 5"" : 7i' : : A' lo' : -.B' :b\ etc., we have P x H:p x h :: A^ . a^ : : ^ : 6', etc. But P X H is the area of the lateral surface of one prism and p x A of thft other, whence the truth of the theorem appeara. OF PRISMS AND CYLINDERS 197 PROPOSITION xiy. 49 7 • Theorem. — The volumes of similar ^^risms are to each other as the cubes of their homologous edges, and as the cubes of their altitudes. Dem.— Let H-ABCDE and h-ahcde be two similar prisms, of which A and a are corresponding triedrals. Placing « so that it will coincide with A, all the faces and edges of one will be parallel to or coinci- dent with the corresponding parts of the other, by definition {493). Let fall the perpendicular FP upon the common base, or its plane produced, so that FP shall equal the altitude of H-ABCDE, and OP, intercepted between the planes of the upper and lower bases of h-abcde^ shall be its alti- tude. Call the former altitude H, and the latter h. Since FP and AF are cut by- parallel planes, we have AF : a/ : : H : A; and AB : aJ : : H : h, since by definition AF : af : : AB : ab^ etc. Call the base of H-ABCDE B, and of Now, as the bases are similar polygons, B : 6 : : A"b' : :^' : : H^ : h\ H : /i : : AB : a6 : : H : ^. B X H : 6 X 7i :: AB' : ^' :: H" : 7i'. Now, as B X H and b x h are the volumes of the respective prisms, and as AB : ah^ as the cubes of any other homologous edges are to each other, the truth of the theorem is demonstrated. Fig. .302. h-abcde b. But Hence, PROPOSITION XV. 408, Theorem, — The convex surfaces of similar cylinders of revolution are to each other as the squares of their altitudes, and as {lie squares of the radii of their bases. Dem.— Let H and h be the altitudes, and R and r the radii of the base^ of two similar C3'linders ; the convex surfaces are 2;rRH and 27crh (481). Now, 27rRH : 27trh : : RH : r^ (?) (1). By hypothesis, H : 7i : : R : r, or H h R .. .R r - and ^ = T r HA. Multiplying the terms of the second couplet of (1) by these cqunls, we have, 27rRH : "litrh : : H^ : ^^ and 2;rRH : 2;rr74 : : R"' : r'. Q. E. D. 198 ELEMENTARY SOLID GEOMETRY. PROPOSITION XVI. 499. Theorem. — The volumes of similar cylinders of revolu- tio?i are to each other as the cubes of their altitudes, or as the cubes of the radii of their bases. Dem. — Using the same notation as in the last demonstration, the student should be able to give the reasons for the following steps. R : r : : H : A (?), whence 7tW : Tti-^ : : W : A' (?). Multiplying the last proportion hy H : h : : H : ^, we have ttR^H : itt^h : : H* : h\ or as R' : ? ^ since H' ://=*:: R" : 7'^ (?). Now, ;rR''H and 7t)'='h are the volumes of the cylinders (?) ; hence the volumes ai-e to each other as the cubes of xAe altitudes, or as the cubes of the radii of the bases, q. e. d. ScH.— It is a general truth, that the surfaces of similar solids, of any form, are to each other as the squares of homologous lines ; and their volumes are as the cubes of such lines. EXERCISES. 1. A farmer has two grain bins which are parallelepipeds. The front of one bin is a rectangle 6 feet long by 4 high, and the front of the other a rectangle 8 feet long by 4 high. They are built between parallel walls 5 feet apart. The bottom and ends of the first, he says, are " square " (he means, it is a rectangular parallelo- piped), while the bottom and ends of the other slope, i. e., are oblique to the front. What are the relative capacities of the bins ? 2. How many square feet of boards in the walls and bottom of the first bin mentioned in Ex. 1? 3. An average sized honey bee's cell is a right hexagonal prism, .8 of an inch long, with faces -^ of an inch wide. The width of the face is always the same, but the length of the cell varies according to the space the bee has to fill. Are honey bee's cells similar ? Is a honey bee's cell of the dimensions given above, similar to a wasp's cell which is 1.6 inches long, and whose face is .3 of an inch wide? How much more honey will the wasp's cell hold than the honey bee's? "i/ 0.- 4. How many square inches of sheet-iron d6es it take to make a joint of 7-inch stovepipe 2 feet 4 inches long, allowing an inch and a half for making the seam ? 5. A certain water-pipe is 3 inches in diameter. How much water is discharged through it in 24 hours, if the current flows 3 feet per OF PYRAMIDS AND CONES. 199 minute ? How much through a pipe of twice as great diameter, at the same rate of flow ? 6. What is the ratio of the length of a hogshead holding 125 gal- Ions, to the length of a keg of the same shape, holding 8 gallons ? 7. What are the relative amounts of cloth required to clothe 3 men of the same form (similar solids), one being 5 feet high, another 5 feet 9 inches, and the other 6 feet, provided they dress in the same style? If the second of these men weighs 156 lbs., what do the others weigh ? 8. If a man 5^ feet high weighs 160 lbs., and a man 3 inches taller weighs 180 lbs., which is the stouter in proportion to his height ? 9. I have a prismatic piece of timber from which I cut two blocks both 5 feet long measured along one edge of the stick; but one block is made by cutting the stick square across (a right section), and the other by cutting both ends of it obliquely, making an angle of 45° with the same face of the timber. Which block is the greater ? Which has the greater lateral surface ? 10. HoAV many cubic feet in a log 12 feet long and 2 feet 5 inches in diameter? How many square feet of inch boards can be cut from such a log, allowing ^ for waste in slabs and sawing ? / SECTION IV, \i\ , ;- ,\ , OF PYRAMIDS AND CONES. ; 500* A pyramid is a solid having a polygon for its base, and triangles for its lateral faces. If the base is also a triangle, it is called a triangular pyramid, or a tetraedron {i.e., a solid with four faces). The vertex of the polyedral angle formed by the faces is the vertex of the pyramid. 501, Tlie Altitude of a pyramid is the perpendicular dis- tance from its vertex to the plane of its base. 502, A Might Pyramid is one whose base is a regular 200 ELEMENTARY SOLID GEOMETRY. polygon, and the perpendicular from whose vertex falls at the middle of the base. This perpendicular is called the axis. 503, A Frustum of a pyramid is a portion of the pyramid intercepted between the base and a plane parallel to the base. If the cutting plane is not pai-allel to the base, the portion intercepted is called a Truncated pyramid. 504, Tlie Slant Weight of a right pyramid is the altitude of one of the triangles which form its faces. The Slant Height of a Frustum of a right pyramid is the portion of the slant height of the pyramid intercepted between the bases of the frustum. Fig. 303. Ill's. — The student will be able to find iUustrations of the definitions in the accompanying figures. 505. A Conical Surface is a surface traced by a line which passes through a fixed point, while any other point traces a curve. The line is the Generatrix, and the curve the Directrix. The fixed point is the Vertex. Any line of the surface corresponding to some position of the generatrix is called an Element of the surface. 506. A Cone of Hevolution is a solid generated by the revolution of a right angled triangle around one of its sides, called the Axis. The hypotenuse describes the Convex Surface of the cone, and corresponds to the generatiix in the preceding definition. The other side of the triangle describes the Base. This cone is right, since the, perpendicular (the axis) falls at the middle of the base. The Slant Height is the distance from the vertex to the circumfer- ence of the base, and is the same as the hypotenuse of the generating triangle. 507. The terms Frustum and Truncated qiq applied to the cone in the same manner as to the pyramid. OF PYRAMIDS AND CONES. 201 S08^ A pyramid is said to be Inscriled in a cone when the base of the pyramid is inscribed in the base of the cone, and the edges of the pyramid are elements of the surface of the cone. The two solids have a common vertex and a common altitude. S09. If the generatrix be considered as an indefinite straight line passing through a fixed point, the portions of the line on oppo- site sides of the point will each describe a conical surface. These two surfaces, which in general discussions are considered but one, are called Nappes, The two nappes of the same cone are evidently alike. Ill's. — In the figure, {a) represents a conical surface which has the curve ACB for its directrix, and SA for its generatrix. The figures indicate the suc- Fio. 304. cessive positions of the point A, as it passes around the curve, while the point S remains fixed. (&) represents a Cone of Bewlution, or a right cone with a cir- cular base. It may be considered as generated in the general way, or by the right angled triangle SOA revolving about SO as an axis. SA describes the convex surface, and OA the base. The figure (c) represents the Frustum of a cone, the portion above the plane ahc being supposed removed. Figure {d) rep- resents the two nappes of an oblique cone. v!- PROPOSITION I. 510. TJieorein,—Any section of a pyramid made ly a plane parallel to its hasc is a polygon similar to the hase. 202 ELEMENTARY SOLID GEOMETRY. Dem.— The section abcde of the pyramid S-ABCDE, made by a plane parallel to ABCDE, is similar to ABCDE. Since AB and db are intersections of two parallel planes by a third plane, they are parallel (?). So also he is parallel to BC, cd to CD, etc. Hence, angle 6 = B, c = C, etc. (?), and the polygons are mutually equiangular. Again, aJ : AB : : S6 : SB, and 5c : BC : : S5 : SB (?). Hence, ab : be :: AB ■ BC (?). In like manner, we can show that be : cd :: BC : CD, etc. Therefore, abcde and ABCDE are mutually equiangular, and have their corre- sponding sides proportional, and are consequently similar. Q. E. d. PROPOSITION n. Sll* Tlieorem, — IfUuo pyramids of the same altitude are cut hy planes equally distant from and 'parallel to their bases, the sections are to each other as the bases. Dem.— Let S-ABC and S'-A'B'C'D'E' be two pyramids of the same altitude, cut by the planes abc and a'b'c'd'e', parallel to and at equal distances from their bases ; then is abc : a'b'c'd'e' : : ABC : A'B'C'D'E'. For, conceive the bases in the same plane. Let SP = S'P' be the common alti- tude, and Sp = S'p' the distances of the cutting planes from the vertex. We have '-ycs- Also, A'B'C'D'E' : a'b'c'd'e' : : A'B'' : a'6'' : : SP'' : Sy' (?). ^- y^ Whence, as SP = S'P', and Sp = S'p' (?), we have ^ -^ abc : a'b'c'd'e' : : ABC : A'B'C'D'E' (?). q. e. d. S12. Cor. — If the bases are equivalent, the sections are also equivalent. Fig. 306. ABC : abc -.'. AB ab SJ' : Sp (?) I PROPOSITION III. 513* Tlieorem. — TJie area of the lateral surface of a right pyra7nid is equal to the perimeter of the base m.ultijMed by one-half the slant height. ^. Dem.— The faces of such a pyramid are equal isosceles triangles (?), whose common altitude is the slant height of the pyramid (?). Hence, the area of OF PYRAMIDS AND CONES. 203 these triangles is the product of one-half the slant height into the sura of their bases. But this is the lateral surface of the pyramid. (See the third cut in Fig. 303.) 514i, Cor. — The area of the lateral surface of the frustum of a right pyramid is equal to the product of its slant height into half the sum of the perimeters of its bases. The student will be able to give the proof. It is based upon {325) and definitions. Fig. 301 PROPOSITION IV. 51S, Theorem. — TJie area of the convex surf ace of a cone of revolution {a right cone with a circular base) is equal to the product of the circumference of its base and one-half its slant height, i. e., ttKH', R being the radius of the base, and H' the slant height. Dem. — In the circle which forms the base of the cone, conceive a regular polygon inscribed, as dbcde. Joining the vertices of the angles of this polygon with the vertex of the cone, there will be constructed a right pyramid inscribed in the cone. Now, if the arcs subtended by the sides of this polygon are bisected, and these again bisected, etc., and at every step a right pyramid conceived as inscribed, it will always remain true that the lateral surface of the pyramid is the perimeter of its base into half its slant height. But, as the number of faces of the pyramid is increased, the perimeter of the base approaches the circumference of the base of the cone, the slant height of the pyi-amid approaches the slant height of the cone, and the lateral surface of the pyramid approaches the convex surface of the cone. Hence, at the limit we still have the same expression for the area of the convex surfiice, that is, the circumference of the base multiplied by half the slant height. Finally, if R is the radius of the base, its circumference is 2;rR, and H' being the slant height, we have for the area of the convex sur- face 2;rR x iH', or ttRH'. 310, Cor. 1. — 77ie area of the convex surface of a cone is also equal to the product of the slant height into the circumference of the circle parallel to the base, and midway between the base and vertex. This follows directly from the flict that the radius of the circle midway between^the base and vertex is one-half the radius of the base, i. e., ^R, whence its circumference is ttR. Now, ttR x H' is the area of the convex surface, by the proposition. 204 ELEMENTARY SOLID GEOMETRY. 517, Cor. 2. — Tlie area of the convex surface of tlie frustum, of a cone is equal to the product of its slant height into half the sum of the circumferences of its bases; i. e., ?r (R + r) H', R and r being the radii of its bases, and H' its slant height. Pmm the corresponding property of the fi-ustum of a pyramid, the student will be able to deduce the fact that i (2;rR + 2itr) H', or -n- (R + r) H', is the area of this surface. 518, Cor. 3. — The area of the convex surface of the frustum of a cone is equal to the product of its slant height into the circumference of the circle midway between the bases. The radius of the circle midway between the bases is ^ (r + R), whence its circumference is n {r + R). Now, tt (r + R) x H' is the area of the convex surface of the frustum, by the preceding corollary. PROPOSITION V. 519. TJieorem, — Two pyramids having equivalent bases and the same altitudes are equivalent, i. e., equal in volume. Dem.— Let S-ABCD and S'-A'B'C'D'E' be two pyramids having the same altitudes, and base ABCD equivalent to base A'B'C'D'E', i. e., equal in area; then is pyramid S-ABCD equivalent to S'-A'B'C'D'E', i. e., equal in volume. For, conceive the bases to be in the same plane, and a plane to start from coincidence with the plane of the bases, and move toward the vertices, remaining all the time parallel to the bases. At every stage of its progress the sections are equivalent, and as the plane reaches both vertices at the same time, by reason of the common altitude, it is evident that the volumes are equal. Or, if desired, we may consider the two pyramids as divided into an equal number of infinitely thin lam^.ncB parallel to the bases. Each lamina in one l^as its corresponding equivalen*^ lamina in the other ; hence the sum of all the lamince in one equals the sum of all the lamince in the other ; i. «., the pyramids are equivalent. OP PYRAMIDS AND CONES, 205 PROPOSITION VI. 520. Uieorem. — TJie volume of a triangular pyramid is equal to one-third the product of its base and altitude. Dem. — Let S-ABC be a triangular pyramid, whose altitude is H* ; then is the volume equal to ^ H X area ABC. For, throui^h A and B draw Aa and Bh paral- lel to SC; and through S draw Sa and Sh parallel to CA and CB, and join a and h\ then Sdb-kBC is a prism with its bases equal to the base of the pyramid. Now, the solid added to the given pyramid is a quadrangular pyramid with obBK as its base, and its vertex at S. Divide this into two triangular pyramids by draw- ing aB and passing a plane through SB and aB. These triangular pyramids are equiva- lent, since they have equal bases aAB and abB and a common altitude, the vertices of both being at S Fio. 310. Again, S-ahB may be considered as having abS (equal to ABC) as its base, and the altitude of the first pyramid (equal to the altitude of the prism) for its altitude, and hence as equivalent to the given pyramid. Therefore S-ABC is one third of the prism Sdb-kBC. But the volume of the prism is H x area ABC. Therefore the volume of the pyramid S-ABC is ^ H x area ABC. q. e. d. 521. Cor. 1. — The volume of any pyramid is equal to one-third the product of its base and altitude. Dem. — Since any pyramid can be divided into trian- gular pyramids by passing planes through any one edge, as SE, and each of the other edges not adjacent, as SB and SC, the volume of the pyramid is equal to the sum of the volumes of several triangular pyramids having the same altitude as the given pyramid, and the sum of whose bases is the base of. the given pyramid. Hence the truth of the corollaiy. 322, Cor. 2. — Pyra^nids having equivalent bases are to each other as their altitudes; such as have equal altitudes are to each other as their bases; and, in general, jjyramids are to each other as the products of their bases and altitudes. * Not drawn in the figure, leet it might confuae. 206 ELEMENTARY SOLID GEOMETRY. Fi6. 312 PROPOSITION TIL 523, Tlieorein, — Tlie volume of the frustum of a triangular pyramid is equal to the volume of three pyramids of the same altitude as the frustum, and lohose bases are the upper base, the lower base, and a mean proportional betioeen the two bases of the frustum, Dem. — Let a6c-ABC be the fnistiim of a triangu- lar pyramid. Through a6 and C pass a plane cutting off the pyramid C-abc. This has for its base the upper base of the frustum, and for its altitude the altitude of the fnistum. Again, draw A6, and pass a plane through A6 and bC, cutting off the pyramid 6-ABC, -which has the same altitude as the frustum, and for its base the lower base of the frustum. There now remains a third pyramid, b-ACa, to be ex- amined. Through b draw bD parallel to aA, and draw DC and aD. The pyramid D-ACa is equiva- lent to 6-ACa, since it has the same base and the same altitude. But the former may be considered as Jiaving ADC for its base, and the altitude of the fnistum for its altitude, i. e., as pyramid a-ADC. We are now to show that ADC is a mean proportional between abc and ABC. ABC : abc : : Also, ABC : ADC whence ABC* : ADC' By equality of ratios, ABC : abc :: ABC' : ADC*; whence ADC'' = abc x ABC, i. e., ADC is a mean proportional between the upper and lower bases of the frustum. 524. Cor. — The volume of the frustum of any pyramid is equal to the volume of three pyramids having the same altitude as the frustum, and for bases, the upper base, the loioer base, and a mean proportional between the bases of the frustum. For, the frustum of any pyramid is equivalent to the corresponding frustum of a triangular pyramid of the same altitude and an equivalent base (?) ; and the bases of the frustum of the triangular pyramid being both equivalent to the corresponding bases of the given frustum, a mean proportional between the triangular bases is a mean proportional between their equivalents. AB' AB ab : : AB : AD (?). AD (?); AB : AD (?). PROPOSITION vm. S2S. TJieorem, — The volume of a cone of revolution is equal to one-third the product of its base and altitude ; i. e., ^;rR'H, R being the radius of the base and H the altitude. OF PYRAMIDS AND CONES. 207 Dem. — This follows from the volume of a pyramid, by a course of reasoning precisely the same as in {515). The volume of a pyramid being equal to one- third the product of the base and altitude, and the cone being the limit of the pyramid, the volume of the cone is one-third the product of its base and altitude. - Now, R being the radius of the base of a cone of revolution, the base (area of) is ttR", whence i^rR^H is the volunae, H being the altitude. 526 » Cor. 1. — The volume of any cone is equal to one-third the prodilct of its base and altitude. S27, Cor. 2. — The volume of the frustum of a cone is equal to the volume of three cones having the same altittide as the frustum, and for bases, the tipjjer base, the lower base, and a mean propor- tional between the two bases of the frustum. The truth of this appears from the fact that the frustum of a cone is the limit of the frustum of a pyramid. PROPOSITION IX. S28, TJieorein, — The lateral surfaces of similar right pyra- mids are to each other as the squares of their homologous edges, their slant heights, and their altitudes ; i. e., as the squares of any two homologous dimensions. Dem. — Let A and a be homologous sides of the bases of two similar right pyramids, H' and h' their slant heights, H and h their altitudes, and P and p tJie perimeters of their bases ; then — (1) P : _p : : A : a, because the bases are similar polygons ; (2) A : a : : H' : h\ because the faces are similar triangles ; (3) H' : A' : : H : h (?). (Vhence, P : p : : H' : A' ; and, as ^H' :^h' :: H' : h', multiplying, we have ^P x H' : \px h' : : H'^ h"" : : A'^ : a^ : : H' : li". But tP X H' and ^pxN are the areas of the lateral surfaces. PROPOSITION X. 529* TJieoretn, — The convex surfaces of similar cones of revo- lution are to each other as the sqicares of their slant heights, the radii of their bases, and their altitudes ; i. e., as the squares of any two ho- mologous dimensions, '^ Dem.— Let H' and h' be the slant heights of two similar cones of revolution, R and r the radii of their bases, and H and h their altitudes; their convex surfaces are ;rRH' and Ttrh'. Now, since the cones are similar R : r : : H' : /t'. 208 ELEMENTARY SOLID GEOMETRY. Multiplying the terms of this proportion by the corresponding terms of nW • ich' : : H' : h\ we have — ;rRH' : TtrJi' : : H" : 7t'\ Hence the convex surfaces are as the squares of the slant heights, and since B: r :: R' '. h' :: H : h (?), R» : r» : : H* ; h"" : : H^ : h' ; and consequently ;rRH' : xrh' : : R^ : r' : : H^ : h\ PROPOSITION XL Y?^ ^ S30» TJieorem. — The volumes of similar pyramids are to each other as the cubes of their homologous dimensions. Dem. — Letting A and a be homologous sides of the bases of two similar pyramids, B and b their bases, and H and h their altitudes, the student should be able to give the reasons for the following proportions : ^H : ^A : : A : a : : H : 7^. Whence iBH : \hh : : A.^ : a^ : -. W : K\ Q. E. D. PROPOSITION xn. j^ ^ 531, Tlieoretn, — Tlie volumes of similar cones are to each other as the cubes of their altitudes, or as the cubes of the radii of their bases, Dem. R and r being the radii of their bases, and H and h their altitudes, R» ^: r* :: ff : A» (?), and R" : r^ : : H^* : A^ Also, ^,7rR:^7ih::R:h. Multiplying, ^tzR-'R : ^itr'h : : H' : h\ or as R' : J'\ Q. e. d. EXERCISES. 1. What is the area of the lateral surface of a right hexagonal pyramid whose base is inscribed in a circle whose diameter is 20 feet, the altitude of the pyramid being 8 feet ? What is the volume of this pyramid ? 2. What is the area of the lateral surface of a right pentagonal pyramid whose base is inscribed in a circle whose radius is 6 yards, the slant height of the pyramid being 10 yards ? What is the vol- ume of this pyramid ? 3. How many quarts will a can contain, whose entire height is 10 inches, the body being a cylinder 6 inches in diameter and 6 J inches OF THE SPHERE. 209 high, and the top a cone ? How much tin does it take to make such a can, allowing nothing for waste and the seams ? 4. If very fine dry sand is piled upon a smooth horizontal surface, / without any lateral support, the angle of slope {i. e., the angle of inclination of the sloping side of the pile with the plane) is about 31°. Suppose two circles be drawn on the floor, one 4 feet in diameter and the other 3, and sand piles be made as large as possible on these cir- cles as bases, no other support being given. What is the relative magnitude of the piles ? 5. In the case of sand piles, as given in the last example, the ratio of the radius of the base to the altitude of the pile is f. How many cubic feet in each of the above piles ? 6. The frustum of a right pyramid was 72 feet square at the lower base and 48 at the upper ; and its altitude was 60 feet. What was the lateral surface ? What the volume ? The student should furnish a synopsis of each section at- ITS close. SECTION V. OF THE SPHERE* 532, A Sphere is a solid bounded by a surface every point in which is equally distant from a point within called the Centre. The distance from the centre to the surface is the Radius, and a line passing through the centre and limited by the surface is a Diameter. The diameter is equal to twice the radius. * a epherical blackboard is almost indispensable in teaching this section, as well as in teaching Spherical Trigonometry. A sphere about 2 feet in diameter, mounted on a pedestal, and having its surface elated or painted as a blackboard, is what is needed. It can be ob« tained of the manufacturers of echool apparatus, or made In any good tumiug-shop. 210 ELEMENTARY SOLID GEOMETRY. CIRCLES OF THE SPHERE. PROPOSITION L 533. Tlieorem. — Every section of a sphere, made hy a plane, ti a circle. Dem.— Let AFEBD be a section of a sphere •whose centre is 0, made by a plane ; then is it a circle. For, let fall from the centre O a pei-pendicular upon the plane AFEBD, as OC, and draw CA, CD, CE, CB, etc., lines of the plane, from the foot of the perpendicular to any points in which the plane cuts the surface of the sphere. Join these points with the centre, O, of the sphere. Now, OA, OD, OB, OE, etc., beini^ radii, are equal; whence, CA, CD, CB, CE, etc., are equal ; i. «., every point in the line of intei-section of a plane and surface of a sphere is equally distant from a point in this plane. Hence, the intersection is a circle, q. e. d. 534. Def. — A circle made by a plane not passing tli rough the centre is a Small Circle ; one made by a plane passing through the centre is a Great Circle. 535. Cor. 1. — A perpendicular from the centre of a sphere, upon any small circle, pierces the circle at its centre ; and, conversely, a perpendicular to a small circle at its centre passes through the cefitre of the sphere. 536. Def. — A diameter perpendicular to any circle of a sphere is called the Axis of that circle. The extremities of the axis are the Poles of the circle. 537. Cor. 2. — The pole of a circle is equally distant from every point in its circumference. The student should be able to give the reason. 538. Cor. 3. — Every circle of a sphere has two poles, lohich, in case of a great circle, are equally distant from every point in the cir- cumference of the circle ; but, in case of a small circle, one pole is nearer any poiiit in the circumference tha^i the other pole is. DISTANCES ON THE SURFACE OP A SPHERE. 211 SS9, Cor. 4. — A small circle is less as its distance from the cen- tre of the sphere is greater. For, its diameter, being a chord of a great circle, is less as it is farther fi'om the centre of the great circle, which is also the centre of the sphere. S4:0. Cor. 5. — All great circles of the same sphere are equal, their radii being the radius of the sphere. PROPOSITION n. 541. Theorem, — Any great circle -divides the sphere into two equal parts called Hemispheres. Dem. — Conceive a sphere as divided by a great circle, i. «., by a plane passing through its centre, and let the great circle be considered as the base of each portion. These bases being equal, reverse one of the portions and conceive its base placed in the base of the other, the convex surfaces being on the same side of the common base. Since the bases are equal circles, they will coincide, and since every point in the convex surface of each portion is equally distant from the centre of the common base, the convex surfaces will coincide. There- fore, the portions coincide throughout, and ai-e consequently equal, q. e. d. PROPOSITION m. 542. Ttieorem. — TJie intersection of any two great circles of a sphere is a diameter of the sphere. Dem. — The intersection of two planes is a straight line ; and in the case of the two great circles, as they both pass through the centre of the sphere, this is one point of their intersection. Hence, the intersection of two great circles of a sphere is a straight line which passes through the centre. Q. e. d. 543. Cor. — The intersections on the surface of a sphere of two circumferences of great circles are a semi-circumference, or 180°, cepart, sifice they are at opposite extremities of a diameter. DISTANCES ON THE SURFACE OF A SPHERE. 544. Distances on the surface of a sphere are always to be under- stood as measured on the arc of a great circle, unless it is otherwise stated. 212 ELEMENTARY SOLID GEOMETRY. (S PROPOSITION IT. S4S, TJieorem, — The distances, measured on the surface of a sphere, from a pole to all p>oints in the circumference of a circle of which it is the pole, are equal. Dem.— Let P be a pole of the small circle AEB ; then are the arcs PA, PE, PB, etc., which measure the distances on the surface of the sphere, from P to any points in the circumference of circle AEB, equal. For, by (537), the straight lines AP, PE, PB, etc., are eqnaA, and these equal chords subtend equitl arcs, as arc PA, arc PE, arc PB, etc., the great circles of which these lines are chords and arcs being equal {540). Thus, for like reasons, arc P'QA = arc P'LE = arc P'RB, etc. 54:6. Cor. — TJie distance from the pole of a great circle to any point in the circumference of the circle is a quadrant {a quarter of a circumference). Since the poles are 180° apart (being the extremities of a diameter), PAQP' = PELF' = a semicircumference. But, in case of a great circle, chord PL = chord .P'L (= chord PQ — chord P'Q), whence arc PEL = arc P'L = arc PAQ = arc PQ. Hence, each of these arcs is a quadrant. 547 • ScH. — By means of the facts demonstrated in this proposition and corollary, we are enabled to draw arcs of small and great circles, in the surface of a sphere, with nearly the same facility as we draw arcs and lines in a plane. Thus, to draw the small circle AEB, we take an arc equal to PE, and placing one end of it at P, c^use a pencil held at the other end to ti*ace the arc AEB, etc. To describe the circumference of a great circle, a quadrant must be used for the arc. By bend- ing a wire into an arc of the circle, and making a loop in each end, a wooden pin can be put through one loop and a crayon through the other, and an arc drawn as represented in the figure. Fig. 315. PROPOSITION T. 548. JProhlem, — To pass a circumference of a great circU through any two points on the surface of a sphere. DISTANCES ON THE SURFACE OF A SPHERE. 213 Solution". — Let A and B be two points on the sur- face of a sphere, through which it is proposed to pass a circumference of a great circle. From B as a pole, with_ an arc ^gual to a quadrant, strike an arc on^ as nearly wliere the pole of tne circle passing through A and B lies, as may be determined by inspection. Then, from A, with the same arc, strike an arc %t intersecting on at P. NowJPjs J,he pole of the great circlej^assing^^hrmj^h^ A an d B. Hence, from P as a pole, with a qua drant arc *"ttfSw a circle; itwiJiJL pass through A and B, and will Fig. 316. be a great circle, since its i)ole is a quadrant's distance frOnTTtrcircumference. [The student should make the construction on the spherical blackboard.] *5^4i>. Cor. 1. — Throuyh any t wo points on the surface of a sphere, one great circle* can always be made to pass, and only one, except luhen the two points are at the extremities of the same diameter, in tohich case an infinite nuniber of great circles can le passed through the tivo points. Since the arcs on and st are arcs of great circles, the circumferences of which they form parts will intersect also on tlie opposite side of the sphere, at a dis- tance of a semicircumference from P. But these two points are poles of the same great circle. Now, as the two great circles can intersect at no other points, there can be only one great circle passed through A and B. But if the two given points were at the extremities of the same diameter, as at D and C, the arcs s^and on would coincide, and any point in this circumference being taken as a pole, great circles can be drawn through D and C. [The student should ti'ace the work on the spherical blackboard.] 550, SCH. — The truth of the corollary is also evident from the fact that three points not in the same straight line determine the position of a plane. Thus A, B, and tlie centre of the sphere, fix the position of one, and only one, great circle passing through A and B. Moreover, if the two given points are at the extremities of the same diameter, they are in the same straight line with the centre of the sphere, whence an infinite number of planes can be passed through them and the centre. The meridians on the earth's surface af- ford an example, the poles (of the equator) being the given points. 551. Cor. 2. — If two points in the circumference of a great circle of a sphere, not at the extremities of the same diameter, are at a quadranfs distance from a point on the surface, that point is the pole of the circle. * The word circle may be understood to refer either to the circle proper, or to its cir- cumference. The word is in constant use in the higher mathematics, in the latter sense. 214 ELEMENTARY SOLID GEOMETRY. PROPOSITION Tl. Fig. 31'; 552. Theorem, — TJie shortest distance on the surface of a sphere, hetiveen any two points in that surface, is measured on the arc less than a s&inicircumference of the great circle which joins them. Dem.— Let A and B be any two points in the sur- face of a sphere, AB the arc of a great circle joining them, and AmC^tB any other path in the surface be- tween A and B ; then is arc AB less than kmCiiB. Let C be any point in A7nCnB, and pass the arcs of great circles through A and C, and B and C. Join A, B, and C with the centre of the sphere. The angles AOB, AOC, and COB form the facial angles of a trie- dral, of which angles the arcs AB, AC, and CB are the measures. Now, angle AOB < AOC + COB (454); whence arc AB < arc AC + arc CB, and the path from A to B is less on arc AB than on arcs AC, CB. In like manner, joining any point in iKmC with A and C by arcs of great circles, their sum would be greater than AC. So, also, joining any point in C/iB with C and B, the sum of the arcA would be greater than CB. As this process is indefinitely repeated, the path from A to B on the arcs of the great circles will continually increase, and also continually approximate the path kmCnB. Hence, arc AB is less than the path A/wCtiB. q. e. d. 553. Cor. — Hie least arc of a circle of a sphere joining any tivo points in the surface, is the arc less than a semicircumference of the great circle pass- ing through the points ; and the greatest arc is the circumference minus this least arc. Dem.— Let kniBn be any small circle passmg through A and B, and ABDoC the great circle. As shown above, A^^B < kmS. Now, circumference ABDoC > circumference kmBn{539). Subtracting the former inequality from the latter, we have BD<>CA > BnA. q. e. d. Fig. 318. PROPOSITION vn. 554. T1ieorein.—The shortest path on the surface of a hemi- sphere, from any point therein to the circumfereiice of the great circle f mining its base, is the arc less than a quadrant of a great circle per- pendicular to the base, and the longest path, on any arc of a great circle, is the supplement of this shortest path. SPHERICAL ANGLES. 215 Fig. 319. Dem. — Let P be a point in the surface of the hemi- sphere whose base is ACBC, and DPmD' an arc of a great circle passing through P and perpendicular to ADCBC ; then is PD the shortest path on the surface from Pto circumference ADBC, and PmD' is the longest path from P to the circumference, measured on the arc of a great circle. For, the shortest path from P to any point in cir- cumference ADBC is measured on the arc of a great circle {552). Now, let PC be any oblique arc of a great circle. We will show that arc PD < arc PC. Pro- duce PD until DP' — PD ; and pass a great circle through P' and C. Draw the radii OP, OD, OC, and OP'. The triedrals 0-PDC and 0-P'DC have the facial angle POD = P'OD, they being measured by equal arcs, and the facial angle DOC com- mon. Hence, as the included diedrals are equal, both being right, the triedrals are equal or symmetrical (440). In this case they are symmetrical, and the facial angle POC = P'OC ; whence the arc PC = arc P'C. Finally, since PC -i- P'C > PP', PC, the half of PC + P'C, is greater than PD, the half of PP'. Secondly, PinD' is the supplement of PD, and we are to show that it is greater than any other arc of a great circle from P to the circumference ADBC. Let PtiC be any arc of a great circle oblique to ADCBC. Produce CwP to C. Now CPiiC is a seniicircumference and consequently equal to DPmD'. But we have before shown that PD < PC, and subtracting these from the equals CP?iC and DPmD', we have PmD' > P;iC'. SS5. Cor. — Fro77i any point m the surface of a liemispliere there are two pe7pe7idiculars to the circumference of the great circle ^vhich forms the base of the hemisphere ; one of which i^erpendiculars measures the least distance to that circumference, and the other the greatest, on the arc of any great cii'cle of the S2)here. Thus PD and PmD' are two perpendiculars from P upon the circumference ADBC. SPHERICAL ANGLES. S36, The angle formed by two arcs of_ Jfcircles of a sphere is conceived as the same as the angle included by the tangents to the arcs at the common point. III. — Let AB and AC be two arcs of circles of the sphere, meeting at A ; then the angle BAC is conceived as the same as the angle B'AC, B'A being tangent to the circle BADm, and CA to the circle CAEn,. Fig. 330. 216 ELEMENTARY SOLED GEOMETRY. Fig. 321. 557. A Splierical Angle is the angle included by two arcs of grea t circles. III. — BAC, Fig. 321, is a spherical angle, and is conceived as tlie same as the angle B'AC, B'Aand C'A being tangents to ihe great circles BADF and CAEF. [The student should not confound such an angle as BAC, Fig. 320, with a spherical angle.} Fig. 3^2. PROPOSITION Tin. 558, Uieorem, — A spherical angle is equal to the measure of the diedral included hy the great circles whose arcs form the sides of the angle. Dem. — Let BAC be any spherical angle, and BADF and CAEF the great circles whose arcs BA and CA include the angle ; then is BAC equal to the measure of the diedral C-AF-B. For, since two great circles intersect in a diameter {542), AF is a diameter. Now B'A is a tangent to the circle BADF, that is, it Hes in the same plane and is per- pendicular to AG at A. In like manner C'A lies in the plane CAEF and is perpendicular to AG. Hence B'AC is the measure of the diedral C-AF-B {4:25). Therefore the spherical angle BAC, which is the same as the plane angle B'AC, is equal to the measure of the diedral C-AF-B- q. e. d. 559. Cor. 1. — If one of two great circles passes through the pole of the other, their circumferences intersect at right angles. Dem. — Thus, P being the pole of the great circle CABm, PO is its axis, and any plane passing through PG is perpendicular to the plane CAB;n {427). Hence, the diedral B-AG-P is right, and the spheri- cal angle PAB, which is equal to the measure of the diedral, is also right. 560* Cor. 2. — A spherical angle is meas- ured by the arc ofja great circle intercepted heticeen its sides, and at a quadrayifs dis- tance from its vertex. Thus, the spherical angle CPA is measured by CA, PC and PA being quad- rants. For, since PC is a quadrant, CG is perpendicular to PG, the edge of the diedral C-PG-A, and for a like reason AG is perpendicular to PG. Hence, CGA is the measure of the diedral, and consequently CA, its measure, is the measure of the spherical angle CPA. SPHERICAL ANGLES. \l 217 S61* Cor. 3. — The angle included by two arcs of ^nqj l circles is I the same as the angle included by two arcs of great circles passing \i through the vertex and having the same tangents. Thus BAC = B"AC". For the angle BAC is, by definition, the same as B'AC, B'A and C'A being tangents to BA and CA. Now, passing planes through C'A, B'A, and the centre of the sphere, we have the arcs B"A, C'A, and B'A, C'A tangents to them. Hence, B"AC" is the same as B'AC, and consequently the same as BAC. 562, ScH. — To draio an arc of a great circle which shall be perpendicular to another ; or^ what Fig. 324. M the same thinrj, to construct a right spher ical angle. Let it be required to erect an arc of a great circle perpendicular to CAB at A, Fig. 323. Lay off from A, on the arc CAB, a quadrant's distance, as AP', and from P' as a pole, with a quad- rant describe an arc passing through A. This will be the perpendicular required. In a similar manner we may let fall a perpendicular from any point in the surface, upon any arc of a great circle. To let fall a perpendicular from P" upon the arc CAB, from P" as a pole, with a quadrant describe an arc cutting CAB, as at P'. Then from P' as a pole, with a quadrant describe an arc passing through P" and cutting CAB, and it will be perpendicular to CAB. [The stu- dent should have practice in making these constructions on the sphere.] PROPOSITION IX. 5GS, Problem* — To pass the circumference of a small circle through any three points on the surface of a sjyhere. Solution. — Let A, B, and C be the three points in the surface of the sphere through which we propose to pass the circumference of a circle. Pass arcs of great circles through the points, forming the spherical triangle ABC. Thus, to pass an arc of a iireat circle through B and C, from B as a pole, with a quadrant strike an arc as near as may be to the pole of the required circle ; and from C as a pole, with the quadrant strike an arc intersecting the former, as at P ; then is P the pole of a great circle passing through B and C (?). Hence, from P as a pole, with a quadrant pass an arc through B and C, and it will be the arc re- quired {551). In like manner pass arcs through A and C, A and B. Now, bisect two of these arcs, as BC and AC, by arcs of great Fig. 325. 218 ELEMENTARY SOLID aEOMETRY. circles perpendicular to each. [The student will readily perceive how this is done.] The intersection of these perpendiculars, o, will be the pole of the small circle required (?). Then from t>, as a pole, with an arc oB draw the circum- ference of a small circle : it will pass through A, B, and C (?), and hence is the circumference required. OF TANGENT PLANES. 564z, A Tanffcnt I^lane to a curved surface at a given point is the plane of two lines respectively tangent to two plane sections through the point. III. — Let P be a point in the cui*ved surfiice at which we wish a tangent plane. Pass any two planes through the surface and the point P, and let OPQ and MPN represent the intersections of these planes with the cui*ved surface. Draw UV and ST in the planes of the sections, and tangent to OPQ and MPN, at P. Then is the plane of UV and ST the tangent plane at P. Fig. 326. PROPOSITION X. 565. TIieorein.—A tangent plane to a sphere is perpendicular to the radius at the point of tanrjency. Dem. — Let P be any point in the surface of a sphere ; pass two great circles, as PaA, etc., and PwAR, through P, and draw ST tangent to the arc 7nP, and UV tangent to the arc aP ; then is the plane SVTU a tangent plane at P, and perpendicular to the radius OP. For, a tangent (as ST) to the arc mP is perpendicular to the radius of the circle, i. e.y to OP, and also a tangent (as VU) to the arc aP is perpendicular to the radius of this circle, i.e., to OP. Hence, OP is perpendicular to two lines of the plane SVTU. and conse- quently to the plane of these lines (?). q. E. D. 566. Cor. l.~Every point in a tangeiit plane to a sphere, except the point of tangency^ is without the sphere. OF SPHERICAL TRIANGLES. 219 For, OP, the perpendicular, is shorter than any line which can be drawn from to any other point in the plane (?), hence any other point in the plane than P lies farther from the centre of the sphere than the length of the radius, and is, therefore, without the sphere. 567, Cor. 2. — A tangent through p to any circle of the sphere passing through this point, lies in the tangent pilane. Dem. — Thus MN, tangent to the small circle PriRh through P, lies in the tangent plane. For, conceive the plane of the small circle extended till it in- tersects the tangent plane. This intersection is tangent to the small circle, since it touches it at one point, but cannot cut it; otherwise the tangent plane would liave another point than P common with the surface of the sphere. But there can be only one tangent to a circle at a given point. Hence this intersec- tion is MN, which is consequently in the tangent plane. OF SPHERICAL TRIANGLES. 568, A Spherical Trianffle is a portion of the surface of a sphere bounded by three arcs of great circles. In the present treatise these arcs will be considered as each less than a semicircumfer- ence. The terms scalene, isosceles, equilateral, right angled, and oblique angled, are applied to spherical triangles in the same manner as to plane triangles. PROPOSITION XI^ 569, Theorem,— Tlie sum of any t200 sides of a spherical tri- angle is greater than the third side, and their difference is less than the third side. Dem.— Let ABC be any spherical triangle; then is BC < BA + AC, and BC - AC < BA ; and the same is true of the sides in any order. For, join the vertices A, B, and C, with the centre of the sphere, by drawing AO, BO, and CO. There is tlius formed a triedral 0-ABC, whose facial angles are measured by the sides of the triangle (208). Now, angle BOC < BOA + AOC {4S4), whence BC < BA + AC : and subtracting AC from both members, we have BC — AC < BA. Fig. 3-28. 220 ELEMENTARY SOLID GEOtLTRY. PROPOSITION XIL 570» Tlieoretn, — The sum of the sides of a spherical triangle may he anytliing between and a "circumference. Dem. — The sides of a spherical triangle are measures of the facial angles of a triedral whose vertex is at the ccntj-e of the sphere. Hence their sum may be anything between and the measure of 4 right angles, as these are the limits of the sum of the facial angles of a tiledral {436). 571* ScH. — As the sides of a spherical triangle are arcs, they can be meas- ured in dagrees. Hence, we speak of the side of a spherical triangle as 30°, 57", 1x5° 10', etc. In accordance with this, we say that the limit of the sum of ;Le sides of a spherical triangle is 360°. PROPOSITION xin. 572* Hieorem* — The sum of the atigles of a spherical triangle may be anything between tivo and six right angles. Dem. — The sum of the angles of a spherical triangle is the same as the sum of the measures of the diedrals of a triedral having its vertex at the centre of the sphere, as in {569). Now the limits of the sum of the measures of these diedrals are 2 and 6 right angles {439). Hence the sum of the angles of any spherical Uiangle may be anything between 2 and 6 right angles. Q. e. d. 573, ScH. — It will be observed, that the sum of the angles of a spherical triangle is not constant, as is the sum of the angles of a plane triangle. Thus, the sum of the angles of a spherical triangle may be 200°, 290°, 350°, 500°, any- thing between 180° and 540°. 574:. Def. — Spherical Excess is the amount by which the sum of the angles of a spherical triangle exceeds the sum of the angles of a plane triangle; i. e., it is the sum of the spherical angles -180°, or 7t. III. — It is not difficult to observe the occasion of \hm excess in the case of the equilateral spherical triangle. Thus.let ABC be such a triangle. Conceive the plane triangle formed by the chords AB, AC, and CB. The sum of the angles of this plane triangle is 180°. Bat each angle of the spherical triangle is larger than the corresponding angle of the plane triangle. Thus, the spherical angle BACis the same as the plane angle CAB', included be- tween the tangents C'A and B'A, which are per- pendicular to the edge of the diedral C-AO-B,and Fia. 329. include its measuring angle. Now, CA and BA OF SPHERICAL TRIANGLES. 221 being different line^ iro^i C'A and B'A are oblique to the edge AO, and in- clude an angle less than its measure, and consequently less than CAB. For a like reason the plane angle ACB < the spherical angle ACB, and plane angle ABC < spherical angle ABC. Moreover, it is easy to see that the inequality between -any plane angle and the corresponding spherical angle increases as the chords BA and CA deviate more from the tangents. Whence we see why the sum of the angles of the spherical triangle is not a fixed quantity. 575, Cor. — A spherical triangle may have one^ tioo, or even three right angles ; and, in fact, it may have one, t2uo, or three obtuse angles ; since, in the latter case, the sum 'of the angles luill not neces- sarily he greater than 540°. 576. Def.— ^ Trirectangular Spherical Triangle is a spherical triangle which has three right angles. PROPOSITION XIV. 577, Theorein, — The trirectangular triangle is one-eighth of the surface of a sphere. Dem. — Pass three planes through the centre of a sphere, respectively per- pendicular to each other. They will divide the surface into 8 trirectangular triangles, any one of which may be applied to any other. Thus, let ABA'B', ACA'C, and CBC'B' be the great circles formed by the three planes, mutually perpendicu- lar to each other. The planes being perpendicular to each other the diedrals, as A-CO-B, C-BO-A, C-AO-B, etc., are right, and hence the angles of the 8 triangles formed are all right. Also, as AOB is a right angle, AB is a quadrant; as BOC is a right angle, CB is a quadrant, etc. Hence, each side of every triangle is a quadrant. Now any one triangle may be applied to any other. [Let the student make the application.] Hence the trirectangular triangle is one-eighth of the surface of a sphere. Q. e. d. , 57 S» Cor. — Tlie trirectangular triangle is equilateral and its sides are quadrants. PROPOSITION XV. 579, TJieorem* — In an isosceles spherical tria^igle the angles opposite the equal sides are equal j and, conversely. If two angles of a spherical triangle are equal, the triangle is isosceles. 222 ELEMENTARY SOLID GEOMETRY. Dem. — Let ABC be an isosceles spherical triangle in which AB = AC ; then angle ABC = ACB. For, draw the radii AO, CO, and BO, forming the edges of the triedral O-ABC. Now, since AB = AC, the facial angles AOC and AOB are equal, and the triedral is isosceles. Hence the dic- drals A-OB-C and A-OC-B are equal {442), and con- sequently the spherical angles ABC and ACB are equal {558). Again, if angle ABC = angle ACB, side AC = side AB. For in the triedral O-ABC, the die- drals A-OB-C and A -OC-B are equal, whence the facial angles AOB and AOC are equal {443), and conse- quently the sides AB and AC which measure these angles. Fig. 331. 580, Cor. — An equilateral spherical triangle is also equiangular ; and, conversely, If the angles of a splierical triangle are equal the triangle is equilateral. ^ PROPOSITION XVI. SSI. Hieorem. — On the same or 07i equal spheres tivo isosceles triangles having two sides and the included angle of the one equal to two sides and the included angle of the other, each to each, can he superimposed, and are consequently equal, Dem.— In the triangles ABC and AB'C, let AB = AC, AB' = AC ; and let AB = AB', BC = B'C, and angle ABC = ABC ; then can the triangle AB'C be superimposed upon ABC. For, since the triangles are isosceles, we have angle ABC — ACB, AB'C = ACB', and, as by hypothesis ABC = AB'C, these four angles are equal each to each. For a like reason AB = AC = AB' = AC. Now, applying AC to its equal AB, the extremity A at A and C at B, with the angle B' on the same side of AB as C, the con- vexities of the arcs AC and AB being the same, and in the same direction, the arcs will coincide. Then, as angle ACB' = ABC, CB' will take the direction BC, and since these arcs are equal by hypothesis, B' will fall at C. Hence B'A will fall in CA. as only one arc of a great circle can pass between C and A, and the triangle AB'C is super- imposed upon ABC; wherefore they are equal. [Let the student give the application when other parts are assumed equal.] 582, Symmetmcal Spherical Triangles are such as have the parts (sides and angles) of the one respectively equal to the parts of the other, but arranged in a dijfferent order, so that the tri- angles are not capable of superposition. OF SPHERICAL TRIANGLES. 223 Fia. :WJ. III.— In Fig. 333, ABC and A'B'C represent sj-mmetiical spherical tri- angles. In these triangles A = A', B = B', C = C, AC = A'C^ AB = A'B', and BC = B'C ; neverthe- less we cannot conceive one triangle superimposed upon the other. Thus, were we to make the at- tempt by placing A'B' in its equal AB, A' at A, and B' at B, the angle C would fall on the opposite side of AB from C. Now, we cannot revolve A'C'B' on AB (or its chord), and thus make the two coincide, for this would bring their convexities together. Nor can we make them coincide by reversing A'B'C, and placing B' at A, and A' at B. For, although these two arcs will thus coincide, as the angle B' is not equal to A, B'C will not fall in AC ; and, again, if it did, C would not fall at C, since B'C and AC are not equal. But, considering the triangles ABC and A'B'C in Fig. 334, in which A =r A', B = B', C = C, AC = A'C, AB =: A'B', and BC = B'C, we can readily conceive the latter as superimposed upon the former. [The student should make the application.] Now, the two triangles are equal in each case, as will subsequently appear of the former. Such triangles as tliose in Fig. 333 are called symmetHcally equal, while the latter are said to be equal by superposition. Fig. 335 represents the same triangles as Fig. 334, and exhibits a complete projection* of the semicir- cumferences of which the sides of the triangles are arcs. The student should become perfectly familiar with it, and be able to draw it readily. Thus, a^Bh is the projection of the semicircumference of which Fig. a35. AB is an arc, aACc of the semicircumference of which AC is an arc, etc., etc. Fig. 334. PROPOSITION XTII. S83, Theorem, — Symmetrical spherical triancjUs are equiva- lent. '* To understand what is meant by the projection of these lines, conceive a hemisphere irith its base on the paper, and represented by the circle abc, and all the arcs raised up from the paper as they would be on the sui-fe,ce of such a hemisphere. Thus, considering the arc aABft, the ends a and b would be in the paper jutt where they are, but the rest of the arc would be off the paper, as though you could take hold of B and raise it from the paper while a and b remain fixed. The lines in the figure are representations of lines on the surface of such a hemisphere, as they would appear to an ej'e situated in the axis of the circle abc, and at an infinite distance from it; that is, just as if each point in the lines dropped 7>er/)««(/ic«^ar/y down upon the paper. Arcs of great circles perpendicular to the base are projected in straight lines passing through the centre, and oblique arcs are projected in ellipses. See Spherical Trigonometry {97-/09), 224 ELEMENTARY SOLID GEOMETRY. Dem. — Let ABC and A'B'C be two symmetrical spherical triangles, with AB = A'B', AC = A'C\ BC = B'C, A = A', B = B', and C = C ; then are they equivalent For, pass circumferences of small circles through the vertices A, B, C and A', B', C, as abc and a'b'c\ of which o and o' are the poles. [The student should execute this on the spherical blackboard.] Now, by reascm of the mutual equality of the sides, the chord AC = chord A'C, chord AB = cJiord A'B\ and chord BC = cJiord B'C, and as the small circles are circumscribed about the equal plajie triangles ABC and A'B'C, these circles are equal. Hence, oA = o'A' = oB = o'B' = oC = o'C. The tri- angle AoB is therefore equal to At^'B', BoC = B'o'C, and A A' ; then is BC > B'C. For, join the vertices with the centre, forming the two triedrals 0-ABC and O-A'B'C. In these triedrals AOB = A'OB', AOC = A'OC, being measured by equal arcs ; and C-AO-B > C'-A'O-B', having the same measures as A and A' {558). Hence COB > COB' {449). Therefore CB, the measure of COB, is greater than CB', the measure of COB'. In like manner, the same sides of the triangles, and con- Fig. 339. sequently the same facial angles of the triedrals, being granted equal, and 15 226 ELEMENTARY SOLID GEOMETEY. BC > B C, A > A'. For, BC being greater tlian B'C, COB > COB' ; whence B-AO-C > B'-A'C-C {450), or A is greater than A'. PROPOSITION XXI. SS9, Hieoreni, — On the same, or on equal spheres, iwo spheri- cal triangles having the sides of the one respectively equal to the sides of the other, or the a?igles of the one respectively equal to the angUs of the other, are equal, or symmetrical and equivalent. Dem. — The sides of the triangles being equal, the facial angles of the triedrale at the centre are equal, whence the triedrals are equal or symmetrical {451). Consequently the angles of the triangles are equal, and the triangles are equal, or symmetrical and equivalent. Again, the tiiangles being mutually equiangular, the triedrals have their diedrals mutually equal ; whence the triedrals are equal or symmetrical {452). Therefore, the sides of the triangles are mutually equal, and the triangles ai'C equal, or symmetrical and equivalent. (See Figs. 333, 334) PROPOSITION xxn. S90, Tlieorem, — On spheres of different radii, mutually equi- angular triangles are similar (not equal). Dem. — Let be the common centre of two un- equal spheres ; and let ABC be a spherical triangle on the surface of the outer. Draw the radii AO, BO, and CO, constructing the triedral 0-ABC. Now, the intersections of these faces with the surface of the inner sphere will constitute a ti-iangle which is mutually equiangular with ABC. Thus, A r= «, B = 6, and C = c, since in each case the correspon- ding diedrals are the same. From the similar sec- tors aOh, AOB, we have aZ> : AB : : aO : AO; and, Fis. 340. in like manner, oc : AC : : aO : AO. "Whence, ab : AB : : flrc : AC. So, also, ab : AB : : iO : BO, and be: BC:: 50 : BO ; whence, ab : AS : -. be : BC. Thus we see that ABC and abc, having their angles equal each to each, have also their sides proportional: therefore they are similar. POLAR OR SUPPLE3IENTAL TRIANGLES. S91» One triangle is polar to another when the vertices of one are the poles of the sides of the other. Such triangles are also POLAR OR SUPPLEMENTAL TRIANGLES. 227 called supplemental, since the angles of one are the supplements of the sides opposite in the pther, as will appear hereafter. PROPOSITION XXIII. 592, I^rohlem^ — Having a spherical triangle given, to draw its i^olar. Solution.— Let ABC be the given triangle * From A as a pole, with a quadrant strike an arc, as C'B', From B as a pole, with a quadrant strike the arc C'A' ; and from C, the arc A'B'. Then is A'B'C polar to ABC. 59H, Cor. — If one triangle is polar to anotlier, conversely, the latter is polar to the former ; i. e., the relation is reciprocal. Fig. 341. Thus, A'B'C being polar to ABC ; reciprocally, ABC is polar to A'B'C; that is, A' is the pole of CB, B' of AC, and C of AB. For every point in A'B' is at a quadrant's distance from C, and every point in A'C is at a quadrant's dis- tance from B. Hence, A' is at a quadrant's distance from the two points C and B of CB, and is therefore its pole. [In like manner the student should show that B' is the pole of AC, and C of AB.] «5i>4. Sen. — By producing each of the arcs struck from the vertices of the given triangles suflaciently,/ lune AIEB, whereas BL < Bl. In a similar manner we can show that surface of sphere is not to lune ACEB : : BCmDn ; any arc greater than BC. Hence, as the fourth term can neither be less nor greater than BC, it must be equal to BC, and we have surf ace of sphere : lune ACEB :: BCmD/i : BC, ». e.j as 4 right angles, to the angle of the lune. €10, SCH, 2. — To obtain the area of a lune whose angle is knoicn, on a c/iven ipliere, find the area of the sphere, and multiply it by the ratio of the angle of the lune (in degrees) to 360°. Thus, R being the radius of the sphere, ^TtW is the surface of the sphere ; and the lune whose angle is 30° is -3^% or •h the surface of the sphere, i. e., -h of 47rR^ = i^rR^ PROPOSITION XXIX. Oil, Hieorein, — Tf Uvo semicircumferences of great circles intersect on the surface of a hemisphere^ the sum of the two opj^osife triangles thus formed is equivalent to a lune whose angle is that included hy the semicircumferences, Dem. — Let the semicircumferences CEB and DEA intersect at E on the surface of the hemi- sphere whose base is CABD ; then the sum of the triangles CED and AEB is equivaleut to a lune whose angle is AEB. For, let tlie semicircumferences CEB and DEA rbe produced around the sphere, intersecting on the opposite hemisphere, at the extremity F ot the diameter through E. Now, FBEA is a lune whose angle is AEB. Moreover, the triangle AFB is equivalent to the triangle DEC: since angle ^ig 350. AFB = AEB - DEC, side AF — side ED, each being the supplement of AE; and BF =: CE, each being the supplement of EB. Hence, tlie sum of the triangles CED and AEB is equivalent to the lune FBEA. Q. £. D. 234 ELEMENTARY SOLID GEOMETRY. PROPOSITION XXX. 612, Theorem, — The area of a spherical triangle is to the area of the surface of the hemisphere i7i which it is situated, as its spheri- cal excess is to four right angles, or 360°. Dem. — Let ABC be a spherical triangle wliose angles are represented by A, B, and C ; tlien is area ABC : surf, of hemisphere : : A + B + C — 180° : 4 right angles, or 360°. Let lune A represent the lune whose angle is the an- gle A of the triangle, i. e., angle CAB, and in like man- ner understand lune B and lune C. Now, triangle AHC+ AED - lune A {6H\ BHI + BEF = lune B, CCF + CDI = lune C . Adding, 2ABC + hemisphere = lune (A + B + C)*, (1) since the six triangles AHC, AED, BHI, BEF, CCF, and CDI, make the whole hemisphere and 2ABC be- sides, ABC being reckoned three times. From (1), we have by transposing and remembering that a hemi- sphere is a lune whose angle is 180°, and dividing by 2, ABC = ^lune (A + B + C - 180^). But, by {607\ i lune (A + B + C - 180°) : surf, of hemisph. : : A + B + C - 180° : 4 right angles. Therefore, ABC : surf, of liemisph. : : A + B + C — 180° : 4 right angles. 613, ScH. \.—To find the area of a spherical tiiangle on a given sphere, the angles of the triangle being given, we have simply to multiply the area of the hemisphere, i. e., 27rR', by the ratio of the spherical excess to 360°. Thus, if the angles are A = 110°, B = 80°, and C = 50°, we have AD/- O TDS. A+B + C-180° „ ^3 60 ,^T„ area ABC = 27rR'» x ^— ^ = 2ii^ ^ S60 ~ ^ ^^ ' 614L, ScH. 2.— This proposition is usually stated thus: The area of a spherical triangle is equal to its sph^?ical excess multiplied by the trirectangular triangle. When so stated the spherical excess is to be estimated in terms of the right angle ; i. e., having subtracted 180° from the sum of its angles, we are to divide the remainder by 90°, thus getting the spherical excess in right angles. In the example in the preceding scholium, the spherical excess estimated in this 110° + 80°j^^50°-180° ^ i. and the area of the triangle would way would be 90= * This signifies the lune whose angle is A + B + C, which ie of course the sum of tho three lunes whose angles are Ai B, and C- YOLUME OF SPHERE. 235 be I of the trirectangular triangle. Now, the trirectangular triangle being i of the surface of the sphere {577) is i of 47rR-', or ^nlC. This multiplied by | gives ^7tW, the same as above. The proportion, ABC : surf, of Tiemisph. : : A + B + C — 180' : 360°, is readily put into a form which agrees with the enunciation as given in this scholium. Thus, surf.ofhemisph. = 27rR'', whence A + B + C-180° , ^, A+B+C-ISO" ABC=r 2;rR'^ x- 360= ^7tW 90° VOLUME OF SPHERE. PROPOSITION XXXI. 6 IS, TJieorem. — The volume of a spliere is equal to the area of its surface multijflied hy \ of the raditis, that is, JttR', E deing the radius. Dem. — Let OLrrR be the radius of a sphere. Conceive a circumscribed cube, that is, a cube whose faces are tangent planes to the sphere. Draw lines from the vertices of each of the polyedral angles of the cube, to the centre of the sphere, as BO, CO, DO, AO, etc. These lines are the edges of six pyramids, having for their bases the faces of the cube, and for a common altitude the radius of tJie sphere (?). Hence the volume of the circumscribed cube is equal to its surface multiplied by ^R. Again, conceive each of the polyedral angles of the cube truncated -by planes tangent to the sphere. A new circumscribed solid will thus be formed, whose volume will be nearer that of the spbere than is that of the circumscribed cube. Let abc represent one of these tangent planes. Draw from the polyedral angles of this new solid, lines to the centre of the sphere, as aO, 50, and cO, etc. ; these lines will form the edges of a set of pyramids whose "bases constitute the surface of the solid, and whose common altitude is the radius of the sphere (?). Hence the volume of this solid is equal to the product of its surface (the sum of the bases of the pyramids) into ^R. Kow, this process of truncating the angles by tangent planes may be con- ceived as continued indefinitely ; and, to whatever extent it is carried, it will always be true that the volume of the solid is equal to its surface multiplied by iR. Therefore, as the sphere is the limit of this circumscribed solid, we have the volume of the sphere equal to the surface of the sphere, which ia 4;rR'', multiplied by ^R, i. e., to ^3 7rR'. q. e. d. Fig. 352. 236 ELEMENTARY SOLID GEOMETRY. 616. Cor. — Tlie surface of the spliere may he conceived as con- sisting of an infinite numler of infinitely small plane faces, and the volume as composed of an infinite number of jjyrajnids having these faces for their bases, and their vertices at the centre of the sphere, the common altitude of the pyramids being the radius of the sphere. 617 • A Spherical Sector is a portion of a sphere generated by the revohition of a circular sector about the diameter around which the semicircle which generates the sphere is conceived to revolve. It has a zone for its base ; and it may have as its other sur- faces one, or two, conical surfaces, or one conical and one plane surface. III. — Thus let db be the diameter around which the semicircle aCh revolves to generate the sphere. The solid generated by the circular sector AOa will be a spherical sector having a zone (AB) for its base ; and for its other surface, the conical surfoce gene- rated by AO. The spherical sector generated by COD, has the zone generated by CD for its base ; and for its other surfaces, the concave conical surface generated by DO, and the convex conical surface generated by CO. The spherical sector generated by EOF, has the zone generated by EF for its base, the plane generated by EO for one surface, and the concave conical s-orface generated by FO for the other. 618. A Spherical Seg^nent is a portion of the sphere in- cluded by two parallel planes, it being understood that one of the planes may become a tangent plane. In the latter case, the seg- ment has but one base ; in other cases, it has two. A spherical segment is bounded by a zone and one, or two, plane surfaces. PROPOSITION XXXU. 619. Tlieorem. — The volume of a spherical sector is equal to the pi'oduct of the zone which forms its base into one-third the radius of the sphere. Dem. — A spherical sector, like the sphere itself, may be conceived as con- sisting of an infinite number of p\Tamids whose bases make up its surface, and whose common altitude is the radius of the sphere. Hence, the volume of the sector is equal to the sum of the bases of these pyramids, that is, the spherical surface of the sector, multiplied by one-third their common altitude, which is one-third the radius of the sphere. Q. E. D. EXERCISES ON THE SPHERE. 237 620, Cor. — The volumes of spherical sectors of the same or equal spheres are to each other as the zones which form their hases ; and, since these zones are to each other as their altitudes (604), the sec- tors are to each other as the altitudes of the zones which form their bases. PROPOSITION xxxm. 621. Theorein, — The volume of a spherical segment of one hase is 7rA'^(R — -JA), A leing the altitude of the segment, and R the ra- dius of the sphere. Dem. — Let CO = R, and CD = A ; then is the volume of the spherical seg- ment generated by the revolution of CAD about CO equal to ;rA^(R - \L). For, the volume of the spherical sector generated by AOC is the zone generated by AC, multiplied by ^R, or 27rAR x ^R = ^ttAR^. From this we must subtract the cone, the radius of whose base is AD, and whose altitude is DO. To obtain this, we have DO = R — A : whence, from the right angled triaugle ADO, AD = y'R^ - (R - Ap" = v^2AR - A^ Now, the volume of this cone is iQD X ttAD", or ^;r(R - A) (2AR - A^) = ^7r(2AR2 Subtracting this from the volume of the spherical sec- tor, we have \TtKW - i;r(3AR2 - SA^R + A^O = ^(A^R - lA') = TT A2(R - lA). Q. E. D. Q22, ScH.— The volume of a spherical segment with two bases is readily obtained by taking the difference between two segments of one base each. Thus, to obtain the volumes of the segment generated by the revolution of Z>CAc about «0, take the differ- ence of the segments whose altitudes are ac and ah. Fig. 354. 3A-^R + A^'). Fig. 355. EXERCISES. 1. What is the circumference of a small circle of a sjohere whose diameter is 10, the circle being at 3 from the centre? Ans., 25.1328. 2. Construct on the spherical blackboard a spherical angle of 60°. Of 45°. Of 90°. Ofl20S Of 250°. 238 ELEMENTARY SOLID GEOMETRY. Sug's. — Let P be the point where the vertex of the required angle is to be situated. With a quadrant strike an arc from P, wliich shall represent one side of the required angle. From P as a pole, with a quadrant, strike an arc from the side before drawn, which shall measure the required angle. On this last arc lay off from the first side the measure of the required angle,* as 60°, 45°, etc. Through the extremity of this arc and P pass a great circle {548). [The stu- dent should not fiiil to give the reasons, as well as do the work.] 3. On the spherical blackboard construct a spherical triangle ABC, having ab = 100°, AC = 80°, and A = 58°. 4. Construct as above a spherical triangle ABC, having AB = 75°, A = 110°, and B = 87°. 5. Construct as above, having AB = 150°, BC = 80°, and AC = 100°- Also having AB = 160% AC = 50°, and BC = 85^ 6. Construct as above, having A = 52°, AC = 47°, and CB = 40°. Sug's. — Construct the angle A as before taught, and lay off AC from A equal to 47°, with the tape. This determines the vertex C. From C, as a pole, with an arc of 40°, describe an arc of a small circle ; in this case this arc will cut the opposite side of the angle A in two places. Call these points B and B'. Pass circumferences of great circles through C, and B, and B'. There are two tri- angles, ACB and ACB'. Note. — The teacher can multiply examples like the three preceding at pleas- ure. This exercise should be continued till the pupil can draw a spherical tri- angle as readily as a plane triangle. 7. "What is the area of a spherical triangle on the surface of a sphere whose radius is 10, the angles of the triangle being 85°, 120°, and 150°? Aiis., 305.4 +, 8. What is the area of a spherical trfkngle on a sphere whose diameter is 12, the angles of the triangle being 82°, 98°, and 100° ? 9. A sphere is cut by 5 parallel planes at 7 from each other. What are the relative areas of the zones ? What of the segments ? 10. Considering the earth as a sphere, its radius would be 3958 miles, and the altitudes of the zones, North torrid = 1578, North temperate = 2052, and North frigid = 328 miles. What are the relative areas of the several zones ? SuG. — The student should be careful to discriminate between the width of a zone, and its altitude. The altitudes are found from their widths, as usually given in degrees, by means of trigonometry. * For this purpose a tape equal in length to a semicircumference of a great circle of the »phere used, and marked off into 180 equal parts, will be convenient. A strip of paper ma^ be need. EXERCISES ON THE SPHERE. 239 11. The earth being regarded as a sphere whose radius is 3958 miles, what is the area of a spherical triangle on its surface, the angles being 120°, 130°, and 150° ? What is the area of a trirectan- gular triangle on the earth's surface ? 12. Construct on the spherical blackboard a spherical triangle ABC, having A = 59°, AC = 120°, and AB = 88°. Then construct the triangle polar to ABC. 13. Construct triangles polar to each of those in Examples 3, 4, and 5. 14. In the spherical triangle ABC given A = 58°, B = 67°, and AC = 81° ; what can you affirm of the polar triangle ? 15. ■ What is the volume of a globe which is 2 feet in diameter ? What of a segment of the same globe included by two parallel planes, one at 3 and the other at 9 inches from the centre ? 16. Compare the convex surfaces of a sphere and its circumscribed cylinder and cone, the vertical angle of the cone being 60°. 17. Compare the volumes of a sphere and its circumscribed cube, cylinder, and cone, the vertical angle of the cone being 60°. 18. If a and l represent the distances from the centre of a sphere whose radius is r, to the bases of a spherical segment, show that the volume of the segment is ;r[r' (b — a) — \{¥ — a')]. See (62 1, 622). ^Y3 PART III. AN ADVANCED COURSE IN GEOMETRY. OHAPTEE I. EXERCISES IX GEOMETRICAL INVENTION, SECTION I. THEOREMS IN SPECIAL OR ELEMENTARY GEOMETRY. S2S. Tliis chapter will afford a review of Parts I. and II., while it will greatly extend the student's knowledge of geometrical facts. Great pains should be taken to secure good habits as to neatness of execution in the construction of figures, orderly and proper arrange- ment of thought, and in style of expression. The practice of con- structing every figure upon geometrical principles — guessing at nothing — cannot be too strongly commended. As to the form of a geometrical argument, observe the following order : 1st. The enunciation of the theorem or problem in general terms. 2d. The elucidation of the general statement, by reference to the particular figure which it is proposed to use. 3d. A description of the figure, with reference to any auxiliary construction which is used in the demonstration or solution. 4th. The demonstration proper. S24z. If two adjacent sides of a quadrilateral are equal each to each, and the other two adjacent sides equal each to each, the diago- nals intersect at right angles. SuG's.^st. Draw a quadrilateral having such sides as the data require, and draw its diagonals. 2d. State the proposition witli reference to the figure. 2U EXERCISES IN GEOMETEICAL INVENTION. 3d. [In this case the regular third step is not required, as no auxiliaiy lines are necessarj',] 4th. Prove that the diagonals are at right angles to each other. The demonstration is based upon a corollary in Section I., Part II., Chapter I. 62o. Cor. — One of the diagonals is bisected. [State which one, and show why.] 626. If a parallelogram has one oblique angle, all its angles are oblique ; and if it has one right angle, all its angles are right angles. Sug's. — Let the student be careful to follow the order as heretofore given. No auxiliary construction is needed. The demonstration is based upon the doctiine of parallels. 627' The sum of three straight lines drawn from any point within a triangle to the vertices is less than the sum,, and greater thau the half sum of the three sides of the triangle. Sug's. — The first statement is proved from (270) and the second from {274:.) 628. A line drawn from any angle of a triangle to the middle of tlie opposite side, is less than the half sum of the adjacent sides, and greater than the difference between this half sum and half the third side. Sug's. — 1st. Draw a triangle, as ABC, bisect one side, as AC, and draw BD. 2d. Make the statement with reference to the figure. 3d. Produce BD until DE = BD, and draw AE and EC. 4th. The first step in the proof is to show the triangle ADE equal to CBD, and ADB equal to DCE ; whence AE = BC, and EC = AB. Fig. 356. Fig. 357. 629. If hues be drawn from the extremities of either of the non-parallel sides of a trapezoid to the mid- dle of the opposite side, the triangle thus formed is half the trapezoid. StjG's. — Tlie third step, or construction, consists in drawing HE parallel to AD and hence to BC (?), and FG through E parallel to AB. 630. Any line drawn through the centre of the diagonal of a parallelogram bisects the figure. THEOREMS IN SPECIAL GEOMETRY. 245 031. Prove that the sum of the angles of a triangle is two right angles, by producing two of the sides about an angle and through this angle drawing a line parallel to the third side. Prove the same by producing one side of the triangle and drawing a line through the ex- terior angle parallel to the non-adjacent side. Fig. 3o8. 632. If any point, not the centre, be taken in a diameter of a circle, of all the chords which can pass through that point, that one is the least w^hich is at right angles to the diameter. ® 033. If from any point there extend two lines tangent to a cir- cumference, the angle contained by the tangents is double the angle contained by the line joining the points of tangency and the radius extending to one of them. 034:. The angle included by two lines drawn from any angle of a triangle, the one bisecting the angle and the other perpendicular to the opposite side, is half the difference of the other two angles of the triangle. Sug's. ABD = 90° - A, whence ABD - EBD = 90° - A - EBD. Also, DBC = 90° - C, whence EBC = 90^ .-= 90° - A - EBD. etc. C + EBD 035. If three lines be drawn from the angled triangle — two bisecting these angles, and a third a perpendicular to one of the bisecting hues — the triangle included by these lines will be isosceles. acute angles of a riijht "Sug's. — It is to be proved that OD COD = OAC + ACQ = 45°, etc. CD. 030, If one circumference be described on the radius of another as a diameter, any straight line extending from their point of contact to the outer circumference is bisected by the inner. SuG.— The demonstration is based upon {159, 211). 246 EXERCISES IN GEOMETRICAL INVENTION. 637 • Prove that the sum of the angles of a regular five point star (Fig. 101) is two right angles. Show, also, that the figure formed by the iuterceiDted portions of the lines is a regular pentagon. 038. If the sides of a regular hexagon are produced till they meet, show that the exterior figures will be equilateral triangles. 639» If from two given points on the same side of a given line, two lines be drawn meet- ing in the line, their sum is least when they make equal angles with the line. Fig. 361. 640. If from tT\'o given points without a circumference, two lines be drawn meeting in the circumference, their sum is least when they make equal angles with a tangent at the common point, the points being on the opposite side of the tan- gent from the circle. 641. The side of an equilateral triangle inscribed- in a circle is equal to the diagonal of a rhombus, whose other diagonal and each of whose sides are equal to the radius. 642. If two circumferences intersect each other, and from either point of intersection a diameter be drawn in each, the other ex- tremities of these diameters and the other point of intersection are in the same straight line. 643. If any straight line joining two parallels be bisected, any other line through the point of bisection and included by the par- allels, is bisected at the same point. 644. If the sides of any quadrilateral are bisected, the quadri- lateral formed by joining the adjacent points of bisection is a par- allelogram. SuG*s. — 1st. Draw a quadrilateral, bisect its sides, and join the adjacent points of bisection. 2d. State the proposition, with reference to the figure. 3d. Draw the diagonals. 4lh. Give the proof. It is based on the similarity of triangles. 645. Cor. L— The parallelogram is one- half the trapezium. Prove it. What figure is formed by joining the centres of EF, FC, Fig. ^n. and FC, HG, etc. ? THEOREMS IN SPECIAL GEOMETRY. 247 64G. Cor. 2. — Lines joining the middle points of the opposite sides of any trapezium bisect each other (?). 6^7' If two straight lines join the alternate ends of two parallels, the line joining their centres is half the dif- ference of the parallels. a,^ ^- Sug's.— We are to prove that EF = i (CD — AB). CH = EF = i (CD - AB). 64:8. In any right-angled triangle the line drawn from the right angle to the middle of the hypotenuse is equal to one-half the hypotenuse. 04:9. The perpendiculars which bisect the three sides of a triangle meet in a common point. Sug's. — Eirst show that the intersection of two of the perpendiculars is equally distant from the three vertices of the triangle. Then that a line drawn from this 'point to the middle of the third side is perpendicular to it. 650. The three perpendiculars drawn from the angles of a tri angle upon the opposite sides intersect in a common point. ■;N Sug's. — Draw through the vertices of the triangle lines parallel to the opposite sides. The proposition may then be brought under the preceding. 651. Cor. — The following triangles ''•../ are similar — viz., BOE, BDC, AOD, and w 'aec, each to each ; also BOF, BDA, DOC, and CFA. Prove it. Fig. 364. 652. If from a point without a circle two secants be drawn, mak- ing equal angles with a third secant passing through the same point and the centre of the eircle, the intercepted chords of the lirst two are equal. SuG. --Prove by revolving one part of the figure. 248 EXERCISES IN GEOMETRICAL INVENTION. 6ij3. The Slim of the alternate angles of any hexagon inscribed in a circle is four riffht ano:les. 634:. If two circles intersect in A and B, and from P, any point in one circumference, the chords PA and PB be drawn to cut the other in c and D, CD is parallel to a tangent at P. Fig. 365. pendicular to each other. Goi>. If two lines intersect, two lines which bisect the opposite angles are per- 656. The angle included by two lines drawn from a point within a triangle to the vertices of two of the angles, is greater than the third angle. Sug's.— The demonstration may be founded on {219) or {231). 657. In a triangle whose angles are 90°, 60°, and 30°, show that the longest side is twice the shortest. 658. Lines which bisect the adjacent angles of a parallelogram are mutually pei-pendicular. 659. If from any point in the base of an isosceles triangle lines are drawn parallel to the sides, a parallelogram is formed whose peri- meter is constant and equal to the sum of the two equal sides of the triangle. Fio. 3G6. 660. If from any point in the base of an isosceles triangle perpendiculars be drawn to the sides of the triangle, their sum is constant and equal to the per- pendicular from one of the equal angles of the triangle upon tlie opposite side. 661. If from any poin« within an equilateral tri- angle, throe perpondiculL s be let fall upon the sides, their sum is constant and equal to the altitude of the Fig. 367. triangle. THEOREMS IN SPECIAL GEOMETRY. 0^ 602. If from a fixed point without a circle two tangents be drawn terminating in the circumfer- ence, the triangle formed by them and any tan- gent to the included arc has a constant perimeter equal to the sum of the first two tangents. 249 Fig. 3fi8. 063, The sum of two opposite sides of a quadrilateral circum- scribed about a circle, is equal to the sum of the other two. 004. If two opposite angles of a quadrilateral are supplemen- tary, it may be circumscribed by a circumference. 005. The square described on the sum of two lines is equivalent to the sum of the squares on the lines, ^;Z?^5 twice the rectangle of the lines. Sug'r. — Be careful to give the construction fully, and show that the parts are rectangles, etc. a b 6' "* a b Fig. 309. 000. The square described on the difference of two lines is equivalent to the sum of tlie squares on the lines, minus twice the rectangle of the lines. a-b 007. The rectangle of the sum and difference of two lines is equivalent to the difierence of the squares described on the lines. Sen.— The three preceding propositions are but geo- metrical conceptions and demonstrations of the algebraic farmul(B, {a + hf = a" + 2ab + b\ {a - bf = «' - 2ab + b\ and {a + b) {a - b) ■= «* - b\ r b' B 3 a-b b ft Fig. 371. 250 EXERCISES IN GEOMETRICAL INTENTION. VARIOUS DEMONSTRATIONS OF THE PYTHAGOREAN PROPOSITION. 668. The square described on the liypotenuse of a right-angled triangle is equivalent to the sum of tiie squares described on the other two sides. 1st Method. — Let ABC be the given trianiilo, and ACED the square described on the hypotenuse. Complete the construction. Show that the four triangles are equal. The square HF is (AB - BC)^ The student can complete the demonstration. Fig. 372. 2d IVIethgd. — Let ACED be the square on the hypotenuse. Let fall the perpendiculars EF, DC, etc. Show that the three triangles are equal, and that FD and LB are the squares of the two sides AB and BC. 3d Method.— Let BL and BH be the squares on the sides. Produce FL and HG till they meet in K. Draw DA and EC perpendicular to AC, and draw DE and KB. Prove that ACED is a square, and also that the triangles ABC, CLE, BFK, KBC, DKE, and AHD are all equal to each other. The demonsti*ation is then readily made. 4th Method.— This is the demonstration usually given in our text-books. Drawing the squares on the three sides, let fall Bl y>erpen- dicular to AC and produce it to K. Draw BD, BE, HC and AF. Show that the triangle HAC = BAD, and that the former is half the square AG, and the latter half the rectangle AK. Hence AG = AK. In like manner show that LC == CK. THEOREMS IN SPECIAL GEOMETRY. 251 We will now give a few other figures by means of which the demonstration can be effected, and leave the student to his own resources in effecting it. r.^ ^ 5th Method. 6th Method. 7th Method. 8th "Method. Fig. 376. ' 9th Method. — The truth of the theorem appears also as a direct consequence of {360). U" Fig. 377. 669' In an oblique angled triangle the square of a side opposite an acute angle is equivalent to the sum of the squares of the other two sides diminished by twice the rectangle of the base, and the distance from the acute angle to the foot of the perpendicular let fall upon the base from the angle opposite. SuG's.— It is to be shown that AB^ = BC' + AC^ - 2AC x DC. Observe that AD' = (AC - DO' = AC + DC' - 2AC x DC. Whence, by a simple application of the preceding theorem, the truth of this becomes apparent. 670' In an obtuse angled triangle the square of the side opposite the obtuse angle is equivalent to the sum of the squares on the other two sides, increased by twice the rectangle contained by the base and the distance from the obtuse angle to the foot of the perpendicular let fall from the augle opposite upon the base produced. SuG. — The demonstnition is analogous to the preceding, C being made obtuse in this case ; whence AD = AC + DC, etc. 252 EXERCISES IN GEOMETRICAL INVENTION. 671. The following is an ontline of a general demonstration cov- ering the three preceding propositions: Letting AE, BD, and CF be the three perpendiculars from the 'angles upon the opposite sides, and observing that a circumference described on any side as a diameter passes through the feet of two of the perpendiculars, {350) and {355) readily give the following : AB X AF =^ AC X AD rr AC' ± AC X CD, and AB x BF = B C x BE ^ BC " ± BC x CE j^ adding. AB^ = AC= + BC' ± 2AC x CD (or 2BC x CE), the + sign being taken when C is obtuse, and the — sign when C is acute. If C is right CE and CD become 0, whence AB"' = AC'^ + BC*. 672. Def. — The line drawn from any angle of a triangle to the middle of the opposite side is called a medial line. 673' The sum of the sqnares of any two sides of a triangle is equivalent to twice the square of the medial line drawn from their included angle, plus twice the square of half the third side. SuG.— Proved by applying {669, 670). 674. The three medial lines of a triangle mutually trisect each other, and hence intersect in a common point. Sug's.— To prove that OE = -^BE, draw FC parallel to AD until it meets BE produced. Then the triangles AEO and FEC are equal (?); whence EF = OE. Also, BO = OF (?). Having shown that OE = ^BE, by a similar construc- tion we can show that OD = ^AD- Finally, we may show that the medial line from C to AB cuts off ^ of BE, and hence cuts BE at the same point as does AD- Another Dem. — Lines through O parallel to the sides trisect tlie sides, etc 67S. In any quadrilateral the sum of the squares of the sides is equivalent to the sum of the squares of the diagon- als, plus four times the square of the line joining the centres of the diagonals. 676. Cor.— The sum of the squares of the sides of a parallelogram is equivalent to the sum of the Fig. osi. squares of the diagonals. THEOREMS IN SPECIAL GEOMETRY. 253 677. In any quadrilateral which may be inscribed in a circle, the product of the diagonals is equal to the sum of the products of the opposite sides. 678. In any triangle the rectangle of two sides is equivalent to the rectangle of the perpendicular let fall from their included angle upon the third side, into the diameter of the circumscribed circle. SuG. — This proposition is an immediate consequence of the similarity of two triangles in the figure. Fig. 382. 670. Cor. — The area of a triangle is equivalent to the product of its sides divided by twice the diameter of the circumscribed circle. 680. If there be an isosceles and an equilateral triangle on the same base, and if the vertex of the inner triangle is equally dis- tant from the vertex of the outer one and from the ends of the base, then, according as the isosceles triangle is the inner or the outer one, its base angle will be J of, or 2J times the vertical angle. 681, Of all triangles on the same base, and having the same vertical angle, the isosceles has the greatest area. Sug's. — Describe a segment on the given base, which shall contain the given angle. The triangle on this base and having its vertex in the arc of the seg- ment is the triangle to be considered. 682. Two triangles are similar, when two sides of one are pro- portional to two sides of the other, and the angle opposite to that side which is equal to or greater than the other given side in one, is equal to the homologous angle in the other. ALGEBRAIC DEMONSTRATIONS. 683. The difference of the squares on any side of a regular pentagon and any side of regular decagon, inscribed in the same circle, is equivalent to the square of the radius. Sug's.— We will give the outline of what may be termed an Algebraic Demon- stmtion of this proposition. This method is often the most convenient and ex- 254 EXERCISES IN GEOMETRICAL INVENTION. peditious. Letting p represent a side of the pentagon, d a side of the decagon, and r the radius, the student should be able to discover the following relations : (1) r'.d::d:r- d.ovr^ - dr = d^ ; Xju^ <'^^ ^ ^*- (2) Vd^ - ip* + v/r* - ^' = ^•• From (2), 2rVr^ — ^p^ = 2r^ — d* = r^ + dr, by substituting for V - c* = i V{a + b + c) {a + b - c) {a - b + c) {- a + b + c) = v/i«(i{j - c) (is - b) (i« - a). A D 256 EXERCISES IN GEOMETRICAL INVENTION. 090. From auy poiut in the plane of a circle the greatest and least distances to the circumference are measured on the line passing through the centre. Sug's. — There are three cases :— 1st. When the point is with- out the circle. 2d. When the point is within. 3d. When the point is in the chcumference. Fie. 386. 091' From any point except the centre of a circle, two. and only two, equal lines can be drawn to the cir- cumference. SuG. — This is a direct consequence of {181, 182). 092. If two opposite sides of a parallelogram be bisected, straight lines from the points of bisection to the opposite vertices will trisect the diagonal. 093. The feet of two perpendiculars let fall from two given points upon a given line are equally distant from the middle of the line joining the points. Fig. 387. Fig. 38S. 094. Two quadrilaterals are equivalent when their diagonals are respectively equal, and form equal angles. 09o. If, on the hjrpotenuse and sides of a right angled triangle, semicircumferences be described, that upon the hypotenuse passing through the ver- tex, the sum of the crescents thus formed will be equal to the area of the triangle. THEOEEMS IN SPECIAL GEOMETRY. 257 696. The bisectors of any two exterior angles of a triangle meet in a point which is the centre of a circle, to which one side of the triangle and the other two produced are tangents. These circles are called the escribed circles. Fig. 3S9. THE NINE POINTS CIRCLE 697. In any triangle the centres of the THREE sides, the feet of the three perpendiculars from the vertices upon the opposite sides or sides produced, and the three middle points of the distances from the vertices to the common intersection of the perpen- diculars, are nine points in the circum- ference of one and the same circle ; the centre of this circle is at the middle of the line joining the centre of the cir- cumscribed circle and the common intersec- tion of the perpendiculars ; and the radius is half the radius of the circumscribed circle. Sug's. — The student will do well to confine his at- tention in the first instance to the first figure, and after he sees the demonstration in this case — i. e., when the perpendiculars fall within, to trace it in the case of an obtuse angled triangle, in which the perpendiculars fall on the sides produced 1st. To show that the circle which passes through a, b, and c, also passes through a, we show the fol- lowing relations among the angles : cab = cAb = Fig. 300. Fig. 391. 258 EXERCISES IN GEOMETRICAL XNVENTION. cka + akh =■ cah. Hence, the vertices a and a are in the same circumference. In like manner we show that ft and y are in the circumference passing through a, h, and c. 2d. Considering one of the partial triangles as BHC, a, /?, and y are the feet of the three perpendiculars from its vertices upon one of its sides and the prolonga- tion of the other two. Therefore, by the first part r and q are the middle points of CH and BH. Considering either of the other partial triangles we find p the centre of AH, 3d. ace and hft being chords of the nine points circle, is its centre, and let- ting Q be the centre of the circumscnbed circle, we may readily show that is ill QH, and also is at its middle point. 4th. Drawing aO, and producing it, we maj^ show that it intersects AH in j9, and hence ^>H = A^; = Qa, and kOiap is a parallelogram. Therefore 0^; = ^pa = \Qik. 608. Cor. — The middle points of the three lines joining the cen- tres, two and two, of the escribed circles of a triangle, and the middle points of the three lines joining the centres of the escribed circles with the centre of the inscribed circle, are six points in the circum- ference of the circle circumscribed about the same triangle. 090. If one triangle be inscribed in another, the circumferences circumscribing the three exterior triangles thus formed intersect in a common point. SuG. — The demonstration is founded on the property of the opposite angles of an inscribed quadrilateral. The construction lines extend from the vertices of the inscribed triangle to the intersection to be examined. 700. The difference between the hypotenuse and the sum of the other two sides of a right angled triangle is equal to the diameter of the inscribed circle. 701. If from the extremities of any side of a triangle two lines be drawn, one bisecting an interior and the other an exterior angle, these lines will meet if sufficiently produced, and their included angle Avill be half the third angle of the triangle. 702. An inscribed equilateral triangle is one-fourth the circum- scribed equilateral triangle about the same circle. 703. The three altitudes of a triangle are to each other inversely as the sides upon which they fall. THEOREMS IN SPECIAL GEOMETRY. 259 704. The bisectors of the angles included by the opposite sides of an inscribed quadrilateral intersect at right angles. SUG.— By means of {214:) show that FC + CE + HA + AG = 180°. Whence FOE = 00^ 705. Two triangles which have an angle in each equal, are to each other Fig as the rectangle of the sides including the equal angle. Sug's. a and D being equal, we are to show- that ABC : DEF : : AB x AC : DE x DF. Take AE' = DE, AF' = DF, and draw E'F'. Now from the facts tliat the triangles AE'F' and DEF are equal, and that triangles of the same altitudes are to each other as their bases, t-he proposition is proved. Fig. 394 706. If of the four triangles into which the diagonals divide a quadrilateral, two opposite ones are equivalent, the figure is a trapezoid. 707. The difference between the angles which a bisector in a triangle makes with the side to which it is drawn, is equal to the difference of the angles of the triangle including this side. 708. If any number of equal right lines radiate from a common point, making equal consecutive angles, and any line be drawn through the common point, the sum of the perpendiculars upon this line from the extremities of the radiat- fng lines on one side, is equal to the sum of those on the other side. 709. Cor. — In any regular polygon, the ^^^ 3^5 sum of the perpendiculars let fall from the vertices on the one side of any line passing through the centre, is equal to the sum of those let fall from the vertices on the other side. *' 'frT\^ d c' \tl'..\ ^ v \-^ n-/ b a ec:=».,r^ a 260 EXERCISES IN GEOMETRICAL INVENTION. 7 10. If the sum of two opposite sides of a quadrilateral is equal io the sum of the other two opposite sides, a circle may be inscribed in it. Sug's. — Bisect any two adjacent angles, as A and B. Then are the perpendiculars r, r, r, equal (?); and it remains to be shown that the perpendicular p = r. Take AD' = AD, and BC == BC, and draw OD' and OC. Since a + h = c + d, and CD' = h — {b — c) — {p — d), CD' = «, and the triangles DOC and D'OC are equal. Hence p = r. Fig. 3%. ^ 711. If two planes are parallel, any right line which pierces one. pierces the other also. SuG. — Proof based on {410). 712. If two planes are parallel, any plane which intersects one, intersects the other also, and the lines of intersection are parallel. 713. Cor. — Two planes which are parallel to a third, are parallel to each other. 7 14. A plane which is perpendicular to a line of another plane, or to a line parallel to that plane, is itself perpendicular to the latter plane. 7 IS. If a straight line is perpendicular to a plane, any line par- allel to the plane is perpendicular to the first line. SuG. — Two lines in space which are not in the same plane, are said to make the same angle with each other as two lines respectively parallel to them and both lying in one plane. 716. In order that a straight line be perpendicular to a plane, it is sufficient that it be perpendicular to two lines not parallel to each other, and situated in the plane or parallel to it. 717. If two right lines in space are perpendicular to each other (not necessarily intersecting), their projections on a plane parallel to either line are perpendicular to each other. SxjG.— The Pi'qjeciion here referred to is that which is called the OriJio- graphic Projection. The proposition is not generally true of the Perspective Projection, i. e., the spaces which the lines (considered as material) would appear to occupy if they were placed between the eye and the plane. (See Ex. 8, page 174, Part II.) THEOREMS IN SPECIAL GEOMETRY. 261 718. The angle of inclination {392, Part II.) of a line oblique to a plane, is less than the angle included be- tween this line and any line of the plane, except its projection, which passes through the point' in which the first line pierces the plane. SuG. BD being the projection ABD', BD' being any line other thi"ough B. 719' Between any two lines not in the same plane, one line, and only one, can be drawn, which shall be perpendicular to both the given lines. Sug's. — Pass a plane through one of the lines parallel to the other; and through the other line pass a plane perpendicular to the first plane. of AB, ABD < / than BD, passing Fig. 3t)T. 720. In a warped quadrilateral, i. e., one whose sides do not all lie in the same plane, the middle points of the sides are in one plane, and are the vertices of the angles of a parallelogram. SuG. — Conceive the planes of two opposite angles of the quadrilateral, the intersection of which will be a diagonal of the given quadrilateral. 721. A line being given in a plane, one plane can be drawn in- cluding the given line and perpendicular to the first plane, and only one. Hence all right diedrals are equal. SuG. — Demonstration similar to {390). 722. The plane angle formed by drawing two lines in the faces of a diedral, from a common point in the edge, is less than tlie measure of the diedral if the angle is acute, and lines lie on the same side of the plane of the measure and are equally inclined to it, and greater if they lie on opposite sides. Li general, if the diedral is acute, the limits of the varying angle are and 90° when both the Imes lie ^-^^ o A p>^ >^^^^ K^D-::::; >^ r ker '6' ■^^*^A («) ■■••.. ^a' \ Fig. 398. 262 EXERCISES I2f GEOMETRICAL I>rVE2fTI0:N". on the same side of the plane of the measuring angle, and 180° when they lie on opposite sides. If the diedral is obtuse the limits are and the angle itself when the lines lie on the same side of the measure, and 90° and 180° when they lie on opposite sides. 0^\., '23. If the projections of a line in the two faces of a diedral are straight, the line is a straight line. SuG.— Proof based on {386). 724. If from the vertex of a triedral a line be drawn at pleasure within the triedral, the sum of the plane angles formed by this line and any two edges is less than the sum of the facial angles formed by the other edge and these two. 72o. If through a point in space two lines be drawn parallel to a given plane, and through the same point two planes be passed re- spectively peq^endicular to the two lines, the intersection of these two planes will be perpendicular to the given plane. -^ 726. The three planes which bisect the three diedrals of a trie- dral intersect in a common line. 727. In any convex polyedral, the sum of the diedrals is greater than the sum of the angles of a polygon having the same number of sides that the polyedral has faces. Srr;.— Proof based upon {T22). 728. Def. — A JPoIyedroii is a solid bounded by plane sur- faces. A RegyJar Convex Pohjcdron is a polyedron whose faces are all equal regular polygons, and each of whose solid angles is con- vex outward, and is enclosed by the same number of faces. 729. There are five and only five regular convex polyedrons — viz. : The Tetraedron, whose faces are four equal equilateral triangles ; The Hexaedron, or Cule, whose faces are six equal squares; The Ocfaedro?i, whose faces are eight equal equilateral triangles ; The DodecaedroUf whose faces are twelve equal regular pentagons ; and Tlie Icosaedron, whose faces are twenty equal equilateral triangles. THEOREMS IN SPECIAL GEOMETRY. 263 Dem. — We demonstrate this proposition by showing — 1st, that such solids can be constructed ; and 2d, that no others are possible. The Regular Tetraedron. — Taking three equal equilateral ti'iangles, as ASB, ASC and BSC, it is possible to enclose a solid angle, as S, with them, since the sum of the three facial angles is (what ?) (Part II., 480). Then, since AC = AB = CB (?), consid- ering ACB the fourth face, we have a regular polyedroQ whose four faces are equilateral triangles. The Reguhr Hexaedron or Cuhe. — This is a familiar solid, but for purposes of uniformity and completeness we may conceive it constructed as follows : Taking three equal squares, as ASCB, CSED, and ASEF, we can en- close a solid angle, as S, with them (?). Now, conceive the planes of CB and CD, AB and AF, EF and ED pro- duced. The plane of CB and CD being parallel to ASEF (?) will intersect the plane of EF and ED in HD parallel to FE (?). In like manner FH can be shown parallel to ED, BH to CD, and HD to BC. Hence the solid has for its faces six equal squares. Fig. 399. A / ! /H 7 c Fig. 400. Fig. 401. Those triedrals are The Regular Octaedron. — At the intersection P, of the diagonals of a square, ABCD, erect a perpendicular SP to the plane of the square, and making SP = AP (half of one of the diagonals) draw SA, SD, SC, and SB. Making a similar construction on the olhei" side of the plane ABCD, we have a solid having for faces eight equal equilateral triangles. The Regular Dodecaedron. — Taking twelve equal regu- lar pentagons, it is evident that we may group them in two sets of six each, as in the figure. Thus, around we may place five, forming 5 triedrals at the vertices of O. possible, since the sum of the facial angles enclosing each is 3| right angles (?) —i. e., between and 4 right angles (Part II., 436). In like manner the other 6 may be grouped by placing 5 of them about 0'. Now, conceiving the convexity of the group O in front and the con- cavity of group O', we may place tl»e two together so as to inclose a solid. Thus, placing A at h, the three faces 5, 6, 1, will enclose a tri- edrul, since the diedral included by 5 and 1 is the diedral of such a tri- edral. Then wnll vertex B fliU at c, and a like triedral will be formed at that point, and so of all the other vertices. Hence we have a polyedron having for faces 12 equal regular penta- gons. Fig. 402. 264 EXERCISES IN GEOMETRICAL INVENTION. Fig. 403. Tlie Regular Icosaedron. — Taking 20 equal equilateral triangles, they can be grouped in two sets, as in tlie figure, in a manner alt6getber similar to the preceding case. The solid angles in this case are included by 5 facial angles whose sum is 3^ right angles (?), which is a possible case (Part II., 436). *As before, conceiving the convexity of group O in front, and the concavity of O', we can place them together by placing A at a, tlms enclosing a solid angle with 5 faces, whence B will fall at 6, etc. Thus >ve obtain a solid with 20 equal equilateral U'iangles for its faces. That there can be no other regular polyedrons than these 5 is evident, since we can form no other convex solid angles by means of regular polygons. Thus, with equilateral triangles (the simplest polygon) we have formed solid angles witli 3 faces (the least number possible), as in the tetraedron ; with 4, as in the octaedron ; and with 5, as in the icosaedron. Six such facial angles cannot enclose a solid angle, since their sum is four right angles (?), and much less any gi-eater num- ber. Again, with squares (the next most simple polygon) we have formed solid angles with 3 faces as in the hexaedron, and can form no other, for the same reason as above. With regular pentagons we can only enclose a triedral, as in the dodecaedron, for a like reason. With regular hexagons we cannot enclose a solid angle (?), and much less with any regular polygon of more than six sides. ScH.— Models of the regular polyedrons are easily foi-med by cutting the fol- lowing figures from cardboard, cutting half-way through the board in the dotted lines, and bringing the edges together as the forms will readily suggest. Fifi. 404. THEOEEMS IN SPECIAL GEOMETRY. 265 730. Any regular polyedron is iuscriptible and circumscriptible by a sphere. Sug's.— From the centres of any two adjacent faces, as c and c', let fall per- jjendiculars upon the common edge, and they will meet it in tlie same point o (V). The plane of these lines will be per- pendicular to this edge (?), and perpendiculars to these faces from their centres, as cS, c'S, will lie in this plane (?), and hence will intersect at a point equally distant from these faces. In hke manner c"S = c'S, and the point S can be shown to be equally distant from all of the faces, and is therefore the centre of the inscribed sphere. Joinmg S with the vertices, we can readily show that S is also the centre of the circumscribed sphere. Fiu. 405. 731. Show that a, being the edge of a regular tetraedron, its volume IS — ^— 732. r>EF. — A Truncated Prism is one whose upper and loAver bases are not parallel. 733. The volume of a truncated triangular prism is equal to the sum of the volumes of three pyramids -' whose common base is the lower base of the prism, and Avhose vertices are the angles of the upper base. Sug's. — Let bD, cD', and «D" be perpendicu- lar to the lower base. Volume of 6-ABC is ^ 6D X ABC. Volume «-6Cc : volume 6-ABC :: cb'C : bBC : : cC : b's : : cD' : bb. .'. Volume a-bC(}' — ^cD' X ABC. In a similar manner volume b-a'AC = iaD" x ABC. Fig. 406. ^ 734. Cor. — The volume of a prism, one of whose bases is a right section and the other an oblique section, is the product of the right section into the arithmetical mean of its edges. Sug's.— The volume of a&c-ABC is as shown above ABC I j. But if ABC is a right section, bD = bB, cD' = cC, and aD" = aA. Hence the volume is ABC ^ ^B + cC + gA J 266 EXERCISES IN GEOMETRICAL INVENTtON. 73o. The volume of any polyedi'on having for its bases any two polygons whatever, situated in parallel planes, and for lateral faces trapezoids, is the product of -J- the distance between the bases into the sum of the two bases plus -i times a section midway be- tween the bases; or v = ^ (b + B' + 4b"), in which H is the dis- tance between the bases, B and B' the bases, and b" a section mid- way between the bases. l' ,' ^ Dem. — Let Li Ml Ni Pi Qi be the section of such a p, polyedron midway bet-ween its bases, and S any point \ in this section. Joining S with the vertices of the ^-^-Ni polyedron, we divide the solid into as many pyramids as it has faces. The volumes of the two which have B and B' for their bases are evidently ^H x B, and ^H X B'. It remains to find the volume of the others. J, ^- Let LML'M' be a lateral face corresponding to LiMi and SO a perpendicular from S upon this face. Draw ri through perpendicular to LM, and consequently to L'M'. Take I'Ki per- pendicular to-the plane section, whence I'Ki — |H. Kow the volume of the pyramid having L'M'LM for its base and S for its vertex is LiMi x 2rii x ^SO. But I'll X SO = Sli X I'Ki (?) ; whence the volume of this pyramid is f LiMi x SI. X I'Ki = I X 2SLiM, x I'Ki = ^ I'K, x 4SLiMi = ^H x 4SLiMi. In like manner the volume of the pyramid having for its base the face in which M^ N^ is situated, can be shown to be ^H x 4SMi Ni and similarly of all the others. Whence the whole volume is ^ H (B + B' + 4B"). 736. Cor. — The proposition is equally true when some or all of the lateral faces are triangles ; i. e., wheu one base has more sides than the other. ScH. — The preceding propositions are of much value in calculating earth- work. 737. If we cut a pyramid by a plane parallel to its base, a second pyramid is formed similar to the first. 738. Two triangular pyramids are similar wiienever they have an equal diedral angle contained between faces, similar each to each, and similarly placed. 739. Two polyedrons composed of the same number of tetrae- drons, similar each to each, and similarly disposed, are similar. PROBLEMS IN SPECIAL OR ELEMENTARY GEOMETRY. 267 740. All regular polyedrous of the same number of faces are similar solids. 74:1. The intersection of the surfaces of two spheres is the cir- cumfer'euce of a circle whose plane is perpendicular to the line which joins their centres. 742. Through any four points not in the same plane one sphere may be made to pass, and only one. Sug's.— The four points may be considered as the vertices of a tetraedron. Conceive perpendiculars drawn to the triangular faces from the intersections of lines drawn in these faces perpendicular to the sides at their middle points. These perpendiculars will meet at a common point (?), which is the centre of the circumscribed sphere (?). [The student should show why only one sphere can be circumscribed.] 743. OoR. 1. — The four perpendiculars erected at the centres of the circles circumscribing the faces of a tetraedron intersect at a common point. 744. Cor. 2. — The six planes, perpendicular to the six edges of a tetraedron at their middle points, intersect at the centre of the circumscribed sphere. 745. One sphere and only one may be inscribed in any tetraedron. SuG. — Bisect the diedrals with planes. 746. The angle included by any two curves intersecting on the surface of a sphere, is equal to the angle included by the arcs of two great circles passing through tlif point of intersection, and whose planes produced include the tangents to the curves at their inter- section. SECTION IL PROBLEMS IN SPECIAL OR ELEMENTARY GEOMETRY. 747. To bisect the angle formed l)y two lines whose intersection is inaccessible. Suo. — M and N are points in the bisector. Fig. 40S. 268 EXERCISES IN GEOMETRICAL INVENTION. 748. To pass a circumfereDce through three points, not in the same straight line, when the radius is so long as to render the ordi- nary method impracticable. SuG. — Let A, B, and C be the three points ; then are M and N other points in the same circumference. Fig. 409. ■N 749. From two given points on the same side of a line given in position, to draw two lines which shall meet in that line and make Ej /tfK/g equal angles with it. i A i / SuG. — K a and/? are equal, what is the relation of !/ MEtoEF? Pig. 410. 750. To construct an isosceles triangle with a given base and vertical angle. Sua.— See Prob. 4, p. 103. 7S1. To trisect a right angle. Sua.— What is the value of an angle of an equilateral triangle ? 7S2. Given the perpendicular of an equilateral triangle, to con- struct the triangle. 7o3. Given the diagonal of a square, to construct it. 7o4. To construct an isosceles triangle, so that the base shall be a given line, and the vertical angle a right angle. PROBLEMS IN SPECIAL OR ELEMENTARY GEOMETRY. 269 755' Given the sum of the diagonal and a side of the equare, to construct it. SuG.— What are the values of a and (5 respectively? 756. To construct a triangle when the altitude, the vertical angle, and one of the sides are given. Fig. 411. 757. To construct a triangle when the sum of the three sides and the angles at the base are given. Sug's.— MN being the sum oia.h, and c, what are the angles M and N as com- pared with the given angles a and fi ? Fig. 412. 758. In a right angled triangle the perimeter, and the perpen- dicular from the right angle upon the hypotenuse being given, to construct the triangle. Bug's.— DE is equal to the perimeter, DBE is an angle of 135°, and FE is the perpendicular on the hypotenuse. ABC is the required triangle. Let the student give the solution in full, and the proof 759. From two given points on the same side of a given line, to draw two equal straight lines which shall meet in the same point of the line. 700. To pass a circumference through two gfv'en points, which shall have its centre in a given line. 76*1. To construct a quadrilateral when three sides, one angle, and the sum of two other angles are given. Sug's.— What is the fourth angle ? When two sides and their included angle are^nown, there will be two cases, according as the two angles whose sum ia known are adjacent to each other or opposite. In the latter case we have to describe a segment on a diagonal, which will contain the fourth angle. For the third case see Ex. 13, page 136. 270 EXZEasES rs geometpjcal inyen-tion. 762. To construct a quadrilateral when three angles and two opposite sides are given. 763. To bisect a trapezoid by a line drawn from one of its angles. Fig. 416. 764. In a given circle, to inscribe a triangle equiangular with a given triangle. SuG. — How does an angle at the centre compare with one inscribed in the same segment ? 76o. To describe three circles* of equal diameters which shall touch each other. 766. In an equilateral triangle, to inscribe three equal circles which shaU touch each other and the three sides of the triangle. 767. To describe a circle of given radius touching the two sides of a given angle. SuG.— How far is the centre from each line ? 768. To describe a circumference which shall be embraced be- tween two parallels and pass through a given point within the par- allels. SuG.— In what hne Is the centre ? How far from the given point ? 769. To describe a circle with a given radius, which shall pass through a given point and be tangent to a given line. 770. To find in one side of a triangle the centre of a circle which shall touch the other two sides. 771. Through a given ]X)int on a circumference, and another PROBLEMS IN SPECIAL OR ELEMENTARY GEOMETRY. 271 given point without, to describe a circle touching the given circum- ference. SuG. — Consider in what two lines the centre must h'e. 772. In the diameter of a circle produced, to determine the point from which a tangent drawn to the circumference shall be equal to the diameter. SuG. — What is the relation between the radius, the required tangent, and the distance from tlie centre to the intersection of the produced diameter and the required tangent? 775. To describe a circle of given radius; touching two given circles. 774. In a given circle, to inscribe a right angle, one side of which is eriven. - , •\ . /;?] 77 5. In a given circle, to construct an inscribed triangle of given altitude and vertical angle. 776' To inscribe a square in a given right- angled isosceles triangle, one side being in tlie hypotenuse. Fig. 4r 777. To inscribe a square in a given quadrant of a circle, the vertex of an angle being at the centre. 778. To find the centre of a circle in which two given lines meeting in a point shall be a tangent and a chord. 779. To describe a circumference which shall pass through a given point and be taugent to a given line at a given point. 780. To bisect a quadrilateral by a line drawn from one of its angles. SuG. — The demonstration is based upon the prin- ciple that triangles having equal bases and equal [^ altitudes are equivalent. *'^ U pj^ ^^q 272 EXERCISES IN GEOMETRICAL INVENTION. 781. Through a given point situated between the sides of an angle, to draw a line terminating at the sides of the angle, and in such a manner as to be bisected at the point, SuG. — Conceive the point as situated in the third side of a triangle of which the two given lines are the other two. 782. To draw a line parallel to the base of a triangle so as to divide the triangle into two equivalent parts. SuG. PB' = DB' = iAB'. See {344,362). Fig. 419. 783. To construct a square when the difference between the diagonal and a- side is given. SuG. — Consider the angles. Fig. 420. 784. To determine the point in the circumference of a circle from which chords drawn to two given points shall have a given ratio. SuG. — Draw a chord dividing the chord joining the given points in the re- quired ratio, and bisecting one of the subtended arcs. 78S. To bisect a given triangle by a line drawn from one of its angles. 786. To bisect a given triangle by a line drawn yi/A from a given point in one of its sides. Fig. 421 787. In the base of a triangle find the point from which lines extending to the sides, and parallel to them, will be equal. 788. To construct a parallelogram having the diagonals and one side given. 789. To construct a triangle when the three altitudes are given. PROBLEMS IN SPECIAL OR ELEMENTARY GEOMETRY. 273 SxjG.— What is the relation of the perpendiculars to the sides upon which they fall? If a triangle can be formed with the perpendiculars as sides, how will it compare with the first triangle ? How proceed when the perpendiculars will not form a triangle ? 790' What is the area of the sector whose arc is 50°, and whose radius is 10 inches? 701- To construct a square equivalent to the sum, or to the dif- ference of two given squares. 702. To divide a given straight line in the ratio of the areas of two given squares. 703. To construct a triangle, when the altitude, the line bisecting the vertical angle, and the line from the vertex to the middle of the base are given. SuG. — The centre of the circle circumscribing the required triangle is in the perpendicular to the base at its middle point ; and the intersection of this per- pendicular and the bisectrix is a point in this circumference. Show that the bisector always lies between the perpendicular and the medial line. 704:. Through a given point, draw a line such that the parts of it, between the given point and perpendiculars let fall on it from two other given points, shall be equal. "What would be the result, if the first point were in the straiglit line joining the other two ? 705. From a point without two given lines, to draw a line such that the part intercepted between the given lines shall be equal to the part between the given point and the nearest line. SuG. — Produce the lines till they meet, if necessary. Draw a line through the given point parallel to one of the lines, and produce it till it meets the other. 706. Given one angle, a side adjacent to it, and the difference of the other two sides, to construct the trians^le. c Queries.— How if Z> > a? How if B is obtuse? v5 707' To pass a circumference through two given p^^ 422 points, having its centre in a given line. 18 274 EXERCISES IN GEOMETRICAL INVENTION. 798. To draw a line parallel to a given line and tangent to a given circumference. SuG. — Draw a diameter perpendicular to the given line. 799. To draw a common tangent to two given circles. SuG.— 1st Method.— There are two sets of tangents, AC, BD, and A'D', B'C. For the first, observe that if PE = AP - OC, OE is parallel to AC, etc. Fig. 4-23. Fig. 424. 2d Method. pO, p'O' being parallel to each other, pp'T gives the intersection of the tangent with the line passing through the centres, since OT : OT, or pO —p'O' : p'O' : : 00' : O'T. Also, PO - P'O' : P'O' : : 00'^ OT. Hence O'T is constant for all positions of the parallel radii. Prove that if the parallel radii are on different sides of the line joining the centres, T' is the point where the internal tangent cuts 00'. Queries. — How many tangents can be drawn — 1st. "When the ch'cles are ex- pO : p'O' ternal one to the other; they intersect each other : within the other ? 2d. When they are tangent externally; 3d. When 4th. When tangent internally; 5th. When one lies 800. To describe a circle tan- gent to a given circumference and also to a given line at a given point. Sug's. — There may be two cases — 1st. When the given circle is exterior to the one sought ; and 2d. When it is interior. In either case the centre of the required circle is in the perpen- dicular AO'. In the former case, 0, the centre of the required circle, is at Fig. 425. r + r' from C ; and in the latter 0' is at r — r' from C. AO = r, and CD = AB = AB' = r'. PROBLEMS IN SPECIAL OR ELEMENTARY GEOMETRY. 275 801' To construct a trai^ezoid when the four sides are given. SuG. — Knowing the difference between the two parallel sides, we may construct the triangle AEC, and hence the trapezoid. Fig. 426. 802. On a given line, to construct a polygon similar to a given polygon. Sug's. — One method may be learned from (.90). Ex. 8, page 152, furnishes another method. The following is an elegant method : To construct on A' homologous with A, a polygon similar to P. Place A' parallel to A, and the figure will suggest the construction. Fig. 427 803. To pass a plane through a given line and tangent to a given sphere. Sug's. — Pass a plane through the centre of the sphere and perpendicular to the given line. Through the point of intersection and in this secant plane draw tangents to the great circle in which the secant plane intersects the surface of the sphere. The points of tangency will be the points of tangency of the re- quired planes (?), of which there are thus seen to be two. 804. Def. — A Tanr/ent I^lane to a cylindrical or conical surface is a plane which contains an element of the surface, but does not cut the surface. The element which is common to the surface and the plane is called the Element of Contact. 80S. To pass a plane through a given point and tangent to a given cylinder of revolution. Stjg's. — 1st. When the point is in the surface of the cylinder. Through the point draw an element of the cylinder, by passing a line parallel to the axis, or to any given element. Through the same point pass a plane perpendicular trTthis element, making a right section (a circle). To this circle draw a tan- gent. The plane of the element and tangent is the tangent plane required. [The student should prove that any point in the plane affirmed to be tangent, not in the element passing through the given point, is without the cylinder.] 2d. When the given point is without the cylinder. Pass a plane through the given point perpendicular to the axis of the cylinder, thus making a right section of the cylinder (a circle). In this secant plane draw tangents to the section. Through the points of contact of these tangents draw elements of the cylinder. These elements are the elements of contact of the tangent planes: Hence planes passing through them and the given point are the tan- 276 EXERCISES IN GEOMETPJCAL INVENTION. gent planes required. [The student should remember that this is but an ouU line, and be careful to fill it up, giving the proof.] 806. To pass a plane through a given point and tangent to a conical surface of revolution. 807. To find, with the compasses and ruler, the radius of a ma- terial sphere whose centre is inaccessible. Sug's. — With one point of the compasses at anj'- point in the surface, as A, trace a circle of the sphere, as a.cb. The chord ka is measured by the distance between the compass points. In like manner measure three other chords, as ac, ab, and he. Draw a plane triangle having these chords for its sides, and circumscribe a circle about it. Thus «D is found. Knowing 2/t', the quantity under the radical sign is negative, and the radical becomes imagi7iary. This means, that 110 t7'iangle can be formed under these circumstances. This case appears in the geometrical solution also, for then h < —^s, or less than \2 /C, and consequently the arc struck from C as a centre, with radius h, will not touch DO, and we get no triangle. * This part of the construction should not be allowed on the figure till it is wanted— i.e., till this stage of the discussion. APPLICATIONS OF ALGEBRA TO GEOMETRY. 279 813. ScH. — This problem is discussed thus at length as an illustration of what may be done by such methods. Of course, all problems are not equally fruitful; but the student should not rest satisfied with a mere determination of the values of the unknown parts in known terms, when anything farther is revealed either by the process or result of the algebraic solution. Especially should he desire to become expert in seeing what geometrical relations are indicated by iha form of the answer obtained. 814:, Given the lengths of the medial lines from the acute angles of a right angled triangle, to determine the triangle, i. e., to find the base and perpendicular. SuG's.— Let AD = a, CE = 5, AB = 2x, and CB = 2y; then 4i;2 + 2^2 ^ a\ and 4/ + x'' = b'{^). .: 2x = AB =3|/i^^lzii!, ' 15 and 2y = CB =2^!^ A E B The form of these results indicates that CB sustains the same relation to CE and AD that AB does to AD and CE — a Fig. 431. fact which is evident from the nature of the case. Again, if 4rt"^ < b'\2x^is imaginary; and if 46- < a^, 2y is imaginary. In either case the triangle cannot exist. So also if 4a^ = ¥, 2x = / and if 46"^ = a'\ 2y = 0, and there can be no triangle. This may be seen from the figure by conceiving AB, for example, to diminish. As A approaches B, AD ap- proaches equality with DB, and CE with CB. Hence the limit is AD = ^CE. Thus we see thai either medial line must be more than half tJie other, — a propo- sition which is proved by this solution. 815. The hypotenuse and radius of the inscribed circle of a right angled triangle being given, to determine the triangle. Results. — Calling the hypotenuse 7i, the radius r, the base x, and the per- pendicular 3^, we have, x _ 2r + 7i ± /y/ h? - 4hr - - 4r-' and 2 2r + hT ^h? - 4hr - Ar' y = The results being the same in other respects, the double sign before the radical indicates that the base and perpendicular are interchangeable — a flict which is evident from the nature of the case. If the radical is 0, i. e., if h^ — 4hr — ir^ = 0, x = r + l\ and y = r + ^h, and the base and perpendicular are equal. Let the student show the same thing geometrically (from a figure). Also, if ^2 _ 4Jir — 4;'2 = 0, 7i = 2r (1 ± ^2). In this result the negative 280 EXERCISES IN GEOMETRICAL INVENTION. sign is to be rejected, since it would make h negative, as ■x/2 > 1. The value U = 2r (1 + ^2) is readily seen from the figure when AB = CB. Thus AC = 2DB = 2 (DO + OB) = 2 (r + r V^) = 2/- (1 + V2) (?). 816. A tree of known height standing perpen- dicular on a horizontal plane, breaks so that its top strikes the ground at a given distance from the foot, while the other end hangs on the stump. How high is the stump ? That is, given the base and the sum of the perpendicular and hyj^otenuse of a right angled triangle, to determine the perpendicular. Result. — Let a be the height of the ti-ee, & the distance from the foot to the point where the top strikes, and x the height of the stump ; then x = a'- - h^ 2a ' a- — b' b^ b^ Since = ^ — — , ^r- is the distance below the middle, at which 2a ' 2a 2a the tree breaks. 817. In a rectangle, knowing the diagonal and perimeter, to find the sides. 818. Knowing the base, b, and altitude, «, of any triangle, to find the side of the inscribed square, x. Result, X = r. a -\- b 819. In an equilateral triangle, given the lengths, a, b, c, of the three perpendiculars from a point within upon the sides, to deter- mine the sides. ^g's. — Find an expression for the altitude in terms of the sides; and then get two expressions for the area of the whole triangle. Equate these. Result, each side = V3 820. In a right angled triangle whose hypotenuse is h, and differ- ence between the base and perpendicular d, to find these sides. „ ,^ - d± \/W - (P . , d^ ^/2¥ - d' Residts, X = :r , x -\- a = ^ . APPLICATIONS OF ALGEBRA TO GEOMETRY. 281 Queries. — Why must the minus sign of the radical be omitted in the geo- metrical interpretation of these results ? What is the least possible value ofd? What are the sides of the triangle for these values? What is the superior limit of the value of d ? What do the sides become at this limit ? 821, In an equilateral triangle given the lines a, h, c, drawn to its three vertices from a point within or without, to find the sides. Result. — Each side = i ft' + y + c' d= a/6 {a'b'' + b'c' + cW) - 3 (a' + b' + c*) ) ^ The radical is + when the point is within, and — when it is without. 822. The perimeter of a right angled triangle and the perpen- dicular from the right angle upon the hypotenuse being given, to determine the triangle. Sug's. — Let 8 be the perimeter, p the perpendicular upon the hypotenuse, and X + y, X — y the two sides about the right angle. Then the hypotenuse = s — 2x, and we readily form the tAvo equations p {s — 2x) = x'^ — y^, and s(s 4- 2?)^ {x + yf + {x — yf ={8 — 2xy (?). Hence x = ^ — and this value substi- tuted in either equation will give y. 823. The base of a plane triangle is b and its altitude a, required the distance from the vertex at which a parallel to the base must cut the altitude in order to bisect the triangle. Result, —pr V2. Query. — What does the fact that h does not appear in the result show ? 824:. Having given the area of a rectangle inscribed in a triangle, can the triangle be determined .^ Can it, if the rectangle is a square ? If the rectangle is a square and the triangle right angled ? If the rectangle is a square and the triangle equilateral ? 825. The sides of a triangle being «, b, c, to find the perpendicu- lar upon c from the opposite angle. 1 Result, 2^ = ^V2c' {a'-\-b') + 2a'b' - a* - b* - c\ Sug's. — Observe that a and b are similarly involved in the result, but c is diffei-ently involved from either. This is evidently as it should be, since a and b are the sides about the angle from which ;> is let fall ; and are thus similarly related to p. But c, the side on which p falls, is differently related to p from 282 EXERCISES IN GEOMETRICAL INVENTION. either of the others. The student should be able to write the value of the per- pendiculars upon each of the other sides, from this one. Thus, that on a is 826. The sides of a triangle are a, h, c, to find the side of an in- scribed square one of whose sides falls in c. Sug's. — The altitude may be found from the preceding, hence may be as- sumed as known. Call it p. Then the side of the requu'ed squai-e is c + p What is the side of the square standing on a? On 6 ? Query. — Will the square be the same on whichever side it stands ? Observe that though the values here found are apparently different, they may not be so really, since p is different in each case. But let the student decide. 827' Having the area of a rectangle inscribed in a given triangle and standing on a specified side, to determine the sides of the rectangle. Besult, h being the base on which the rectangle stands, j9 the alti- tude from this base, and s the given area, we have for the sides h /W sb , p . /// sp ^ = 3 ^|/ 1 - ? ^'"^ ^ = 2 ^|/i -y Sro's. — The ± and t signs indicate that, in general, there can be two equal rectangles inscribed standing on the same base. The student will do well to illustrate it with definite numerical values, as p = 10, i = G, s = 10. Again, — must be greater than — , J^nd j- > '^, i. e., s must be less than ^pb. That is, the greatest rectangle is half the area of the triangle, since i ^ is the area of the triangle. 828. The Algebraic solution of a problem often enables us to effect a geometrical construction. We will give a few examples. Through a given point within a circle, to draw a chord of a given length. SoLUTiox.— Let 8 be the length of the required chord, and P the given point. Since P is a known point, call AP = a, PB = b, AB being the diameter througJi P. Let CD represent the required chord, and calliug CP, r, PD = s — x. Then sx — x- — ah ; whence x — Isi ± Vi* — ''<^- Fig. 4:33. APPLICATIONS OF ALGEBRA TO GEOMETRY. 283 To effect the geometrical construction, let 8 be the length of the given chord, and P the point in the given circle. Draw the diameter through P, and erect PE perpendicular to it. Make EH = is; then since PE^ =: ab, PH = -y/i s^ — ab. Now take H\ = is, and from P as a centre, with a radius PI = is + /^i«'^ — ab, strike the arc Dl intersecting the circum- ference. DPC is the chord reqmred. Fig. 4S4. From the radical Vi*" ~ ^^^ we see that, if ab > ^s-, x is imaginary, as we say in algebra. In such a case the problem is geometrically impossible, as will appear from the construction, for then PE is greater than EH, which makes HP, the representative of -^/l^^ X has but one value, and the segments are equal. ab, impossible. If \8'^ = ab^ 829. To find a point in a tangent to a circle from which, if a secant be drawn to the extremity of the diameter passing through the point of tangency, the external segment shall have a given length. Solution. — Let AB = d he the diameter of the given circle, DX = a the external segment of the re- quired secant, and the whole secant BX = x. Then x'^ — ax = cP, and x = ia ± ■\/lV + \a;K To effect the geometrical construction, construct the radical by taking AC = \a\ whence BC =: ■yjd^ + \cC\ Now make CY = \a, and with B as a centre, and BY as a radius, strike an arc cutting the tangent, as in X. Then is Fig. 435. BX^.x = ia+ ^/cP + \a\ The negative value of the radical is inapplicable in this elementary, geomet- rical sense, since as -y/cP + W > i«> this would make x a negative quantity. Again we see that no real value of a can render x imaginary. We can observe the same things from the geometrical construction. Thus, if the negative value of the radical were taken, x would be numencally less than BC, by ia,.or AC. But BC — AC < BA. Hence an arc struck from B with the required radius would not cut the tangent. "We see also that a may have any value between and oo . 8S0. Given the hypotenuse and area of a right angled triangle, to construct the triangle. Sug's. — Let li be the hypotenuse, s"^ the area, and x the perpendicuhir from the right angle upon the hypotenuse. Then hx = 2s'^, or ih : s : : s : x, and 7i : 2s : : 2s : 2x. The figure will suggest the construction. 284 EXERCISES IN GEOMETRICAL INVENTION. 831. Through a point between two lines which intersect, to draw a line which shall cut off a triangle of given area. Sug's. — Let AY = x, and the requhed area = «'. We have h :H :: x - h : x. .'. H = -^. X — And Hx = 2sK .: H 2*2 X ' Thus = ^-/K^4 To construct this, find c = ^, i, e., construct a third proportional to h and ."*. Then construct yc (c — 26), i. e., find a mean proportional between c and c — 26 ; let this be m. Whence x = c ■±: m. In gen- eral, there may be two solutions, if any, since there are two values of re. [This. should also be observed from the figure.] But if 26 > t there is no solution. If y = 26, there is but one solution. In the latter case where is the given point ? What is the geometrical difl3culty when 26 > - ? Can m be numerically III greater than c ? 832. To construct the four forms of the affected or complete quadratic equation, viz., (1.) 0, (3.) x' - px + q equations. 0, (4.) + 2)X — q = 0, (2.) X* — px — q = '^ + px -f (^ = 0, without solving the First Form:, x- + px — q = 0. — Draw any two lines as LM, NP, intersecting in some point O. Resolve q of the equation into two factors, as r and q\ so that we have x'^ + px — r x q' = 0. Take OA = ;?, OB = r, OC = q'. Bisect CB and AO by perpendiculars, and from their in- tersection F as a centre, with a radius FB, draw a circle. Then DO, or AE, is x, the positive root. For x {x + p) = rq', or x"^ + px — rq' = 0. The negative root is OE. Thus, let OE = (— x). Theu DO = AE = (- a; - p). Hence (- x) ^0. This construction is evidently always possible irrespective of the relative magnitudes of p, r,q' ; a fact which agrees with the statement in algebra that this form always has real roots. Second Form, x^ —px — rq' = 0. — The construction is the same as for the first form; only, in this case OE is the positive, and DO the negative root. Thus for OE = x (positive), we have DO x OE = (x — p) x = rq', or x^ — px — APPLICATIONS OF ALGEBRA TO GEOMETRY. 285 rg' = 0. For DO = (- x), we have DO x OE = DO (OA + AE) = DO (OA + DO) = (— -i) {p — ^) = ^'Q\ 01' -^"^ — P-^ — ^'?' = 0. Observe that in the first case the negative root is numericall}' greater than the positive ; while it is the reverse in tliis form. This agrees with tlie conclu- sions of algebra (See Complete School Algebra, 104). Thikd Form, x- — px + rq' =0. — Draw any two lines, as OM, OP, meet- ing at 0. Take OA = ;?, OB = r or q', and OC = q' or r. Erect perpendiculars at the middle points of OA, and BC ; and from their intersection F as a cen- tre, with a radius FB, strike a circum- ference. Then OE and OD are the values of x. For OE = .t, OE x OD = OE X EA = OE (OA -OE.) = x{p- .?) = rq\ or x'^ — px + rq' = 0. For OD = X, OD X OE = OD (OA - AE) = OD (OA — px + 7'q' = 0. Observe that the former value of x is greater than the latter, but that neither is negative. So also, we may readily see that the roots ma)- become equal, and also, im- aginary. Tlius if the circle were tangent to OA, the roots would be equal, and if it did not touch OA they would both be imaginary. (See Algebra, as above.) Fig. 439. OD) = X ip — x) = rq\ or a;- FouRTH Form, x'^ + px + rq' = 0.— The constniction is the same as the last, only both values of x are negative. Thus, (— x) [p (p + x) = rq', — px — X- — 7'q' = 0, or x- + px + rq' = 0. (- ^0] = (- ^O ScH. — Thus we see that we can construct any equation of the second degree containing but one unknown quantity, which has real roots. Hence, if the al- gebraic solution of a geometrical problem requires only the resolution of such an equation, the algebraic solution will lead to the geometrical construction. 833. "We have now given sufficient illustrations of this most in- teresting and important subject, so that the student should have caught the spirit of this method of using algebra to subserve the •^iurposes of geometrical investigation. We shall simply append a list of problems, upon which the student can put in exercise both his algebraic and geometric knowledge. But we cannot refrain from repeating the advice, that the learner should not rest satisfied with the mere algebraic resolution of the problem. ITc should be ambitious to trace, as fully as possible, the wonderful relations which exist between the abstract operations of algebra, and *he more con- crete relations of geometry. 286 EXERCISES IN GEOMETRICAL INVENTION. EXA3IPLES. 834:. Given the perimeter of a right angled tri- y angle and the radius of the inscribed circle, to de- termine the triangle. 835. Given the hypotenuse of a right angled triangle and the side of the inscribed square, to de- termine the triano:le. 836. In a right angled triangle, given the radius of the inscribed circle, and the side of the inscribed square, the right angle of the triangle constituting one angle of the square, to determine the triangle. Sug's. — Letting x and y be the sides, z the hypotenuse, r the radius of the inscribed circle, and s the side of the inscribed square, we have « = — ^—* X -\-y xy = r {x + y + z), and x + y = z + 2r. TThence z = 2r l~ 1, etc 837' In any triangle whose sides are «, J, c, to find the radius of the inscribed circle. 838. Show that the area of a regular dodecagon inscribed in a circle whose radius is 1, is 3. 839. Find the area of a regular octagon a\;®*^ whose side is a. Result, 2 (V2 4- 1) a\ 840. Find the radii of three equal circles de- FiG. 441. scribed in a given circle, tangent to the given circle and to each other. 841. The space between three equal circles tangent to each other is a ; what is the radius ? 842. In a triaugle, given the ratio of two sides, and the segments of the third side made by a perpendicular let fall from the angle opposite. APPLICATIONS OF ALGEBRA TO GEOMETRY. 287 843' In a triangle, given the base and altitude, and the ratio of the other sides, to determine the triangle. 844' Given the base, the medial line, and the sum of the other sides of a triangle, to determine the triangle. 84S. To determine a right angled triangle, knowing the perim- eter and area. Sug's. x"^ + y^ = z\ X + y + z = 2p, and xy = 2s\ give y + x = 2p — '^, x'^ + 2xy + y^ = 4p"^ — 4pz + 2^, 2^ + 4^-2 — ^f — 4pz + z'^ ; whence z = ' . Now use y + a; = 2» — z = , and xy = 2s^. p if 1^ P ' "^ 846. To determine a right angled triangle, knowing the perim- eter, and the sum of the hypotenuse, and the perpendicular upon the hypotenuse from the right angle. Sug's. x"^ -^ y'^ = z'^, x -^ y -{■ z = 2p, z + u = a, xy = zu. Then a;"^ + 2xy + y'^ = 4:p'^ — 4pz + s^ ; whence 2xy = 4p'^ — Apz, and hence 2s (a — 2) = 4p — 4^)2, etc. 847' The volume, the altitude, and a side of one of the bases of the frustum of a square pyramid being known, to determine a side of the other base. 848. To determine a right angled triangle, knowing the perim- eter, and the perpendicular let fall from the right angle upon the hypotenuse. 849. To determine a triangle, knowing the base, the altitude, and the difference of the other sides. 8S0' To determine a triangle, knowing the base, the altitude, and the rectangle of the other sides. 851' To determine a right angled triangle, knowing the hypote- nuse and the difference between the lines drawn from the acute amgles to the centre of the inscribed circle. Sug's.— Let fall CD a perpendicular upon AO produced. Now, since the the angles BAG and ACB are bisected, and COD = QAC + OCA, and ICD = lAB, they being complements of the equal angles CID, lAB, we have, COD = OCD, and CD = OD = -y/^ CO. Hence, putting AC = A, CO = X, and ^O = x + d, we have {x + d + ^i xy + (-y/^ x)' = h^. From this a; is readily found. The student should then be able to complete the solution. / 288 INTRODUCTION TO MODERN GEOMETRY. 852. Given two sides of a triangle and the bisector of their in- cluded angle, to determine the triangle. 853. Given the three medial lines, to determine a triangle. 854:. Given the three sides of a triangle, to determine the radius of the circumscribed circle. 855. Four equal balls whose radius is r are placed on a plane so that each is tangent to the other three, thns forming a pyi'amid; what is its altitude ? 856. Given the base of a triangle, the bisector of the opposite ano-le, and the radius of the circumscribing circle, to determine the Sug's.— First to fiud ED = x. Since EM = r — -y^r^ — b'\ it may be considered known and put equal to c. We then have DM = -^/c^ + x'^ ; and also, 59 _ 3.2 DM X a :=: AD X DB = ^/-^ — x-, or DM = . Whence Vc^ + £^ = ' ~ ^ , and x is readily found. Calling ED = s, the student will have no difficulty in proceeding with the solution. CHAPTEE 11. . INTBOnUCTION TO MODERN GEOMETRY.* SECTION I. OF LOCI. 857. The term Locus t, as used in geometry, is nearly synony- mous with geometrical figure, yet having a latitude in its use which the other does not possess. The locus of a point is the line (geo- * With strict propriety only the latter sections of this chapter helong to the Modem Geome- tnj. technically so called. Bnt, as the entire chapiter is composed of matter which has not hitherto found place in our common text-books, and the relative importance of which is be- coming more fully appreciated in modern times, the author has ventured to embrace the whole under this title. t The word Locus is the Latin tor place. OF LOCI. 289 metrical figure) generated by the motion of the point according to some given law. In the same manner, a surface is conceived as the locus of a line moving in some determinate manner. Ill's.— The locus of a point in a plane, which point is always equidistant from the extremities of a given right line, is a straight line perpendicular to the given line at its middle point. C Thus, suppose AB a fixed line, and tlie locus of a point equidistant from its extremities is required ; that point may be anywhere in a perpendicular to AB at its mid- dle point, and cannot he anywhere else in this plane. a This perpendicular is the locus (place) of a point subject to the given law. Again, a boy on the green is required to keep at just ^ 20 feet from a certain stake ; where may he be found? P^^ ^^ i. e., what is his locus (place) ? Evidently, the circum- ference of a circle whose radius is 20 feet. Thus, the locus of a point in a plane, equidistant from a given point, is the circumference of a circle. This is the place of such a point. What is the locus in space of a point equidistant from a given point ? "What is the locus of a point in space equidistant from the extremities of a given line ? A plane. What is the locus of a line moving so that each point in it traces a right line ? In general, a plane ; if it move in the direction of its length, a straight line. What is the locus of a right line parallel to and equidistant from a given line? What is the locus of a right line intersecting a given line at a constant angle ? * A conical surface of revolution. What is the locus of a semicircle revolving on its diameter ? PROPOSITIOIVS AXD PROBLEIIS Ii\ DETERMIXIXG PLANE LO€L [Note.— The student should be required to give every demonstration in form, and in detail. Frequent exercise in writing out demonstrations, is almost the only method of securing a good, independent style in demonstration] 858. Theo. — The locus of a 2:)oint in a plane, equidistant from the extremities of a given liiie, is a per2)encUcular to that line at its middle point. SuG. — To prove tiiis we have simply to show that every point in such a per- pendicular is equidistant from the extremities of the given line, and that no other point has this property (Part II., 129). * That is, an angle which remains of the same size. 19 290 INTRODUCTION TO MODERN GEOMETEY. SS9. JProb. — Fi7id the locus of a jjoint at atiy constant distance m from a straight line. Of what proposition in Part II. is this the converse ? Sug's. — To prove the proposition which the answer to this question asserts, it will be necessary to show that every point in the affirmed locus is at the same distance from the given line and that no other point is at that distance. We aflBrm that the locus is tieo right lines parallel to the given line and at a distance m therefrom. The formal demonstration is as follows : Let AB be the given line, and OE, OE', perpendiculars thereto, each equal to m. Through E and E' draw CD and CD' parallel to AB ; then is CD, CD', the Fig. 445. locus required.* For, by Part II. {156), every point in CD, CD', is at the distance m from AB ; and we may readily show that any other point, as P or P', is at a distance greater or less than m fromAB. Hence CD, CD', is the locus required. P E A P' :/n Oim C t:' SGO. Theo. — In a circle, the locus of the centre of a chord ^jar- allel to a given line is a diameter. Dem. — Let mn be any circle, and AB a given line. Then is the locus of the centre of a chord parallel to AB, a diameter of the circle. For, let DH be any chord parallel to AB. Through the centre of tlie circle C, and P, the middle point of DH, draw EL. Now EL is perpendicular to DH (?), and con- sequently to AB (?). Then will EL be perpendicular to any and eveiy chord parallel to DH (?), and hence will bisect such chord (?). Therefore the locus of the centre of a chord parallel to AB is a diameter. Again, any point in the circle and out of the line EL is not the middle point of chord parallel to AB. Thus, letting P' be such a point, draw a chord through P' parallelto AB. As there can be but one such chord (V), and as EL bisects it (?), P' is without the diameter (?). Fig. 446. 861. Tlieo. — Tlie locus of the centre of a circumference passing through tiuo given poi)its is a straight line. SuG.— Consult Part tl. {159, 168, 197), The student should put the argument in form. 862. Tlieo. — The locus of the centre of a circle lohich is tan- * It is important that the student think of these two lines a? one locns, or as parts of one and the same locus, if this will aid the conception. A locus may consist of any number of detached parts; all that is necessary being that the given conditions be fulfilled. In this respect the word locus has a more enlarged meaning than the term (jeoineh^ai figure. OP LOCI. 291 gent to a given circle at a given j)oint, is a straight line passing through the centre of the given circle. Dem. — Let C be the centre of thegiveu circle, and B the point in the circumference to which the circle * shall be tangent, the locus of whose centre is required. Through B drawTL tangent to the given circle. Now, a circle passing through B, and tangent to the given circle, will have TL for its tangent (?), and as a radius is perpendicu- lar to a tangent at its extremity, and only one perpendicular can be drawn to TL through B, the centre of a circle tangent to the given circle at B must be in this straight line. Moreover, as the given circle is tangent to the right line TL at B, its centre is in the perpendicular AX. Hence AX is the locus required. 863' Theo. — The locus of the centre of a circle of given radius R, and tangent to a given straight line, is two parallels to this line at a distance R therefrom, on each side. Give proof in form. 864. JProb. — Fi7id the locus of the centre of a circle of given radius R, whose circumference ptdsses through a given point. Give proof in form. 86^. Tlieo. — The locus of the centre of a lirie of constant length, having its extremities in two fixed lines luhich cut each other at right angles, is the circimifere?ice of a circle. Sug's.— Let MN be the length of the given line, and CD, and AB, the two lines intersecting at right angles, in which the extremities of MN are to remain. Now, in whatever position MN may be placed, its middle point, P, is at the same distance (|MN) from O (?). To show that any point not in this circumference, as MN, if Avithiu, less. Hence, the required locus is a circumference whose centre is 0, and whose radius is iMN. Fig. 44S. * Observe the form of expression. We say " tlie circle," and not. "the circles." usin:; the term in a generic sense, as including all which have the required property, i. e., all which arc tangent to the given circle at B. 292 INTRODUCTION TO MODEEN GEOMETEY. 866' JProh. — Find the locus of the centre of a chord of constant length, in a given circle. itl SuG. — We say, once again, always give the proof in form. 867. I^roh. — Find the locus of the vertex of the right angle of a right angled triaiigle of a constant hypotenuse. 868. JProb. — Find the locus of the middle poi7it of the chord in- ' tercepted on a line through a given point, hy a given circumference, when the given point is without the circumference, when it is in, ■ and when it is within the circumference. 869. JProh.— Find the locus of a point the sum of whose dis- tances from ttoo fixed intersecting liiies is constant, i. e., is equal to a given line. Solution.— Let AB and CE be the fixed lines, and m the constant distance. Draw MN parallel to AB, and at a distance m ^ from it. Bisect the angle CPN. Then is LR (a part of) the locus required. For [the student will here show that the sum of the distances from any point in LR to AB aud CE, as PD, P"D" + P"d", P'D' + P'd', P"'d"', P'^D'' + P'V^ is con- stant and equal to w], observe that when one of the perpendiculars measuring the distance from a point in the locus, changes from one side to the other of the line on which it is let fall, its sign changes. Thus P"D", P"d" being considered +, P'd' and P"D'' are to be con- sidered — . This is a general principal in mathematics. See Part II . {215), and foot note. Finally, LR is only a part of the locus, since there is another line on the op- posite side of AB, obtained by drawing the auxiliary MN on that side, which fulfills the same condition. The student should show what the result is when we draw the auxiliary parallel to CE, and on either side of it, also that any point not in oue of these lines cannot fulfill the required condition. The complete locus is four indefinite right lines intersecting each other at right angles, so as to inclose a rectangle. Fig. 449. 870. I*voh. — Find the locus of a point such that the sum of the squares of its distances from two fixed points shall he equivalent to the sqtiare of the distance between the fixed ptoints. 871- JProb. — Find the locus of the intersection of two secants OF LOCI. 293 drawn through the extremities of a fixed diameter in a given circle, one of the secants heing alivays perpendicular to a tangent to the circle at the point where the other cuts it. Sug's. P being the point, show that PB = AB, for any position of AP and BP. Hence, any point in the circumference having B for its centre, and AB for its radius, fulfills the conditions. Show that any point out of the ciroumference does not fulfill the conditions. 872. JProb. — Find the locus of the infrrsection of tioo lines draimi from the acute angles of a right angled triangle, through the points vjhere the po'pendicular to the hgjjotciiuse cuts the op- /\ ■. \b posite sides, or sides produced. Sug's. — The locus of P is required. Prove that APC is always a right angled triangle, wherever the perpendicular EF to the liypoteuuse AC is drawn. 873. JPvoh. — Find the locus of a point luhich divides a line draionfrom a fixed point to a fixed line i?i a fixed ratio. Sug's. — Most problems in finding loci such as are treated in Elementary Plane Geometry, viz., right lines and circles, are readily solved by constructing a few points according to the given conditions, whence we can determine by inspection whether the required locus is a right line or the circumference of a circle; and, having discovered this fact by inspection, it will remain to show why it sliould be so. Thus, in the present problem, O being the fixed. point, and AB the fixed line, drawing a few lines, OC, according to the requirements, and dividing them i^ the same ratio (in the figure 3:2), we find a few points P in the locus. We then discover at once that the locus is a right line parallel to AB, and can easily see w7iy it should be so. Fig. 452. 874. I^rob. — Two fixed circumferences intersect: to find the locus of the middle point of the line drawn through one of the points of intersection and terminated ly its other intersections ivith the cir- cumferences. 294 INTRODUCTION TO MODERN GEOMETRY. FUr. 45:3. Bug's. — TTe will first give an example of the course TNhicli the miud of the student might take iu his eflforts to discover tlie solution. He would naturally draw two iinequal * circles, as M and N, and, through one of the points of intersec- tion, as A, draw BC,and bisect it at P. It is the locus of P that is desh-ed. Now, sup- pose the line BC to revolve about A, B passing towards B', and C towards A. It is the path of the middle point that he seeks. When C reaches A, the line becomes tangent to N, and P is the middle point of the chord AB'. In a similar manner, he sees that the middle point of the chord AC, tangent to M at A, is also a point in the locus. Again, he observes that as B moves towards M, and C towards N, P moves to- wards A, and when AC = AB, P is at A. It now appears probable that the locus of P is a circumference. Proceeding on this hj-pothesis, he reasons, that, if this is true, AP' and AP" are chords of the locus, and, bisecting them with perpendicu- lars, he will have the centre of the locus. Locating thus, he observes tliat it appears to be in the line joining the centres 0' and O", and about midwaj' be- tween them. This leads him to see, whether, by assuming the middle point of ' as the centre of a circle, and OA as a radius, he can prove that any such line as BC drawn through A is bisected by this circumference, as at P. This jic can readily prove by means of the perpendiculars OD, O'D', and O'D", which bisect the chords AP, AB, and AC. For, since these perpendiculars are parallel, and 00 =: 00", D'D = DD" ; whence DP = AD", and, adding AP to each, DA = PD", or DB = PD". Adding to BD', DP, and D"C (= AD' = DP), there results BP = PC, and the hypothesis is true. Bat, in (jii'ing this problem as a recitation, the student will proceed as follows: Letting M and N be the two fixed circumferences, intersecting at A, join their centres 0' and 0", and bisect 00" as at O. With O as a centre and OA as a radius, describe a circle. Then is this circumference the locus required. For, let BC be any secant line passing through A, we may show that P is the middle point of BC- [Having done this, as above, and shown that any point not in this circumference is not the middle of the secant line passing through A, his solution is complete.] ^■) 87 o. I^roh. — If the line AB is divided at Cjfind the locus of P, so that angle apC = angle BPC S^^g's. — In seeking for t1i£ solution, the following would be a natural process. Drawing any line, as AB, in the lower part of the figure, taking C, any point in it. and conceiving BP and AP drawn so as to md^ie equal angles vj'Wh PC, we would naturally discover that, if a circle were circumscribed about BPA, PC produced would bisect the arc below AB. Thus we discover a ready method of locating P ; i. e., in the main figure, bisect AB by a perpendicular, as ED, and * Equal circles would probably have sj)€Cial relations. OF LOCI. 295 with any point on ED as a centre, pass a circumference through A and B. Through D and C draw a line, and P is a point in the locus (?). Any number of poinds can be found in this way ; and, having found a few, as P, P', P", | P'", etc., the situation of these will suggest that, probably, the locus is the circumference of a circle whose centre is in A B pro- duced. If this should be the fact, CP is a chord of that circle, and, erecting a perpendicular at the middle point of CP, its inter- section with AB produced, as O, will be the centre of the locus (?). "We will now endeavor to prove that any point in this circumference, as P', is so situated that BP'C = CP'A, and that no point out of this circumfer- ence has this property. We can readily show that the angle OPB - OCR — BPC = OCR — CRA (?). But PAC = OCR - CRA (?). . • . Triangle ORB is similar to OPA (?), and OA ■: OR : : OR : OB, or OA : OC : : OC : OB. Now, for any^omX in this cir- cumference, as R', we shall have OA : OR' : : OR' : OB, since OR' = OC, and OA, OC, and OB are constant. Hence, wherever R' is taken (in this circumference), the triangle ORB is similar to ORA, angle OR'B = P'AB, and BR'C = CP'A. Finally, that no point out of this circumference possesses this property is evi- dent, since the distance of such a point from O would not equal OC, and the angle OPiB (Pj being such a point) would not equal PiAB. SKG. Prob. — In a fixed circle^ any two chords intersect at right -^ angles in a fixed point ; find the locus of the centre of the chord join- ^ying their extremities. Give the proof. 877. Prob. — Fi7id the locus of the point in space equidistant from three given p>oints. Give the proof. 878. Prob. — Find the locus of the point in space equidititant from tico given points. Give the proof. 879. Prob. — Find the locus of the point in a plane such that the difference of the squares of the distances from it to two fixed points tuithout the plane shall he constant. Sug's. — Conceive the two points without the plane joined by a right line, and a perpendicular to this line drawn from either extremity of it ; the point where this perpendicular pierces the plane is a point in the locus (?). The re- quired locu* is two parallel straight lines. 296 INTRODUCTION TO MODERN GEOMETRY. 880. Proh. — Find the locus of the middle point of a straight line of constant length, whose extremities remain in two lines at right angles to each other, lut which are not in the same plane. Give '^ \ the proof. 881. I*roh. — Find the locus of the point equidistant ffom two fixed p la n es. Give proof. Sug's.— Consider, 1st, When the fixed planes are parallel ; and 2d, when they intersect. 882. JProh- — What locus is the intersection of a plane and the surface of a sjihere 9 Give proof. 883. I^rob. — Wliat locus is the intersection of the surfaces of tico given spheres 9 884. JProb. — Find the locus of the point in space such that the ratio of its distance from a given right line to its distance from a fixed point in that line is constant. SECTION IL OF SYMMETRY. 885. Def. — Two points are said to be symmetrical with resjoect to a third, when the right line joining the two points is bisected by the point of reference, called the Centre of Symmetry. 886. Def. — Two loci, or two parts of the same locus, are sym- p, ^^ metrical luith respect to a p>oint, when / \/^"^~^^ every point in one has its symmetrical point ■'^ 'C^^^^^^^:—^ ^^^ ^^ other. Xa) b^C^^.^-'^^B^ Ill's. — In (a) P' is symmetrical with P in re- m' spect to S, if S is the middle point of PP'. In (&) C*) we observe that the semi-circumference A^ti'B is ^ ' symmetrical with the semi-circumference A7/iB, p/^]\,.^^ ^-<ect to a plane, is an equal p>lane angle. 300 INTRODUCTION TO MODERN GEOMETRY. Dem. — Let AOB be any plane angle, and MN the plane of symmetiy. Let P be any point in AO, and Pi any point in OB. Let 0' be the symmetrical of O, P' of P, and Pi' of Pi ; then is AG' the symmetrical of AO, O'B' of OB, and angle A'O'B' of AOB. Now by the preceding proposition the two triangles POPi,and P'O'Pi', are mutually equilateral, whence AOB = its symmetrical A'O'B'. Query. — When will the triangle 'pop' exist, when not ? and 008. TJieo. — Any lilane polygon has for its symmetrical, loith reference to a iMne, an equal lylane lyolygon. Sug's.— ABODE being any plane polj'gon, and MN the plane of symmetry, by constructing A',B',C',D',E' sj^mmetrical with A, B, C, D, E, we have by the preceding propositions A'B'C'EyE' equilateral and equiangular with ABODE ; whence it only remains to show that A'B'C'D'E' is a plane (not a warped) surface. Let F be any point in the angle AED, draw HI, and let H' and I' be the sym- metricals of H and I {895). Draw H'l'. Then is the symmetrical of F in H'l' (?), as at F'. Now, every point in HF within the angle BAE has its symmetrical in H'F' {^05\ Thus, by taking three points, not in a straight line, in the angle BAE, we can show that their syrametricals are in the plane B'A'E', and also in A'E'D'. In like manner, all the angles of A'B'C'D'E' can be shown to be in the same plane. 000. Cor. — If two x>lcines intersect, their symmetricals intersect, and the two intersections are symmetrical right lines. The student should show how this grows out of the proposition. 010. TJieo. — The symmetrical of a diedral is an equal diedral, Dem. AOB being the measure of the diedral A-OC-B, and A'-O'C'-B' the symmetrical diedral, and 0' the sym- metrical of O, the symmetrical of AO being A'O', the angle A'O'C is right, and in like manner B'O' being the symmet- rical of BO, B'O'C is right. But BOA = B'O'A' (?), whence the diedrals are equal. Fig. 466. OF SYMMETRY. 301 Oil. HieO' — Two 'pohjedrons, symmetrical with respect to a plane, have their faces equal, each to each, and their homologous solid angles symmetrical. Sug's. — This is an immediate consequence of preced- infi^ propositions. Thus E' being the symmetrical solid liomologous with E,the homologous plane foces includ- ing them are equal {908). Again, the facial angles being equal, but not similarly disposed, the solid angles are symmetrical. ,^ //' ri }A^ __ ^ M i i :B ; /n 11/ 'H ■i-irf^ /f \~\ ' l- H' V Fig. 467. 912. Cor. — Two symmetrical 2)olyedrons can he decomjjosed into the same nnmher of tetraedrons, symmetrical each to each. For we can decompose one of the polyedrons into tetraedrons having for their common vertex one of the vertices of this ]3olyedron, and each of these tetraedrons will have its symmetrical in the other. 913' Theo. — Two symmetr iced polyedrons are equivalent. Dem.— From the last corollary it will appear that it is sufficient for the de- monstration of this proposition to show that two symmetrical tetraedrons are equivalent (?). Let S-ABC, and S-ABC be two tetraedrons symmetrical with respect to their common base. They have a common base and equal alti- tudes (?), hence they are equivalent. 914:. General Scholium.— We may speak of two loci, or two parts of the same locus, as symmetrical with respect to a line or plane, whenever all the points in one have symmetrical points in the other, even though the line joining the symmet- rical points be not 'perpendicular to the axis, or the plane, of symmetrj"- ; observing, however, that this line is always bisected by the axis or plane. Thus, tlie ellipse in the figure is symmet- rically divided by the line X'X, since every point in one portion has a symmetrical point in the other, as Ps = P'«, for every point in the curve. In such a case the parts cannot be brought into coincidence by simple revolution : one part must be reversed. Fig. 468. 302 INTRODUCTION TO MODERN GEOMETRY. SECTION III OF MAXIMA A>D MINDIA. 915. Def. — A Maxiniinn value of a magnitude conceived to vary continuously in some specified way, is a value ffi^^ "which is greater than the preceding and succeeding values of the magnitude. Ill's. — Thus, suppose in a given circle, a chord passing through a fixed point, P, revolves so as to take successively the positions 1«, 26, Ac, 3^, 4«, etc. It is at a maximum when it passes through the centre, as Ac. The chord is the magni- tude which is conceived to vary in the way specified, and Ac is a value greater than the preceding and the succeeding values. Again, conceive a circle to be compressed or ex- tended, as in the direction mn, so as to take the forms in- dicated by the dotted lines, its area will be diminished, the perimeter remaining the same. That is, of all figures of a given perimeter, the circle has the maxi- mum area. Fig. 470. 910' Def. — A Mininiuni value of a magnitude conceived to vary continuously in some specified way, is a value which is less than the preceding and succeeding values of the magnitude. Ill's. — Thus, conceive the varying magnitude to be a straight line from the fixed jwint P to the fixed line X'X ; that is, suppose such a line to start from some position PI. and move through the successive positions P2, PA, P3, P4. PA is a minimara, since it is less than the preceding and succeeding values. PROPOSITIOXS COXCERXIXG MAXDIA AM) MIXDIA. 917* Axiom. — Tlie minimum distance between tico poi?its is a straight line. 918. TJieo. — The minimum distance from a point to a line is a straight line 2)erpendicular to the given line. Student give proof. 919. Tlieo. — Tiie maximum line which can he inscribed in a given circle is a diameter. OF MAXIMA AND MINIMA. 303 Proof based on the fact that the hypotenuse of a right angled triangle is the greatest side. 920. Theo^ — The sum of the distances from tiuo points on the same side of a line, to a poi?it in the line, all heing in the same plane, is a mini- mum lohen the lines measuring the distances make equal angles with the given line. Student prove AP + BP < AP' + BP'. Fig. 472. 921. TJieo. — If a triangle have a constant base and altitude, its vertical angle is a maxi- 7num iuhen the triangle is isosceles. SXFG. — By what is the vertical angle measured ? Fig. 473. 922. TJieo. — The lase and area of a triangle being constant, its perimeter is a minimum johcn the triangle is isos- celes. A Sug's. — The area and base being constant, the vertex remains in a line parallel to the base, for all values of the other sides. The figure will suggest the demonstration, which is based on the fact that any side of a triangle is less than the sum of the other two. Fig. 414, 923. TTieo. — The difference between the distances from tico points on opposite sides of a fixed line toa p)oint in that line, is a maximum, when the lines 7neasuring these distances make equal angles ivith the fixed line. Sug's. P'O = AP - AP'; but PO > A'O (= A'P) — A'P'. Fig. 475. Query.— Having the points P, P', and the fixed line given, how is the point A found by geometrical construction ? 924. Theo. — The lengths of two sides of a triangle being constant, the area is a maximum lohen the iiicluded angle is right. Fio. 476. 304 INTRODUCTION t6 MODERN GEOMETRY. 925. Tlieo. — The sum of tico adjacent sides of a redaitgle being constant (AB), the area is a maxi- mum when the sides are eqiial. ISOPEREttETRY. 026' Isoperinietric Figures are such as have equal perim- eters, i. e., bounding lines of equal length. Problems in isoperimetry are a species of problems in Maxima and Minima. Thus, of all figures whose perimeters are m (say 10 inches), to find that which has the greatest area, is a problem in isoperimetry. Again, what must be the form of a pentagon whose perimeter is m, in order that its area may be a maximum ? 027' Theo. — Of isoperimetric triangles with a constant base, the isosceles is a maximum. Sug's.— By means of the figure to Theorem {921), we can readily show that any triangle having the same base as the isosceles triangle, and its vertex either in or beyond the line through the vertex of the isosceles triangle and par- allel to its base, has a greater perimeter than the isosceles triangle. Hence, the isoperimelric triangle on the given base has its vertex below this parallel, ex- cept when isosceles ; and consequently the isosceles is the maximum. 028. Cor. — Of isoj^erimetric triangles, tlie equilateral has the maximum, area (?) 020. JProb. — Given any triangle with a constant lase, to con- struct the maximum isoperimetric triangle. 030. ProT). — Given any triangle, to construct the maximum isoperimetric triangle. 031. Theo. — Of isoperimetric quadrilaterals, the square has the maxinmm area. Dem.— Let ABCD be any quadrilateral. If AD is not equal to DC, ADC can be replaced by the isosceles isoperimetric triangle AD'C, and the area of the quadri- lateral increased. So ABC can be replaced by AB'C. There- fore AB'C'D > ABCD. In like manner if AD' is not equal to AB' , D'AB' can be replaced by the maximum isoperi- metric triangle DAB'. So also D'CB' can be replaced by D'C'B'. Therefore AB'C'D' > AB'CD' > ABCD. Now, A'B'C'D' is a rhombus (?), and the student can show that the square on A'B' is greater than any rhombus with the same side. Fig. 478. ISOPERIMETRY. 305 932. JProb. — Having given a quadrilateral, to constrtict the maxivium isoijerimetric quadrilateral. 933. Theo. — Of isoperir}ietric quadrilaterals luitli a constant base, the maximum has its three remaining sides equal each to each, and the angles ivhich they include equal. Dem. — Let ABCD be the maximum isoperimetric quadrilateral on the base AD, then AB = BC = CD, and angle ABC = BCD. For, if AB is not equal to BC, draw AC, and replacing the triangle ABC with its isoperimetric isosceles triangle, we shall have a quadrilateral isoperimetric with ABCD, and greater than ABCD, i. e., greater than the maximum, which is absurd. Again, if angle ABC is not equal to BCD, let ABC < BCD, whence BCE < EBC, and BE < EC. Take EF = EC, and EG = EB, whence the triaugles FEG and BEC are equal, and FC = BC. Also, since AB + BC + CD = AE + ED - (EB + EC) + BC, and AF + FG + CD - AE + ED - (FE + EC) + FG, it follows that AFGD and ABCD are isoperimetrical, and, since ABCD = AED - BEC, and AFGD = AED - FEG, that AFGD and ABCD are equal. Therefore, AFGD is a maximum, and by tlie preceding part of the demonstration AF = FG = BC = AB, which is absurd; and there can be no inequalit}'- between angles ABC and BCD. Fig. 479. 934. Theo. — Of isoperimetric polygons of a given number of sides, the regular polygon has the maximum area. Dem. — First, the polygon must be equilateral ; for, if any two adjacent sides, as AB, BC, are unequal, the triangle ABC can be replaced by its isoperimetric isosceles triangle, and thus the area of the polygon be increased. Second, the pol3^gon must be equiangular; for, if any two adjacent angles, as B and C, are unequal, the quadrilateral ABCD can be replaced by its isoperime- tric quadrilateral with B = C, and thus the area of the polygon be increased. Fig. 480. ^ 93 S. TJieo. — Of isoperimetric regular polygons, the one of the greater number of sides is the greater. Dem. — Let ABC be an equilateral (regular) triangle. Join any vertex, as A, with any point, as D, in the opposite side. Replace the triangle ACD with the isosceles isoperi- metric triangle AED. Then is the quadrilateral ABDE > the triangle ABC. But, of isoperimetric quadrilaterals, the regular (the square) is the greater. Hence, the regular quadrilateral (the square) isoperimetric with the trianulc ABC, is greater than ^20 306 IXTEODUCTION TO MODERN GEOMETRY, the triangle. In the same manner the regular pentagon isoperimetric with the square can be shown greater than the square ; and tJius on, ad libitum. 936. Cor. — Of plane isojjerimetric figures, the circle has the maximum area, since it is the limiting form of the regular polygon, as the number of its sides is indefinitely increased. .:,v^ SECTION IV, OF TRAXSYERSALS. ^R f>,?7. Def. — A Transversal is a line cutting a system of lines. A transversal of a triangle is a line cutting its sides ;* it either cuts two sides and the third side produced, or the three sides produced. In speaking of the transversal of a triangle (or polygon), the distances on any side (or side produced) from the intersection of the transversal with that side to the angles, are Segments. Of these there are six. Adja- cent segments are such as have an ex- tremity of each at the same point. Xon- adjacent segments are such as have no extremity common. Ii.l's. TR is a ti-ansversal of the triangle P,g 4^ ABC ; rtA, aC, bC, bS, ^A, cB are adjacent seg- ments two and two ; aC, 6B, cA, and aA, bC, cB are the two groups of non-adjacent segments. 938. The two Fukdamextal Propositions of the Theory OF Transversals. 939. Theo. — Tlie product of three non-adjacent segments of the sides of a triangle cut hy a transversal, is equal to the j^roduct of the other three. Dem. — ABC being cut by the transversal TR, aA x 6C x cB = aC x Z/B x ck. Draw BD parallel to^AC, and from the similar triangles we have DB aC rtA cA Bb=^'''^"^^7;B bO whence, multiplying, rtA _ rtC X cA hB ~ bt x cB' or aA X 6C x cB «C X 6B X cA. Fio. 483. * Or. sides produced— this expression bein? usually omitted in higher Geometry are to be considered indefinite unless limited in the problem. all lines OF TRANSVERSALS. 307 940. Cor. — Conyerselj, If tliree i^oints le tahen in the sides of a triangle (as a, b, c) such that the product of three non-adjacent seg- ments equals the product of the other three, the points are in the same straight line. For, passing a line through a and h, let it cut the third side in c . Then, by the proposition, aA x hC x c'B = aC x bB x c'A. But, by hypothesis cB cA aA X bC X cB = aC X bB X cA. Whence -—- = -—-, and c and c' must coincide. cB c A ' ScH. — This theorem is known among mathematicians as 77ie Ptolemaic TJieo- rem, and is usually attributed to Claudius Ptolemy, an Egyptian mathematician and philosopher who flourished in Alexandria during the first half of the second century. But it is thought to be more properly due to Menelaus, who lived a century before Ptolemy. 041. TJieo. — The three angle-transversals* of a triangle, 2)assiug through a common point, divide the sides into seg- 7ne7its such that the product of three non-adjacent segments equals the product of the other three. ^/Ln\ ^-?-\ Dem. — From the triangle ACc cut by the transversal r«B, we have aA x CO x cB =: AB x aC x Oc ; and from CBc cut by *A, Oc X iC X AB — CO x bB x cA. Multiplying, we obtain aA x 6C x c B = aC x Z*B x cA. 942. Cor. 1. — Conversely, If the three angle- transversals of a triangle divide the sides into seg- ments such that the product of three non-adjacent " fig. 4^. segments equcds the product of the other three, the transversals pass through a com^non p)oint. For, the sides being divided at «, b, and c, so that aA x &C x cB = ^tC x bB x cA, draw Cc, and A&, and let O be their intersection. Now, let a' be the point in which BO cuts AC. Then, by the proposition, a' A x Z'C x cB == rt'C x //B x cA. Whence -rr = -77^, and a and a' coincide. a'A a'Q ' 943. Cor. 2. — If any one of the sides is bisected, the line joining the other p>oints of division is parallel to this side. For, let 6C = bB. Then «A x 6C x cB = aC x bB x cA, becomes «A X cB = aO X cA ; or Aa : «C : : Ac : cB. Query. — How does this apply to the second figure ? 944. Cor. 3. — If the line joining two points of division is j^^r- allel to the third side, the latter side is Usected. * The transversals passing' through the angles. 308 INTRODUCTION TO MODERN GEOMETRY. For, if ah is parallel to AB, aC :bC: :ak : 6B, -whence aC x JB = 6C x aA. And, since rtA x 6C x oB — aC x6B xcA, cA = cB. .9d?J. We will now give a few problems to illustrate the use of the theory of transversals. 04:6. J^rob. — To show that the medial lines of a triangle j^ciss through a common 'point. Solution. — Since aA = (tC, bC = JB, and cB = cA, by mul- tiplying, we have a A x bC x cB = aC x bB x cA ; -whence by the last corollary these transversals pass through a com- mon point. 047. JProh. — To shoio that the bisectors of the angles of a tri- angle pass through a common point. Solution. — In the last figure let «B, 5A, cC be the bisectors. ^, rtA AB bO AC cB CB , . , . «A x 5C x cB T^^^ ^=CB'Z;B=AB'^=AC' "^"It^Ply^^g' aZxbBx cA = ^' ^^ rtA x &C X cB =rtC X 6B x cA. Therefore these transversals pass through a common point. 048. JProb. — To shoio that the altitudes of a triangle yass through a common point. Sug's. — In the last figure, if rtB, bk, cC, -were the perpendiculars, there would ^ . ., . , . . rtA AO JC CO cB OB be three pairs of similar triangles givmg ^ =|^' M = AO' ^ == CO ' whence, as in the last. / » cc A X t L X c£ £, vp v aJ^^x ,7 , 040. Proh. — To shoiu that the angle-transversals terminating in the points of tangcncy of the sides of the triangle with its inscribed circle, 2)ciss through a common jwint. Sug's.— In the last figure, if a, b, c were the points of tangency we should have rtA = cA, bC = aC, cB =bB ; -whence rtA x ^>C x cB = aC x JB x ck. AVhich shows that the transversals pass through a common point. OoO. Theo. — If two sides of a triangle are divided proportion- c ally, starting from the vertex, the angle-trans- versals from the extremities of the other side to the corresponding points of division, in- tersect in the medial line to this third side. Dem. — Since AC and CB are divided proportion- ally at a and a\ rtA x a'C = aC x rt'B ; and as DB = DA, rtA X rt'C X DB = rtC x rt'B x DA, the angle-transversals Art', Brt intersect in CD. The same may be sho-wn of any other angle-transversals from A and B, dividing CB and CA proportionally. OF TRANSVERSALS. 309 9S1' Cor. — In any iraj^ezoid the transversal j^^^ssing through the intersection of the diagonals, and the intersection of the non- parallel sides, bisects the parallel sides. SuG. — Joiniug cut' iu the last figure, CD is such a transversal. The student will readily see the connection with the proposition. [%'^ OS 2. JProb.— Through a given point to draiu a line which shall meet two given lines at their intersection in an invisible, inaccessible 2^0 int. Solution. — Let Mm, N?i be the two given lines which meet in the invisible, inaccessible point S, and P the given point through which a line is to be lo- cated which will meet Mw, Hn in S. Through P draw any convenient trans- versal, as BF, and any other meeting this, as AF. Now, considering MS as a trans- versal of the triangle CDF, we have AF X BC X SD = AD X BF X SC ; whence SD _ AD X BF SC HP _SD_ AD PC AF X BC* parallel to BF, we have But, HD being drawn Fio. BF or HD = AD BF PC SC AF X BC AF X BC whence HD is known, as AD, BF, PC, AF, BC can be measured. Tlie points P and H determine the required line. 953. Def. — The Complete Quaclrilateral is the figure formed by four lines meeting in six points. The complete quadrilateral has three diagonals. III. — ABCDEF is a complete quadrilat- eral, and its diagonals are CF, BD, and AE, the latter being spoken of as the third or exterior diagonal. Fig. 488. 954. Tlieo. — The middle points of the three diagonals of a complete quadrilateral are in the same straight line. Dem.— m, n, o being the centres of the diagonals of the complete quadrilat- eral, in the preceding figure, are in the same straight line. Bisect the sides of the triangle FDE, as at I, N, L, and draw IN, IL, LN. Since IN is parallel to BE, and bisects DF and DE, it also bisects DB (?) and hence passes through n. For like reasons IL passes through m, and LN through o. Now, AC being a trans- versal of the triangle FDE gives CD x BE x AF = CE x BF x AD. Therefore, 310 ES'TRODUCTION TO MODERN GEOMETRY. noticing that iCD = ;/J, ^BE = ;jN, {AF = oL, ^CE = mL, ^BF = nl, and MO = oN, we have ml x 71N x oL = mL x fi\ x oN. Hence these thiee points m,n,o lie in a transversal to the triangle ILN. SECTION V. HARMONIC PROPORTION AND HARMONIC PENCILS. ^oo» Def. — Three quantities are in Harmonic Projyortion when the difference between the first and second is to the difference be- tween the second and third, as the first is to the third. III. — 6, 4, 3 are in harmonic proportion, since 6 — 4:4— 3::6:3. In gen- eral, a,b, c are in harmonic proportion, \i a — b : h — c : : a \ c. 9o6. Theo. — If a given line he divided infernally and externally in the same geometric ratio, the distance beticeen thepoitits of division is a harmonic mean between the distances of the extremities of the given line from the point not included between them. Dem. — Let AB be the given line ; and let O and 0' be so taken that AO : BO : : AO' : BO' ; then is 00' a har- -— ~ — g ^7 — monic mean between AO' and BO'. For AO = AO'-00', and BO = 00' - BO' ; Fig- 490. whcDce AO', 00', and BO' are such that AO'— 00' : CO' —BO' : : AO' : BO', that is, tliey are in harmonic proportion. 9J7. Cor. 1. — AO, AB, and AO' are in harmonic proportion^ i.e.; AB is a harmonic mean between AO and AO'. For AB - AO (= BO) : AO' - AB (= BO') : : AO : AO'. OoS. Cor. 2. — ]]lien AOfOO', BO' are in harmonic proportion, AO X BO'=r BO X AO'. U^O. Cor. 3. — Conversely, }Ylien a line is divided i?ito three parts K((ch that the rectangle of the extreme parts equals the rectangle of the mean part into the whole line, the line is divided harmonically. Thus, let AO' be the line, and AO x BO* = BO < AO'; then AO: BO : :A0' : BO', whence, b}' the proposition, 00' is a harmonic mean between AO' and BO'. Def. — The points and O' are called Harmonic Conjugates. HARMONIC PROPORTION AND HARMONIC PENCILS. 311 060. Theo. — Iftivo lines M clratvn, one hiseding the interior and the other the adjacent exterior angle of a triangle, and oneeting the op- posite sidef^ they divide this line har- monically. SuG.— By means of {358, 359, Part II.) the student will be enabled to establish the relation AO : BO : : AO' : BO', wlience, by the last proposition, AO',00', BO' are in harmonic proportion. 961' Tlieo. — In the complete quadrilateral, any diagonal is Q divided harmonically hy the other two, Dem. — Thus, AFH is divided harmoni- cally at C and H. For, considering BH as '\., a transversal of the triangle AGF, we have -■^H HF X DC X BA = HA X DF x BC. And CC, AD, FB being angle-transversals of the same triangle, we have CF x BA x DC = Whence, dividing, ~ = ^, i.e.^ AH is divided harmonically Gr GA G FrG. 489. CA x BC X DF. Again, if CH is drawn, CA, CG, CF, CH constitute a harmonic pencil, and BH, a transversal of it, is cut harmonically at B, I, D, H. Finally, if F and I be joined, FH (or FA), FB, Fl, FD constitute a harmonic pencil, and hence CG is cut har- monically at C, I, E, G. 962. Cor. — An angle-transversal of a triangle, and a line passing through the feet of the other uiigle-transversals, divide the third side harmonicall}-. 963. JProb.— Given a right line to locate ttvo harmonic coiijugate points. Solution.— Let AB be the line. O may be taken at pleasure between A and B. We are then to find 0', so that AO : BO : ; AO' : BO', Taking this by di- vision, we have AO - BO : BO : : AO' - BO' (= AB) : BO'. The first three terms being known, the other can be constructed. Or, we may first locate 0' at pleasure, and then find O. 964. Theo. — If from the given point C in a line the distances CO, CB, CO' he taken in the same di- rection, so that CO X CO' = CB* ? and if CA = CB bo taken in the opposite di- rection, AO' will he divided harmonically at O and B. Dem. — From CO CB -f- CO(=AO): CB O B Fig. 492. X CO' = CB^ we readily Avrite CO - CO (= BO) : : CO' + CB (= AO') : CO'- CB::CB: CO', CB {= BO'). * The bieector of the exterior nn«,'le nuetf the side prodMed ; but in higher geometry, as it i^i always understood that lines are iudefiuite unless limited by hypothesis, such specifications are deemed unnecessary. .:^-; /^' 312 DsTEODUCTION TO MODERN GEOMETRY. i'jy'Kl OGo. Con, 1. — Conversely, If a line AO' be cut liarmonically at V- b and B, and either of the harmonic means be bisected^ as AB at c. the three segments CO, CB, CO' icill be in geometric proportion. For, since AO' : BO' : : AO : BO, AO' + BO' : AO' - BO' : : AO + BO : AO - BO, or 2C0' : 2CB : : 2CB : OCO, and CO' : CB : : CB : CO. 066. Cor. 2. — lu a given line, as ab» as o approaches the centre C, 0' recedes, and when is at c, 0' is at infinity, since CO' = ^• 067- TJieo. — The geometric mean between two lines is also the geometric mean between their arith- metic and harmonic means. Dem.— Let AO' and BO' be the two lines. Ou their difference, AB, draw a semicircle, draw the tangent O'T and let fall the perpen- dicular TO. Then O and O' are harmonic con- jugates, since CO x CO' = CB' (?), CO' is the arithmetic mean (that is, \ the sum) of AO' and BO' (?) and TO' is tlie geometric mean (?). .'. CO' : TO' : : TO' : 00' (?). Queries. — Which is the greatest, in general, the arithmetic, geometric, or har- monic mean between two quantities ? Are they ever equal? 968. ScH. — This proposition affords a ready method of finding either of the harmonic conjugates O or O', when the other is given. The student will show how. \^ 060. Cor. 1. — The rectangle of the harmonic means and the S2im of the extremes f is equivalent to twice the rectangle of the extremes. For, CO' X 00' = TO"'- - AO' x BO', whence 2C0' x 00' = 2A0' x BO'; and, since 2C0' =r AO' + BO', (AO' + BO') x 00' = 2A0' x BO'. "v. 070. Cor. 2. — Tlie rectangle of the harmonic mean and the dif- ference of the differences of the \d and 2nd, and the 2nd and 3rd, is equivalent to twice the rectangle of these differences. That is, 00' x [(AC - 00') - (00' - BO')] = 2 (AO' - 00') (00' - BO'), or 00' X (AO — BO) = 2A0 x BO. Let the student give the proof. ^ 071. Cor. 3. — If three quantities are in harmonic proportion their reciprocals are in arithmetic proportion {i.e., the difference be- tween the 1st and 2nd equals the difference between the 2nd and 3rd). For, from AO', 00', BO, we have the reciprocals ^^» =r^/ ^^,. Now J 1_ _ AO' - 00 ' _ AQ J !_ _ 00' - BO' 00' AO' ~ 00' X AO^ ~ 00' X AO' ' BO' 00' '~ 00' x BO' HARMONIC PENCILS. 313 BO But 1 " 00' AO BO OO'xAO' OO'xBO' AO AO' BO BO' (?). 1 00' - OO'xBO" 1 1_ AO' ~ BO' 072, JProb, — Give7i the harmonic mean and the difference be- tween the extremes, to find the extremes. SuG's.— We have 00' and fiiB,{Fig. 493, Art. 967) given. Then CO x CO' = CT' =r iAB", and CO' - CO = 00', whence CO'" - 00' x CO' = iAB". From this equation CO' can be constructed {832)y and the problem solved. 073* Theo, — When two circles cut each other orthogonally (i. e., so thiit the tangents at the common point are at right angles), any line passijig through the centre of one, and cutting the other, is divided har- moniccdly ly the circumferences. Dem.— The tangents being perpendicular to each other pass through the centres, hence CO x CO' = CT'. But CB - CT. Therefore AO' is cut hai-monically. Fig. 494. 974. Proh, — To find the altitude of a triangle in terms of th radii of the escribed circles touching the adjacent sides. Solution. — Let r and r' be the radii of the escribed circles, and p the altitude. Now RT, AT, and QT are in harmonic proportion ; since, considering the triangle ACT, CQ bisects its in- terior and RC its exterior angle (?), we have QT : QA : :, RT : RA. But r, p, r', sustain the same relation to each other as RT, AT, QT; hence ?•, p, r' are in harmonical proportion. Therefore, by ^G9) p (r+r') =2rr'; or ;?= -?!ll. m Fig. 495. e.6 >- HARMONIC PENCILS. £>7tT. Def. — A JPencil of lines is a series of lines diverging from a common point. Def. — A Harmonic JPeiicil is a pencil of four lines cutting another line harmonically. III.— In the following figure OA, OB, OC, OD constitute a Ediinonic Pencil, if they divide the line mn harmonically at A, B, C, D. 314 INTRODUCTION TO MODERN GEOMETRY. 976. Tlieo. — A harmonic pencil divides harmonically every line which cuts it. Dem.— OA, OB, OC, OD being a harmonic pencil, that is, AD, BD, CD, being in harmonic proportion, A'D', any other line cutting the pencil, is divided harmonically, so that A'D', B'D', CD', are in harmonic propor- tion. Through C and C draw parallels to OA, as LK and L'K'. Now, from similar triangles, AB : BC : : AO : CK, and AO : CL : : AD : CD. But AD : CD : : AB : BC, since AD is harmoni- cally divided. Hence AO : CK : : AO : CL, and CK = CL. Hence from similar triangles CK' = CL'. Again A'B' : B'C : : A'O : CK' (?), Fig. 49G. and A'D' : CD' : : A'O : CL' {= CK') (?), whence A'B' : B'C : : A'D' : CD', or A'D', B'D', CD) are in harmonic proportion. Sen. — If the line through C cut the prolongation of AO beyond O, it is still harmonically divided ; and, in fact, it is scarcely necessary to make this state- ment, since in all general discussions lines are to be considered indefinite, un- less limited by hypothesis. 977. Def. — The alternate legs of a harmonic pencil are called conjugate, as OA and OC, OB and OD. \d7^. Theo. — If tico conjugate legs of a harmonic imicil he at right angles, one of them bisects the angle included by the other ixiir, and the other the suyplement of this angle, SxjG.— This is the converse of {962), remembering that the bisectors of two adjacent supplemental angles are at right angles. SECTION VI. ANHARMOMC RATIO. 979. Di^F.—TJie Anhavmonic Batio of fonr points in a right line is the ratio of the rectangle of the distance between the first and fourth into the distance between the second and third to the rectangle of the distance between the first and second into the distance between th§j^^^wi^ and fourth. III. — The anharmonic ratio of the four points A, B, C, D is AD x BC : AB x CD. B Fig. 497 AN HARMONIC RATIO. 315 980. The relation AD x bc : AB x CD is expressed for brevity [abcd]. III.— Thus [ABCD] means AD x BC : AB x CD ; [ADBC] means AC x DB: AD X BC; [BACD] means BD x AC : BA x CD; etc. The ratio [ABCD], or AD X BC : AB X CD is evidently the same as ^^^ •' ^^^r^. dC \*\j 981. ScH. — The appropriateness of the term anharmonic (;^^^harmonic) will be seen when we observe that, if AD is harmonically (\\\\(\it(\, ^^ equals j:—-. dC CO If, therefore, g^ is not equal to ^, which is the general case of division, irre- spective of the position of the points B and C, we may consider the ratio of BC ^^CD' '^^^^ general ratio, or, what is the same thing, AD x BC : AB x CD, is called the anharmonic ratio. 982. Theo. — Tlie anharmonic ratio of four x>oints is not clianged by interchanging tiuo of the letters, 2^^'ovided the other two be inter- changed at the same time. Dem. [ABCD] = [DCBA] = [BADC] = [CDAB], /. e., AD x BC : AB x CD = DA X CB : DC X BA = BC x AD : BA x DC = CB x DA : CD x AB, which are evidently identical. [The student should notice the different segments of the line indicated by the different forms.] 983. ScH.— But [ACBD] is a different anharmonic ratio from [ABCD] ; since AD X CB : AC X BD is not necessarily equal to AD x CB : AB x CD. Now, as there can be twenty-four permutations of four letters, there may be formed six different anharmonic ratios from four given points in a line. 984z. TJieo. — If a ^^encil of four lines is cut hy any transversal, the anharmonic ratio of the four points of intersection is constant. Dem. SL, SM, SN, SO, or, as we may read it, S-L,M,N,0, being such a pencil, and AD any transversal, draw through C NP parallel to SO. Then, AD X BC:AB X CD:: jf^'^^h: -;^N:AS) ^^^^^p, < AD : CD J ^ AS : CP ) But CN : CP is constant for all positions of C on SM. There- fore AD X BC: AB X CD is constant for any transversal. 985. Sen. — Other constant ratios may be written from the preceding prop- osition and scholium. Tlic anharmonic ratio [ABCD] is called the anharmonic ratio of the pencil. The angles of the pencil are the six angles included by the rays. 316 INTRODUCTION TO MODERN GEOMETRY. 086. — Cor. 1. — //' tiuo 2)C7icils are mutuaUy equiangular their anharmonic ratios arc equal. Query. — Is the converse of this corollary true ? IV.^ • 987. CoR. 2. — If two pencils have their ijiter- sections in the same right line, their anharmonic ratios are equal. Fig. 499. 988. Def. — The anharmonic ratio of four points on tlie circumference of a circle is the anharmonic ratio of the pencil formed by joining these points Avith any point in the circumference. III. — Thus, the anharmonic ratio of the points A, B, C, D is the anharmonic ratio of the pencil 0-A,B.C,D, it being im- material where in the circumfei-ence the point O is taken, since by Cor. 1, preceding, the ratio is the same for any posi- tion of O (?). BC Fig. 500. 989. TJieo. — If four fixed tangents to a circle are cut dy a fifth, the anharmonic ratio of the four ])oints of intersection, called the anharmonic ratio of the tangents, is constant. — V Fig. 501. Dem. a, B, C, D being the fixed points of tangenc}'', any transversal, as TV, cutting the tangents, has the anharmonic ratio [LMNP] constant. For the pencil 0-L,IVI,N,P lias its angles constant. Thus LOM is measured by \ arc (AX — BX) = ^AB, which is con- stant. And in like manner MON is measured b}'- \ arc BC, and NOP is measured by I arc CD. Hence, by the first of the preceding corollaries, the anharmonic ratio [LMNP] is constant. 990. The theory of anharmonic ratio is applied with great facility to the demonstration of theorems showing that several points are in a right line, and that several lines intersect in a common point. We give three specimens of each class. 991. TJieo. — If tico pencils have the same anharmonic ratio and a homologous rag common, the intersection of the other homolo- gous rays are in the same right line. Dem.— Let S-A,B,C,D and S'-A,B',C',D' Pjq 5Q.2. ^^ ^^'<^ pencils having the same anharmonic AN HARMONIC RATIO. 317 .ratio, and the rays SA,S'A coincident; then the intersections E,F, H are in the same right line. Let the line passing through E and F intersect SA iu K, and sup- pose it intersect SD in H', and S'D' in H". Then, since the anharmonic ratios of the two pencils are equal [KEFH'] = [KEFH"]; whence H' and H" are the same point, and must be the intersection of the two lines SD, S'D', that is, H. 992. Theo.— If in two right tines four poi?its in the one have the same anharmonio ratio as four points in the other, and one ho- mologo2ts 2)oint in common^ the three lines X)0.8sing through the other pairs of homologous points meet in a common point. Dem.— Let A be common, and [ABCD] = [A 6'C'D'J. Draw SA and SD'. Call the point in which SD' cuts AL D" (for the time being). Then [AB'C'D'J = [ABCD"]. But by hypothesis [AB'C'D'J = [ABCD]. Therefore [ABCD] = [ABCD"], and D and D" are one and the same point. Hence the three lines which pass through B and B', C and C, D and D' meet in a common point S. Fig. 503. 993. TJieo. — If the lines p>assing through the corresponding ver- tices oftiuo triangles meet in a common point, the intersections of their homologous sides lie in the same right line. Dem.— Let ABC and A'B'C be two triangles so situated that the lines AA', BB', CC meet iu the com- mon point S ; then L, M, N, the in- tersections of the homologous sides, are in a right line. For the pencil S-L, B, A, C being cut by the two transversals LD, LD', gives [LBAD] = [LB'A'D'] {984:). But C-L,B,A,D, and C'-UB',A',D', have these anharmonic ratios, hence C-L, 0,M,N, and C'-L,0,M,N, their equiv- alents, tmd having a common ray 'CC',have equal anharmonic ratios, and consequently L, M, N are in the same right line {991). Fig. m\. \ ^ J 994. TJieo. — If the intersection of the correspo7iding sides of two triangles are in the same right line, the lines jjassing through their corres2)07iding angles meet in a common 2)oint. Dem.— In the last figure, if AB and A'B', AC and A'C, BC and B'C have their 318 INTRODUCTION TO MODERN GEOMETRY. intersections in tlie same right line, as LN, the lines passing through B and B', A and A', C and C meet in a common point, as S. By {987) C-L,0,M,N has the same anharmonic ratio as C'-L,0,M,N, ^vhence [LBAD] = [LB'A'D'], and the truth of the theorem follows from (992). OOo. Theo. — The opposUc sides of an inscriled hexagon have their intersections in the same straight line. Dem. — The anharmonic ratios of the pencils B-A,E,D,C,* and F-A,E,D,C be- ing equal {988), LGDE, which intersects the first, is divided in the same anhar- monic ratio as NHDC, which cuts the sec- ond, or [LGDE] = [NCDH]. But these lines have a common homologous point D, hence the lines joining the other pairs of homologous points, as LN, CC, EH, meet in a common point, as M. Therefore L, M, N are in the same right line. 996. Sen. — This theorem is due to' Pascal, whose wonderful achievements in his brief life of thirty-nine years (1623- 1662) have been the admiration of all suc- ceeding generations. Fig. 505. 907. Theo. Fig. 506. The diago7ials joining the opposite vertices of a cir- cumscribed hexagon intersect in a common ptoint. Dem.— Consider AB. BC, CD, EF four fixed tangents cut by ED and FA. Then [PNDE] = [AQMF] {989). Hence the an- harmonic ratios of B-P,N,D,E,* and C-A,Q,M,F are equal (985); and since they have a common ray (CQ, BN) the in- tersections A, 0, D, of their homologous raj's, are in the same right line. Therefore the diagonals pass through a common point. * The student can conceive the rays BE, BDi elc„ as drawn, without encumbering the figure with them. POLE AND POLAR IN RESPECT TO A CIRCLE. 319 SECTION VIL POLE AND POLAR IN RESPECT TO A CIRCLE. ^^S. Def. — If a secant to a circle be revolved about a fixed point in the plane of the circle, the locus of the harmonic conjugate of the fixed point, in reference to the in- tersections, is the I^olav of the fixed point. The fixed point is the JPole of the Polar Line. The terms pole and polar as here used are correlative and neither has any significance without the other. III. — Let AP be a secant revolving about tke fixed point P, and let C be so taken that (in every position) AC : CB : : AP • BP, then is the locus of C the Polar of P, and P is Fio. 507, the Pole of the locus of C. yOO' Tlieo. — Tlie Polar of a given point in respect to a circle is a rigid line. Dem. — Let P be the pole, AP any secant passing through P, and a point in the polar. The locus of C is required. Draw PL through the centre, and let fall the perpendicular CC. Draw AL, AH, ACT, and CB. Since AC :CB:: AP: BP, C'P bisects the angle BC'F, the exterior angle of the triangle ^^' AC'B (?) ; hence, as LAH is a right angle, AL (^./ bisects NAC, the exterior kngle of the tri- ' angle CAP (?). Therefore, PL is harmonically divided at C, and H ; and, C being a fixed 'point, and C any point in the locus, the locus is the perpendicular TCCV. Fig. 508. 1000. Cor. 1. — Si7ice, as the secant revolves, the points a a?id B will vanish in C", C" is the point at which a tangent from the pole p touches the circle. 1001. CoR. 2.—Draioing OC", C'P, we see that OC"' (or oh') = 0C X OP. 320 INTRODUCTION TO MODERN GEOMETRY. 1002. CoR. 3. — The jjolar of a point hi ilic circumference is a tangent at that point. For, as OC x OP is constant and equal to OH", OP diminislies as OC in- creases, and when OP=OH,OC'=:OH also. 1003. JProh. — To draiv the ptolar to a givenp)ole in respect to a given circle. Cor. 1 effects the solution. 1004. JProb. — To find the piole of a polar to a given circle. Through the centre draw a perpendicular to the polar. [The student should be able to complete the solution.] 1005. Def. — The point c wliere the polar cuts the line passing through the pole and the centre of the circle is called the Polar Point. 1006. TJieo. — The pole and polar 2mnt are inter changeable. Dem. — TV being the polar to P, we are to show that T'V, parallel to TV and pass- ]15=:^>^3^ ing through P, is polar to C ; i.e.^ that any ^""^cant, as AC'C", passing through C, is di- vided harmonically in the intersections with the circumference, C, and the intersection with T'V. Drawing AP, since P is the pole of TV, we have, as in the last demon- stration, angle APB bisected by PC ; and consequently RPB bisected by PC". Therefore AC : CB : : AC": BC". Q. E. D. Fig. 509. 10011 * Tlieo. — The polars of all the points in a right line pass through the pole of that line; and, con- versely, The poles of all straight lines luhich pass through a given point are in the polar of tliat point. Dem. — 1st. TV being a given line and P its pole, we are to show that the polar to an}' point, as N , passes through P. Draw through P a perpendicu- lar to ON ; then P' is the polar point to N. For, OP : ON : : OP' : OB (?) : whence ON x OP' = OPxOB = OA^- Therefore, T'V is the polar of N (?). 2nd. P being any point and TV Fig. 510. RECIPROCAL POLARS. 321 its polar, the pole of any line, as T'V passini^ through P, is in TV, as at N. Draw ON perpendicular to T'V. Then, as before, ON X OP = OP X 0B= OA^ and N is the pole of TV. lOOS. Cor. — The j^ole of a straight line is the intersection of the ])olars of any tivo of its lioints ; and, conversely, The i^olar of any point is the straight line joining the j)oles of any tivo straight lines passing through that point. RECIPROCAL POLARS. a ^ 1009 . Def. — If two polygons be constructed, one within, or inscribed in, a circle, and the other without, or circumscribed about the same circle, such that the vertices of the one are the poles of the sides of the other, the two polygons are called Hecii^rocal Polars ; and the circle is called the Auxiliary Circle. The possibility of constructing such polygons is apparent from the last the- orem. When the points P, P', P", P'" Fig, 51o are in the circumference, TV, TT', T'V, VV become tangents, as appears from (1002). 1010. JProb. — Raving given one of two reciprocal polars, to con- struct the other. The student should be able to make the construction. 1011. By means of the relation between reciprocal polars a large class of propositions relating to the relative positions of lines and points, become, as it were, double ; i.e., one proposition being proved, another can be inferred. The process by which the inference is made is called reciprocation. We will give an example. 322 INTRODUCTION TO MOLELN GEOMETE: 1012. JProb. — To deduce the reciprocal of PascaVs theorem (995). Solution. — Draw tangents at the six ver. tices of the inscribed hexagon. Thus, a cir- cumscribed hexagon is formed whose sides are the polars of the vertices of the inscribed hexagon, through the vertices of which they respective]}' pass (1002). Now, drawing the diagonals PM, NQ, OL, they are the polars of the intersections of the opposite sides of the inscribed hexagon, as PM, polar to the intersection of DE and CB (?); and hence they pass through a common point, as V. Thus we have Brianchon's theorem, viz. ; The lines joining the opposiie angles of a cir- cu7nscn'bed hexagon pass through a common point. Fig. 513. 1013. — The three following theorems are of frequent use in ap- plying the theory of reciprocation. 1014. TJieo. — The angle included by tiuo straight lines is equal to the angle included hy the lines joining their 2Joles to the centre of the auxiliary circle. Dem. — The pole of a line being in the perpendicular from the centre of the auxiliaiy circle upon the line (1000), O' is the supplement of O ; hence o = 0. Fig. 514. WIS. T/ieo. Tlie distances of any tiuo points from the centre of the auxiliary circle are to each other as the distances of each point from the polar of the other. Dem. P, P' being the points, and TV, T'V their polars respectively, we are to show that CP : CP' : : PD" : P'D". By UOOl) R' = CP X CD = CP' x CD', R bemg the radius of the auxiliary circle. Whence CP : CP' : : CD' : CD. But CP : CP' : : CF : CE (?), and there follows CF : CE : : CD' : CD, CD' - CF (= PD") : CD - CE (= P'D'") : : CF : CE : : CP : CP. ScH.— -This is known as Salmon's Theorem. RADICAL AXIS AND CENTRES OF SIMILITUDE OF CIRCLES. 323 1016. Theo. — The anharmonic ratio of four points in a straight line is equal to that of the pencil formed by the four polars of these points. Dem. — 1st. The polars pass through a common point and thus form a pencil (?). 2cl. The angles included by the lines joining the four points with the centre, and those included by the polars are equal (?), hence the two pencils have the same anharmonic ratio. Fig. 516. SECTION VIII. RADICAL AXES AND CENTRES OF SBIILITUDE OF CIRCLES. 1017. Def. — The JPower of a I^oint in the plane of a circle is the rectangle of the distances from the point to the intersections of the circumference by a line passing through the point. ^ — .A . III. — Thus, the power of a point P, without the circle, is PA X PB ;— the power of a point within, as P', is P'A' X P'B' ; the power of a point in the circumference is zerOy since one of the distances is then ; — the power of the centre is the square of the radius. yiq. 517. 1018. Cor. — Tlie 2)0iver of a given p>oint zuith respect to a given circle is a constant quantity. Thus PA X PB = the square of the tangent from P to the circle, in whatever position PB lies, so long as it passes through P. So also P'A' x P'B' = the rectangle of the segments of any other chord passing through P'. 1019. T)E¥.—The Hadical Axis of Two Circles is the locus of the point whose powers ^vitb respect to tlie two circles are equal. 324 INTRODUCTION TO MODERN GEOMETRY. 1020. I*rojh — Tlte Jiadical Axis of two circles is aright line. Fig 518. Dem. — Join the centres of the two circles O, 0', and take a point R on this line such that OR^ - OR' = OT" - OT', or OR*' - OT' = OR^ - OT'', and erect PR perpendicular to 00'. Then P being ariy point in this perpendicular, OF* — or' = OP' — or?'. Adding this to the preceding equation, we have OP- - or" = O^P' - 0T'^ or PT^ = PT'". .-. PT = P T'. PT and PT' being tangents to the circles from any point in PR. Hence PV is the radical axis of the two circles. 1021. Cor. — WIie7i the circles are exterior to each other, the Radi- cal Axis lies between them, touching neither ; tuhen they are tangeni, either externally or inter7ially, the radical axis is the common tan- gent J when they cut each other, the axis ^ the common chord produced, 1022» ScH. — When the circles intersect it might seem that the above de- monstration fails for points within, as in the common chord. But, the powers of any point in this chord are still equal. Thus, at the intersections the powers are zero ; and at any other point in the chord, as a, ab x ae = ad x ac, since each is equal to ao x as. 1023. Cor. There ^ an infinite number of circles having their cejitres in the same right line, luhich have the same radical axis as any two given circles. Thus, in the first figure, PV being the radical axis of the circles O, 0', letting circle remain fixed, 0' may vary indefinitely so that O'R^ — O'T'' remains constant, and equal to OR — OT . 1024. I*roh. — Given two circles, to draw their radical axis. Solution. — Draw a common tangent, bisect it, and through the point of bisection draw a line perpendicular to the line joining the centres. "When the circles are tangent to each other, the distance between the points of tangency is ; hence the perpendicular is erected at this point. When they intersect, produce the common chord, or use the first method. CENTRES OF SIMILITUDE. 325 1025. JProj). — Wheji two circles cut each other orthogonally, that is, at right angles, the square of the radius of either is equal to the poiuer of its centre luith resj^ect to the other. Dem.— The power of with respect to circle O' is Qa X On — OP^, and of 0' with respect to circle O, O'b X O'm = O'P^ ; since, as the circles cut each other orthogonally, their tangents are at right angles, and the tangent to either passes through the centre of the other. Fig. 519. 1026. IProp.— Tlie radical axes of a system of three circles ivhose centres are not in the same straight line, intersect at a common 2)oint v Dem. — Since 0, O', 0" are not in a straight line, the radical axes of O, O', and 0, 0", as PV" and PV intersect. Let P be their common point. Now the power of P with respect to 0' is equal to its power with respect to O", since each is equal to its power with respect to O. Hence P is a point in the radical axis of O', 0". 1027. Cor. — If the centres are in the same straight line, their radical axes are parallel, and the common point is at infinity. 1028. Per — The intersection of the radical axes of three circles is called their Hadical Centre, Fig. 5-20. CENTRES OF SIMILITUDE. 1029. Def. — If the line joining the centres of two circles be divided externally, as at c, ai^d internally, as at c, in the ratio of the radii, these points are respectively the External and the In- ternal Centres of Sim- ilitude of the two cir- cles. III.— If CO : CO' : : EO : E'O', C is the external centre of similitude ; and, if CO : CO' : : E^b : E'O', C is the internal centre of similitude. Fig. 521. 326 INTRODUCTION TO MODERN GEOMETRY. The student should construct the figure when the circles are tangent exter- nally, — when they are tangent internally, — and when one is wholly within the other. Query. — How are the centres of similitude situated in the three different relative positions of the circles ? 1030. I^rop. — In Uvo circles the line passing through the ex- tremities of two liarallel radii on the same side of the line passing through the centres, intersects this line in the external centre of sim- ilittide, and if the radii are on ojyposite sides of this line the inter- section is the internal centre of similitude. The proof consists in showing that the line passing through the centres is divided as above. Let the student show it for the three diflFerent positions of the circles. 1031. Cor. 1. — Conversely, If any transversal he draion from either centre of similitudey the radii drawn to the inter sect io7is are parallel. Thus in the last figure, since CO : CO' '• EO : E'O', and the triangles have the angle C common, EO and E'O' are parallel. 1032. Cor. 2. — Tangents draion at the alternate intersections of a transversal through the external centre of similitude are parallel ; also, those at the mean intersections, and those at the extreme inter- sections, if the transversal he draion through the internal centre of similitude. This follows as a consequence of the parallelism of the corresponding radii, to which the tangents are perpendicular. Thus, tangents at E and E' are par- allel, as are those at F and F'. So, also, tangents D and D', and at E' and E" are parallel. 1033. Def. — The extremities of two parallel radii on the same side of the line joining the centres are called Homologous Points, and those of non-parallel radii where the transversal cuts the circumferences, as E, F', are called Anti-Homologous Points. 1034. CoR. 2. — The distances of a centre of similitude from two homologous points are to each other as the radii. 1035. CoR. 3. — The cejitres of similitude and the centres of the circles are four harmonic points. 1036. JProp. — If a circle touch two others, the line joining their points of contact imsses through the external centre of similitude of CENTRES OF SIMILITUDE. 327 the latter if the contacts ttre both external or both internal ; and through the internal centre of similitude if the contacts are the one external and the other internal. Fig. 522. Dkm.— In either case let 0" be the circle tangent to 0, and O' ; and tlirouffh the points of tangency draw E'C. The angle 0"ET = 0"FE' = EFO = FEO ; whence OE and O'E' are parallel, and the similar triangles CEO, CE'O' give CO : CO' : : OE : O'E'. [The student should make the otlier constructions.] 1037. Froh. — To draw a li7ie 2-)arallel to a given line so that the distance betiveen the extreme intersections with two ^iven circles shall be a maximum. SoLUTiON.-^Draw a line through the internal centre of similitude and par- allel to the given line. Now, at the extreme intersections draw tangents, and it will become evident that the line first drawn is a maximum. [The student should make the figure and fill out the proof] If the circles are wholly exterior to each other, the distance between the mean intersections is a minimum. 1038. Concluding Note. — Our limits preclude our pursuing these topics farther. We have given enough to make the language of the Modern Geome- try intelligible, and to afford some insight into its character. One of the best elementary resources for the Euglish student who wishes to pursue the subject at greater length, is Mulcahy's PHnciples of Modern Oeometry, Dublin, 1862. It is, however, much to be regretted, that there is no English treatise which presents the elements of this subject with the philosophic elegance of the French. The best of the latter is Rouche and Comberousse's Treatise on Ei^ementary Geometry. For a more extended view- of the subject, Salmon or Whitworth will furnish the Euglish student ; but he who would be proficient must read the works of Cuasles and Poncelet, who are the irreat authorities. 3"^ OLNEY'S MATHEMATICAL SERIES. ELEMENTS OP TEIGON-QMETEY, PLANE AND SPHERICAL. Bt EDWARD OLNEY, PROFESSOR OP MATHEMATICS IN THE rNIVIIRSlTT Of MICHIOAX. NEW YORK: SHELDON & COMPANY, No. 8 MURRAY STREET. 1877. Stoddard's lathematical Series. -MM- STODDARD'S JUVENILE MENTAL ARITHMETIC STODDARD'S INTELLECTUAL ARITHMETIC .... STODDARD'S RUDIMENTS OF ARITHMETIC STODDARD'S NEW PRACTICAL ARITHMETIC .... SKORT AND FULL COURSE FOR GRADED SCHOOLS. STODDARD'S PICTORIAL PRIMARY ARITHMETIC STODDARD'S COMBINATION ARITHMETIC STODDARD'S COMPLETE ARITHMETIC .... The Combination Scliool Aritlimetic being Mental and Written Arithmetic in one book, will alone serve for District Schools. For Academies a full high course is obtained by the Complete Arithmetic and Intellectual Arithmetic. OLNEY'S HIGHER MATHEMATICS. A COMPLETE SCHOOL ALGEBRA in one volume, 390 pages. Designed for Elementary and higher classes in Schools and Academies. By Prof. Edward Olney, University of Michigan. A GEOMETRY AND TRIGONOMETRY in one vol. By Prof. Edward Olney. One vol. 8vo, A GENERAL GEOMETRY AND CALCULUS in one vol. Entered according to Act of Congress, in the year 1870, Br SHELDON A COMPANY la the Office of the Librarian of Congress at Washiu^tco. CONTENTS PART IV.— TRIGONOMETRY. CHAPTER I. PLANE TRIGONOMETRY. SECTION I. Definitions and Fundamental Relations between the Trigonometri- cal Functions op an Angle (or Arc). page. Definitions 1-5 Fundamental Relations of tlie Trigonometrical Functions of an Angle 5-7 Signs of the Functions 7-10 Limiting Values of the Functions 10-1-4 Functions of Negative Arcs 15-16 Circular Functions 1 G Exercises 17-30 SECTION II. Relations between the Trigonometrical Functions of different An- gles (or Arcs). Functions of the Sum or Difference 20-25 Functions of Double and Half Angles 26 Exercises 26-29 SECTION III. Formula for rendering Calculable by Logarithms the Algebraic Sum of Functions 29-30 Exercises 30-31 SECTION IV. Construction and Use of Tables. Definitions 31-32 To Compute a Table of Natural Functions 32-33 To Compute a Table of Logarithmic Functions 33-34 Exercises in the Use of the Tables 34-39 Functions of Angles near the Limits of the Quadrant 39-42 Exercises 43 IV CONTEXTS. SECTION V. Solution op Plane Triangles. page. Of Right Angled Triangles 44-45 Exercises and Examples 45-48 Practical Applications 48-49 Of Oblique Angled Plane Triangles 49-51 Exercises 51-55 Oblique Triangles Solved by means of Right Angled Triangles . . . 55-56 Exercises 56-57 Functions of the Angles in Terms of the Sides 58-59 Exercises 59-60 Area of Plane Triangles 60-61 Practical Applications 61-64 CHAPTER 11. SPHERICAL TRIGONOMETRY. IXTRODUCTIOX. Projection of Spherical Triangles. Definitions and Fundamental Propositions 65-66 Projection of Right Angled Spherical Triangles 66-71 Projection of Oblique Angled Spherical Triangles 71-73 SECTION I. Solution op Right Angled SruERiCAL Triangles. Definitions 73-75 Exercises 75-76 Napier's Rules 76-78 Determination of Species 79-81 Exercises in Solution of Right Angled Spherical Triangles 81-84 Quadrantal Triangles 84-85 SECTION XL Op Oblique Angled Spherical Triangles. To find the Segments of a Side made by a Perpendicular let fall from the opposite angle 85-86 The Relation of the Sides and Opposite Angles 87 Solution of Oblique Spherical Triangles by Napier's Rules, in Three Problems 87-90 Exercises • - 90-95 CONTENTS. V SECTION III. General Formula. page. Angles as Functions of Sides, and Sides as Functions of Angles. 96-100 Gauss's Equations 100-101 Napier's Analogies 103 Exercises in the Use of these Formulae 102-107 SECTION IV. Area of Spherical Triangles 108-110 Practical Applications of Spherical Trigonometry 110-110 TABLES. Introduction to the Table of Logarithms 1-10 Table of Logarithms of Numbers 11-28 Table of Natural Sines and Cosines, and of Logarithmic Sines, Cosines. Tangents, and Cotangents 29-74 Table for Precise Calculation of Functions near their Limits 75-78 Table of Tangents and Cotangents 79-88 PART IV. TRIGONOMETRY, CHAPTER L PLANE TBIGONOMETRY. SECTION L DEFINITIONS AINT) FUNDAMENTAL RELATIONS BET^ITIEN THE TRIGONOMETRICAL FUNCTIONS OF AN ANGLE (OR ARC). 1* Trigonometry is a part of Geometry which has for its sub- ject-matter, Angles. It is chiefly occupied in presenting a scheme for measuring and comparing angles, by means of certain auxiliary lines called Trigonometrical Functions, in investigating the relations between these functions, and in the solution of triangles by means of th3 relations between their sides and the trigonometrical functions of their angles. 2. I^lane Trif/ononietri/ treats of plane angles and triangles^ in distinction from Spherical Trigonometry, which treats of spherical angles and triangles. 3. A FuiictiOil is a quantity, or a mathematical expression, conceived as depending upon some other quantity or quantities for its value. Ill's. — A man's wages /(>r a given time is a function of tlie amount receivetl per day ; or, in general, his wages is a function of boili the time of service and the amount received per day. Again, in the expressions y = ^ax^^y ='x* — 2i».T + 5,y = 21ogaa;, y = a^,2/ is a function of a; ; since, the numbers 2, 5, a andd being considered fixed or constant, the value of y depends upon the value we assign to x. For a like reason such expressions as V^a" — x*, and Zax^ — 2>/l^^ may be spoken of as functions of x. Once more, the area of a triangle is a func- tion of its base and altitude. 4. Angles as Functions of Ares. — We have learned in Geometry (Part II., Sec. VI.), that angles and arcs may be treated as functions of each other; and that, if the angles be taken at the 1 2 PLANE TRIGONOMETRY. centre of the same or equal circles, tlie arcs intercepted nave the same ratio as the angles themselves, and hence may be taken as their measures or representa-tives. For trigonometrical purposes, an angle is considered as measured by an arc struck with a radius 1, from the angular point as a centre. o. A Degree being the ^^ part of the circumference of a circle, becomes the measure of -5^ of a right angle; and, for convenience, it is customary to speak of such an angle as an angle of one degree, of four times as large an angle as an angle of four degrees, etc., apply- ing the term directly to the angle. A small circle written at the right and a little above a number indicates degrees (°). 6. A Minute is -^ part of a degree. Minutes are designated by an accent ('). A Second is -^ part of a minute. Seconds are indicated by a double accent ("). Smaller divisions of angles (or arcs) are most conveniently represented as decimals of a second, though the designations thirds, fourths, etc., are sometimes met with, and signify further subdivisions into 60ths. 5° 12' 16" 13'" is read, " 5 degrees, 12 minutes, 16 seconds, and 13 thirds." Ill's. — In Fig. 1 AOB is an angle of 35°, because the measuring arc db contains 35 of the 360 equal parts mto which the circumference whose radius is Oa, could be divided. In like manner BOC is an angle of 7°. BOC = iAOB = iAOC- Hence, it becomes evi- dent that we may use the numbers 35, 7, and 42, to represent the respective angles AOB, BOC. and AOC, or the corresponding arcs o^, 6c, and ac. 7. A Ouadt^ant is an arc of 90°, and is the measure of a right angle ; hence, a right angle is called an angle of ^- 1- 90°. Thus arc ad, Fig. 1, = 90°; or angle AOD = 90°. S, TJie Complement of an angle or arc is what remains after subtracting the angle or arc from 90°. The Supplement of an angle or arc is what remains after subtracting the angle or arc from 180°. Ill's. — In Fig. 1, the angle BOD is the complement of AOB, and the arc bd 18 the complement of arc ab. The complement of 35° is 90° — 35°=- 55°. The supplement of 35° is 180°- 35°= 145°. DEFINITIONS AND FUNDAMENTAL RELATIONS. 3 ,9. A Quadrant is often represented by I'T, since -?. is the semi- circumference when the radius is unity. When this notation is used, 180° the Unit ^rc becomes = 57°.29578 nearly, or 57° IT' 44".8 +, '71' which is an arc equal in length to the radius. 10, For trigonometrical pui-poses, an angle is conceived as g^T. erated by the revolution of a line about the angular point, ana hence may have any value luhatever, not only from 0° to 180°, but from 0° to 360°, and even to any number of degrees greater than 300°, as 1280°, etc. An angle of 45° is generated by \ of a revolu- tion, 90° by } of a revolution, 180° by 4 a revolution, 270° by } of b revolution, 360° by one revolution, 450° by 1} revolutions, 1280° by 3| revolutions, etc., etc. 11, In accordance with the conception of an angle as generated by a revolving line, the measuring arc is considered as orifjinating at the first position of the revolving line- {i, e., with one side of the angle), and terminating in the line after it has generated tlie angle under consideration (*. e., with the other side of the angle). The first extremity is called the Origin of the arc, and the other the Termination. Ill's. — In Fig. 1, let the angle AOB be considered as generated by a line starting from the position OA, and revolving around the point 0, from right to left,* till it reaches the position OB. Oa being taken as unity, the arc ab is the measuring arc of the angle AOB ; a is its origin, and b its ierminaiion. 12. In the generation of angles by means of a revolving line, the normal motion is considered to be from right to left, and the quad- rants are numbered 1st, 2d, 3d, and 4th, in the order in which they are generated. 13. The Trig onometr leal Functions are eight in num- ber; viz., sine, cosine, tangent, cotangent, secant, cosecant, versed- sine, and cover sed- sine. These lines are functions of angles, or. what amounts to the same thing, of arcs considered as measures of angles, and are the characteristic quantities of trigonometry. 14z. Hie Sine of an angle (or arc) is a perpendicular let fall from the termination of the measuring arc upon the diameter passing through the origin of the arc. Thus in Fig. 2, M is in each case the sine of the angle AOB, or of the arc axh. * The pnpil will understand that, if he imagines himeelf ptanding at the centre of moti -u. as the moving body or point passes before him, the distinctions " from right to left." and " from left to right," are easily made. PLANE TRIGONOMETRY. lo. Hie Tri{/ono2}ietrical Tanf/ent of an angle (or arc) is a tangent drawn to the measuring arc at its origin, and limited Fig. 2. by tlie prodrbced diameter passing throngli the termination of the arc. Thus in Fig, S, ac is in each case the tangent of the angle AOB, or of the arc axb. 16, The Secant of an angle (or arc) is the distance from the angular point, or centre of the measuring circle, to the extremity of the tangent of the same angle (or arc). Thus in Fig. 2, Oc is in each case the secant of the angle AOB, or of the arc axb. 17, The Versed-Sine of an angle (or arc) is the distance from the foot of the sine of the same angle (or arc) to the origin of the measuring arc. Thus, in Fig. 2, da is in each case the versed- sine of the angle AOB, or of the arc axh. IS. The prefix co, in the names of the four trigonometrical func- tions in which it occurs, is an abbreviation for the word comjyUment. Thus cosine means complement-sine, i. e., the sine of the comple- ment; cotangent means tangent of the complement; etc. The co- sine of 40° is the sine of 90° — 40°, or 50° ; the cosine of 110° is the sine of 90° - 110°, or — 20°; the cotangent of 30° is the tangent of 60° ; the cosecant of 200° is the secant of - 110°. 19. Construction of the Coniplefuentary Functions. — Let us now see how the complementaiy functions are constructed with refer- ence to their primitives, premising that all arcs in Fig. 3, reckoned from A, are DEFINITIONS AND FUNDAMENTAL RELATIONS. to he reckoned around from right to left in this discussion. 1st. Let AP be any arc less than 90° ; then 90° — AP = aP is its complement. Now considering a as the origin and P the termination of this complementary arc, Pd is its sine, at its tangent, Ot its secant, and ad its versed- sine. Hence, Pd^ at, Ot, and ad are respect- ively the cosine, cotangent, cosecant, and coversed-sine of the arc AP, or the angle AOP. 2d. Letting APP' be any arc between 90° and 180°, its complement is 90° — APP' or — aP', the — sign signifying that the arc is reckoned backward from P' to a. But as the values of the functions will be the same whether the origin be taken at P' or at a, we may take a as the origin of this comple- mentary arc, and P' as its termination, whence P'c?' becomes its sine, rt^' its tan- gent, 0^' its secant, and ad' its versed-sine. Therefore P'd', at'. Of, and ad\ are respect- ively the cosine, cotangent, cosecant, and ^^^- 3- coversed-sine of the arc APP', or the angle AOP'. 3d. In like manner, aP" is shown to be the complement of arc APP'P " ; and as P"d",at, Ot, and ad" are respectively the sine, tangent, secant, and versed-sine of this complement, they are the corresponding cofunctious of the arc APP'P", or the salient angle AOP"* 4th. In the same way, it appears that P"'d"', at', Ot\ and ad"' are the cosine, cotangent, cosecant, and coversed-sine of the arc APP'P" P", or the salient angle AOP"'. Observe that a.^ a point on the measuring arc 90° from the pritmlive origin, is tJie oHgin of all the complementary fiuictions. ScH. — It will readily appear from the figure that the cosine of an angle (or arc) is always equal to tJie distance from the foot of the sine to tJie vertex of the angle (or the centre of the measuring arc). This is the more convenient prac- tical definition. Thus the cosine of AP is PfZ = DO; the cosine of APP' is P'd' = D'O, etc. 20. Notation, — Letting x represent any angle (or arc), the several trigonometrical functions of it are writteL sin a:, cosx, hxnx, cot a;, sec a;, cosecT, vers.T, and covers a;. They are read -^sine-r," " cosine x" " tangent a;," " cotangent ic," etc. FUNDAMENTAL RELATIONS BET^^TEN THE TRIOONOMETRICAI. FUNCTIONS OF AN ANGLE (OR ARC). [Note.— These fundamental relations must he made perfectly familiar. 'XTiey must i>e meaiorized, and be as familiar as the Multiplication Table. The student can do nothing in trigonometry without them.] The discussions in this treatise all proceed upon one general plan; viz., — First obtain the j^ai'ticiilar x^^^operty of the 6 PLA>iE TRIGOXOMETEY. sineund cosine, and from this deduce all the others according to the dependencies shown in tlie foUoiv- ing proposition* 21, I*rop. — TJie Fundamental Relations which the Trigono- metrical Functions sustain to each other are: 1 (1) sin' a; + cos' a; = 1 (2) sm X tan X = ; cosa; (3) cos a; cot X = -. — ; sma; (4) 1 cot X = . ; cos a; 1 (5) seca; = (6) coseca: ^ ' sma;' (7) sec' a: = 1 + tan' x ; (8) cosec'a; = 1 4- cot' a; (9) vers a; = 1 — cos a; ; (10) covers a; = 1 — sin a;. (The forms sin'a", sec'a;, etc., signify tlie square of the sine, the square of the secant of x, etc., and are read " sine square a;," " secant square x" etc. The student should distinguish between sin'a;, and gin a;'.) Dem.— In Fig. 4, let x represent any arc as AP, less than 90°. Then PD = sin x, OD or Pd = cos J-, AT = tana;, OT = secx, at = coti, Ot = cosec j, AD = yei-sin x, and ad = coversin x. Fig. 4. (1). In the right-angled triangle POD, pd' + OD' = 0P'> or sm^'a; + cos' a; = 1, since OP = radius = 1. (2). From the similar triangles POD and TOA, AT PD sin a; £J- = ~^, or tan x = OA OD' cos a-. (3). From the similar triangles POcf and iOa, at Pd cos a; = ^^.. or cot X — —. — . sma; ^ = 7^' 01" cot X Oa Od (4). Multiplj-ing (2) and (3) together, sinajcosar . ^ 1 : — = 1, 01 tan X — — — cosajsma; cot* tan a; cot 2- = (5). From the similar triangles OTA asd CRD, 1 ^ = ^ ; but OP and OA each = 1, .*. sec x = DEFINITIONS AND FUNDAMENTAL DELATIONS. i (6). From the similar triano:les Ota and OPd, Ot OP 1 "TT- = 7^, or cosecic = -; — . Oa Od sma* (7). From the right-angled triangle OAT, or" = oa' + AT", or sec'jj = 1 + tan'a;. (8). From the right-angled triangle Oat^ oT = Oa' + "a^y or cosec^^ = 1 + cofa;. (9). AD = AO - OD, or vers a; = 1 - cos aj. (10). ad = aO — Od, or covers aj = 1 — sin «. Thus the fundamental relations of the functions are established for an arc less than 90°. But it will readily a])pear that the relations are the same for any other arc. For example, let x = AP' be any arc between 90° and 180^ Theu the triangle P'D'O gives sin^'a; + cos'* a; = 1, since P'D' = sin x, and OD' = cos x. The similar triangles P'D'O and OAT' give =rr- = frj^, or tan x = ; and the '^ * OA D'O' cos a;' cos X similar triangles P'd'O and t'aO give cot x = - — . In like mianner let the student observe the relations when x = APP'P", or an arc between 180° and 270°. So also when x = APP'P"P"'. or an arc between 270° and 360°. 22, CoK. 1. — The tangent and cotangent of the same angle are reciprocals of each other ; so also are the secant and cosine, and the cosecant and the sine. Thus, if tan x = Z,Qotx = \] since cot a; = , . If sec 2; = 2, cos a; = :|: sinceseca; = , or cosa; = . tanrc cos a; seca; 23. Cor. 2. — Sines and cosines cannot exceed 1. Tangents and cotangents can have any values from to ao» Secants and cosecants can have any values hetioeen 1 and co . Versed-sines and cover sed- mies can have any values 'between and 2. These conclusions will readily appear from the definitions, and an inspection of Fig, \, SltJNS OF THE TRIGONOMETRICAL FUNCTIONS. 24:, I*rop. — Angles (or arcs) considered as generated from right to left being called positive * and marked +, those considered as gcH" eratedfrom left to right are to be called negative and marked — . • This is purely an arbitrary convention. We might equally well reTerse ii PLA^'E TRIGONOMETRY. Dem. — Tliis is a direct application of the significance of the + and — signs. Bee Complete School Algebra, pp. 20-23.) Thus, in Fig. 5, if the angle AOP, considered as generated by the revolution of a line trom the position OA in the direc- tion of the arrow-head (from right to left), is called positive and marked + , an angle generated by the motion of a line from the position OA in the opposite direction (from left to right), as the angle AOP" thus gen- erated, is to be considered negative and marked — . Let it be carefully observed that it is the assumed direction of the motion of the generatrix that determines the sign of the angle (or arc). Two lines meeting at a common point may be con- sidered as designating either a -f or a — angle, according to the direction of motion assumed. Thus the lines OA and OP', Mg. 5, may form the positive angle measured Fig. 5. by the arc APP', or the negative, salient angle measured by the negative arc AP"'P"P'. q. e. d. 2S, JPvoj)* — Radius being considered as alioays extending in the same direction, viz., from the centre toward the circumference, is always positive. 26, JProj), — T/te sign of the sine of an angle between 0° and 180° being +, that of an angle between 180° and 360° is — . Dem. — In Fig. 5, we observe that the sines of all angles terminating in the 1st and 2d quadrants, i. e., between 0° and 180°, may be considered as measured from the primarj' diameter AB, npu>ard,v;\n\Q those of angles terminating in the 3d and 4th quadrants, i. e., between 180° and 360°, are reckoned downward irom the same line; hence, the former being called +, the latter shojild be — , .w t7i€ two species are estimated in opposite directions, q. e. d. A more elegant conception is to consider the sine as projected upon the diam- eter vertical to that passing through the origin, as aC ; whence Od is the sine of AOP (or arc AP). Now this line evidently is when the angle is ; and as the angle increases, the sine increases, being generated from O vpicard, and hence is called + . This is the same conception as we use in the case of th^ cosine. Adopting it, we see that sines reckoned from upward are +, and downward — . Cosines reckoned from to the right, are + , and to the left, — . 27» Cor. — T7ie cosecant of an arc has the same sign as its sine, aince coseca; = -. — ; and as 1, being the radius, is -f , the sign of sma; %\nx is the same as the sign of sin x. DEFINITIONS AND FUNDAMENTAL RELATIONS. 9 28, IProp. — The sign of the cosine of an angle letween 0' and 90°, and between 270° and 360°, is +, lohile that of an angle hetiucen Dem. — In Fig. 5, we observe that the cosines of all angles terminating in the 1st and 4th quadrants, may be considered as estimated from the centre towarc the right, as OD, OD'" ; while correspondingly, the cosines of angles terminating in the 2d and 3d quadrants will be estimated from the centre toward the left, as OD', OD". Hence, by reason of this opposition of direction, the former are called + , and the latter — . q. e. d. 29, Cor. — TJie secant of an angle has the same sign as its cosine, Biuce these functions are reciprocals of each other, (See 2*^*) 30. JProp. — The sign of the tangent of an angle hettveen 0° and 90°, and also letioeen 180° and 270°, is + ; while that of an angle hetiveen 90° and 180°, and betiueen 270° and 360°, is — . Dem. — Since tan x = , when sin a; and cos x have like signs, tan a; is + , by cos X the rules of division ; and when sin x and cos x have different signs, tan a; is — . Now, in the 1st and 3d quadrants* the signs of sin a; and cos x are alike, hence in these quadrants tana; is plus; but in 2d and 4th quadrants sin a; and cos.e have unlike signs, and consequently in these tan x is — . q. e. d. 31, Cor. — The sign of the cotangent is the same as the sign of the tangent of the same angle j since cot x = . tan 3/ 32, JProp, — Versed-sine and coversed-sine are always +. Dem. — Vers a; = 1 - cos a; ; and as cos a; cannot exceed 1, 1 — cosa; is al ways +. In like manner, covers a: = 1 — sin a;; and as sin a; cannot exceed 1, 1 — sin a; is always + . q. e. d. ScH. 1. — It is essential that the law of the signs, as explained above, be well understood, and the facts fixed in memory. Fig. 6 will aid the student in fixing the law in the memoiy. Having this constantly be- fore the mind, and remembering that tan and cot are + when sin and cos have like signs, and — when they have unlike, and that cos and sec have like signs, as also sin and cosec, or, more simply, that 1 Cin = — , cot = : — , sec = — , and cosec = cos tan cos sm the student cannot fail to know the sign of a func- tion at a glance. It will be of service to remember that versed-sine and coversed-sine, and all the functions of angles of the 1st quadi-aut, are + ; but tliat of the other functions than the versed-sine and coversed-sine, of Fig. 6. angles terminating in the other quadrants, but Uco are -f- in each quadrant * This is a convenient elliptical form for iBt quadrant," etc. an angle whose measuring; arc terminates in the 10 PLANE TRIGONOMETRY. ScH. 2. — The signs of functions of angles greater than 360* are readily deter- mined by observing in what quadrant the measuring arc terminates. Thus, Bin 570° is — , since an arc of 570° terminates in the 3d quadrant. In any given case, the sign of the function is the same as the sign of the same function of tho remainder left after dividing the arc by 360% or 2;r. Thus tan 1180° is the same as tan 100° ; i. e., it is — . The same may also be said of the value of the func- tion. LDHTENG YALUES OF THE TRIGONOMETRICAL FUSCTIOXS. 83. Proiy.—Sin 0° = if 0, sin 90° = 1, sin 180° = ±0, «i?i 270° = —1, sin 360°= tO, and the limits within which the sines of all angles are comprised, are + 1 and —1. Deit,— Let the point P be supposed to start from A and move around the circumference from right to left, and let x represent the angle (or arc) generated. Now, when P is at A, a; = 0, and PD = 0. Moreover, if we consider the sine P'" D'" as reaching its limit, by the moving of P'" from some point in the fourth quadrant to the origin, the sine of 0° becomes — 0, since what is true of a taryinrj quantity all the way as it appi'oaches its limit, is assumed true at tJie limit. But, if the sine reaches its limit by the passage of the point P from some point in the first quadrant to the origin, the sine of 0° is to be considered as + , since the function is + as it approaches its limit .-. sin 0° = T 0.* As x increases from 0° to 90° (/. e., as P passes from A to a), PD is + and increases from to 1 (+ to + 1): .*. sine 90° = 1. As a; continues to increase from 90°, sin x diminishes and becomes at Pig. 7. 180°. To determine the sign of sin 180°, we notice that it is + M'hen the point P approaches this limit from the second quadrant, and — when it approaches it from the third quadrant, .'. sin 180° = ± 0. Conceiving x to go on increasing from 180°, sin a; appears below AB aud is — , and beginning at — diminishes (a numerical increase of a negative quantity being considered an absolute decrease) till at 270° it becomes — 1. .". sin 270° = — 1. As x passes from 270° to 360°, sin x increases (see above) from — 1 to 0. The sign of this limit is ambiguous, as appears by regarding tlie limit as reached by the angle passing from something less than 360' to 360°, whence we have — ,»nd also as reached by the angle passing hack from something greater than 360° to 360°, whence the sign is + . .-. sin 360° = T 0. Finally, as it is evident that these values would recur in the same order as the point P passed around again, the above comprise all real values of sines of angles. Q. e. d. * It has been customary to disregard the sign of 0. in such a series as — m . . — 3, — 2, —1, T 0, + 1, + 2, + 3, .... + m, whereas it should be regarded as ambiguonfl. as appears above. DEFINITIONS AND FUNDAMENTAL RELATIONS. 11 34. JProjy.—Cos 0° = 1, cos 90° = db 0, cos 180° = - 1, cos 270° = ^0, cos 360° = 1, and the cosines of all angles are comprised hetween + 1 and — 1. Dem. — Using the same figure and the same conception as in the last demon- stration, it is evident that as P approaches A, from either direction, that is as « approaches 0°, the cosine OD approaches to equality with the radius and reaches it at a; = 0, being + in either case. .*. cos = 1. As a; approaches 90° from a less value, i. e., from the first quadrant, cos a; is +, and approaching + ; but as X approaches 90° from some greater value, i. e., from the second quadrant, cos a; is — and approaching — 0. .*. cos 90° =: ± 0. In like manner \ve obseiTC that as X increases from 90° to 180°, cos x decreases (see Dem. of 33) from to — 1, which it reaches at 180°, and this, whether the point is reached in one way or the other. .-. cos 180° = - 1. Again, cos 270° = T ; since it is — if aj passes to 270° from some value less than 270°, and + 0, if it passes from some value greater than 270. While x passes through the fourth quadrant, cos x passes from to + 1, reaching the latter value when x = 360°. .-. cos 360° = 1. Finally, as it is evident that the above values would recur in the same order as the gen- erating point passed around again, this discussion comprises all real values of a cosine. 3S. JPro2).—Tan 0° = zp 0, tan 90° = ± oo, tan 180° = ^ 0, tan 270° = ± go , tan 360° = =f 0, and the tangents of all angles are comprised between -t-co and—a^. cosO 1 cos 90 ± TanlSO- = ": = ^= ^ 0. Tan 270° = ^-1^ = ::ii = ± ». Tau360- cos 180 —1 cos2<0 TO =■ ^jTTT^ = -— = T 0. From these results it appears that the tangent may COS OOU J. Lave any value whatever from +00 to — c» ; and as subsequent revolutions of the generatrix would evidently only repeat these values, these are all the real values of a tangent. (Moreover, all real values are comprised between + oo and — 00). Q. E. D. It is easy to observe the direction of the change in the tangent (whether it is sin increasing or decreasing) by observing the fraction — ;. As the arc increases in the first quadrant, the sine increases and the cosine decreases, for both of which reasons the tangent increases, and hence changes more rapidly than either sine or cosine. The student should obseiTC the change in each of the four quad- rants in the same manner. ScH.— These values of the tangent are illustrated by Fig. 7. Thus AT, which is +, becomes + when P returns to A, or a; = 0. Also AT', which is — , and may be considered as the tangent of AP'", a negative arc, becomes — 0, whenAP"' = 0. Again, if AP passes to Art, AT passes to + x . But, if P' passes hack to a, so that AaP' p;isses to Aa, or 90°, AT' passes to — oo. .*. We see that tan may be considered =F 0, and tan 90° = ± 00 . In like manner thf other limits are illustrated. 12 PLANE TEIGONOMETRY. SO. Prop.— Cot 0° = ip o), co/J 90° = =t 0, cot 180° = :?:«>, cot 270° = ± 0, co^ 3G0° = =p co, and the limits of the cotangent are + 00 , and — oo . Dem.— These values are the reciprocals of the coiTesponding values of the tan- cos gent ; or they may be deduced from the relation cot = -r-, in a manner altogether analogous to those of the tangent. Fig. 7 also affords geometrical illustraticns of them. The student should not fail to deduce and illustrate them, and also to observe the law of change. 37. JProp.—SecO° = 1, sec 90° = ± od, 5ecl80° = - 1, sec270'* = ip CO, sec 360° = 1, a7id all real values of this function are com- prised hetiueen 1 and ± oo , and — 1 and =f oo . Dem. — These values are the reciprocals of the corresponding values of the cosine. The student should obtain them from the cosine, and illustrate them from Fig. 7, observing the law of change. Thus beginning at OA = 1, which is the secant of 0°, the secant increases till x = 90°, when the secant OT becomes + oc, if we consider this limit as reached thus ; but — oo , if we consider such an arc as AaP', whose secant is — OT', which becomes — oo, when the point P' passes back to a. Sen. — The series which represent the values of secants are, for the first and second quadrants, + 1, + 2, + 3, + . . . . + tw, ± oo , - w, — . . . . — 3, — 2, — 1; for the third and fourth quadrants, — 1, — 2, - 3, — ....— w, T oo, + 771, + .... + 3, + 2, + 1, understanding the numbei*s in these series as rep- resenting a few terms of series which have an infinite number of terms of all values exten(ung, in the first case, from + 1 to ± oo , and thence to — 1. It will be a good exercise for the pupil to write the series representing the values of each of the ti'igonometrical functions. Thus, for the sine, we have T 0, + i, + i, + a^ + 1^ 4- a^ + ^, + ^, ± 0, for the first and second quadrants, understand- mg that all values intermediate between those represented are included. For tlie second and third quadiants, we have, + 1, + f , + i, + i, ± 0, - i, — i. 38. JPro2)'—Oosec 0° = qr oo , cosec 90° = 1, cosec 180° = =t oo , cosec 270° = — 1, cosec 3G0° = :^ oo , and all the real values of this function are comprised letween 1 and ± oo , and — 1 and z^ cf^, Dem.— Let the student demonstrate and illustrate as in the preceding article. Do not neglect to go through the whole in detail; it is an important and excellent exercise. 39. Prop.— Versin 0° = 0, versin 90° = 1, versin 180° = 2, ver- sin 270° = 1, versin 360° = 0, a7id the real limits of the function aro and 2. DEFINITIONS AND FUNDAMENTAL RELATIONS. 13 Dem. — The student will readily deduce these results from the relatiou versa : 1 — cos X. Thus when a; = 0, vers = 1— cos = 1 — 1 =0, etc. 4z0, Prop* — Covers 0° = 1, covers 90° = 0, covers 180° = 1, covers 270° = 2, covers 360° = 1, and the limits of the real values of this fmction are and 2. Dem. — The student should be able to give it 4:1» General Scholium. — It is important to observe that in the case ol each of the above functions it cJianges its sign by passing through 'S BEr\>T:E> THE TRIGOXOXETRICiX FU>CTIO>S OF DrFFERE>T 1>GLES (OR ARCS). (a) FUKCTIOXS OF THE SUM OR DIFFERENCE OF ANGLES (OR arcs). 4,8, I^rop, — Tlie sine of the sum of tivo angles {or arcs) is equal to the sine of the first into the cosine of the second, phis the cosine of the first into the sine of the second. Thus letting x and y represent any two angles (or arcs), sin {x + y) = sin a; cos ^ + cos x sin y. Dem.— Let AOB and BOC be the two angles represented respectively by x and y. Draw the measuring arc aP\ and the sines PD, aud P'E of the angles. AOC = AOB + BOC, is the sum of the two angles. Draw P'D', the sine of the sum of the two angles. Then PD = sin x, P'E = sin y, OD = cos a:, OE = cos y, P'D' = sin (.» + y), aud CD' = cos{x-\-y). Now sin(j: + y) = P'D' = EF + P'L. But from the similar ti'iangles EOF, EP _ sin x cos y 1 D' F D ^ A Fio. 11. ana POD, we have ;|^^ = ??, or OE OP the similar triangles P'EL and POD, we have p,^ EF = sin J cosy. Also, from P'L OD ... P'L C0S2 ODor P^ OP smy 1 .-. PL = cos a; sin y. Substituting these values of EF and P'L, we have sill (.? + y) = sin a; cosy + cos a; siu y. q. e. d.* * This demonstration may seem defective, eince the snm of the angles z and y, as represented in the diagram, is less than 90* ; nevertheless, in the General (Analytical) Geometry, we con- Ftautly proceed in a manner entirely analogous ; viz., first produce the equation of a locua from some particular fi^Tire, and then make it general in application. The demonstrations in cases in which the sum"of the angles is greater than 90=", etc., are similar, and some of them will be given in the Exercises at the close of the section. It is not thought best to cumber thtr purely theoretical part of the subject with such matter. FUNCTIONS OF THE SUM OR DIFFEEENCE OF ANGLES. 21 Cor. Si?i{90° + x)=cosx. Sin{lSO°+x) = -smx. Sin{270° i-x) = — cosx. Si7i{SQ0° -i- x) = sinx. Dem.— From the proposition we have, sin (90° + x) — sin 90° cos x 4- cos 90° sin x = cos x ; since sin 90° = 1, and cos 90° = 0. In like manner, sin (180° + a:) = sin 180° cos a; + cos 1 80° sin a; = — sin a? ; since sin 180° = 0, and cos 180° = — 1. [The student will readily produce the other forms.] 49. JProj^. — T/ie sine of the differencG iettueen two angles (or arcs) ts equal to the sine of the first into the cosine of the second, viinus the cosine of the first into the sine of the second. Thus, letting x and y represent the angles, sin [x — y) = sin x cos y — cos x sin y. De-m.— In sin(:z; + V) = sin a; cosy + cos.r siny, substituting —y for y, Ave have, sin {x — y) — sin x cos (— y) -^ cos x sin (— y) = sin ^ cos y — cos a; sin y ; since cos (— y) = cosy, and sin (— y) = — siny. (45, 4:4.) Cor. Sin(^0'' -x) = cos x. Sin(lSO°-x) = sin x. Sin{270°-x) = — cosx. Sin (360° — x) = — sinx. Dem. — This is simply an application of the proposition, as the preceding corollary is of its proposition. (The student will make it.) Or we may-deduce the results from the corollary under the preceding proposition by merely sub stituting — X for x. Thus, sin (90° — x) = cos (— x) = cos a; ; sin (180°— x) — — sin {— x) = — (— sin x) = sin x ; etc. 50. JProj), — TJie cosine of the sum of two angles (or arcs) is equal to the rectangle^ of their cosines, minus the rectangle of their sines. Thus, letting x and y represent the angles, cos (a; + y) = cos a; cos ?/ — sin a; sin ?/. Dem. — Taking the formula sin {x — y) = sin a; cosy — cos x siny, and substi- tuting 90°— X for X, we have, sin (90° — x —y)= sin (90° — x) cos y — cos (90' - X) sin y. Now 90° - a; - y = 90° - {x + y) ; and sin [90° - (a? + y)] = the sine of the complement of {x -f- y), or cos {x + y). Also sin (90° — a;) = cos ar and cos (90°— x) = sin x. .'. Substituting, cos {x -f- y) = cos a; cos y — sin x sin y. CoR. Cos(dO° + x)=-si7ix. Cos(lSO° + x)=-cosx. Cos('270°+z) = sin X. Cos(3G0'' + x) = cos x. Dem. — Apply the proposition. 22 PLANE TKIGONOMETRY. 51, JProp. — Tlie cosine of the dijference ofUvo angles {or arcs) is equal to the rectangle of their cosines, j^lns the rectangle of their sines. Thus, letting x and y represent the angles, cos {x — y) = cos X cos y + sin x sin y. Dem.— In cos {x+ y) = cos a; cos y — sin x sin y, substituting — y for y, wo have, cos {x — y) = cos x cos (— y) — sin a: sin (— y), or cos {x — y) ~z cos x cos » + sin a; sin y ; since cos (— y) = cos y, and sin (— y) = — sin y. [Nctice that the last term becomes — sin x (— sin y), which equals + sin x sin y.] Cor. a?5(90°-rc)=5i/ia:. Cos{lSO''-x)= - cosx. Cos(270°-x) = — si';i it*. Cos (360° — ic) = cos a:. Dem. — Apply tlie proposition. 32, ^rop. — Tlie tangent of the swn of ttvo angles {or arcs) it equal to the sum of their tangents, divided by 1 mimis the rectangh of their tangents. Thus, x and y being the angles, we have, tan X + tan ii tan {:c + y) = -^ : — ^ ^ ^^ 1 — tan.^ tun^ sin (x + v) sin x cos v + cos x sin y ,^. . ,. Dem Tan (x + v) = = ^—- — ■. ^- Dividmg numer- UEM. 1 an v^ -h y) ^^^ {x + y) cos xcosy- sin x sin y ator and denommator of the last fraction by cos x cos y* we have, tan {x + y)z= sin X sin y sin x sin y ?;;^ "^ ^ ^ ^^^ coflj/ ^ tana; + tan y, ^^^^ !!^ = tanar, etc., and sin a; sin y sin a; siny 1 — tan a; tan y cosa; cos X cos y cos x cos y sin X sin y _ sin x siny ^ j^ p, cos X cos y c^)s a; cos y' ' Cor. Trt?i(90°+rc)= -6-0/^2:. Tan{lSO° + x) = tanx. Ta7i{270°-hx) = - cotx. ra7i(360°+ x) = tan x. * The three following fiorms may readily be obtained by dividing reepectively by sin x sin y, cot ?/ + cot X ^ , ^ 1 + cot a; tan y , sin X cos y, and cos x sin y : viz., tan (x + y) = '^^^J^—'i: *^° fa? + y) = ^^^^_^^j^y > and tan (X + ?/) = ^^" ^ '^^ ^ — — . Any one of these may be reduced to the one given in the prop- cot y — tanx osiiion, bv substituting in it — for cot, and reducing. Notice that the form in the Prop, is ic tan terms oJ the tangents. Also observe tchy dividing by a certain term gives a particular form, ciid by which to divide to get a required form. FUNCTIONS OF THE SUM OR DIFFERENCE OF ANGLES. 23 Dbm. Tan (90° + x) = 'i^i^^51jl5i! = _5£i^ = _ cot a:. Tan (180^ + x) =.- ' cos (90 + «) — sinaj ^ ' sin (180° -V X) — sin a; ^ — r-r-- -L = . = tan X, etc., etc. cos (180 + a?) — cos aj 53, I*rop, — The tangent of the difference of two angles {or arcs) is equal to the difference of their tangents^ divided by 1 j^lfis the rect- angle of their tangents. Thus, x and y being the angles, , . tan X — tan ii tan (x — y) = : — - . ^ ''' 1 + tan x tan y rx m / . sin (o; — y) sin x cos y — cos x sin y Demonstkation. Tan (a; — j^) = \ ^ = "- ,- zJtM ^ cos [x — y) cos X cosy + sin x sin y Bin X cos y cos a; sin y sin a; _ sin y cos a; cosy cos a; cos v cos* cosy tana;— tan, y -1 . __^ — ; JL — — ^_. Q, Y^ D^ (gj^e COS. a; cos y sin .csiny . sin a; sin y 1 + tan a; tan y cos ar Cos y cos a; cos y cos x cos y foot-notes to preceding proposition.) This proposition is also readily deduced from tlie preceding by substituting in tlie formula tan {x + y)=z ^Q ^ + J^" V ^ 1 — tan X tan y — y for y, and remembering tliat tan {— y) = — tan y. Cor. Tan{^^°-x)= cotx. Tan{1^0°-x)= -tanx. ^«?^(270°-.^•) = co^.T. Trt?i (360°— x) = — tan x. DKM. Taa (90= - .) = '^^:^, = |^ = cot ... Tan (180= - ., ?i^-^=.^^ = -tau.. Tan(270»-^)=^i5^„;^=:^^2ii cos.(180 — a;) —cos a; cos (270 — a-) — sin.c m •o/.Ao X sin(3G0°— a;) — sin.i; cot a?. Tan (360°- x) = — )--^ { = = - tan x. ' cos (800*^— a)) cos a; * These and the kindred formula maybe produced by a direct application of the proposition. n,u ^ mn« ^ tan 90° + tan x tan 90" 1 Thus, tan (90o -h x) = --^^^^^^ = ^:^^^^^ = ZI^^ = " <^^' ^' . otherwise, tan (90° + a;) = ;^"»0°-^tanx ^ "■ ^K^^^ "" "^ = _1_ .j^^e a ^ 1 — tan 90° tana; 1 1 —tana:' ^ * — t^f:^ — tan a; tan a; tan 90° 00 finite divided by an infinite equals 0. Again, tan (180° + x) = i£!Li^±i^.^_ = 1^E£ .-, 1 — tan 180° tan x 1 tana;, since tan 180° = 0, etc., etc. 24 PLANE TRIGONOMETRY. o4z. Projy- — Tlie cotangent of the sum of two angles {or arcs) u equal to tlie rectangle of their cotangents mimis 1, divided ly their sum. Thus, X and y being the angles, , , . cot X cot ?/ — 1 cot (x + y) = ^ , . . ^ ^^ coty + cot a; COS.T cosy sinx siny cos(^ + ?/) _ co&r cosy — sin.r siny _ sin.c siny sin.r sin^ Sin (^ + y) sinxcosy 4.cos^ smy sin^cosy cos^smy siiLC siny sulc siny cos X cos y sin.r siny cota:coty— 1 _ i i -^ xi = = . Q. E. D. Or we may deduce it tlms, cosy cosi; coty + cot a; siny sin x 1- ' 1 _ 1 — tan X tan y cot x coty _ cot x cot y — 1 tan(i; + y) ~ tan ^ + tany "~ 1 _1 "" coty + cota; cot J coty Q. E. D. Cor. Cot{dO°-h x) = ^ tanx. Cot{lSO°+ x) =cotx. Cot{270°-\-x) = - tan X. Cot (360°+ x) = cot x. DEir. Divide cosine by sine, or take tlie reciprocals of the corresponding tauircnts Thus, cot (90° + «) = - — j^r^^ r = — = — tan a;, etc. ' ^ ' tan (90 + a) — cotx 5o, ^vop, — Tlie cotangent of the difference of ttuo angles (or arcs) is equal to the rectangle of their cotangents plus 1, divided by their difference. Thus, x and y being the angles, , . _ cot X cot y + 1 ^ ^^ ~ coty — cot a: Dem. Substitute — y for y, in the preceding formula ; or, divide cos [x — y) by sin (.c — y) and reduce ; or, take the reciprocal of tan {x — y), and substitute — for tan. cot Cor. Cot (90° - x) = tanx. Cot (180°~a:) = - cotx. Cot (270°- x) = tan X. Cot(dC)0°—x) = —cotx. Dem. Same as above. ScH. 1. The farmulm for the secant and cosecant of the sum and of the difference of two angles (or arcs) are not of sufficient importance to waiTant their introduction here ; some of them will be given in the exercises, as also the extension of those already given to the case of the sum of three or more angles, or arcs. FUNCTIONS OF THE SUM OR DIFFERENCE OF ANGLES. 25 Ben. 2. The results reached in this discussion are so important that we will rollect them into a (A) sin {x + y) (B) sin {x - y) -. (C) C(w(.t! + y) - (D) cos(.^ -y)- (E) tan {x + u): (F) tan (^-2^): (G) cot {x + y) (H) cot {X - y) TABLE : sin X cos y + cos x sin y. sin X cos y — cos x sin y. cos X cos y — smx sin y. cos- a; cosy + sln.c siny. tan X + tan y 1— tanaj ismy tana; — tan j/ 1 + tana; tan y* cot a; coty — 1 coty + cot a; cot a; cot?/ + 1 coty — cot.c (I) X 90°— X 90° +x ; 180°— a; 1 180° + x 270°— X 270° + X 360°- a; 360° +x tfhie cos a; cos a; sin a; — sin X — cos a; — cos a; — sin X sin a; uoeine sin a; — sin a; — cos a; — cos X — sin X sin X CO:? a; cos a; tangent cot a; — cot a; — tana; tanx cot a; — cot a; — tana; tana; cotangent tan X — tana; — cdt x cot X tan X — tan X — cot a; cot a; It will not be found difficult to memorize and extend set (I), if the student observes, that, when the number of whole quadrants is odd (as 90°, 270°, etc.), the function changes name (as from sin to cos, from cos to sin, etc.) ; but, when the number of whole quadrants is even, the function retains the same name. The sign of the sine and cosine is readily deteiTiiined according to fundamental principles by observing where the arc ends, assuming x < 90°. Thus 180°+ x ends in the third quadrant ; hence its sine (which in numerical value is sin a*) is — , and its cosine is also — . As the signs of the tangent and cotangent of the sOi^ne arc are alike, we have only to observe whether the sine and cosine, in any tjiven case, have like or unlike signs, in order to determine the sign of the tan- gent and cotangent. For example, what is cot (630° + a; ) equal to ? The num- ber of quadrants being odd (7), the function changes naj)ie, and since the arc ends in the fourth quadrant, its sine is — , and its cosine + ; therefore cot (630° + x) = — tan x. If in any given case x > 90°, determine the character of the function as above, on the hypothesis x < 90°, and then modify the result for the partic- ular value of X. Thus in the last case, if x was between 90° and 180", its tan- gent would be — , and for such a value cot (630° + x) = tan x. Or, we may consider at first where the arc ends, taking into consideration the given value of a;. 26 PLANE TRIGONOMETRY. (b) FUXCTIOXS OF DOUBLE AND HALF ANGLES. 30, Pr02>' — Letting x represent any angle {or arc), (K) siu 2x = 2sin x cos x ; (L) cos 2x = cos' a; — sin'' a: = ^^ 2cos" X — 1, or 1 — 2siu'' x ; (M) tim2x (N) cot2.'c 2 tan x 1 - tan' a cot' X — 1 2cot X Dem. These results are readily deduced from (A), (C), (E), aud (G) Thus, in siu {x + y) = sin x cos y + cos x sin y, if we make y —x^ we have sin 2x = Bin X cos a; + cos x siu x = 2siu x cos a;. (In like manner produce the others.) o7. J^rox), — Letting x rejjresent any angle (or arc), we have, (0) sinfr = ±Vi(l-cosa;); (P) cos \x = ± Vi (i + cos a;) ; (Q) tan i. = ± /l^^^^ 1 + COS a; (R) cot J a; •^^ cos a; cos a; Dem. From (L), 2sin"''a; = 1 - cos 2^-, or sin a; — ± /y/i (1 — cos 2a;). Puttiug ix for X, this becomes siu ia; = ± -y/i (1 — cos x). In like manner, from the same formula (L), 2cos^a; = 1 + cos 2x ; whence, cos \x = ± ^\ (1 + cos x\ Again, . „. sini.r /l — cosa; . . 1 , /l + cos x tan \x = = ±i/ ; and cot^a; = — = ± i/ . Q. E. D. cosic ' 1 + cosa; tan ^a; r i — cosa; ScH. The sign of the function in the case of each of these is + if a; < 180" ; but can only be determined b}' the value of a; in any given case. EXERCISES. 1. Prove from Fig. (a) that sin {x -f- y) = sin x cos y + cos x sin y, when X and y are each < 90°, but x + y > 00°. 2. Same as in £Jx. 1, from (b), when X < 90°, X -\- y> 90°, and < 180°, and y > 90° and < 180°. SuG. hi this case, sin (x -^ y) = P'D' = P'L - EF. Ill other respects the demonstration is identical with the preceding. This gives sin(a; + y) = cos a; sm y — sin a; cos y. But the F D Ci^A. (a). ^u^"CTIONS of the sum or difference of angles. 27 0). — sign is accounted for in the general formula, sin {x + y) = sin a; cos y + cos x sin y, by notic- ing that cosy is — , when y > 90° and < 180°, 3. Same as in the preceding, when X > 90°, y < 90°, and (x + y) < 180°. Sua. Here sin (.2; + 2/) = P'D' = EF - P'L In all other respects the demonstration is identical with the otlier cases. The — sign in this case arises from x being between 90' and 180", whence cos a; is — . [Note. A number of other cases may be devised, but the ri)ove illustrate the varieties.] 4. From i'V^. 11,Art. 48, demonstrate geometrically the formula cos {x ■{■ y) = cos X cos ?/ — sin a; sin y. The same for each of the cases in Ex's 1, 2, and 3, above. SuG. In Fig. 11, cos {x + y) = OD' = OF - LE. g| = ^^, or OP = cos x cos 2/- -^ = q^, or LE = sin x sin y, 5. Prove geometrically the relation sin [x —y) = sin X cos y — cos x sin y. SuG. Let aP =x, and since y is to be subtracted we measure it back from P, and y = PP'. Now sin {x — 2^) = P'D'=r EF - P'L. 0. Prove from Fig. (d) that cos {x — y) = cos X cos y + sinx sin y. DaA 7. Given sin 45°= Vi^ and sin 30°= J to find sin 75°, and sin 15" Also tan, and cot. Result^ sin 75° = .97, sin 15°= .2G, nearly. SuG. Use formula (A , . . . H). 8. Given sin 30°= J, to find sine, cosine, tangent, and cotangent, of 15°, 7° 30', 3° 45', and 1° 52' 30". Results, sin 15°= .2588, cos 15° = .97, nearly. SuG. Use the formulm in (57). Compare results with those fcnmd in tha Table of Natural Sines, etc. 9. Of what angles may the trigonometrical functions be found from sin 45°= jV2, by means of the fornuihe in (J7) ? How ? 28 PLANE TRIGONOMETRY. 10. Prove that sin (x -hy + z) = sin x cos y cos z + cos x sin y cos z + cos a; cos y sin 2 — sin x sin y sin 2;. Also., cos {x -\- y + z) = cos a; cos y cos 2; — sin x sin ycosz — sin a; cos ?/ sin 2; — cos x sin ?/ sin z. . , , , , ^ / . tan X + tan y + tan z — tan a: tan y tan 2 Also that tan (x-{- y +z)=: —^ —^ . ^ 1 — tan X tan y — tan x tan z — zany tan z SuG. Sin {x + y + z) = sin [{x + y) + z] = sin (.2; + 2/) cos z + cos (a; + y) sin z. ScH. Since if {x + y + z)= Tt^ tan (^ + y 4- 2) = 0, we have from the last form, tana; + tan y + tans = tan x tan y tan z; i. e., if a semicircumference be divided into any three parts, the sum of the tangents of the three parts equals the products of the tangents of the same. ^ ^ T^ . , , , V sec a: sec y cosec x cosec y 11. Prove that sec (x + y) = —, ^ ' cosec X cosec y — sec x sec y 12. Prove that sin dx = 3sin x — 4sin^ x. Also that cos 3x = 4:Cos^ X — 3cos X. Also, tan 3.?; = — :; ^z — '-^ — -» Also, cot 3a; = 1 — 3 tan a; cot' a; — 3cota; 3cot' a; - 1 SuGS. Sin 3^ = sin {2x + x) = sin 2x cos x + cos 2x sin x = 2sin x cos x cos a; + (1 — 2sin'^) sin x = 2sin x cos' a; + sin a; — 2sin' x = 2sin a; (1 — sin'' x) + sin aj — 2sin' X = Ssin x — 4sin'' x. tan '^ c + tan a^ Tan dx = tan (2.c + a;) = :; ^^^— . In the latter substitute for tan 2a; its 1 — tan 2a; tan x value iu terms of tan x. 13. Prove that sin 4a: = 4(sin x — 2sin''a;) cos a;. lA r> i-i r • 2tan4a; 2 14. Prove that sm x = T—^-r- = z — r-. 1 + tan ^x cot ^x + tan ^x sin hx „ „. r. • , , 2smir cos^a; 2tan i.r 2tan i.?; ^ SuG. Sm X = 2sm ^c cos ^v = f- = p- = — ^-^ = , — — -4-r . To sec i.c sec i-v sec' i a; 1 + tan' i^x produce the last form divide numerator and denominator of t-^^^ It- by 1 + tanHa; "^ 1 - T> i-i i. i. 1 1 — cos a; . - , ^ 1 4- cos a; lo. Prove that tan la; = — ■. . Also cot la; = : . sin a; " sm x SuG. From (L), (56*), 2sin'' ^x = l — cos x, and from (K), 2sin \x cos \x = sin x. Divide the fornior by the latter. FORMULiE ADAPTED TO LOGAEITHAIIC COMPUTATION. 2^ 16. Find the trigonometrical functions of 18°. Solution.— Letting x = 18°, 2x = 3G°, and 3x = 54°, hence sin 2x =■- cos Sx But sin 2x = 2sin x cos x ; and cos Zx = 4005" x — 3cos x ; hence 2sin a; cos ^ = 4cos' X — 3cos a;, or 2sin x = 4cos' x — 3 = 4t — 4siu' a; — 3. From which 4sin' x + 2sin x = l. Solving this quadratic, we have sin x, or sin 18° = -^^— j — neglecting the — root, since sin 18° is +. From this, cos 18° = \\/^^ + 2/y/5. These may be put in approximate decimal fractions. 17. Having given the functions of 18°, and 15° (Ex. 8), find those of 3° ; then of 6°, 12°, 24°, etc. Compare the results obtained with the values as given in the Table of Natural Sines, etc., obtaininG; all the values in decimal fractions. SUCTION III FORMULIE FOR RENDERING CALCULABLE BY LOGARITHMS THE ALGEBRAIC SUM OF TRIGONOMETRIC.iL FUNCTIONS. S8, Since multiplication, division, involution, and evolution are the only elementary combinations of number which we can effect by means of logarithms, if we l^isli to add or subtract trigonometrical (or other) quantities, we have first to discover what products, quotients, powers, or roots, are equivalent to the proposed sums or difierences. SO* IPvop, — To render sin x ± sin?/, and cos a; db cos y calculable by logarithms. Solution. From {55, Son. 2) we have sin (2; + y) = sina; cosy + cos a: siny, and sin {x — y) — sin x cos y — cos x sin y. Adding these formulas sin {x + y) + sin {x — y) = 2sin x cos y. Now putting x + y = x\ and x — y = y' • whence x = ^{x' + y'), and 2/ = i {x' —y') ; we have sin x' + sin y' = 2sin l{x' + y') cos^(a;' — y);-or, dropping the accents, as the results are general, (A') sin X + s'my = 2sin ^{x + y) cos \{x — y). Again, by subtracting formula B {55, Son. 2) from formula A, and makuig 'he same substitutions, we have, (B') sin X — siny = 2cos ^{x + y) sin i{x — y). 30 PLA^'E TRIGONOSrETET. In like manner adding cos {x + y) = cos a: cos y — sin x sin y, and cos {x — y) = cosx cosy + sinar siuy, and making the same substitutions, we have, (C) cos X + cosy = 2cos \{x + y) cos \{x — y). Finally subtracting formula C (55, ScH. 2) from formula D, and making the same substitutions, we have, (DO cos y — cosx = ' 2sin ^{x + y) sm \{x — y\ or cos x — co5y= — 2sin l{x + y) sin i{x — y). 00. Cor. l. — T7te su?yi of the sines of two angles is to tlieir differ- ence, as the tangent of one-half the sum of the angles is to the tangent of one-half their diff'erence. Dem. — Dividing A' by B', we have, sin X + sin y _ sin \{x + y) cos \{x — y) _ sin \{x + y) cos i{x — y) _ sin « — sin y ~ cos \{x + y) sin \{x — y)~ cos \{x + y) sin \{x — y)~ 1 tin ^(x -\- v) tanKx + y)coti(.r-y) = tani(^ + y) x __^_-_^ = -1-| ___ q. e. d. 61» Cor. 2. — TJie difference of the cosines of two angles divided by tlieir sum is numerically equal to the product of the tangent of one- lialf the Slim of the t2co angles into the tangent of one-half their difference. Dem. — Dividing D' by C, we have, cos a; — cosy _ — sin \{x + y) sjn ^x — y) _ _ sin \{x + y) sin-^(3; — y) cos a: + cos y ~ cos \{x + y) cos l{x — y) ~ cos \(x + y) cos \{x — y) ~ • tan \{x + y) tan \{x — y). q. e. d. (Observe the opi)osition in signs.) 02, JProb, — To render tan x db tan f/ calciilahle by logarithms. _ _ ^ sin.?; sin V sin t cos y ± cos a? sin y sin(a:±y) Dem.— Tan a; ± tan y = ± ^ = — — cos a: cosy cos a,- cosy cos a; cosy q. E. D. EXERCISES. Let the student deduce the following relations : -. i-. i. . ^ sin {x + ?/) 1. Cot X + coty = -T— 5^ — 7-^. ^ sin X sm y « o 2 cos i(.T + y) cos i(:c — y) 2. Sec a; + sec y = ^ ^ ^ -. ^ cos X COS y « o 2 sin4-(a: + ?/) sini(.'C — y) 3. Sec a; — secy = — — -- ^ COS a; cosy COXfeTRUCTION AND USE OF TIlIGONOMETRlCAL TABLES. 31 4. 1 + COS a; = 2cos'ia;. (See S7.) 5. 1 — cos a; = 2sin'Jrc. 6. £!L^-iHll^ = tani(a; + y). . (Divide A'byC^^.9.) cos re + cos?/ ^^ ^/ V J 7 / ^ sin a: — sin y , , , . 7. ; = tan i(3;- t/). cos a: + cosy ^v .// _ sin ic + sin ?/ , ^ . . 8. = — cot i(a; — y). cos a; — cos?/ *^ ^^ ^ sin a: — sin V ' , , , 9. ^ = — cot -Ka; + y). cos a;— oosy '^^ '" SECTION IV. CONSTRUCTION AND USE OF TRIGONOMETRICAL TABLES. [Note. — In order to read this and the subsequent sections, the student needs a knowledge of the nature of logarithms, and the method of using common logarithmic tables. If he is familiar with the last chapter in The Complete School Algebra of this series, he is j^repared to go on. If lie has not this knowledge, he should read the inti'oduction preceding the table of Logarithms before reading this section.] 63, A Table of Trigonometrical Functions is a table containing the valnes of these functions corresponding to angles of all different values. In consequence of the incommensurability of an arc aAd its functions, these results can be given only approxi- mately; yet it is possible to attain any degree of accuracy which practical science requires. 04:, There are two tables of trigonometrical functions in common use, the Table of Natural Functions, and the Table of Logarithmic Functions. 05, A Table of JS'atiiral Trigonometrical Functions is a table in which are written the values of these functions for angles of various values, the radius of the circle being taken as the measuring unit, and the function being expressed in natural num- bers extended to as many decimal places as the proposed degree of accuracy requires. 00, A Table of Logarith^nic Trigonometrical Func- tions is the same as a table of natural functions, except that the logarithms of the values of the functions are written instead of the functions themselves, and to avoid the frequent occurrence of nesja- 32 % PLANE TTJGONOMilTEY. tive characteristics, the characteristic of each logarithm is increased by 10. For example, sines and cosines being always less than unity, except at the limit (33), and tangents of angles less than 45° and cotangents of angles greater than 45° being also less than unity, the logarithms of all such functions have negative characteristics. To obviate the necessity of writing these with their sign, the charac- teristic of each logarithm is increased by 10. 67, JProb, — To compute a table of natural trigonometrical func- iionsfor every degree and minute of the quadrant. Solution. — It is evident that an arc is longer than its sine, but that this disparity diminishes as the arc grows less. Thus, in a circle whose radius is 1 inch, the length of the sine of an arc of 1° would not difer appreciably from the arc. Much less should we be able to distinguish between the sine of 1' and the arc. Now, since when the radius is 1, a semicircumference = 7t = 3.141592G, and also = 180', or ISO X GO = 10800', we have the length of an arc of V = "^ = 0.0002908882 approximately. Assuming this as the sine of 1', we lObOO obtain the cosuie thus. cos 1' =\/l - sin'^ r =\/(l + sinl') X (1-sinl') = v/l.0002908882 x .9997091118 = 0.9999999577. Having thus obtained sufficiently accurate values of sin 1' and cos 1', we can continue the operation as follows: from the formula sin (a: + y) + sin {x — y) = 2 sin X cos y, and cos {x + y) + cos {x — y) = 2 cos x cos y, we have sin (x + y) = 2 sin ar cos y — sin {x — y), ' cos (x + y) = 2 cos a; cos y — cos [x — y). Now letting y remain constantly equal to 1', and letting x take successively the values 1', 2', 3', etc., we have _ j sin 2' = 2 cos 1' sin 1' - sin 0' = 0.0005817764 For a: _ 1 , ^ ^^^ ^^ _ ^ ^^^ ^, ^^^ ^ _ ^^^ q, _ 0.9999998308 _ ^, j sin 3' = 2 cos 1' sin 2' - sin 1' = 0.000872664G For ar - ~ , ^ ^^^^ 3' = 2 cos 1' cos 2' - cos 1' = 0.999999G193 _ , I sin 4' = 2 cos 1' sin 3' — sm 2' = 0.001 1G35526 For ar _ 3 , ^ ^os 4' = 2 cos 1' cos 3' - cos 2' = 0.9999993232 _ , ( sin 5' = 2 cos 1' sin 4' - sin 3' = 0.0014544407 For X - 4', -^ cos 5' = 2 cos 1' cos 4' - cos 3' = 0.9999989425 etc., etc. These operations present no difficulties except the labor of performing the numerical operations. Of course GO operations are required for ever}' degree, and for 30°, 1800. But having computed the sines and cosines for every degree and minute up to 30° CONSTRUCTION AND USE OF TRIGONOMETRICAL TABLES. 33 we can complete the work by simple subtraction of values already fouml For example, letting x = 30°, the first formula used above becomes sin (30° + y) = cos y - sin (30° - y\ and from cos {x + y) — cos {x — y)= — 2siD x sin y, we have cos (30' + y) = cos (30° — y) — sin y. Now making y successively = 1', 2', 3', etc., these give j sin 30* V = cos 1' — sin 29° 50 ( cos 30° 1' = cos 29° 59' - sin 1' j sin 30° 2' = cos 2' - sin 29° 5S' i cos 30° 2' = cos 29° 58' - sin 2' j sin 30° 3' = cos 3' - sin 29° 57' ( cos 30° 3' = cos 29° 57' — sin 3' etc., etc. All of these values which occur in the second members having been deter- mined in reaching sin 30' and cos 30°, those in the first members can be found by performing the requisite subtractions. Proceeding in this way till we reach 45°, the numerical values o^ all sines and cosines become known, since the sine of any angle between 45° and 90°, being the cosine of the complementary angle, will have been computed in reaching 45°. And so also the cosines of angles between 45° and 90° will have been computed as sines of the complementary angles l^elow 45°. The sines and cosines being computed, the corresponding tangents, cotan- gents, and, if need be, the secants, cosecants, versed-sines, and coversed- sin 2 1. cos 7! sines, can be calculated from the relations tan x = , cot x = or -: — ~, cos X tan c sm x 1 1 . , . . seca; = , cosecx = - — , vers x = 1 — cosa*, and covers .c = 1 — sm .r. cos X sm j; 68, Sen. — If it is desired to obtain the natural functions of angles esti- mated to seconds, it is necessary that the values in the ta])les computed as above be extended to 7 decimals at least. From such a table we may make iiiteri)o- lations for seconds with suflScient accuracy for most practical ends, except for values near the limits, where the disparity between the variation of the aic and that of the function changes very rapidly. For example, let it be required to find sin 34° 24' 12 " from the data sin 34° 24' = .5G49G70, and sin 34° 25' = .5652070. We observe that an increase of 1' upon the angle of 34° 24' makes an increase of .5652070 — .5649670 = .0002400 in the sine. Hence an increase of 12", or i of 1', makes an increase of i of .0002400, or .0000480, approxi- mately. Adding, we have sin 34° 24' 12" = .5650150. The student must be careful to notice whether an increase of the angle makes a numerical increase or a de- crease of the function, and add or subtract as the case may require. 69. JProb, — To constrtict a table of logarWimic trigonometrical functions. Solution. — Compute the natural sines and cosines as in the preceding prob- 'em. Take the h^garithms of the values tlius obtained, and add 10 to each 3 34: PLANE TRIGONOMETRY. cLaractcristic. The results are the ordinary tabular logarithmic sines and co- sines. For example, we find from the table of^natural functions that sin 34° 24' r- .5049670. The logarithm of this number is 1.752023. Adding 10 to the char- acteristic, Tre have log sin 34' 25' = 9.752023, as usually'- given in the tables. In like manner the cosines are obtained. To obtain the tabular logarithmic tangents, we have from tan x — cos X log tan X = log sin x — log cos x. If we now take the log sin x from the table as computed by the preceding part of this solution, and from it subtract the cor- /espondlng log cos ar, the result is tlie true log tan x^ since the extra 10 in the tabular log sin and log cos is destroj^cd by the subtraction. Therefore, to thi« difference we must add 10 to get the tabular log tan, as above explained. For example, the tabular log sin 34° 24' = 9 752023, and log cos 34° 24' = 9.916514. Hence, the tiibular log tan 34° 24' = 9.752023 - 9.916514 + 10 = 9.835509. In like manner the tabular log cot x = log cos x — log sin x + 10. If the logarithmic secants are required they can be obtained from the relation sec x = , which gives log sec :c = — log cos x. In applying this by means cos.^ of tlie tabular functions, it must be observed that the log cos x, as we get it from the table, is 10 too great ; hence,- the tnte log sec a; = — log cos a; + 10. In tabulating log secants and cosecants, it is not necessary to add 10, since, as these functions are never less than 1, their logarithms are never negative. 70. Sen. — The interpolations for seconds are usually made in the same way when using the logarithmic functions, as explained above for the natural fimc- tions. But to facilitate the operation, the approximate change of the logarithm for a change of 1" of the angle is commonly written in the table, in a colmnn called Tabular Differences, and marked D. EXERCISES. 1. Find from the tables cat the close of the vohime the natural trigonometrical functions of 25° 18'. Solution. — To find (he sine and cosine w^ look in Table II., and find 25° at the top of the page. In the extreme left-hand column we find the minutes, and passing down to 18, find opposite, in the column headed N. sin (natural sine) 42736 ; also in the column N. cos, we find 90408. Now, as these are the lengths of the sine and cosine as compared with radius, we know they are fractions. .-. Sin 25° 18' = .42736, and cos 25° 18' = .90408. To find the tangenty^e turn to Table IV., and finding 25° at the top of the page, pass down the column of minutes, on the left-hand of the page, to 18, opposite which, and under the column headed 25°, we find 2698. To this we prefix the figures 47, which stand in the same column, opposite 11', and belong to the tan- gents of all tlie angles from 25° 10' to 25° 19', and are omitted in the table sim- ply to relieve the eye and to economize space. Thus we find tan 25° 18' = .472698, tha number being laiown to be a fraction because the angle is less tlian 45° CONSTRUCTION AND USE OF TRIGONOMETRICAL TABLES. oO Tojincl the cotangent we look at the bottom of the page in the same table till we find 25*, and tlien passing up the minutes column at the right hand, find c:ot25° 18' = 2.11552. If the aecant were required we should be obliged to obtain it by dividing 1 by the cosine, as our tables do not include this function. Thus sec 25° 18' = cos 25° 18' ^ ~Mm ^ ^••^^^^• [Note. — Tables of secants and cosecants are sometimes given, but they are not of sufficient importance to justify their introduction into an elementaiy text-book.] 2. Show that sin 37° 43' = .61176 ; cos 37° 43' = .79105 ; tan 37° 43' = .773353; cot 37° 43' = 1.29307; sec 37° 43'= 1.2G4142; cosec37° 43' = 1.634628 ; yers 37° 43' = .20895 ; covers 37° 43' = .38824. 3. Find that sin 64° 36' = .90334; cos 64° 36' = .42894; tan 64° 36' = 2.10600 ; cot 64° 36' = .474835 ; sec 64° 36' = 2.331328 ; coseo 64° 36' = 1.107003 ; vers 64° 36' = .57106; covers 64° 36' = .09666. SuG. — In looking for sines and cosines of angles above 45°, seek the degrees at the bottom of the page, and be careful to observe that the columns of sines and cosines, as named at the top, change names when read from the bottom. The foundation of this arrangement will be readily perceived. Thus, turning in Table II. to 24° 32', we find sin 24° 32':^ .41522. But sin 24° 32' = cos (90 - 24°V2') = cos 65° 28' = .41522. Thus the degrees and minutes read from the bottom of the page are the complements of those read from the top. 4. Find that sin 42° 27' 12" = .67499; cos 42° 27' 12" = .73783; tan 42° 27' 12" = .914834; cot 42° 27' 12" = 1.09309. SuG.— Sin 42° 27' = .67495, and sin 42° 28' = .67516. .'. An increase of 1' in the angle makes an increase of 21 (hundred-thousandths) in the sine, and 12" will make ^^ or \ as great an increase, approximately. Observe that in the case of cosine an increase of the arc makes a decrease of the function. 5. Find that sin 143° 24' = 0.596225; cos 151° 23' = .877844; tan 132° 36' = 1.08749; and cot 116° 7' = .490256. SuG.— Sin 143" 24' = sin (180° —143° 24') = sin 30° 36'. Also the trigone- metrical function of any angle is numerically equal to the same function of its iiupplement {56). 6. Find the logarithmic trigonometrical functions of 32° 15' 2; from the tables at the end of the volume. 36 PLANE TRIGONOMETRY. SOLUTION.— Turning to Table II. we find 32° at the top of the page, and opposite 15', and in the column L. sin (logarithmic sine), we get 9.727228 ; i. e., log sin 32° 15' = 9.727228. Kow from the column of differences, D. 1", we learu that an increase of 1" of the arc at this point makes, approximately, an increase of 3.34(raillionths) in the logarithm of its sine. Hence, we assume that an in- crease of 22" makes 22 x 3.34 = 73 (millionths). .-. log sin 32° 15' 22" = 9.727228 -f .000073 = 9.727301. In a sunilar manner we have log cos 32° 15' = 9.927231. An increase of 1" in the arc makes a decrease of 1.33 (millionths) in the log cos. .-. an increase of 22" makes 29 (millionths) decrease in the log cos, and log co? 82= 15' 22" = 9.927202. Log tan 32° 15' 22" = 9.800100 ; and log cot 32' 15' 22" = 10.199901. 7. Find rliat log sin 24° 27' 34" = 9.617051 ; log cos 26° 12' 20" = 9.952897 ; log tan 26° 12' 20" = 9.692125 ; log cot 126° 23' 50" = 9-8675;9. Sufe.— Observe cot (126^ 23' 50") = cot (180° - 126° 23' 50") = cot (53° 36' 10"). Also that angles above 45' are found at the bottom of the table ; and remember to subtract the correction for co-functions, if an increase of arc is assumed. 8. Given the natural sine .45621, to find the angle from the tables. Solution. — Looking for this sine in the table of natural sines, we find the next less sine to be .45606, and the angle corresponding, 27° 8'. Now, at this point, an increase of V in the arc makes an increase of 20 (hundred thousandths) in the natural sine. But the given sine .45621 is only 15 (hundred thou- sandths) greater than .45606, the sine of 27" 8'. Hence the required angle is but H of V or 60" = 35", greater than 27' 8'. .-. sin-».45621 = 27° 8' 35", and its supplement 152° 51' 25", which has the same sign, and these arcs in- creased by every multiple of 27r. 9. Find sin-\62583; cos-\34268; tan-M:68531; cot-\876434. Results. Sin--^62583 = 38° 44' 35", and 141° 15' 25"; and these arcs increased by every multiple of 2'7r. cos-\34268 = 69° 57' 36", and 360° - 69° 57' 36" = 290° 2' 24", and these arcs increased by every multiple of 27r. tan~'.468531 = 25° 6' 16", and 180° + 25° 6' 16" = 205° 06' 16", and these arcs increased by every multiple of 2;r. cot-\876434 = 48° 46' 3", and 180° + 48° 46' 3" = 228° 46' 03", and these arcs increased by every multiple of 27r. CONSTRUCTION AND USE OF TEIGONOMETRICAL TABUIS. 37 SuG. — Observe that an increase of the arc makes a decrease of its co-functions. In the table of tangents as given, Table IV., the proportional parts given at the bottom of each column are the approximate changes which the functions undergo for a change of 1" in the function. Thus, in finding cot-*.87G434, we find cot-* .876462 = 48° 46' ; and at the bottom we find that a change of 8.64 (million ths) in the function makes a change of 1" in the angle. Hence, as the given cotangent is 28 (millionths) less than the cotangent of 48° 46', the angle required is 28 -;- 8.64 = 3 (seconds), greater than 48° 46'. 71» Sen. — It is usually best lo take from the table that function which is nearest in value to the given function, and then increase or diminish the corresponding arc as the case may require. If wx always take from the table the next less function than that given in the example for sine, tangent, and secant, and the next greater for the cosine, cotangent, and cosecant, coiTections for seconds will require alwaj^s to be added. If we always take from the tables the func- tions next less than the one given, the corrections for seconds must be added for sine, tangent, and secant, and subtracted for the co-functions. If we were always to take from the tables the next greater function than the one given, the seconds corrections would be added for the co-functions, and subtracted for the others. [Note. — It is very important that the pupil become so familiar with the nature of these tables as to use them intelligently, and not mechanically. For this reason we refrain from giving the usual specific, mechanical directions for their use, and substitute illustrations showing how they are used in accordance with the principles upon which they are constructed.] 10. Find sin-^- .34256); cos-^- .62584); tan-^(-- 3.41621) ; cot-^(- 1.21648). Results. sm-X-.34256):=200°l'58",and339°58'02", and these arcs increased by every multiple of 2ir. cos-'(-.62584) = 128° 44' 38", and 231° 15' 22", and these arcs increased by every multiple of 2-^'. tan-H- 3.41621) = 106° 18' 57", and 286° 18' 57", and these arcs increased by every multiple of 2t. cot-^- 1.21648) = 140° 34' 42", and 320° 34' 42", and these arcs increased by every multiple of 2-^. 8uG. — To obtain these results the pupil will need to recall the principles m the corollaries to {4S—55). Thus, to find cot- \ - 1.21648), we find from the table that cot-'(1.21648) = 39° 25' 18"; and from (55) Cor., we learn that cot (180°- X) :- - cot X. .-. Cot-»(- 1.21648) = 180° - 39° 25' 18" = 140° 34' 42". Again, from the same corollary, we learn that cot (360° — x) = — cot x. \ Cot-'(- 1.21468) = 360° - 39° 25' 18" = 320° 34' 42". 11. Gi-^en the logarithmic sine 9.451234, to find the corresponding Angle. 38 PLAN'S TEIGONOMETRY. Solution. — The next nearest log sin found in Table II., is 9.451204 = log sin 16° 2o'. Now we learn from the table that an increase of 1" in the angle at this point, makes an increase in its log sin of 7.14 (millionths). But the given log sin, 9.451234, is 30 (millionths) greater than log sin 16° 25'. /. The required angle is 30 h- 7.14 = 4 (seconds) greater than 16° 25' ; and we have sin 16° 25' 4" = 0.451234. Again, as sin 16° 25' 4" = sin (180° - 16° 25' 4") = sin 163° 3^' 50 the latter angle has for its log sin 9.451234. Finally, either of these augles in, creased by any multiple of 2Tt has the same logarithmic sine. 12. Show from the table that the angle whose log cos is 9.778151, is 53° 7' 49", and also 306° 52' 11", and each of these angles increasec^ by any multiple of '2ir. 13. What angles correspond to the logarithmic cosines 9.246831. and 9.889372? 14. Find from the table what angles have for their logarithmic tangents 9.895760, 10.531054, and 11.216313. Results. The first two are the log tans of 38° 11' 20", and 73° 35' 43", and also of 180° + either of these angles, and each increased by any multiple of 2t, 15. Find the angles corresponding to the logarithmic cotangents 10.008688, 9.638336, and 9.436811. Results, The first two are the log cots of 44° 25' 37", and 66° 29' 54", and also of 180° + either of these angles, and each increased by any multiple of 2-^. 72, ScH.— Strictly speaking, negative numbers have no logarithms; since no base can be assumed, such that all negative numbci-s can be represented by said base affected with exponents. It is therefore customary to say that nega- tive numbers have no logarithms. Nevertheless, ue do apply logarithms to nega- tite trigonometriMl functions. Thus, if we have — cos r, the — sign is inter- preted as simply telling in what quadrants x may end; while, in other respects, the function is treated exactly like + cos x. 16. Given log (- cos a;) = 9.346251, to find x. Solution. — The logarithmic cosine 9.346261, considered independently of its sign, corresponds to 77" 10' 35". But the — sign requues that the arc shall end in the 2d or 3d quadrant, for such angles, and such only, have negative cosines. •. The angles required are 180° T 77° 10' 35" = 102° 49' 25". and 257° 10' 35", and these increased by entu-e circumferences, as all these angles have loga- rithmic cosines, which are numerically equal to 9.3462G1, and the cosines them- selves are negative. 17. What angle less than 180° has a negative cosine whose tabu lar logarithmic value is 9.653825 ? Ans. 116° 47' 4", CO^■STEUCTION AND USE OF TEIGONOMETRICAL TABLES. o9 18. What angle lebj than 180° has a negative tangent whose tabu lav logarithmic value is 9.884130 ? A?is. 142° 33' 15" 19. What angle less than 180° has a negative sine whose tabular logarithmic value is 9.341627 ? 20. What angle less than 180° has a negative cotangent whose tabular logarithmic value is 9.564299 ? A7is. 110° 8' 15". 21. Find values of ic < 180° which fulfil the following conditions : log (- cos x) = 9.562468 ; log (- tan x) = 10.764215 ; log (- sin x) = 8.886432; log (- cot x) = 11.152161. Eesults, 111° 25'; 99° 45' 54"; none; 175° 58' 14". 22. Having at hand only the common logarithmic tables of trig- onometrical functions, and the table of logari thins of numbers, I wish to find the number of degrees, minutes, and seconds corre- sponding to the natural tangent 2.16145. How is it done, and what is the result? Ansiuer: Find the logarithm of 2.16145, to this add 10, and fii^d the angle corresponding to this tabular logarithmic tangent. The angle is 65° 10' 20". 23. From the same tables as above find the natural cosine of 35° 23'. Also what angle corresponds to natural tangent 2. 24. From the same tables as above find the angle corresponding to natural tangent — 1.82645. Also to natural cosine — .42536. 25. Why is it in the table of logarithmic functions that the sine of an angle minus its cosine + 10 gives the tangent ? Why that cosine — the sine + 10 gives the cotangent ? Wliy that the sum of the tan- gent and cotangent of any angle = 20 ? Why is but one column of tabular differences needed for tangents and cotangents, while the sines and cosines require each a separate column ? (7] FUISCTIONS OF ANGLES NEAR THE LliMITS OF THE QUADUANT. TABLE III. [Note. — This subject may be omitted in an elementary^ course, the first time going over, if thought best.] 73, Failure of Table II. — The method which has been given in the preceding pages for finding the logarithmic functions of angles involving seconds, by means of the Tabular Differences, Table II., in Bufliciently accurate in most cases for practical purposes, but ia 40 PIANE TRIGONOMETRY. entirely too rude for the siues, tangents, and cotangents of angles near the beginning of the quadrant (those less than 2° or 3°), and for cosines, tangents, and cotangents of angles near the close of the quadrant (those between 87° or 88° and 90°). An example will render this clear. Suppose we wish to find log sin 1' 12". We find from Table II., log sin 1' = 6.463726 ; and also that the average increase of the log sin between 1' and 2' is 5017.17 (million tLs) for every second increase of the angle. But this average rate of increase ©f the function dui-ing the minute is much less than its real rat^ of increase ^;^ the first part of the minute, as from I'to 1'12", and much greater than the real rate of increase in the latter part., as the angle approaches 2'. In fact, we see from this table, that we should use 2934.85 as the increase of log sin 2' for 1" increase of the arc. Now, in our proposed example, we want the increase of the log sin while the angle is passing from 1' to 1' 12". This, as shown above, is con- siderably more than 5017.17 (millionths) for every second. TJie cosine being the sine of the complement is subject to the same law of change near the close of the quadrant, that governs the sine at the beginning. The case of the tangent of a small angle is similar to that of the sine ; and since the cotangent is the reciprocal of the tangent, it has the same laiu of change, only that the one increases as the other decreases. Thus, since doubling a small arc, as 1", doubles its tan- gent (approximately), it divides its cotangent by 2. Finally, while the law of change in the sine is very different neai the close of the quadrant from what it is near the beginning, the sine changing very rapidly at the beginning and very slowly at the close, and the cosine is just the opposite, the tangent, and cotangent have the same law of change at both extremities of the quadrant. Thus, if near the beginning of the quadrant a certain small increase of the arc increases the tangent at a particular rate, it decreases the cotangent at the same ra/e,- since these functions are reciprocals of each other. Moreover, since tan 1° = cot (90° — 1°) = cot 89°, cot 89° changes according to the same law as tanl°; and tan 89 changes reciprocally with cot 89°. 74. JDescrijytion of Table III, — The first page of the table enables us to find the sines of angles less than 2° 36' 15" (and con- sequently the cosines of angles between 87° 23' 45" and 90°) with Very great accuracy. The columns headed Angles contain the degrees, minutes, and seconds of the proposed angles, and the columns at their right give the same angles in seconds. The columns headed CONSTEUCTION AND USE OF TEIGONOMETRICAL TABLES. 41 Diff. contain tlie corrections to be used according to th6 following problems. The second and third pages answer a similar purpose with reference to tangents and cotangents of arcs within 2° 36' 20" of the limits of the quadrant. 75* ^rop. — Letting x represent any number of seconds less than 2° 36' 15", lue have, log sin x" = 4.685575 + logx — Diff. Dem.— The length of 1" of an arc to raduis unity is 3.14159265358979 (the length of the semicircumference) -f- 648000 (the number of seconds in 180"), and = .00000484812. FOr practical purposes this fraction may also be taken as the sine of 1". Though, actually, the sine is less than the arc, the expressions for arc 1" and sine 1" agree to as many places of decimals as we have here. Again, for these small arcs the sine increases at nearly the same rate as the arc, so that sin 3" = .00000484812 x 3 nearly; sin 102" = .00000484812 x 102 nearly; these results being slightly in excess of the true values. It is the correction for this excess that is furnished by Table III. in the columns marked Diff. But this table is adapted to logarithmic computation; hence we have log sin a;" = log sin 1" + loga; — Difif. In this expression log sin x" is the logarithm of the natu- ral sine of x" (not increased by 10, as each function in Table II. is) ; log sin 1" + logic, that is, the logarithmic sine of 1" plus the logarithm of the number of seconds, corresponds to multiplying the sin 1" by the number of seconds, and gives the Icgarithm of the product, or strictly, the logarithm of the length of the arc of re". Now, the sine o(x" being less than the arc, its logarithm is less than the logarithm of the length of the arc. Just how much less the table tells. This difference, therefore, between the logarithm of the arc and the logarithm of its sine, which is given in the table, is -to be subtracted. Finally, to make this result agree with Table II. we must add 10 to the result. Now, log sin 1" = log .00000484812 = 6.685575, and adding 10, we have 4685575. .-. log sin a;" = 4685575 + \ogx — Diff., a result which agrees with the logarithmic functions in Table II. q. e. d. 76. Cor. l.— To obtain the log cos of an angle between 87° 23' 45' •%nd 90'', from this table, take the log sin of its coinplement. 77. Cor. 2. — To obtain the log tan of an angle less than 2° 36' 20", from this table, use the formula, log tan x" = 4.685575 + logrc + Diff. Dem.— For as small an arc as 1", sine, arc, and tangent are practically equal ; hence, log tan 1" = log sin 1'' = 4685575 (10 being added). Moreover, for these small arcs the tangents increase (like the sines) in nearly the same ratio as the arcs;. hence, we add log a;. Finally, the tangent is a little in excesg jf the arc, which excess is given in the table, and is to be added, q. e. d. ^ PLANE TRIGONOMETRY. 78. Cor. 3.— To oUain the log cot of an angle less than 2° 36' 15', frojn this table, use the formula, log cot a;" = 15.314425 - logo; - Diff. DEMONSTRATioN.-Since cot X" = ^^^' log cot X" = log 1 - log tan x" = 20 - (4.685575 + log x + Diff.) = 15.314425 -\ogx- Diflf. The 20 arises from adding 10 twice to log 1 (= 0). One 10 is added because 4.685575 is 10 in excess of the true log tan 1" ; aud the other 10 is added in order to make the log cot x" agree with the ordmary tabulated value, as in Table II. Q. e. d. 79, Cor. 4.— To obtain the log tan of an angle between 87° 23' 45" and 90°, take the log cot of its C07npleme7it ; and to obtain the log coi of an angle between the same values, take the log tan of its complement. 80. JProb. — Having given a log sin less than 8.657397 {the log t>i7i 2° 36' 15''), to find the corresponding angle. Solution.— From log sin x" — 4.685575 + log x — Diflf., we have, log x = log sin 7^' — 4.685575 + Diff. Hence, if from the given log sin, we subtract 4.685575, and then add the proper correction as furnished by Table III., we have the logarithm of the number of seconds sought. But we cannot tell what Diff. to take till we know the number of seconds. To meet this difficulty, find the angle corresponding to the given log sin from Table II,, and reduce it to seconds. This will be sufficiently accui-ate to furnish the required Diff. 81, Cor. 1. — Having given a log cos less than 8.657397 {the log cos of 87° 23' 45"), to find the corresjjonding angle, treat it as if it ivere a log sin, and having found the corresponding angle, take its compile- rneJit. 82, Cor. 2. — For log tan aiid log cot, the formulce in {77, 78), give, logx = log tan x" - 4.685575 - Diffl, and,log:c = 15.314425 — log cot a;" — Diflf. These are applied as iu {80) ; tliat is, the Diflf. to be subtracted 13 found by getting from Table II. the required angle in seconds, a3 near as may be, and then take from Table III. the corresponding Diff. CONSTRUCTION AND USE OF TRIGONOMETRICAL TABLES. 43 EXAMPLES. 1. Fiud the log sin, tan, and cot of 1° 11' 15". Solution.— Log sin 1° 11' 15" = 4.685575 + log 4275 - .000031 = 8.316480. 1° 11' 15" = 4275". Since 4275" is between 4230" and 4300, the Diff. is 31 (millionths). Log tan 1° 11' 15" = 4.685575 + log 4275 + .000062 = 8.316573. Log cot 1° ir 15" = 15.314425 - log 4275 - .000062 = 11.683427. Or, log cot can be found by subtracting log tan fi*om 20. 2. Verify the following by using Table III. : log sin 56' 26" = 8.215242; log tan 56' 26" - 8.215301; logcot56' 26" =11.784699. 3. Verify the following by using Table III. : log cos 88° 17' 44" = 8.473396 ; log tan 88° 17' 44" = 11.526412 ; ■ loa cot 88° 17' 44" = 8.473588. 4. Having given the logarithmic sine 7.246481 to find the angle. Solution.— From Table XL we find 6' 5" — 3G5" as the angle. But this is subject to the inaccuracy exhibited in {73). To obtain the coiTect result from Table III., we have {80\ log X = 7.246481 - 4.685575 + = 2.560906. .-. X = 363.8; or the angle is 6' 3" .8. 5. Given the logarithmic tangent 7.805487, to find the correspond- ing angle. Solution.— Table IL gives 21' 58".3 = 1318".3 as the angle. From Table III. the DLff. corresponding to this is 6 (millionths). Hence, log X = log tan a;" - 4.685575 - Ditf. {82) becomes, log x = 7.805487 - 4.685575 - .000006 = 3.119906. .-. X = 1318 ; and the angle is 1318" = 21' 58". 6. Given the logarithmic cotangent 12.197148, to find the corre- sponding arc. Table II. gives the angle 21' 49".8, but the true angle as given by Table III. is 21' 50". M PLA^•E TRIGONOMETRY. SECTION V, TRIGOOMETRICAL SOLUTION OF PLAXE TRIANGLES. 83, There are six parts in every plane triangle : three sides and three angles ; one side and any other two of which being given, the remaining parts can be found by means of the relations which exist between the sides and tabulated trigonometrical functions. To exhibit these highly important practical operations is the object of this section. We shall treat first of right angled plane triangles, and then of oblique angled plane triangles. OF RIGHT ANGLED TRIANGLES. JProj}. 84, — Tlie relations between the sides and the trigoiimnet- rical functions of the oblique angles of a right angled triangle are as foUoios : side opposite ^ hypotenuse ' side adjacent ^ hypotenuse ' /ox |. f _ si de opposite ^ ^ '^ ' ° ~ side adjacent' (1) sine = (2) cosine = .*. cosecant = .*. secant = cotangent = hypotenuse side opposite ' hypotenuse side adjacent' side adjacent side oj^posite* FiQ. 12. «EM.— Let CAB, Fig. 12, be a tiiangle, right angled at A. Let aM be the mcasuriug arc of the angle B, PD = sin B, and BD = cos B. From the PD similar triangles PDB and CAB, we have gp = since BP = 1. CA BC I. e. . r> Side opposite sin B = i-* , hypotenuse From the same tiiangles BD BP BA BC I. «., cos B = side adjacent . Tangent being equal to sine divided by cosine, we nave tan B = hypotenuse side opposite ^ side adjacent _ 8ide opposite hypotenuse The hypotenuse hypotenuse side adjacent other functions being the reciprocals of these three, are as given in the proposi- tion. Finally, as a similar construction could be made about the other obhque angle, C, this demonstration may be considered general, q. k. d. SOLUTION OF EIGHT ANGLED PLANE TRL^NGLES. 45 ScH. 1. — These formula are so important that it is well to have them fixed in the memory, not oaly as written above, but also as follows: r,. , , . ^J . , side-adj (IV Side-opp =: hy x sm, or , or tan x side-adj, or ; COScC COL mN a- 1 T 1 ^iv ^ ., side-opp (2). Side-adj = hy x cos, or — ^, or cot x side-opp, or -^ ; -' -^ sec ^ tan ' of which the relations side-opp = hy x sin, and side-adj = hy x cos are of the most frequent use. ScH. 2. — The six ratios given in this proposition are frequently made the def- initions of the tiigonometrical functions. Thus, referring to a right angled tri- angle, a sine of an angle may be defined to be the ratio of the side opposite to the hypotenuse ; the cosine as the ratio of the side adjacent to the hypotenuse, etc. Sen. 3. — The student will be aided in remembering these important relations by observing that the side opposite the angle is analogous to the sine, and the ., ,. , -..-r , . its analogue side adjacent to the cosine. Now, the sine = r , and so also the co- sine. Tangent equals sine divided by cosine, and in this case it is the part analogous to the sine, divided by the part analogous to the cosine. One shouM hv not make the blunder of saying that sin = . -, ^ , since tliat would make side-opp tbe sine always more than 1 ; but we have seen that it never can exceed 1. Similar checks against error may be made in the case of the other relations. EXERCISES. [Note. — The first five of these exercises are mainly designed to illustrate the proposition, and familiarize the mind with the relations.] 1. In a right angled triangle whose sides are 3, 4, and 5, what are the trigonometrical functions of the angles ? What are the functions when the sides are 6, 8, and 10 ? 2. In a right angled triangle the hypotenuse is 12, and the angl^ at the base sin~^ J. What are the sides ? What is the sine of th/» other angle ? Suo.— Represent the angles by B, A, and C, A being the right angle ; and the sides opposite by 5, a, and c. Then sin B = -, or ^ = — ; whence, h = ^ Sin C = cos B = ^y^ c = 6 yT 3. In a right angled triangle whose hypotenuse is 12, and the angle at the base tan~^2, what are the other parts ? A71S. Cos-^f\/5; ^Vo7and^^/5, 46 PLANE TRIGONOMETEY. 4. The sides of a right angled triangle are .20 antl 32. What are the angles and hypotenuse? Obtain the hypotenuse by means of the secant. _ Ans. Tan-4,tan-^f, and WS9. 5. The hypotenuse of a right angled triangle is 120, and one side 100. Show that the angle opposite the latter is tau~^^jVTi, the adjacent angle cosec~^3^A/llj aiid the remaining side 20\/ll. Obtain these results in the order given. Use a trigonometrical function to obtain the last. EXAMPLES. (a) BY MEANS OF THE TABLE OF I^ATURAL PUXCTIO:S"S. 1. In a right angled triangle ABC, the hypotenuse BC is 235, and the angle B is 43° 25'. Find the angle C, and the sides AB and AC. C = 46°"35'; AB = 170.7; AC = 161.52. SoLunox — To find C, we have but to remember that the angles of a right angled triangle are complements of each other ; whence, C = 90" — B = 46' 35'. AR To find AB, we have cos B = ^' or AB = 235 x cos 43° 25'. Now, from the table of natural functions we find cos 43° 25' = .72637; whence AB = 235 x .72637 -= 170.7. To find AC, we have sin B = || ; whence AC = 235 x .6873 = 161.52. 2. In a right angled triangle ABC, the hypotenuse AC is 94.6, and the angle c is 56° 30'. Find the angle A, and the sides AB and BC A =33° 30'; AB =78.88 ; BC = 52.21. 3. In a right angled triangle BDF, the hypotenuse BF is 127.9, and the angle B is 40° 10' 30". Find tlie angle F, and the sides BD and DF. F = 49° 49' 30"; BD = 97.72; DF = 82.51. 4. In the triangle CDE, right angled at E, given the side DE 75, the Bide CE 50.59, to find the other parts. Hyi^otenuse = 90.47. 5. In the right angled triangle CDE, given the hypotenuse CD 264, the side CE 135.97, to find the other parts. DE = 226.28. 6. Given the hypotenuse 435, and one of the acute angles 44°, to find the other parts. 7. Given the hypotenuse 64, and the base 51.778, to find the other parts. SOLUTION OF KIGHT ANGLED PLANE TRL^NGLES. 47 8. GiTeu the hypotenuse 749 feet, and the base 548.255 feet, to find the other parts. 9. Given the hypotenuse 125.7 yards, and one of the acute angles 75° 12 23", to find the other parts. 10. Given one side 388.875, and the adjacent angle 27° 38' 50", to find the other parts of a right angled triangle. (b) BY MEAXS OF A TABLE OF LOGARITHMIC FUN"CTIOj;rS. 11. In a right angled triangle, given an oblique angle 54° 27' 39", find the side opposite 56.293, to find the other parts. Solution.— The other oblique angle is 90° - 54° 27' 39" = 35° 32' 21". To find (lie hypotenuse, we have sin 54° 27' 39" ——j- — , or liy = hy ' ' sino4°27'89" Applying logarithms to facilitate computation, log hy = log 56.293 — log sin 54° 27' 39" + 10. The 10 is added since the log sin 54° 27' 39" found in the t^iilar displace neut of the moon due to these different points of observation is horizontal pariillax. 50 PLANE TRIGONOMETEY. 86, Pro2)»—TIie sum of any two sides of aj^^ctne triangle is to iJieir difference, as the tangent of half the sum of the angles opposite is to the tangent of half their difference. Dext. — Letting a and h represent any two sides of a plane triangle, and A and B the angles opposite, we have a : 6 : : sin A : sin B. Taking this both by composition and division, we have a + b : a — b : : sin A + sin B : sin A — sin B. But from (60), sin A + sin B : sin A — sin B : : tan i(A + B) : tan ^A — B). .-. a + J : a — 6 : : tan ^A + B) : tan \{^ — B). q. e, d. 87* JProp. — Tlie tangent of half of any angle of a plane triaiigle equals Ic, divided hy half the perimeter of the triangle minus the side opposite the angle ; in which h is the radius of the inscribed circle, and equals the square root of the contiiiued product of half the peri- meter minus each side separately, divided hy half the perimeter. DEif. — Let ABC be any plane ti'iangle. Represent the angles by A, B, and C, the sides opposite by a, b, and c, the perimeter by Py and the segments of the sides made by the radii of the inscribed circle, by a*, y, and z, as in the figure. Then a + b -\- c = %x + 2y + 2z = p,OT u + y + 2 r= 4^. Whence x = \p — a, y =lp —b, and z Fro. 15. = iP —c ; since y + z = a,x + z = b, and x + y = c. k k Now from the triangles AOD, DOB, and COE, tan ^^=-=-, , tan X ip — a ^B = ^ = -A_ and tan ^0 = - = ^^—. y ip-b z ip-c To find k. ^A + iB + iC =r 90°, or iA = 90° - QB + ^C) ; whence, tan i^ = I = tan [90- - (iB + ^C)] = cot «B + iC) = j"'^"^^^ (54, Dem). Substituting for tan ^B, and tan ^C, -, and -, we have - = ^• y s X k k' whence ** = ^ ^^ -r •Ax-\-y + z)k' = xyz ; and k = |/--^^_. In this value of A:, sub- stituting for a-, y, and z their values, we have k = j/^^^ ~ ^^ ^^^ ~ ^^ ^^^ ~ ^\ SOLUTION OF OBLIQUE ANGLED PLANE TRIANGLES. 51 88, ScH. 1. — These three propositions {85, 80, 87), furnish the most elegant and expeditious means for finding the unknown parts of an oblique angled plane triangle, when a sufficient number of parts are given or known (^.5) 89, Sen. 2.— Important Practical Suggestions, 1st. Two angles of a triangle being given, the third is known by implication, It being the supplement of the sum of the other two. 2nd. When two of the known, or given, parts are opposite each other, the first proposition {85) effects the solution. 3rd. When two sides and the included angle are given, the solution is effected by means of the second proposition {80). 4th. When the three sides are given, the angles are found by the third proposition {87). EXERCISES. 1. In the plane triangle CDE, given the angle D = 15° 19' 51", C = 72° 44' 05'', and the side c, opposite C, 250.4, to find the other parts. ^ d ^^^^^^ Solution.— i^rs^, E = 180°- (D + C) = 91° 56' 04" L ^^-- {89, 1st). ^ Secondy To find side d opposite angle D {89^ 2nd), sin C : sin D : : c : p and DB there is but one triangle which fulfills the conditions, viz., the one with an acute angle at F'. SOLUTION OF OBLIQUE ANGLED PLANE TRIANGLES. 53 The results in the above examples are, for triangle DF'B, angle DFB .— 47" 16' 28", BDF' = 92° 43' 33", and side BF' = 543.89; for triangle DFB, angle DFB = 132° 43' 32", angle BDF = 7° IG' 28", and side BF = 68.94 90, Cor. — Li applying trigonometrical formulce to the solution of triangles, if the part sought is found in terms of its sin"e, the result is amhiguouSf and we are to determine zuhether there really are two solutions to the prollem in a geometrical sense, ty certain geometrical considerations, or else hy tryijig loth values for the angle determined ly its sine. This ambiguity arises only tohen an angle is determined hy its sine, as will appear hereafter. 6. Given two sides of a plane triangle 201 and 140, and the angle opposite the latter 36° 44'. Find the other parts. Results. — There are two triangles. Parts of the first, 120° 49' 49", 22° 26' 11", and 89.34; Parts of the second, 59° 10' 11", 84° 5' 49", and 232.84. 7. Given two sides of a plane triangle 180, 100, and the angle op- posite the former 127° 33', to find the other parts. There is but one triangle, and the parts are 26° 7' 59", 2G° 19' 1", and 100.65. 8. Given two sides of a plane triangle 30.8 and 54.12, and the angle opposite the latter 36° 42' 11", to find the other parts. Why but one triangle ? 9. Given two sides of a plane triangle GOO and 250, and the angle opposite the latter 42° 12'. Find the other parts. SuG. — Attempting to get the angle opposite 600, we find log sin = 10.207400, which is impossible. It is in some such way that a trigonometrical solution shows a geometrical absurdity. 10. Given two sides of a plane triangle 1337.5 and 493, and the angle opposite the former 69° 46'. Find the other parts. 11. In a plane triangle, given two sides 1686 and 960, and the in- cluded angle 128° 04', to find the other parts. c Solution. -Let a — 1686, h = 960, and C = 128° 04'. {See 89, 3rd.) The sum of the angles A and B is 180° - 128° 04' = 51° 56', and i{^ + B) = 25° 58'. From (86) we have, a + b : a —b :: tan J,(A + B) : tan KA — B), or 2646 : 726 : : tan 25° 58' : tan i{A - B). 54: PLANE TRIGONOMETRY. Making the computations by logarithms, we find log tan KA — B) = 9.1258S7 Hence, ^(A — B) = 7° 36' 40", the angle found in the table, or its supplement But ^ the difference of two angles of a ti'iangle is less than 90° ; consequently KA - B) = 7= 36' 40". Now having i(,A + B) = 25° 58', and i(A - B) = 7° 36' 40", we find A = 33° 34' 40", and B = 18° 21' 20". The side c can be found by (So) as two opposite parts are now known. c = 2400. 12. In a plane triangle ABC, the side AB is 304, BC 280.3, and the included angle B is 100°. Find the angles A and C, and the side AC. A = 38° 3' 3" ; C = 41° 56' 57" ; AC = 447.856. 13. In a plane triangle ABC, the side AB is 103, AC 126, and the included angle A is 56° 30'. Find the angles B and c, and the side BC. B = 72° 20' 15" ; C = 51° 9' 45" : BC = 110.267. 14. In a plane triangle ABC, given the three sides, a = 3459, h 4209, and c = 6030.4, to find the angles. Solution. — Applying (57), we have, y ip ' log^ = i{log (ip-a) + log {ip -b) + log (^ - c) - log ip}. k k k Also, tan -iA = , tan |B = -, and tan ^C = , , or log tan ^A = log ^ — log {ip — a), log tim ^ B = log A; — log (^ — b), and log tan ^C - log k - log {ip - c). COMPUTATION. a = 3459 b= 4209 c= 6030.4 p = 13698.4 ip = 6849.2 (ar. comp.) log = 6.164360 ip- a= 3390.2 log = 3.530226 Ip- b= 2640.2 log = 3.421637 ip - c = 818.8 log = 2.913178 2) 6.029401 log k = 3.014700 log tan ^A = log A; - log {^p - a) = 9.484474 .'. A = 33" 56' 10".5 log tan iB = log A;- log (i;?- 6)= 9.593063 .'. B = 42M7' 25'.3 log tan iC = log A; - log {ip - c) = 10.101522 .-. C = 103° 16' 24".2 Proof, A + B + C = 180* 00' 00" SOLUTION OF OBLIQUE ANGLED PLANE TRIANGLES. 55 15. In a plane triangle ABC, the side AB is 95.6, BC is 275, and AC 300. Find the angles A, B, and C. A= 65° 47' 55"; B = 95*» 42' 52"; C = 18° 29' 13". 16. In a plane triangle BDF, the side BD is 500, DF is 403.7, and BF 395.75. rind the angles B, D, and F. B = 52° 0' 3"; D = 50° 34' 45"; F = 77° 25' 12". OBLIQUE ANGLED TRIANGLES SOLVED BY MEANS OF RIGHT ANGLED TRIANGLES. [Note. — Articles 91-94 inclusive, may be omitted in an elementary course, if thought desirable. Or 91 and its applications may be taken instead of 56*- 88, 85 should be included in any course. It is too elementary and important to be omitted.] 91. I*rop, — All cases of oUique angled plane triangles may he solved ly the solution of right angled triangles. Dem.— Of the three given parts we may affirm that they are, 1st, All adjacent; 2nd, Two adjacent and one separated; or 3rd, All separated. 1st. When tlie given parts are all adjacent ; i. d., when they are tico sides and the included anrjle, or two angles and the in- cluded side. To solve the first let fall a perpen- dicular from the extremity of one of the given sides upon the other given side, or upon that side produced. There will thus be formed two right angled triangles which can be computed, and from the parts of which the parts of tlie required triangle can be found. Thus, let A be the given angle, and b and c the given sides. In the right angled triangle ADC there are given A and 5, whence AD, P, and angle ACD, can be computed. Then passing to triangle CDB, we know P, and DB (since we have c given and have computed AD). Hence, we can compute B, Fia. 18. rt, and DCB. Thus, the parts of ACB become known .... When the given parts are two angles and the included side, find the third angle by taking the supplement of the two given. Let fall a perpendicular from one extremity of the given side upon the opposite side. The two right angled triangles thus formed can then be computed. Thus, if A, h, and C are given, having found B, let fall CD. The triangle ACD has the angle A and side b known, whence its parts can be computed. Having computed P we can pass to the triangle CDB, ^nd knowing P and B, can compute it. Thus the parts of ACB be '.ome known. 56 PLANE TRIGONOMETRY. 2nd. W/ien tico of the given parts are adjacent and one separated; i. e , when two angles and a side opposite one are given ; or two sides and an angle oppo- site one are given. The first of these cases is viitually the same as the last given. To solve the other, let fall a perpendicular from the angle between the given sides, and two right angled triangles will be formed which can readily be computed. Thus a, b, and A being given, and CD let fall from C, the triangle ACD can first be computed, and then CDB. This is the ambiguous case, but it is easily determined. Having computed P, if the given side a is less than P there is no solution ; if = to P, on^ solution (a right angled triangle) ; if a > P and < h, there are two solutions, i. e., it will go in between CD and AC, and also beyond CD ; if a > P and also > h there is only one solution, as it will not go in between CD and AC. 3rd. When the three given parts are all separated from each oilier. This is the case in which the three sides are given to find the angles. It is readily solved by letting fall a perpendicular from the angle opposite the greatest side, upon that side, as CD upon AB. Then compute the segments AD (which call m), and DB {n), from the following relation (Part II, Ex. 12, page 162) : m + 71 (or c) : h + a : : b — a:m — n. Knowing m and n, the angles of the two right angled triangles ACD and CDB can be computed, and these make known the angles of ACB. q. e d [Note.— A few additional examples are here given which the pupil can use to illustrate the theory presented in (91). If more are needed the preceding can be used : tliese may also be used to apply the methods before given. Again, a very great variety and number of examples may be made from these by as- signing different parts as known.] EXERCISES. 1. In a plane triangle BDF, the side BF is 123.75, DF 500, and the included angle F 120°. Find the angles B and D, and the side BD. B = 49° 12' 4"; D = 10° 47' 56"; BD = 572.006. 2. In a plane triangle ABC, the angle A is 70° 21', the angle B 54° 22', and the side BC 125. Find the angle C, and the sides AB and AC. C = 55° 17' ; AB = 109.1 ; AC = 107.88. 3. In a plane triangle ABC, the side AB is 98, the side BC 95.12, and the angle C 33° 21'. Find the angles A and B, and the side AC A = 32° 14' 55"; B = 114° 24' 5"; AC = 162.33. 4. In a plane triangle DAC, given AD = 450, AC = 309, and D = 27° 50', to find the other parts. C = 137° 9' 36", or 42° 50' 24"; A = 15° 0'24", or 109° 19' 36"; OC ^ 171.36, 01 624.5. SOLUTION OF OBLIQUE ANGLED PLANE TRIANGLES. COMPUTATION. It Tvill give definiteness to the stu- dent's thought, if he first sketch the figure geometrically. Thus, lay off D = 27° 50', and taking AD = 450, let fall the perpendicular AP. 1st To compute p. p=c sin D = 450 sin 27' 50', log 450 = 2.653213 log sin 27° 50' = 9.669225 log p = 2.322438. .-. p = 210.106, 57 Fig. 19. Knowing p, we see by inspection that AC can lie in both the positions AU 'ind AC, and hence that there are two solutions. 2nd. To compute 0^ from the triangle ACP, in which d and p are now snown. _P 210.106 Sm C - ^ - 3Qt^ log 210.106 = 2.322438 log 309 = 2.489958 log sin C = 9.832480. .-. C'= 42° 50' 24", and C = 137° 9' 33". 3rd. To find tlie angle A. DAC = 180° - (D + ACD) = 180° - 164° 59' 36" = 15° 0' 24". DAC = 180° - (D + ACD) = 180° - 70° 40' 24" = 109° 19' 36". 4th. To find DC. Compute DP and CP from the triangles APD and APC. DP - CP = DC, and DP + CP = DC. 5. In a plane triangle ABC, the side AB is 460, BC is 340, and AC 280. Find the angles A, B, and C. A = 47° 23' 16"; B = 37° 18' 31"; C = 95° IS' 13". 6. The sides of a plane triangle are 40, 34, and 25 feet respect- ively ; required the angles. 38° 25' ;20", 57° 41' 24", 83' 53' 16". 7. The sides of a plane triangle are 390, 350, and 270 feet respect- *vely; required the angles. 42° 22' 06", 60° 52' 33", and 76° 45' 21". 8. Given two sides of a plane triangle 450 and 540, and the in- cluded angle 80°, to find the remaining parts. Angles, 56° 11', 43° 49'; and the side, 640.08. 9. Given two sides of a plane triangle 76 and 109, and the in- cluded angle 101° 30', to find the remaining parts. Angles, 30° 57' 30", 47° 32' 30"; and the side, 144.8. 58 PIAXE TRIGONOMETRY. FOXTI0>S OF THE A>GLES OF A TRIA>GLE IN TERMS OF THE SIDES. 9fl, Tro]}. — A)uj side of a plane triangle equals the swn of the products of each of the other sides into the cosine of the angle lohich it makes with the first side, Dem.— In the first figure AB = AD + DB. But AD = 6 cos A, and DB = a cos B. /. c = & cos A + a cos B. In the second figure AB = AD — DB. But AD = 6 cos A, and DB = a cosCBD = a (— cos CBA)= — a cos B. .-. c = b cos A — (—a cos B) = b cos A + a cos B. In like manner, we have a = b cos C + c cos- B, and b = a cos C + c cos A. Collectmg and ai-ranging. (1) a =b cosC + c cos B ; (2) b = acosC + c cos A ; (3) c =a cos B + b cos A. q. e. d. Tie. 20. 93, Cor. — Tlie square of any side of a plane triangle equals the sum of the squares of the other two, minus twice their rectangle into the cosiiie of their included angle. Dem. — From (3) {92), we have by transposmg and squaiing, a/* cos' B = c' + 6" cos' A — 2Z>c cos A ; and from {85) a" sin' B = 6" sm' A. Adding, a} z=z c^ + 6' — 2Z>c cos A. In like manner, 6' =a^ -\- c^ — 2ac cos B ; and c* = a' + 6' — 2aZ> cos C. Q. e. d. 94, ScH. — These formula affor^l another means for finding the angles of a plane triangle when the sides are given. Thus, (li) cosA = (2i) cos B = (3.)cosC = 2ab J' + c'- ■a' 2bc a' -t-c'- -&« 2ac a' + 6»- c» These foi-muloB give directly the natural cosines of the angles in terms of the sides. To adapt them to logarithmic computation, we transform tliem as follows : t> + c« - rt» Subtracting each member of (li) from unit}', 1 — cos A = 1 — g«-y-c'4- 2hc _ a''-{h-c)^ _ {a^{b-c)] {a-{h-c)] _{a 2be ~ 2bc ~ 2bc ~ 2bc b— c){a + e—b) 26c~ SOLUTION OP OBLIQUE ANGLED PLANE TKIANGLES. 59 But 1 — cos A = 2siiiHA (57, O) ; and Icltini; j> = a + 6 + c, ^(a + 6 — c) = Ip _ Cj and i{a + C'-b) = ip — b. Whence, substituting, 2sin*iA = 2bc (U) sin^A = i/^JL-L^^^ZLhl. la like manner, (20siniB = |/ii^^^E«);and V CIC In a manner altogether similar, by adding each member of (1') t(i unity, and reducing, we get (Is) cos iA = a/ ^^ (^ - ^) ; and from (2,) and (3.), (23) cosiB = |/Si^HS; r ac Dividing (I2) by (I3), (2^) by (2,), and (3^) by (3,), we have. ^^^) -^^=/^^^^ (3J taniC = i/^p|ii^>. ^ ii?(ii>-c) EXERCISES. [Note.— In order to render X\iQ%e, formulcB familiar, and to give the student exercise in applying /<97'??i?i^, a few examples are appended. If necessary, any which precede can be used.] 1. The sides of a plane triangle being 40, 34, and 25, find the angles. Solution. — By natural functions. Let the sides be represented by a, 6, and c in order, and the angles opposite by A, B, and C ; then Cos A = '^^^r^-^ ^1136+623-1600 ^ _jog,,_ _. '^ ^ 33. 53. ^^..^ 2bc 1700 60 PLANE TRIGONOMETRY. There is no ambiguity in this case, since the cosine is +, and hence the angle is < 90'. The same angle is found by logarithmic computation, thus ; a = 40 ... log = 1.602060 6 = 34 ... log = 1.531479 c = 25 . . .log = 1.397940 p = 99 ... log = 1.995635 ^ = 49.5 ... log =: 1.694605 ip-a= 9.5 ... log = 0.977724 ^p -b = 15.5 . . . log = 1.190332 ^p-c = 24.5 ... log = 1.389166 log {hP -c) = 1.389166 \ogiip- 6) = 1.190332 a. c. log b = 8.468521 a. c. log c = 8 .602060 2)1.650079 T.825039 log sin |A = 9.825039. .-. iA = 41° 56' 38" and A = 83° 53' 16". In like manner the other angles may be found. 2. The sides of a plane triangle being 6, 5, and 4, find the angles. The angles are 82° 49' 09", 55° 46' 16", and 41° 24' 35". 3. The sides of a plane triangle being 8601.5, 4082, and 7068, find the angles. The angles are 54° 35' 12", 28° 4' 44", and 97° 20' 4". 4. The sides of a plane triangle being .51238G4, .3538971, and .3090507, find the angles. The angle opposite the last side is 36° 18' 10".2. AREA OF PLANE TRLiXGLES. Oo» JPro2^' — The area of a plane triangle is equal to half the product of any two sides into the sine of the included angle, Dem.— Let ABC be any plane triangle, and b and c any two sides with A as the included angle. From the extremity of one of these two sides remote from A, let fall a perpendicular p, upon the othor side Now, Area ACB = ^pc. Fig. 21. But, from ACD, ^ = 6 sin A. ibc sin A- q. e. d. Area ACB 06. Cor. — Tlic area of a plane triangle is equal to the square root of the continued product of half its periruter into half its perimeter minus each side separately. AREA OF PLAME TRIANGLES. Gl Dem. — From the proposition, and since sin A = 2sin |A cos ^A, we have, ^l^ea = ibc sin A = be smiAcosi A = be ^^^P^^tZ:^^ x |/iME«I^ EXERCISES. 1. Given two sides of a plane triangle 125.81 and 57.05, and tlie included angle 57° 25'. Find the area. Area = 3055.7. 2. Given the sides of a plane triangle 103.5 and 90, and the- included angle 100°, to find the area. Area = 45SG.74. 3. How many square yards are there in a triangle whose sides are 30, 40, and 50 feet? AreazzQG^. 4. Find the area of a triangle whose sides are 20, 30, and 40. Area = 290.4737. 5. What is the area of a triangle whose sides are 30 and 40, and their included angle 28° 57' ? Area = 290.427. 6. What is the number of square yards in a triangle, of which the sides are 25 feet and 21.25 feet, and their included angle 45° ? Area - 20.8694. 7. Find the area of a triangle in which two of the angles are 80° and 60° respectively, and the included side 32 feet. Area = 679.33 square feet. 8. Find the area of a triangular field having one of its sides 45 poles in length, and the two adjacent angles, respectively, 70° and 69° 40'. Area = 1378.411 square poles. 9. Find the area of a triangular piece of ground having two angles respectively 73° 10' and 90° 50', and the side opposite the latter 75.3 poles. Area = 748.03 square poles. PRACTICAL APPLICATIONS. [Note. — The following problems are inserted, not as any part of a treatise npon the subject of trigonometry as pure science, but as affording the student rood mental exercise, and valuable and interesting information.] 62 PLAKE TKIGONOMETRY. 1. To find the length (in miles) of a degree of longitude at Ann Arbor, Mich. Solution. — Let NESQ be a meridian section of the earth, EQ the equatorial diameter, and EL the lati- tude of Ann Arbor, 42' 16' 48".3. A degree of longi tude at L is 3^0 of the circumference of the circle whose radius is LD. CL the radius of the earth at tliis point = 3957.* Now in the right angled ti-iangle LCD, we have CLD = ECL = 42° 16' 48".^3, and CL = 3957; whence, LD = CL x cos 42° 16' 4S".3, and LD = 2927.6. .*. A degree = 51.1 miles. As a degree in longitude makes 4 minutes di£Ference in time, 51.1 miles east or west on this parallel is equivalent to 4 minutes difference in time. Query.— How does it appear from the above solution that the length of a degree of longitude varies as the cosine of the latitude ? 2. To find the distance of a planet from the earth at any par- ticular time. Solution — To render the problem as simple as possible, we will suppose two observatories on the same meridian, at N,and N'; and that when the planet P is on the same meridian, the angles ZNP, and Z'N'P (the zenith distances) are measured. With these data and the radius of the earth, CN, CN', known, the problem comes quite within the scope of the present study. The process is as follows: The arc NN' being known, the angle NCN' is known. Then in the triangle NCN', two sides and tlie included angle are known, whence the other parts can be found. Now, knowing the angles PNC, PN'C, and CNN', CN'N, we can find the angles PNN', PN'N. This affords suflScient parts of the triangle PNN' to determine the triangle, and we find PN, or PN'. Finally, in the triangle PNC, we know PN, NC, and the included angles whence the other parts can be computed. But PC is the distance sought Fig. 23. 3. Suppose in case of the moon, the angles PNZ, and PN'Z', being measured, are found tp be respectively 44° 54' 21", and 48° 42' 57", the distance between the points of observation N and N' is 92° 14', and the radius of the earth is 8956.2 miles; find the distance to the moon. Distance = 237,954.098 miles. • The equatorial radinp of the earth is 3962.8 miles ; but in consequence of the flattening in the direction of the polar diameter it is less here. / PRACTICAL PEOBLEMS. 63 4. Required the height of a hill D above a horizontal plane AB, the distance between A and B being equal to 975 yards, and the angles of elevation at a and B being respectively 15° 36' and 27° 29'. DC = 587.61 yards 5. Find the area of a regular hexagon, and also of a regular octagon, whose sides are each 10 feet. Areas, 259.8, and 482.84 square feet 6. Find the area of a regular pentagon, and also of a regular dec- agon, whose sides are each 12 feet. Areas, 247.74, and 1107.96 square feet. 7 Wishing to know the length of a certain pond of water, 1 measured a line 100 yards in length, and at each of its extremities observed the angles subtended by the other extremity and a couple of trees at the extremities of the pond. These angles were, at one end of the line, 32° and 98°, and at the other, 37° and 118°; what was the length of the pond ? Draw the horizontal line AB equal to 100; make the angle BAD 32% BAC 98°, ABC 37°, and ABD 118°. The intersections of the lines AC and BC, AD and BD, determine the extremi- ties of the pond; the straight line CD is the length of the pond. CD = 161.868 yards. 8. The distances AB, AC, and BC, between the points A, B, and c, are known ; viz., AB = 800 yds., AC = 600 yds., and BC = 400 yds. From a fourth point P, the angles APC and BPC are measured; viz., APC = 33° 45', and BPC = 22° 30'. ■Requir'id the distances AP, BP, and CP. r AP = 710.193 yds. Distances,^ BP = 934.291 yds. t CP = 1042.522 yds. 64 PLANE TRIGONOMETRY. Sug's. — Conceive tlie circumference passed through A, B, and P, and AD and DB drawn. In the triangle ADB, angle DAB = the given angle DPB, and DBA — APD. Ilence, all the parts of triangle ADB can be found. Again, since the sides of the tiiangle ACB are given, its angles can be found. Then, since angle CAB — DAB = CAD, there are two sides and the included angle known in triangle ACD ; whence angle ACD can be found. Thus we reach the ti'iangle ACP , in which there are now known AC and the angles. 9.. From the top of a mountain, three miles high, the angle of depression of a line tangent to the earth's surface is taken, and found to be 2° 13' 27". What is the diameter of the earth, considered as a sphere ? Arts. 7946.28 miles. 10. Taking the sun's mean apparent diameter as 32' 3".4, and his distance from the earth 91,430,000 miles, show that, if his centre were coincident with the earth's, his body would extend in all direc- tions nearly 200,000 miles beyond the moon. (See Ex. 3.) Sun's diameter = 852,574 miles. 11. Assuming the height of the Great Pyramid to be 486 feet, how far off may it be seen across the desert ? Ans., 27 miles. 12. "WTiat was the perpendicular height of a balloon, when its angles of elevation were 35° and 64°, as taken by two observers on the same level, at the same time, both on the same side of it, and in the same vertical plane ; the distance between the two observers being 880 yards ? Ans.y 935.757 yards. 13. Given two sides of a parallelogram 60 and 80, and a diagonal 100. Is this the longer or shorter diagonal ? What is the other ? What are the angles of the parallelogram ? 14. A balloon being directly over one of two towns standing on uhe same horizontal plane, at a distance of eight miles from each ether, the angle of depression to the more remote town was observed by the aeronaut to be 10°. What was the height of the balloon ? Ans., 1.41 miles. 15. The most recent observations make the sun's horizontal par- allax 8 ''.94, and the earth's equatorial radius 3962.8 miles. Show that the distance of the sun from the earth is nearly as given in Ex. 10, instead of 95 millions of miles, as it has been heretofore con- sidered. CHAPTER n. SrHEBICAL TBIGONOMETBY^ INTR OD UCTION. PROJECTION OF SPHERICAL TRIANGLES. 97. To Project a Spherical Triangle on a plane surface is to draw the triangle on that surface so that it will present the same appearance to the eye, situated at a particular point, as when drawn on the surface of a sphere. 98. The Simplest Method of projecting a spherical triangle is to project it on the j^lane of one of its sides, the eye being supposed situated in the axis of the sphere perpendicular to this plane, and at an infinite distance from it. The plane is called the Plane of Pro- jection ; and its intersection with the sphere is called the Primitive Circle, and is the base of the hemisphere on which the triangle is conceived as situated. 99. Fundamental Propositions. — 1st. Wientlie ixirts of a spherical triangle are each conceived as less than 180°, any such triangle can he represented on a hemisphere, 2d. The primitive Circle has its axis, and consequently its pole, projected at its centre. 3d. The semi-circumference of any circle of the sphere, perpendic- ular to the Primitive Circle, is p^rojected in the chord representing the intersection of the circles ; and, if the perpendicular circle he a great circle, its semi-circumference is projected in a diameter of the Primitive Circle. 6G SrHERICAL TRIGONOMETRY. Ill's. — These propositions are direct cousequences of the fundamental con- ception.^ Thus, let ABA'B' represent the base of the hemisphere on which the triangle is conceived as situated. This is the Primitive Circle, and the e3'^e is supposed situated at an infinite distance, and in a line perpendicular to the plane of the paper at P. The pole of the Primitive Circle being in this line is projected (seen as) at P. As all great circles perpendicular to the Primitive Circle pass through its pole and include its axis, the eye is in all such planes, and any lines of these planes, as the semi-circumfer- ences of the great circles in which they intersect the sphere, are projected (appear to the eye) as diameters of the Primitive Circle. Moreover, since the eye is at injinily, it is to be conceived as in the plane of any small circle which is perpendicular to the primitive, and which is therefore projected in a chord, as CC. PROJECTION OF RIGHT ANGLED SPHERICAL TRIANGLES. 100. JProb, 1. — To project a right angled spherical triangle on the plane of one of its sides, luhen tlic tiuo sides about the right angle are given.* Solution. — Let the angles of the tnangle be represented hy A, B, and C, A Let the sides opposite these angles respectively be rep- resented by a, h, and c ; whence b and c are the given sides. Draw the primitive circle and the diameters BB', NN' at right angles to each other. From B lay off BA = c.\ Let the right angle be at A; whence the side b is • perpendicular to the primitive circle, and projected in the diameter A A'. To project the vertex C, conceive the semi-circum- ference, of which A A' is the projection, to revolve on AA' until it falls upon the semi-circumference A'BA, then will the point C f\\ll at d. Hence make kd = b. In like manner revolving the semi-cir- cumference, of which AA' is the projection, until it falls upon A'B'A, the point C will fall at d'. Hence make M' = b. The point will de- scribe the semi-circumference of a small circle perpendicular to the primitive circle, and whose projection is dd'. Now, as the Fig. 25. * In this treatise the discussions embrace only such triangles as have each part less than 180°. t For the purposes for which we shall use these projections, an arc can be laid olT with sufficient accuracy by observing its relation to 90°, 60°, 30°, or some aliquot part of the circum- ference, which is readily obtained on geometrical principles. PROJECTION OF EIGHT ANGLED SPHEEICAL TRIANGLES. 67 projection of the vertex of the triangle C is at the same time in AA' and cld\ it must be at their intersection. Finally, the hypotenuse a is projected in a curve passing through B, C, and B', since two great circles intersect at the extremities of a diameter. This curve is really an ellipse, but for our present purpose it may be considered as the arc of a circle passing through B, C, and B'. Tiiere- fore, BAG is the projection required. Queries. — Will a solution of this problem be possible for all values of J and c ? How does it appear from the projection ? EXAMPLES. 1. Having given h = 110°, and c = 60°, to project the triangle. See Fig. 26. 2. Having given h = 50°, and c = 130°, to project the triangle. 3. Having given I? = 90°, and c = 30°, to \ project the triangle. 4. Haying given b = 90°, and c = 90°, to project the triangle. 101. JProb. 2, — To ^^J'oject a right angled splicricaJ iriangh tolien the liypotemise and one side ar'e given. Solution. — Using the common notation, let c represent the known side. Drawing the primitive circle and the conjugate* diameters BB', NN', layoff BA = c, and draw the diameter AA'. . The projec- tion of b will lie in AA', and the projection of the vertex C will fall somewhere in this line. Now the arc a lies in a semi-circumference passing through B and B'. Conceive this semi-circumfer- ence to revolve on BB', as an axis, till it coincides, first, with.BNB', and then with BN'Bl The point G will trace the semi-circumference of a small circle perpendicular to the primitive circle, and whose projection is dd'. Hence, making Bd = Bd' = a, and drawing dd\ the projection of the vertex C lies at the same time in AA' and dd\ and is there- fore at their intersection. Passing an arc of a circle (strictlj^ an ellipse) through BCB', we have ABC, the projection desired. Fig. 27. * Two diakieters of a circle which are at right angles to each other arc called ConjugaU DiameUrs. 68 SPHERICAL TRIGONOMETRY. Queries. — Will a solution of this problem be possible for all values of a aud e ? If a had been greater than c in the above case, would the solution have been possible ? Will dd' and AA' always intersect, whatever may be the relative values of a and c ? EXAMPLES. 1. Having given c = 75°, and a = 64°, to project the triangle. 2. Having given c = 45°, and a = 136°, to project the triangle. 3. Having given b = 110°, and a = 85°, to project the triangle. See Fig. 28. 4. Having given b = 110°, and a = 120°, to project the triangle. ■ 5. Having given c = 90°, and a = 75°, or a = 120®, to project the triangle. Fig. 28. Query. — If one side of a right angled spherical tiiangle is 90°, what must the hypotenuse be ? Why? 102, JProb, 3. — To 2^J^oject a right angled spherical triangle when an ohlique angle and the hypotenuse, or the oblique angle and the adjacent side are given. Solution. — Draw the primitive circle and the conjugate diameters BB', NN' as usual. To construct the given angle B, we observe that this angle is measured by an arc of the great circle which is projected in NN'. Hence, lay oflf Hd = N^' = B , and draw rfcf ; then is NO the projection of the arc which measm'es B, and the projection of a lies in tlie arc passing through BOB'.* Having the angle B projected, if the hypotenuse a is the other given part, find the projection of C by taking Be = Be' = a, and drawing ee'. Com- plete the projection by drawing AC through P. When the adjacent side c is given, take BA = c, and draw AC as before. [The student will be able to give the reasons.] 103, ScH.— As OP is the cosine of the angle ABC, the point may be found by measuring * A? has been remarked, this arc is really a semi-ellipse. This fact, together with the method of constructing the semi-ellipse, and thus getting the correct projection of the hypot PROJECTION OF EIGHT ANGLED SPHERICAL TRIANGLES. 69 from P (towards N if B < 90% and towards N' if B > 90°) a distance equal to the natural cosine of B. Query. — Is the solution of this problem possible for all values of the hj'pot- enuse or adjacent side, and the angle? EXAMPLES. 1. Having given B = 65°, and the hypotenuse a = 120°, to proiect the triangle. See Fiff. 30. 2. Having given C = 45°, and the adjacent side b = 50°, to project the triangle. Sug's. — Project the angle C as the angle B of the preceding, and lay off 5 = 50° from its vertex on the circumference of the primitive circle. 3. Having given C = 170°, and hypote- nuse a = 160°, to project the triangle. 4. Having given B = 150°, and c = 40°, to project the triangle. 104, JProb. 4, — To 2^^oject a right angled spherical triangle when an ohlique angle and side opposite are giveii. SoLUTiox. — Project the given angle at B, Figs. 31, 32, as in the last problem. Then, from an}' point in the circumference of the primitive circle, as N, take NO', in the diameter passing through that point, equal to the projection of the given side. (This is done by taking Hd = Nd' = b, as in Prob. 1, and drawing del'). Now, with P as a centre, and radius PO', describe a cir- cumference cutting BOB'. One extremity of the "given side h will be projected in this circumference, since this circumference contains the projections of all the points in the surface of the hemisphere which are at a distance h from the circumference BNB'N'. But the vertex C is also projected in the cnuse, belong to a treatise on Conic Sections. In this case, BB' and 20 P are the axes of the ellipse, and the curve can be constructed by takin 90% must b be greater or less tlian B in order to have two solu- tions ? If B < 90% how is it ? If B > 90°, can b be less ? If B < 90% can o > 90=" ? EXAMPLES. 1. Given C = 120°, c = 150, to project the triangle. See Fig. 33. 2. Given c = 80°, c = 60°, to project the triangle. See Fiff. 3-4. Fig. 33. 3. Given B = 70^ 4. Given B = 6-4°, Fig. 34. T0°, to project the triangle. See Fig. 35. 75°, to project the triangle. See Fig. 36. A \ Fig. 35. Fig. 36. 5. Given b = 160°, b = 110°, to project the triangle. PROJECTION OF OBLIQUE ANGLED SPHERICAL TRIANGLES. 71 105, Sen. — When the given parts are the two oblique angles, the projection is most readily effected by first computing one of the sides. The projection in this case will be considered in connection with the numerical solution of the case, in the next section. PROJECTION OF OBLIQUE ANGLED SPHERICAL TRIANGLES. lOG. Proh, 1, — To project a sjylierical trimigle when two sides Project the given angle and the included angle are given. Solution. — Let a, e, and B denote the given parts. B at some point in the circumference of the primi- tive circle, as B. Lay off one of the given sides, as c, from B on BNB'. Let BA = c. Determine the extremity C of the projection of the other given side a, as in Prob. 2, etc., and drawing the diameter AA', pass tlie arc through ACA' ; BCA is the pro- jection sought. Query. — Is this projection always possible, what- ever the relative magnitude of the given parts ? EXAMPLES. 1. Given A = 130°, c = 85°, and b = 100°, to project the triangle. 2. Given c = 40°, a = 37°, and I? = 80°, to project the triangle. 107. JProb. 2, — To project a spherical triangle tv7ie7i tivo sides and an angle opposite one of tliem are given. Solution. — Let the given parts be a, 5, and B. Make the projection upon the plane of tlie iinknowii side c. Thus, drawing the primitive circle and the conjugate diameters BB', NN', conceive c as projected from B on the arc BNB', and project the given angle B as in preceding problems. On the arc BB', take BC = the pro- "jection of the given adjacent side a. To deter- mine the projection of the opposite side 5, describe a circle about P, as a centre, with a radius PC. Through C draw PD, and taking Dd = Dd' = b draw dd\ Tlirough the intersections o, o' of dd! with the circumference of the small circle, draw the radii PA, PA'. Finally, passing arcs througli the points A, C, A", and A', C, A'", BAC, and BA'C are the projections of triangles which fulfill the given conditions. The projections of B an(\ 72 SPHERICAL TRIGONOMETRY. a were made upon principles pre\iously established ; and it only remains to show that AC, and A'C are projections of b. Since by construction DL is the projection of an arc equal to b, the projections of arcs of great circles connect- ing D with and o' are projections of arc^ equal to b. But the figure OoP — ACP, and Do'P = A'CP ; therefore AC = A'C = Do = the projection of b. 108. ScH.— It is evident that this problem has one solution, two solutions, or no solution, according to the value of 6 as compared with a and B. Thus, if the projection of 6 = DC, o and o' coincide, there is but one solution, and the tri- angle is right angled at A (which in this case falls at D). Also, if the projection of b is intermediate in value between DC andon/^ one of the arcs BC, B'C, there is onXjone solution. If, however, as in the figure given, the projection of b is intermediate in value between DC and both BC and B'C, there are tico solutions. Finally, if 6 is given of such value that the chord M does not touch the small circle, there is no solution. The latter case occurs when B < 90% if the projec- tion of Z> < DC ; and when B > 90% if the projection of J > DC, as will appear from Figa. 38, 39. We may observe, also, that there are tico solutions when o and o' both fall on the same side of BB as c ; one solution when o and o' coincide, and when they fall on opposite sides of BB' ; and no solution when o and o' are imaginary, i. c, when dd' does not touch the small circle, or when both fall on the opposite side of BB' from c. EXAMPLES. Fis. 39. 1. Given B = 110°, a = 120°, and 5 = 83°, to project the triangle. See Fig. 39. 2. Given B = 110°, a = 120°, and ^> = 130°, to project the triangle. 3. Given C = 64°, a = 120°, and c = 75°, to project the triangle. 4. Given C = 80°, b = 60°, and c = 40°, to project the triangle. 5. Given = 112°, b = 75°, and c = 150°, to project the triangle. 109. I^rob. 3.— To project a spherical triangle when the three sides arc given. SOLUTION OF BIGHT ANGLED SPHERICAL TRIANGLES. 73 Solution.— Drawing the primitive circle and the conjugate diameters, as nsual, take BA = c, the side on the plane of which it is proposed to project the triangle. Take Be = Be' = a and draw ee' ; then as before shown, the projection of the vertex C lies in ee'. In like manner taking ^d — Ad' = b, and drawing dd' the projection of C lies in dd[. The intersection of the chords ee' and dd' is there- fore the projection of the vertex C. Finally, passing arcs through AC A' and BCB', the projec- tion is complete. Queries.— How does the projection show the impossibility when the sum of the three sides is greater than 3G0° ? How when the sum of two sides is less than the third side ? Fiu. 40. EXAMPLES. 1. Given a = 100°, b = 80°, and c = 68°, to project the triangle. 2. Given a = 108°, b = 120°, and c = 25°, to project the triangle. See Fig. 41. 3. Given a = 120°, I? = 65°, and c = 40°, to project the triangle. 4. Given a = 150°, b = 140°, and c = 170°, to project the triangle. Fio. 41. SECTION I. SOLUTION OF RIGHT ANGLED SPHERICAL TRIANGLES. 110. Splierical Trigononietry treats of the relations be- tween the trigonometrical functions of the sides and angles of spherical triangles, and of the solution of such triangles by means of these relations. 111. ScH.— In the present treatise we shall confine our attention to triangles none of whose sides or angles exceed 180°. 112. The Sioc Parts of a spherical triangle are the three sides and the three angles : any three of these being given, the others may be found. 74 SPHERICAL TRIGONOMETRY. 113, SCH.— The last statement involves the assertion that the three angles of a spherical triangle determine the sides, whereas we are accustomed to say in Plane Trigonometry, that, at least one given part of a plane triangle must be a side, in order that the triangle may he determined. Tliere is really no such difference as these two statements imply. For example, if we have the angles of a plane triangle given, we know the ratios of the sides to each other, since the sides are to each other as the sines of the angles opposite ; but we cannot determine the absolute mines of the sides. This is in accordance with the state- ment that mutually equiangular plane triangles are similar figures (not neces- sarily equal). Now these are exactly the facts in the case of spherical triangles, if ue do not limit them to the same or equal spheres. Thus, the angles of a spheri- cal triangle being given as 4S' 30', 125° 20', and 62° 54', we solve tlie triangle by the rules of spherical trigonometry and find that the sides are 56^ 39' 30", 114° 29' 58", and 83° 12' 6". But, so long as the radius of the sphere is unknown, these results are merely ikQj:£ltUi£^ values of the sides, not their absolute lengths. Moreover, consider two concentric spheres whose radii are m and n. Now, any triangle being constructed on the one, if planes are passed through its sides intei*secting at the common centre, their intersection with the surface of the other sphere will form a triangle mutually equiangular with the first, and any one side of the one triangle is to the corresponding side of the other, as the radii of the spheres ; hence the homologous sides are proportional. We see, therefore, that to determine absolutelj^ a spherical triangle, it is necessary to know one of the sides in linear extent as well as angular measure, or, what is equivalent, the radius of the sphere must be known. 114, TJie Species of a part of a spherical triangle is deter- mined by its relation to 90°. Two parts, both greater or both less than 90°, are of the same S2)erAes ; two parts, one of which is greater and the other less than 90°, are of different sjjecies. Thns, two parts which are 58°, and 63°, respectively, are of the same species ; two which are 160°, and 115°, are of the same species ; two which are 120°, and 48°, are of different species. 1 IS, Napier^ s Five Ci7*ciilar JParts are ^jwo sides of a rij/ht anfjled spherical triangle about the right an^le, and the comjile- ments of the h}i3otenuse and of the oblique angles. These terms are merely conventional, and are applied exclusively to right angled triangles. III.— In a spherical triangle ABC, right angled at A, the sides h and c, the complement of hypotenuse «, and the com- plements of the angles B and C, are the circular parts. We may designate them 5, c, comp a, comp B, and comp C. [It is essential that this nomenclature and the statements of the two following paragraphs be clearly understood, and firmly fixed in the mind, in order that the phraseology of ^iG 42. ^^^ fundamental rules may be intelligible.] SOLUTION OF EIGHT ANGLED SPHERICAL TRIANGLES. 75 116* When five things occur in succession, as it were in a circle, like the circular parts b, c, comp B, comp a, and comp c, in the preceding figure (no account being made of A), it will be observed, that, taking any three at pleasure, one of the three may always be selected which lies adjacent to each of the others, or separated from each of them. Of the three parts thus considered, the 3Iiddle JPart is the one which has the other two adjacent to or separated from it; while the latter are called the Uxtrenies, adjacent or 02)2^osite, as the case may be. Ill's. — Let the three parts under consideration be comp a, b, and c; comp a is tlie middle part, and b and c are the o2yposite extremes. If b, c, and comp C are under consideration, b is the middle part, and c and comp C are the adjacent extremes. 117» Sen. — In solving a right angled spherical triangle, there are always three parts under consideration at once, viz., the two given parts and the part sought, no mention being made of the right angle. Now, tlie first question to be settled in oi-der to a solution is, WJdcJi of the three parts under consideration is the middle part^ and are tlie extremes opposite or adjacent? Beginners are very liable to make mistakes by failing to use the complements of the proper parts ; or by not correctly distinguishing the middle part, and the character of the extremes, as opposite or adjacent. The student should practise upon such simple exer- cises as the following until the questions can be answered instantly^ and with- out mistake. EXERCISES. 1. In Fig. 42, given a and c, to find C. What are the circular parts under consideration ? Which is the middle part ? Are the extremes adjacent or opposite ? Answers. — The circular parts are comp a, c, and comp C. c is the middle part, and the extremes are opposite. 2. Having C, and a given, to find h. What are the circular parts ? Which is the middle part ? Are the extremes adjacent or opposite ? 3. Having c, and h given, to find B. What are the circular parts ? Which is the middle part ? Are the extremes adjacent or opposite ? 4. Having a, and 1) given, to find C. What are the circular parts ? Which is the middle part ? Are the extremes adjacent or opposite? 0. Having B and C given, to find I. What are the circular parts ? Which is the middle part ? Are the extremes adjacent or opposite ? 76 SPHERICAL TRIGONOMETRY. 6. What are the opposite extremes when h is the middle part? What the adjacent extremes ? Which are the opposite and which the adjacent extremes when c is ^the middle part"^ When comp B is the middle part ? When comp c is the middle part ? When comp a is the middle part ? 7. What part is middle part to comp C and c as adjacent ex- tremes? As opposite extremes? Ans., l, and none, 8. In Fig. 43, M being the right angle, what are the circular parts ? Given O and m, to find 0. What are the circular parts under consideration ? Are the extremes adjacent or opposite ? 9. What are the opposite extremes to comp O ? What the adjacent ? To comp m ? To o? Pig. 43. Ton? NAPIER'S RULES. 118, Hule !• ^rop* — In any right angled sjyJierical triangle^ the sine of the middle part equals the pro- duct of the cosines of the opposite ex- j, tr ernes. Dem. — In the spliencal triangle BAC, right angled at A, taking h, c, comp B, comp C, and comp a in succession as middle paits, we are to prove that sin h = cos (comp a) x cos (comp B), . or sin b = sin a sin B ; (1) sin c = cos (comp a) x cos (comp C), or sin c = sin a sin C ; (3) sin (comp B) = cos b x cos (comp C), or cos B = cos 6 sin C ; (3) sin (comp C) = cos c x cos (comp B), or cos C = cos c sin B (4) sin (comp a) = cos b x cos c, or cos a = cos b cos c. (5) To demonstrate these relations, let O be tlie centre of the sphere, and draw the radii OA, OB, and OC. The angles BOC, AOC, and AOB, are measured respectively by «, b, and c, the sides of the trianigle ; hence these angles at the centie and their measuring ai'cs may be used interchangeably. From one of NAPIER S RULES. 77 the oblique anj^les, as C, let fall a perpendicular upon the radius OA. From the foot of this pei'pendicular draw DE perpendicular to OB, and join C and E. Now CDE is a right angle (Part II., 426), CE is perpendicular to OB (Part II., S99), and DEC is equal to angle B of the triangle (Part II., 55S). CD = sin h, OD = cos b, CE = sin a, and OE — cos a. (1) From the triangle CDE, right angled at D, we have CD = CE X sin CED, or sin ft = sin a sin B. Generalized, (1) becomes, TTie sirie of either side about tlie right angle — the sine of the hypotenuse into the sine of the angle opposite the side. Hence, from analogy to (1), we may write sin c = sin ct sin C. (3) Or (2) may be proved in the same manner as (1), by letting fall from B a perpendicular upon OA, from its foot drawing DE perpendicular to OC, and joining E and B. Then BD — sin c, OD = ccs c, BE = sin a, OE = cos a, and angle BED = angle C. From the triangle BDE,'we have BD = BE X sin BED, or sin c = sin a sin C. (2) To prove (3), we have from the triangle CDE, Fig. 44, cos CED = — , or cos B CE But from triangle OED, right angled at E, ED = OD x sin DOE = cos 5 sin c = [from (3)], cos b sin a sin C. Substituting this value of ED, we have _, cos 6 sin a sin C , . ^ cos B = r = cos b sm C. (3) sma ^ We may write (4) from (3) by analogy, as (2) was from (1) ; or, better, let the student produce it from Fig. 45, as (3) was produced from Fig. 44. Finally, to produce (5), consider the triangle ODE, in either figure, right angled at E. This gives ED sin 6 OE = OD X cos DOE, or cos a = cos 6 cos c. (5) 119, JRtile 2. JProjy* — -^^^ any right angled splier- ical triangle, the sine of the middle part equals the i^o- duct of the tangents of the adjacent extremes, Dem.— In the spherical triangle BAC, right angled at A, taking by c, comp B, comp C, and comp a, in succession as middle parts, we are to prove that, 78 SPHERICAL TRIGONOMETRY. sin b = tan c x tan (comp C), or sin b = tan c cot C ; (1) sin c = tan & x tan (comp B), or sin c = tan 6 cot B ; (2) sin (comp B) = tan c x tan (comp a), or cos B = cot « tan c ; (3) sin (comp C) = tan b x tan (comp a), or cos C = cot a tan b ; (4) sin (comp a) = tan (comp B) x tan (comp C), or cos a = cot B cot C- (5) Taking theformiihv of Rule 1st, and ting the value of each factor as found write tlie followius^ sin & = sin a sin B = sin c = sin a sin C = cos B = cos 6 sin C = cos C = cos c sin B = cos a = cosb cos c = in the second member of each siibstitu- in some other of the set, we readily sin c cos C sine cose cosC = tan c cot C ; (1) sin C cos c sin C sin b cos B sin b cos B sinB = tan b cot B ; (2) sin B cos b ~ cosb cos a sin c cos c sm a cos a = -. X sin a sin c COSC = cot a tan c ; (3) CCS a sin b cos bsina' cos a. = -; — *x sma sin b cosb = cot a tan 5 ; (4) cos B cos C sin C sin B cosB &m B cosC sinC = cot B cot C. (5) Q. E. D. 120. Sen. 1. Fig. 47. -It will be a good exercise for the student to demonstrate Rule 2d from the an- nexed figures, as Rule 1st was from Figs. 44 and 4o. The 5th is not as readil}^ obtained from the figure as the others. The stu- dent may trace the fol- lowing relations, some .in one figure, and some in Fig. 4S. the Other. Cos « = gi = '-^ = cos b COS c. But, cot C = ^ = '^\ and cot B = §5 OE sec c AE tauc AE s^^ '^ T. ^ _, , ^ sin 5 sin e ^ . « . /^ = ^ — ; ; whence, cot B cot C = r = cos b cos c. /. cos a = cot a cot C. tan 6 tan 6 tan c 121, ScH. 2. — It is of much importance, especially for the purposes of Spherical Astronomy, that the student observe that the relations expressed in the above formulae, and in fact all the relations between the sides and angles of spherical triangles, are also the relations between the facial and diedral angles of triedrals. Thus, if a, b, and c represent the facial angles, and A, B, and C the opposite diedrals, all these relations can be established, and in exactly the same manner as above, without any allusion to the ep/ierical triangle. [The student should do it.l DETERMINATION OF SPECIES. 79 DETERMINATION OF SPECIES. 122. In the solution of spherical triangles the determination of the species of a part sought becomes of essential importance, since any part of such a triangle may have any value l^etween 0° and 180°. Hence, when we have learned the numerical value of any function of a part, we have yet to determine whether the part itself is less or greater than 90°, i. e., the species of the part. This may always be effected by some one of the following propositions. 123, JProp, — If the part sought is found in terms of its cos, tan, or cot, its S2)ecies can he determined hy the algebraic signs of the functions in the formula used. Dem. — In each oi i\ve formulm arising from Ihe application of Napier's rules, there are three functions, the arcs corresponding to two of which are always known, hence the algebraic signs of their functions are known, and the signs of these two determine the sign of the third or unknown function. Now, when a cos, tan, or cot is + , the corresponding angle is less than 90° (if less than 180°) ; and when one of these functions is — , the corresponding angle is greater than 90° ; i. e., in a triangle, it is between 90° and 180°. 124. When -the part sought is found in terms of its sine, the species cannot be determined by the signs of the formula, since the part being less than 180° its sine is always +. The three following propositions dispose of such cases. 125, ^rop, — An ohlique angle of a right angled spherical tri- angle and its opposite side are always of the same species. ' Dem.— From Napier's first rule we have, cos B = cos J sin C. But sin C is necessarily + ; therefore, cos B and cos h always have the same sign, and B and h are of the same species. In like Dianner, we see from cos C = cos c sin B, that C and c are of the same speaies. q. e. d. 126, JProp, — When the hypotenuse of a right angled spherical iriangle is less than 90°, the other two sides {and consequently the 80 SPHERICAL TRIGONOMETRY. ohUque angles) are of the same species with each other. But tvhen the hypotenuse is greater than 90'^, tJie other two sides {and consequently the oblique angles) are of different species from each other, Dem. — From Napier's first rule we have, cos a — cos h cos e. Now, if a < 90% cos a is + ; lience cos h and cos c must have like si^^s, and b and c be both less or both greater than 90° But if a > 90' (and less than 180% as it is), cosrt is — ; hence, cos b and cos c must have different signs, and b and c be one greater and the other less than 90% Finally, since the oblique angles are of the same species as their opposite sides, they are of the same species with each other when a < 90% and of different species from each other when a > 90% 127. I^vOjy* — TT72f?i a side and its opposite angle are given in a right angled sp^herical triangle, there is xo solution if the sine of the side is greater than the si7ie of its opposite angle; there is one solution and the triangle is hi-rectangular if the sine of the side equals the sine of its opposite angle ; and there are two solutions if the sine oftlie side is less than the sine of its opposite angle. Dem. — We have sin 5 = sin a sin B, or sin a = -: — = . Now, sin 6 > sin B sm B makes sin a> 1, which is manifestly impossible. Sin& = sin B makes 6 = B, since they are of the same species. But when the arc included by tJie sides of an oblique angle of a right angled spherical triangle is equal to the angle, the vertex of the angle is the pole of the arc. Hence, in this case the other sides of the triangle are each 90% Finally, if sin b < sin B, a has two values, one ^eater and the other less than 90°. Hence there are two triangles. 128, ScH. — These relations between an angle and its opposite side may be observed directly from a figure. When B < 90% the measure of it, that is 6 = B, is tlie greatest included perpendicular which can be drawn to one side of the lune BAB'C. Hence, in tliis case, b cannot exceed B, which implies that sin b cannot be > sin B, as when the arcs are less than 90', the greater arc has the greater sine. If sin 6 = sin B, b = B, and BA = BC = 90% and the side b can occupy but one position in the lime, thus giving rise to but one tiiangle BAC, which satisfies the conditions (or two equal triangles BAC and B'AC). If 6 < B, which, when B is less than 90° unplies that sin b < sin B, the side can occupy two positions m the lime, b' and b" giving rise to two triangles, BA'C and BA"C'% both of which fulfill the conditions. EXERCISES IN SOLVING RIGHT ANGLED TRIANGLES. 81 Again, when B > 90", the measure of it, i. e., 6 = B, perpendicuhir that can be drawn to one side of the lune. Hence, in this case we cannot have & < B, which imphes that sin h cannot be > sin B, since the greater arc has the less sine. If sin b — sin B, J = B, and BA = BC r= 90°, and the side h can oc- cupy but one position in the lune, thus giving rise to but one triangle BAC, which satisfies the conditions (or two equal triangles BAC and B'AC). If 6 > B, which implies that sin h < sin B, the side b can occupy two positions in the lune, as b' and b", thus giving rise to two triangles BA'C, and BA"C' conditions. is the least included C A' Fig. 50. both of which satisfy the EXERCISES. 1. In a right angled spherical triangle BAC, A being the right angle, B = 80° 40', and a = 105° 34', to project the triangle and compute the other parts. Projection.* — Projecting the triangle upon the plane of the side c (102), we have, BCA, Fig. 51. [The student should give the process.] Solution.— It is immaterial which of the re- quired parts we seek first. We will seek c. Now the three circular parts under consideration are c, comp«, and comp B. Comp B is middle part, and the extremes are adjacent ; hence, by Napier's second rule we have, cos B = tan c cota. cos B cos 80° 40' or tan c= _ ,^^, ,,,,. cot a cot lOo d4 Now cos 80° 40' is + , and cot 105° 34' is Fig. 51. Therefore tan c is — , and c > 90°. Computing by logarithms, log cos 80° 40' = 9.209992 - log cot 105° 34' = 9.444947 = log tan c = 1.765045 Add 10 for tab. tan 10. 9.765045. .-. c = 149° 47' 37' * It is recommendecl that the projection be given ahvays before the trigonometrical eolution. It is an excellent exercise, and gives clearness of perception. 6 82 SPHERICAL TRIGONOMETRY. To find b, sin b = sin a sin B = siu 105' 34' x sin 80° 40'. This makes b known by means of its sine, whence the signs of the formula do not determine the species of 6. But b is of the same species as B {124)^ and therefore less than 90°. Computing by logarithms, log sin 105° 34' = 9.983770 + log sin 80° 40' = 9.994212 Deducting 10 = 9.977983 = log sin 5. .-. b = 71° 54' 33". rr ^ 1 r^ * B ^ ^ ^ « cos « COS 105° 34' __ To find C, COS a = cot B cot C, or cotC = — — ^ = — . Whence cot C is -, and C > 90°. Computing by logarithms, log cos 105° 34' = 9.428717 - log cot 80° 40' = 9 .215780 Adding 10 = 10.212937 = log cot C. .'. C = 148° 30' 54". SCH. — It is expedient to find each part directly from the parts given in the example, in order that an error in finding one may not extend itself through the -whole solution. 2. Given a = 86° 51', and B = 18° 03' 32", to project the triangle and compute the other parts. c = 86° 41' 14", b = 18° 01' 50", C = 88° 58' 25". 3. Given b = 155° 27' 54", and c = 29° 46' 08", to project the triangle and compute the other parts. See Fig. 52. a = 142° 09' 13", C = 54° 01' 16", B = 137° 24' 21". 4. Given c = 73° 41' 35", and b = 99° 17' 33", to project the triangle and' compute the other parts. C = 73° 54' 46", b = 99° 40' 30", Fxa. 32. « = 9^° 42' 17". 5. Given B = 47° 13' 43", and c =. 126° 40' 24", to project the triangle and compute the other parts. b = 32" 08' 56", a = 133° 32' 26", c = 144° 27' 03". EXERCISES IN SOLVING EIGHT ANGLED TRIANGLES, 83 Projection. — In order to project this case, i. «., when the two oblique angles are given {105), it is most convenient to compute the base before pro- jecting. It is also expedient, when two angles are given, to project the larger at a point in the cir- cumference of the primitive circle, as at C, espe- cially if the smaller be quite small. In this case, projecting the angle C at C, Fig. 53, conceive BA as drawn through P (or, if desired, sketch it hj-po- thetically), and then compute b, from the relation cos B cos B = sin C cos 5, or cos h •■= found h . ^. Having sm C ^ 32° 08' 56", take CA = b, and draw AB through Fig. 53. P. 6. Given B = 100°, and b = 112°, to project the triangle and com- pute the other parts. Projection.— See Fig. 54. Numerical Solution.— r., BCA, Or BC'A'. Si SPHERICAL TRIGOXOMETBY. Thus -we see that each of the two triangles BCA and BC'A' fulfills the con- ditions of the problem. T. Given one side of a right angled si^herical triangle 160°, and the opposite angle 150°, to project the triangle and compute the other parts. Jtesidts. — There are two triangles. The other sides of the first are 136° 50' 23", and 39° 01' 51"; and the angle opposite the latter side is 67° 09' 43". The corres^Donding j^arts of the other triangle are 13° 09' 37", 110° 55' 09", and 112° 50' 17". 8. In the spherical triangle DEF, right angled at E, given an oblique angle 58°, and the side oj^posite 61°, to project the triangle and com- pute the other parts. 9. In a right angled s]oherical triangle given an oblique angle 165°, and the o^Dposite side 112°, to project the triangle and com- pute the other parts. 10. In a right angled spherical triangle given one side 65° 23' 12", and the opposite angle 6b° 23' 12", to project the triangle and com- pute the other parts. 11. Given c = 60° 47' 24".3, B = 57° 16' 20".2, and A = 90°, to project and compute. a = 68° 56' 28".9, c = 51° 32' 32".l, and b = 51° 43' 36".l. 12. Given c = 116°, b = 16°, and the included angle 90°, to pro- ject and compute. QUADRANTAL TKLLXGLES. 129. A QiiadranUil Triangle is a spherical triangle one of whose sides is a quadrant, or 90°. Such a triangle is readily solved by passing to its polar, solving it, and then passing back. The polar triangle to a quadrantal triangle, being right angled, is solved by Napier's rules. Ex. 1. Given a = 90°, B = 75° 42', and c = 18° 37', to compute the other parts. Sug's. — Representing the supplemental parts of the polar triangle by A', B', C, a', b', and c', we have A' = ISO* - a = 90', b' = 180' - B = 101° 18', and OF OBLIQUE ANGLED SPHERICAL TEIANGLES. 85 C = 180° - c = 1G1° 23', from which to find B', a', and c'. This being rii^ht angled, we find, by applying Napier's rules, B' = 94° 31' 21", a' = 76° 25' 11", and c' = 161° 55' 20". Hence in the primitive triangle we have b = 180° - B* = 85° 28' 39", A = 180° - a' = 103° 34' 49", and C = 180° - c' = 18° 04' 40 ". Ex. 2. Given a = 90°, C = 42° 10', and A = 115° 20', to find the other parts. B = 54° 44' 24", b = 64° 30' 40", c = 47° 57' 47". SECTION II, OF OBLIQUE ANGLED SPHERICAL TRLINGLES. ±30, All cases of oblique angled spherical triangles can be solved by Napier's rules and the following proposition. 1.S1, ^rop, — In any spherical triangle, if a perpeyidicular be let fall fiwn either vertex upon the opposite side or side produced, the tangent of half the siim of the segments^ of that side is to the tangent of half the sum of the other tivo sides, as the tangent of half the difference of those sides is to the tangent of half the difference of the segments. Dem. — In the triangle BAC let fall the perpen- dicular p, from C upon the opposite side. Let BD = 8, and DA = s'. By Napier's first rule, cos a = cos p cos s, and cos b = cos p cos s'. Dividing the former by the latter, = — — , ; ^ •' cos b cos s ' whence, by composition and division, cos b — cos a cos s' — cos s cos a + cos b cos s + cos s'' Fig. 55. But by (6i), and cos b — cos a cos a + cos b cos s' — cos = tan i{a + b) tan i (a — J), cos 8 + COS 8 > = tan i {s + s') tan ^{s — s'), * When the perpendicular falls without the hase, a? in Fig. 50, this term is to be understood as meaning; the distances from the foot of the perpendicular to each extremity of the base, as BD and AD- This, in fact, is the general statement— applying as well to the case when the perpendicular falls on the base. 86 SPHERICAL TPilGOXOMETEY. .-. tan i {a + b) tan |(a — 5) = tan i (s + s') tan i (« — «') ; tan i{s + s') : tan ^ {a + b) : : tan ^{a — b) : tan ^{s — «'). q. e. D. 132, ScH. 1. — Since from a point in the surface of a hemisphere two perpendiculars can always be di-awn to the circumference of the great cii'cle which forms its base, and since the feet of thesq perpendiculars are 180° apart, and no side of a spherical triangle can equal 180°, the foot of one perpendicular will always fall within the base or upon one extremity of it, and the other without the base ; or both will fall without the base. K we take the foot of the pei-pendicular which falls within the base, or the nearer one when both fall without, the sum of the distances from the foot of the perpendicular to the extremities of the base is always less than 180°^ t. €., s + s' < 180°. When the perpendicular falls within, 8 + s' makes up one side of the triangle, and hence is less than 180°. If both perpendiculars fall without, let D, Fi{/. 56, be the foot of the nearer one. Now DB + BD' = 180° ; but by hypothesis DA < BD', .-. DB h- DA < 180°. When DA = BD', DB + DA = 180°. 133. ScH, 2. — As in spherical triangles the greater segment is not always adjacent to the greater side, it becomes necessar}' to determine the position of the segments. This can be done by the signs of the proportion tan i {s — s). Fig. 56. tan i (s + s') : tan ^ {a + b) : : tan i {a — b) 1st. Tan ^{s + «') is always + , since, if D falls in the base, 8 + s' < 180° ; and if D falls without, by taking the nearer perpendicular, « + &' is still < 180° (132). .: H« + «') < 90°, and tan H« + *') is + . 2d. When a + b < 180°, tan i (a + 5) is + ; and when a + b> 180°. tan i {a + b) is — . 3d. When a > b, a — b is a. positive arc less than 180°, hence tan ^{a — b) is + ; and when a s\ and — when s < s'. 5th. When 8 + 8' = 180°, tan i{s + s')= oc. Now as a - 6 < 180°, tan i {a-b) cannot be oc, nor can tan ^ {s — s') = when the perpendicular falls without. Hence to make the proportion possible, tan ^{a + b) must be cc,or a + b= 180°. In this case we project on the plane of a or b. If: a + b = 180", aad a + G = 180°, we project on the plane of a. U a + b = 180°, a + c = 180°, and b + c = 180°, the tiiangle is trirectangular. 134, ScH. 3. — If either segment is greater than the whole base, the perpen- dicular falls without the triangle. In this case the shorter segment lies in an opposite direction from its angle to that considered in the demonstration, and hence is to be considered — ; and the algebraic sum of the segments is still equal to the side upon which the perpendicular is let fall. OBLIQUE ANGLED TRIANGLES SOLVED BY NAPIER's RULES. 87 135* Prop. — In a spherical triangle, the sines of the sides are to each other as the sines of their 0])posite angles. Dem. — By Napier's first rule we have from either Fig. 55 or Fig. 56, sin p = sin a sin B, and sin ^ = sin 5 sin A. .'. sin a sin B = sin h sin A, or sin a : sin 5 : : sin A : sin B. q. e. d. ScH. — This proposition is not introduced here because it is necessary for the solution of spherical triangles, but because of its essential importance. It is often convenient to use it in the solution of a triangle, but never necessary, as will appear hereafter. It affords a ready metlwd of determining a part opposite a given part, provided the species of the part he determined by other considerations. SOLUTIOX OF OBLIQUE ANGLED SPHERICAL TRIANGLES BY NAPIER'S RULES FOR RIGHT ANGLED SPHERICAL TRIANGLES. 130, The examples which arise in the solution of oblique angled spherical triangles are all comprised under the three following problems, each of which consists of two cases : — 1. When the given parts are all adjacent to each other. 2. When two of the given parts are adjacent and one separate. 3. When the given parts are all separate from each other. 137, J^voh* 1. — Given three adjacent parts of an ohlique angled spherical triangle, to solve the triangle. Case 1st. — Given tiuo sides and the included angle. Solution. — Project the triangle on the plane of one oftlie given sides {106), and let fall a perpendicular from the angle opposite upon this side or upon Fig. 58. this side, produced, as the case may be. There are thus formed two right angled triangles, as BDC and DC A, each of which can be solved by Napier's 88 SPHERICAL TRIGONOMETRY. rules, by first solving the one containing the given angle. Thus, in the triangle BDC right angled at D, a and B are supposed known ; whence CD, BD, and the angle BCD, can be computed. As BA = c is known, the segment DA can be found, it being the difiference between c and the arc BD. When the solu- tion of tliis triangle gives BD > <•, it is evident that the perpendicular falls without tlie triangle, which will agree with the projection. Passing to the triangle ADC, right angled at D, we now know CD and DA ; whence the other parts can be found. Finally, the angle BCA of the required triangle = BCD + DCA when the perpendicular falls witliin the triangle, and BCD — DCA when the perpendicular falls without. Case 2d. — Given tico angles and the included side. Solution. — The solution of this case is effected by passing to the polar triangle, projecting and solving it by Case 1st, and then passing back. 138. ScH. — A slight saving of labor is effected by usin^ {135) in the solu- tion. Thus, in the triangle BCD, compute CD and BD as before, and (not com- puting angle BCD) then passing to the triangle DCA, compute h and A. Finally, compute C (the entu-e angle) from the proportion sin 6 : sin c : : sin B : sin C. 139. JProb, 2.-^1 an oblique angled spherical triangle, given fivo 2)arts adjacent to each other and one separated from both of them, to solve the triangle. Case 1st. — Given two sides and an angle opposite one of them. Solution. — Project the triangle on the plane of the unknown side, with the given angle at B ; and let fall the perpendicular from the angle C opposite the unknown side. Compute the tiiangle BDC. Having computed this triangle. Fig, 60. compare the side opposite the given angle, as 5, with the perpendicular and the arcs BC and CB', i. e,, with ^, a, and 180° —a. If b = p there is but OBLIQUE ANGLED TRL^GLES SOLVED BY NAPIER's RLXES. 89 one solution and the triangle is right angled, A falling at D. If b is inter- mediate in value between p and both a and 180° — «, it can occupy two positions as in Fig. 59, and there are two solutions. If b is intermediate in value between J) and only one of the arcs a or 180° — a, there is but one solution. When B < 90° the perpendicular is less than any oblique arc; hence in this case, if Z> < j9, there is 710 solution. But if B > 90°, the pei-pendicular is greater than the obhque arcs ; hence in this case, \ib> p, there is no solution. [These results should be obtained independently of the results given by the projection, and one be made a check upon the other.] The solution is now completed by computing the parts of DCA, and adding or subtracting the segments BD and AD, and the angles BCD and A CD, as the case may require. Case 2d. — Given tivo angles and a side opposite one of them. Solution. — ^Pass to the polar triangle ; solve it, and then pass back.* 14:0, ScH. — The relation established in {13o) may also be used in the solution of this problem. Thus, having projected the triangle, computed j9, and determined whether there are one or two solutions, to find A, when «, by and B are given, we have, sin b : sin a : : sin B : sin A. Then computing the third side c (or sides), by means of the right angled triangles BCD and DCA as before, we may use the proportion {135) to find the angle BCA and BCA'. But the use of this proportion gives no advantage except in cases in which there is only one solution. 14:1, JPvoh. 3, — In an oblique angled sp)herical triangle, given three parts all separated from each other, to solve the triangle. Case 1st. — Given the sides to find the angles. Solution. — Project the triangle on the plane of one of its sides, as c. From the proportion, tan|(s 4- s') : tan \{a + b) : : tan i{a — b) : tan i{s—s'), * This case can be projected and solved in a manner altogether similar to the first, without passing to the polar triangle. Thus, let B, «i and A be the given parts. ]^roject the triangle on the plane of c, as in the figure. Project B in the usual way, and make BC = the projec- tion of a. Through C draw DD', and make BDO = the projection of A- Drawing the small circle with radius PC, draw diameters through the intersections and Q\ Then will A and A' ^c the vertices of the triangle re- quired. The student may prove that the figures POD and PCA are equal, and also PQ'D and PCA', and hence that angle BDO = A = A'. 90 SPHERICAL TRIGONOMETRY. Fig. 62. find i{s — s'). Then half the sum of the segments + half the difference gives the greater segment, and half the sum — half the difference gives the less. Determine from the signs of the terms whether s is greater or less than s' : and also determine whether the perpendicular lies wiUiin or without the tri- angle {134). Observe that these results con-espond to those given by the projection. Finally, in each of the two right angled triangles BCD and DCA, there are two sides given ; whence the angles can be found by Napier's rules. If the perpendicular falls wiUiin, C = BCD + DCA, and A, of the re- quired triangle = DAC. If the perpendicular falls without, C = BCD — DCA, and A of the triangle = 180' - DAC. Case 2d. sides. ■Given the angles to find the Solution. — ^Pass to the polar triangle ; solve it, and then pass back. 14:2, ScH.— Here, again, {135) affords a slightly more expeditious solution. Having projected the triangle, found and located the segments, and com- puted one angle, as B, by the methods given above, the other angles may be found from the proportions, sin b : sin a : : sin B : sin A, and sin Z> : sin c : : sin B : sin C. EXERCISES. 1. Given l = 120° 30' 30", c to project and. solve the triangle. 70° 20' 20", and A = 50° 10' 10", Projection. — See Fig. 64. Trigonometrical Solution.— 1st. To sohf tlie triangle ABD, in which the two known parts Qre situated. (a) To find p, sin p = smc sin A. log sin 70' 20' 20" = 9.973913 -f- log sin 50° 10' 10 ' = 9.885329 Rejecting 10 = 9.859241 = log sin p .*. p = 46° 19' 01", tlie species being determined by the opposite angle {125). [Observe that the re- sult corresponds with the projection]. OBLIQUE ANGLED TRIANGLES SOLVED BY NAPIER's EULES. 91 (ft) To find AD, cos A = cote tan AD, or tan AD = -4- . ^ ' ' ' cot c log cos 50° 10' 10" = 9.806532 - log cot 70° 20' 20" = 9.553016 Adding 10 = 10.253516 = log tan AD. .-. AD = 60° 50' 49", the species being determined by the signs of the formula. (c) To find angle ABD, cos c =: cot A cot ABD, or cot ABD =: — —. log cos 70° 20' 20" =: 9.526929 - log cot 50° 10' 10" = 9.921204 Adding 10 = 9.605725 = log cot ABD. .-. ABD = 08° 01' 53" the species being determined by the signs of the formula. 2d. To solve the triangle DBC. [a) To find DC. Since AD < 5, the foot of the perpendicular falls in the base, and DC = AC - AD = J - AD = 120° 30' 30" - 60° 50' 49" = 59° 39' 41". {b) To find a, cos a = cosp cos DC log cos 46° 19' 01" = 9.839270 + log cos 59° 39' 41" = 9.703386 Rejecting 10 = 9.542656 = log cos a. /. a = 69° 34' 56", the species being determined by the signs of the formula. sin T) (c) To find C, sin j; = sin a sin C, or sin C = — — . '' sm a log sin p = 9.859241 - log sm 69° 34' 56" = 9.971820 Adding 10 = 9.887421 = log sin C. .'. C = 50° 30' 08", the species being determined by the side opposite. {d) To find angle DBC, smp = tan DC cot DBC, or cot DBC = -^~. tan uC log sin 2? = 9.859241 - log tan 59° 39' 41" = 1 0.232653 Adding 10 = 9.626588 = log cot DBC. .*. DBC = 67° 03' 36", the species being determined by the signs of the formula. Finally, B = ABD + DBC = 68° 01' 53" + 67° 03' 36" = 135° 05' 29". ScH.— We might have omitted the computation of angle ABD in the first part, and DBC in the second, and have found instead the entire angle B from sin a : sin6 : : sin A : sin B. To compute this requires the looking out of but two logarithms, since sin a is given in the second part (c), and sin A in the first part {a). 2. Given a = 97° 35', h = 27° 08' 22", and A = 40° 51' 18", to project and compute the triangle. Between Avhat limits must the 92 SPHERICAL TRIGONOMETRY. value of a be assigned in order that there may be two solutions ? Be- yond what limiting values of « is a solution impossible ? Projection. See Fig. 65. Trigonometrical Solution.— To find p, sin p = sin b sin A. log sin 27° 08' 22" = 9.659115 + log sin 40° 51' 18" = 9.815675 Rejecting 10 = 9.474790 — log sin p. B .'.p = 17° 21' 40". [Reason for the species.] TVe now observe that there is one and onli/ one solution, since the arc a (97° 35') cannot lie between CD (17° 21' 40") and b (27° 08' 22"), but can lie be- tween CD aud CA' (180° - b = 152° 51' 38"). To find AD, cos A = cot 5 tan AD or tan AD = cos A cot 6" log cos 40° 51' 18" = 9.878733 - log cot 27° 08' 22" = 10.290226 Addinsr 10 = 9.5«8o07 = log tan AD. AD = 21° 11' 30".6. cos 5 To find ansrle ACD, cos b = cotA cot ACD, or cot ACD = . = ' cot A log cos 27° 08' 22" = 9.949340 - log cot 40° 51' 18" = 10.0630575 Adding 10 [Reason for the species. ] smp sin a 9.8862825 = log cot ACD. .'. ACD = 52° 25' 01". To find B, sin p = sm a sin B, or sui B = log sin p (as above) = 9.474790 - log sin 97° 35' = 9.996185 Adding 10 = 9.478605 = log sin B. . '. [Reason tor the species. ] cos a B = 17° 31' 09" To find DB, cos a = cos p cos DB, or cos DB = cos p log cos 97° 35' = 9.120469 - log cos 17° 21' 40" = 9 979750 Adding 10 = 9.140719 = log cos DB. [Reason for the species.] AB = AD + DB = 21° 11' 30".6 + 97° 56' 51".3 = 119° 08' 21".9. DB = 97° 56' 51".3. sin v To find DCB, sin;? = tan DB cot DCB, or cot DCB = — tan DB" log sin p (as above) = 9.474790 — log tan 97° 56' ol".3 = 10.855090 Adding 10 = 8.619700 = log cot DCB. [Reason for the species.] ACB ^ C = ACD + DCB = 52° 25' 01" + 92° 23' 7".7 = 144° 48' 8".7. DCB = 92° 23' 7".7. OBLIQUE ANGLED TELA.NGLES SOLVED BY NAPLER'S RULE. 93 Finally, we observe that, if any value were assigned to a between h (27° 08' 22") and CD (17° 21' 40") there would be two solutions ; since for such values the side a could lie on both sides of CD. But, for any value of a less than CD (17° 21' 40"), there would be no solution ; since CD is the shortest distance from C to the arc ABA'. Also, for any value of a greater than the arc CA' (152° 51' 38"), there would be no solution, as such an arc would fall between CA' and CD' (if not > CD'), and consequently would make c > 180". SCH. — Such examples as this and the preceding can be more expeditiously solved by using p in each equation in solving the triangles ACD and DCB. By this means and using {135) to determine the side c, the solution can be eflFected with only 12 logarithms. Thus in Ex. 2. 1st. To find i?, Bin p = sin & sin Ai require? 3 logarithms 2d. To find ACD, cos ACD = cot & tan 2?, requires 3 " 3d. To find DCB, cos DCB = cot « tan;?, requires 2 " (log tan p being known). 4th. To find B, ^ini? = sin a sin B, requires 2 (log sin 2? being known). 5th. To find c, sin A : sin C : : sin a : sin c, requires _2^ " Gog's of sin A and sin a Total 12 logarithms being known.) •3. Given a = 7G° 35' 3G", 1) = 50° 10' 30", and c = 40° 00' 10", to project and solve the triangle. Projection. — See Fij. 66. TRiGONO>rETRiCAL SOLUTION. — Ist. To find the segments CD and DB, we have, tan i{s + s'), or tan ^a : tan i{b + c) : : tan i(Z> - c) : tan i{s - s'). f \ Ja ^<-^ /\ Computing by logarithms. "'Va \J \\ a.c. log tan ia = log tan 38° 17' 48" = 0.102561 ^ \ ^^^^^^=^ + log tan \{b + c) = log tan 45° 05' 20" = 10.001347 + log tan lib -c) = log tan 5^ 05' 10" = 8.949406 Fig. GG. Rejecting 10 = 9.053314 = log tan i{s - s'). .-. i(s_ 5') = 6°27'02". In order to determine whether s or s' is the greater,* we observe the signs of the proportion, and finding tan h{s — s') positive, know that s > s'. ' Hence, s = \{s + s) + l{s - s') = 38° 17' 48" + 6° 27' 02" = 44° 44' 50", and s' = Ks + s') - K« - «') = 38° 17' 48" - 6° 27' 02" = 31° 50' 46". The angles sought are now readily found by computing the two right angled triangles ADC and ADB, * Though the projection generally determines such facts as this, the species of parts, and the number of solutions in ambiguous cases, the student should not rely upon it, but determine each such fact upon purely trigonometrical considerations, merely using the projection to give clearness to the conception, and as a rough check against errors. 94 SPHERICAL TRIGONOMETRY. Or, having computed C from the triangle ACD, we maj find A and B more expeditiously by using the proportions, sin c : sin 6 : : sin C : sin B, and sin c : sin a : : sin C : sin A. The angles are C = 34° 15' 03", A = 121° 36' 12", and B = 42° 15' 13". 4. Given A = 128° 45', C = 30° 35', and a = 68° 50', to solve the triangle. What values of A give two solutions ? What none ? c = 37° 28' 20", b = 40° 09' 04", and B = 32° 37' 58". 5. Given A = 129° 05' 28", B = 142° 12' 42", and C = 105° 08' 10", to solve the triangle. a = 135° 49' 20", b = 146° 37' 15", and c = 60° 04' 54". 6. Given a = 68° 46' 02", b = 37° 10', and C = 43° 37' 38", to project and solve the triangle. A = 116° 22' 22", B ^ 35° 29' 54", and c = 45° 52' 34". 7. Given a = 40° 16', b = 47° 44', and A = 52° 34', to project and solve the triangle. What values of a give but one solution ? What none ? There are hvo triangles.— Ivl the 1st, c = 53° 19' 20", B = 65° 16' 35", and C = 79° 52' 21". In the 2d, c = 14° 18' 22", B = 114° 43' 25", and C = 17° 39' 22". 8. Given a = 62° 38', b = 10° 13' 19", and C = 150° 24' 12", to project and solve the triangle. A = 27° 31' 44", B = 5° 17' 58", and c = 71° 37' 06". 9. Given a = 56° 40', b = 83° 13', and c = 114° 30', to- project and solve the triangle. A = 48° 31' 18", B = 62° 55' 44", and C = 125° 18" 56". 10. Given A = 50° 12', B = 58° 08', and a = 62° 42', to solve the triangle. What values of A give but one solution ? What none ? There are Uoo soMions.— 1st, b = 79° 12' 10", c = 119° 03' 26", and C = 130° 54' 28". 2d, b = 100° 47' 50", c = 152° 14' 18", and C = 156° 15' 06". 11. Given A = 36° 25', B = 42° 17' 10", and c = 95° 10' 05", to project and solve the triangle. OBLIQUE ANGLED TRIANGLES SOLVED BY NAPIER's RULES. 95 12. Given a = 124° 53', b = 31° 19', and c = 171° 48' 22", to solve the triangle. 13. Given a = 150° 17' 23", h = 43° 12', and c= 82° 50' 12," to solve the triangle. 14. Given a = 115° 20' 10", b = 57° 30' 06", and A = 126° 37' 30", to solve the triangle. 15. Given A = 109° 55' 42", B = 116° 38' 33", and C = 120° 43' 37", to solve the triangle. 16. Given A = 50°, b — 60°, and a = 40°, to solve the triangle. 17. Given a == 50° 45' 20", b = 69° 12' 40", and a = 44° 22' 10", "^o project and solve the triangle. There are fivo solutions.— Ui, B = 57° 34' 51", c = 115° 57' 51", and c = 95° 18' 16". 2d, B = 122° 25' 09", c = 25° 44' 32", and c = 28° 45' 05". 18. Given b = 99° 40' 48", c = 100° 49' 30", and A = 65° 33' 10", to project and solve the triangle. a = 64° 23' 15", B = 95° 38' 04", and c = 97° 26' 29". 19. Given A = 48° 30', B = 125° 20', and c = 62° 54', to solve the triangle. a = 56° 39' 30", b = 114° 29' 58", and c = 83° 12' 06". 20. Given c = 54° 15' 03", B = 40° 18' 13", and a = 70° 30' 30", to solve the triangle. 21. Given A = 47° 54' 21", c = 61° 04' 56", and a = 40° 31' 20", t@ project and solve the triangle. 22. Given a = 50° 10' 10", b = 69° 34' 35", and a = 120° 30' 30", to project and solve the triangle. 96 SPHEPJCAL TilIGO^'OMETRY. SECTION III. GOTHAL formula:. [N'OTE.— This section is designed for such as make matliematics a specialty. The preceding sections are thought suflScieut for the general student] 14:3, Prop. — In a Sj)herical Tri- angle the cosine of any side is equal to the product of the cosines of the other two sides, plus the product of the sines of those sides into the cosine of their included angle j that is, (1) cos a = cos h cose + sin b sin c cos A ; (2) cos I? = cos a cos c + sin a sin c cos B ; (3) cos c = eosa cos b + sin a sin Z> cos C i Dem.— From Fig. 67, we have, cos a = cos {c — x) cos p cos (c — x) cos b r . = ^: smce cos cos X L cos b cos c cos X + c os b sin c sin x ~~ cos X = COS b cos c + cos b sin c tan x i _ cos b "I ^ ~ cos X J [expanding. cos (c — x)] ] sm X since = tan x cos a; cos b tan x~ = cos b cos c + sin J sin c cos A | since cos A = cot 6 tan x= , .. In a similar manner (2) and (3) may be produced. 144, Cor. l.—From set A, hy jjassiiig to the polar triangle, loe have, (1) cos A = — cos B cos c + sin B sin c cos« ; \ (2) cos B = — cos A cos c + sin A sin c cos & ; ^ B. (3) cos C = — cos A cos B + sin A sin B cos c. ) Dem.— Lettmg a', b\ d, A', B', and C represent the parts of the polar triangle, GENERAL FORMULAE. 97 we have a = 180° - A', 6 = 180° — B', c = 180° - C, A = 180' - a\ B = 180° — b\ and C = 180° — e'. Whence, substituting in (1) A, we have, cos (180° - AO = cos (180° - B') cos (180° - C) + sin (180° - B') sin (180° — C) cos (180° - «'), or, cos A' = — cos B' cos C + sin B' sin C cos a', since cos (180° - A') = - cos A', etc. ; and sin (180° - B) = sin B', etc. Finally, dropping the accents, since the results are general, and treating (2) and (8) of set A in the same way, we have set B. 14S, Cor. 2. — From A a7icl B we readily find tlie angles in terms of the sides, and the sides in terms of the angles. Thus, from A, ,^ . cos a — cos h cos c (1) C0SA = : — r—' ; ^ sm sm c cos h — cos a cos c (2) cosB (3) cosC = sm a sm c cos c — cos a cos h sin a sin ?> A'. rrom B, ,^ . cos A + COS B COS c (1) COS a = - . ■ — , (2) COS h = (3) cos c = sm B sm C cos B -I- cos A cos c . : : } sm A sm C cos C + cos A cos B sin A sin B > B'. 146, JProp. — FormtdcB A', and B', adapted to logarithmic coin- 2mtation, lecome. (1) sin^A (2) sin^B (3) sin /sin (is — b) sin (\s - -'^■, t sin b sin c A /'''' iis — a) sin (fsf - -0). V sin a sin c y i /'^ (is — a) sin {^s - -b) sm a sm 7 M\ 98 And (1) sin (2) sin ^b (3) sin ^c SPHERICAL TRIGONOMETRY. in ^a = a/ - ' COS is COS (is - ■A). sin B sin c J COS is cos (is - B). sin A sin c ? COS is cos (is — C ) sin A sin B B" Dem.— Subtracting each member of (1) A' from 1, we have, cos a — cos h cos c cos h cos c + sin 5 sin c — cos a "" 1 — cos A = 1 — •. 2 sin" ^A = sin b sin c cos {b — c) — cos a . sin 5 sin c , since 1 — cos A = 2 sin^ ^A {62, 5), sin b sin c and cos b cos c + sin & sin c = cos (6 — c) (55, D). Now letting y = b — c, and x = «, we see from (59, D') that cos (6 — c) — cos a = 2 sin ^(a + b — c) sin ^(a + c — b). Hence, 2 sin* ^A 2sini(a + b — c) sini(ct + c — b) sin 6 sin c ' or, sin ^ A _ . / sin i{a b — c) sin i(a + c — 6) sin 6 sin c Finally, putting s = a + b + c, whence K« + b —c) = ^s — c, and we have, x{a + c — b) = |s — 6, „•„ lA i A^^ (2-"« - ^) sin (i« - c) sin tA = 4/ : — j-^ . r sm sin c sin b sin c In like manner, (2) and (3) of set A' reduce to (2) and (3) of set A". Again, subtracting each member of (1) set B', from 1, we have, cos A 4- cos B cos C sin B sin C — cos B cos C — cos A 1 — cos a = \ .'. 2sin'^ ia = sin B sin C sin B sin C — cos (B + C) — cos A _ cos (B + C) + cos A sin B sin C s sin B sin C 2 cos i(A + B -i- C) cos i(B + C^,A) ^^^Jt, sin B sin C GENERAL FORMULA. V Now putting S = A + B + C, whence i(B + C — A) = ^S — A, we have, . , , / cos iS cos (iS - A ) sm ia = i/ -. — fi—~^ r sin B sm C In the same manner, (2) and (3) of B" are deduced from (2) and (3) of b . 147. Cor. 1. — Passing to the polar triangle, A" and B" become .^. , , /cos (is — B) cos (is — C). (1) cos Uc = 4/ — — : — —. — — -' ^ ^ '^ 1/ sm B sm C (is — A)cos(-|S - C). 1/ sm A sm C _ /cos (is — A) cos (is — b) sm A sm B And (2) cos i^ (3) cos ic (1) cos (2) cosiB=|/- ,„. , , /sin is sin (3) cos ic = A/ ~. ^ ^ ^ 1/ sm a ^ B'" sm i5 sm(i.s" ~ ^0. sin 5 sin c ^ sin i^ sin (^s — b) ^ sin a sin 6* ' ' " sin b 14S» ScH. — Formulae A'" and B" can be obtained directly from A and B, in a manner altogether similar to that in which A" and B" were deduced, by adding each member of the equations in sets A' and B'to 1, instead of subtract- ing, and observing that 1 + cos.-c = 2cos^ ^x. 149, Cor. 2. — Dividing the formitlm of set k." by the correspond- ing ones of set A'"; and, in a similar manner, those o/ B'" by those ofB", and putting ^^r.(y - a) .m(y-J>),m(is - c) I: and |/ c"MiS-A)co.sas-B)coB(tS-c) ^ ^^.^ ' . — cos is (1) tan (2) taniB sin {\s — b) sin (i-s^ — c) _ l: sin is sin (is —a) ~ sin {\s — a) ' sin (is — a) sin (is — c) _ k sin is sin (is — b) ~ sin (is-Z>j (3)taniC=i/5HLii£zi4l4£lfc ^ ^ ^ •!/ sm is sm (is — -b) _ h c) ~ sin {\s — c)' 100 SPHERICAL TRIGONO^IETRY. (1) COtJrt (2) cotib = A/' / cos (is y -cos B) COS (jS — C) _ ^S cos (^s — A) COS (4s — A) ' K cos (^S — A) COS (Is — C) (3) cotl-c / cos (is 1/ -cos COS is COS (is — B) COS (is - B) K B' a) cos (is — B) is cos (is — C) cos (is — C). ScH. — In these formulcB k is the tangent of the arc with which the inscribed circle is described, and K is the cotangent of the arc with which the circum- FiG. 68. Fig. 69. scribed circle is described. Thus, nsmg the common notation, we have In Fig. 68, AD = AD' = \s — a, and angle PAD = \^\ whence tan PD sin AD = cot PAD x tan PD tan PAD' or tan ^A = tan PD k ., [(l)Ai^]. .:k= tanPD. sin (^s — a) sin {{s — «)' From Fig. 69, we have, AD = ic, and angle PAD = ^S — C. cos(iS -C), Hence, cos (iS — C) = cot AP x tanic, or tan ^c cotAP or cot \c = cot AP cos (is - C) cos(iS — C) , [(3)Bi^]. .'. K = cot A P. GAUSS'S EQUATIONS. ISO, I^rob, — To deduce Gauss's Equations, loliich are sin i(A + b) cosi(rt — V) (1) cos iC" cos \c , . sini(A — B) _ sini(rt — V) ^ cosic ~ siuic ' (3) gauss's equations. 101 cos^(A + b) _ cos-l{a + b) , smTj C "" COS ic ' COS |^(A — B) _ sin ii^i + i) sin ^C "" sin ^g Solution.— From A, page 25, we have, sin (^A + iB), or sin KA + B) = sin ^A cos iB + cos ^A sin ^B. Substituting in the second member the vahies of sin ^A, cos iB, cos ^A, and sin iB , from A" and A'", there results, • -,/« r^x sm{hs — b) . /smissmiis — c) fiin {U — a) /sm ^s sm{is — C) smi(A + B) = 7 4/ : : — 7 — H -. i/ -. : — -. ' sm c r sm a sm b sm c r sm a sm d _ sin (jj? -b) + sin {js—a) / sin js sin {js - c ) ^, . sin c y sin 6t sin 6 ' '■ ^ ' cos iC sm sin (is — b) + sin Qs — a) sm c But sin (is — 5) + sm {is — a)z= sin {^a + |& + §c — &) + sin {^a + ^b + ic — a) = sin [ic + i{a — b)] + sin [^c — -iCa — b)] = 2 sin ic cos ^^(a — 6). (59, A'). Also, sin c = 2 sin |c cos ic. Substituting these values, the preceding becomes, • 1/ A ox 2 sin |c cos Ua — b) , _ sm i(A + B) = r-^ — , ^^ , ^ cos iC . 2 sm ic cos ic ' sin |(A + B) _ cos i{a — b) cos iC ~" cos ^c • w In like manner starting with sin (|A - |B ), or sin |(A — B) = sin |A cos ^B — cos -^A sin -JB, ., ,^ sin i(A — B) sin i{a — b) ,^, there results, -—r?i — - = ^^ ■ (2) cos iO sm ic ' Starting with cos (iA + ^B), or cos \{fK + B) r= cos ^A cos ^B — sin ^A sin ^B, there results, cos «A + B) ^ cosK« ^) sm -JC cos \o ' Starting with cos (^A — IB), or cos \{k — B) = cos M cos iB + sin ^A sm ^B; *u 1* cos \{k — B) sin \{a + J) ,,, there results, ^r— — ' = f^ r— -^ (4) sm iC sm \c ^ ' 102 SPHERICAL TRIGONOMETEY. NAPIER'S ANALOGIES. lol. I^roh, — To deduce Napiefs Analogies, which are (1) (2) (3) tan |( a + b) _ cos ^{a — h) ^ cot -|-c ~ cos \{a + h) ' tan |^(a — B) _ sin ^{a — h) ^ cot \C ~~ sin ^{a +T) ' tan ^ {a + i) _ cos |(A — b). tan |c ~ cos ^(A + B)' tan ^(^—5) _ sin^(A — B) tan Jc "" sin4^(A + b)' Solution.— To deduce (1), divide the 1st of Gauss's Equations by tlie 3d. To deduce (2), divide the 2d of Gauss's by the 4th. To deduce (3\ divide the 4th of Gauss's by the 3d. To deduce (4), divide the 2d of Gauss's by the 1st. 152, ScH. — In using these fonnulse the species must be carefully attended to. Thus in (1), cot iC and cos \{a — h) are necessarily + ; hence tan |(A + B) and cos^(a + h) are of the same sign with each other. In (2), cot ^C and sin \{a + h) are both + ; hence, tan |(A — B) and sin \{a — b) are of the same sign with each other. And similar inspections may be made upon (3) and (4). EXERCISES. 153, The proposition tliat " The sines of the angles are to each other as the sines of their opposite sides" (133), Napier's Analogies (151), and formulae A'% B'"' (140) are sufficient, in themselves, to effect the solution of all cases of oblique spherical triangles; and for practical purposes they generally require less labor than Xapier's Rules. «We give a few solutions and refer the student to the pre- ceding Exercises for further practice. 1. Given a = 100°; c = 5° and h = 10°, to solve the triangle. (Prob. 1, Case 1st, 137.) 1st. To find A and B we have, cos K« + *) : cos K« - b) : : cot ^C : tan i(A + B) ; and sin ^{a + b) : sin ^{a - b) : : cot ^C : tau i(A - B) [15^ (1) (2)] Napier's analogies. 103 Computing by logarithms, we have, ar. CO. log cos [K« + b) = 55°] = 0.241409 + log cos [i{a -b) = 45'] = 9.849485 + log cot [i C = 2° 30'] = 11.35990 7 Rejecting 10 = 11.450801 = log tan i{k + B). .-. ^(A + B) = 87" 58' 18". ar. CO. log sin [K« + b) = 55°] = 0.086635 + log sin [kia - b) = 45°] = 9.849485 + log cot [^C = 2° 30'] = 11 .359907 Rejecting 10 = 11.290027 = log tan ^{^ - B). ... i(A - B) = 87° 06' 16" The signs of all the terms being + , |(A, + B) and ^(A — B) are both less than 90°. KA + B) + i(A - B) = A = 87° 58' 18" + 87° 06' 16" = 175° 04' 34" i(A + B) - i(A - B) = B = 87° 58' 18" - 87° 06' 16" = 0° 52' 02". 2cl. To find c. This may be found from the proportion, sin A : sin C : : sin a : sin c, or from the 3d or 4th of Napier's Analogies. We use the last, though the first is equally expeditious. sin ^A — B) : sin ^(A + B) : : tan i{a — b) : tan ^c. ar. CO. log sin [^A - B) = 87° 06' 16"] = 0.000555 + log sin [KA + B) = 87° 58' 18"] = 9.099728 + log tan [\{a - b) = 45°] = 10.00000 Rejecting 10 = 10.000283 = log tan ^c. .-. c = 90° 02' 14". 2. Given a = 135° 05' 28".6, c = 50° 30' 08".6, and b = G9° 34' 5G".2, to solve the triangle. 1st. To find a and c. The 3d and 4th of Napier's Analogies give, cos KA + C) : cos KA — C) : : tan lb : tan Ka + c) ; and sin KA + C) : sin KA — C) : : tan ^b : tan l{a — c). Computing by logarithms, we have ar. CO. log cos [i(A + C) = 92° 47' 48". 6] = 1.3116286 * ^ + log cos [KA - C) = 42° 17' 40"] = 0.8690535 + log tan [id = 34° 47' 28".l] = 9.8418527 Rejectmg 10 = 11.0225348 = log tan Ua + c). .'. ^{a + c) = 95° 25' 25". i{a + c)> 90% since cos KA + C) is -, cos KA - C) is +, and tan ib is +, making tan K<^ + c) —. * These logarithms are taken from 7-place tables, in order to obtain the tenths of seconds accurately. 104 SPHERICAL TRIGONOMETRY. ar. CO. log sin [KA + C) = 92° 47' 48".6] = 0.0005176 + log sin [KA - C) = 42° 17' 40 " ] = 9.8279768 + log tan lib = 34° 47' 28". 1] = 9.8418527 Rejecting 10 = 9.6703471 = ] ; tan ^{a — c). c) = 25° 05' 05". lia — c) < 90°, since the signs of the terms are all + . K« + c) + i{a - c)=a = 120° 30' 30", and i{a + c) — i{a- c) = c = 70° 20' 20". 2d. To find B. Either of the 1st two of Napier's Analogies will give B. Thus (1) becomes, cos i(a — c) : cos i{a + c) : : tan ^(A + C) : cot ^B ; and (2) sin i{a — c) : sin ^a + c) : : tan i(A — C) : cot ^B- But as i{a + c) is so near 90°, it will be better to use the second of these than the first. Or we may with equal accuracy use, sin c : sin 5 : : sin C : sin B. ar. CO. log sm (c = 70° 20' 20" ) = 0.0260878 + log sm (6 = 69° 34' 56". 2) = 9.9718202 + log sin (C = 50° 30' 08".6) = 9.887421 Rejectmg 10 = 9.8853290= log sm B. .-. B = 50° 10' 10". 3. Given a = 50° 45' 20", b = 69° 12' 40", and A = 44° 22' 10", to solve the triangle. 1st. To find B. sin a ; sin 5 ; ; sin A : sin B. ar. CO. log sin {a = 50° 45' 20'') = 0.1110044 + log sin {b = 69° 12' 40 ") = 9.9707626 + log sm(A= 44° 22' 10") = 9.8446525 Rejecting 10 = 9.9264195 = log sin B. .-. B = 57° 34' 51".4, and 122° 25' 08".6. There are two solutions, since a is intermediate in value between _?; and both b and 180° — b* * The determination of the species of B, or what is the same thing, the number of solutions, can usually be effected by a simple inspection without any computa- tion whatever. Thus, sin j9 = sin 5 sin A. the loga- rithms of which are given above, as ie log sin a. Now, as both a and p are < 90°, and log sin p < log sin a, p <:a. But a < 6, and also Icsa than 180° — b. All this can be seen at a glance. Napier's analogies. * 105 To find C and c of the larger triangle in which B = 57° 34' 51".4 Napier's 1st gives ar. CO. log cos [i(6 - a) = 9° 13' 40"] = 0.0056570 + log cos [i(6 + a) = 59° 59'] = 9.6991887 + log tan [KB + A) = 50° 58' 30".7] = 10.0912464 Rejecting 10 = 9.7960921 = log cot iC .-. C= 115° 57' 50". 7. Napier's 3d gives ar. CO. log cos [|(B - A) = 6° 36' 20".7] = 0.0028928 + log COS [i(B + A) = 50° 58' 30".7] = 9.7991039 + log tan [i(6 + a) = 59° 59'J = 10.23826 89 Rejecting 10 = 10.0403656 = log tan ^. .-. c = 95° 18' 16".4, 3d. To find C and c of the smaller triangle in which B = 122° 25' 08".6. Using the same formulm as before. ar. CO. log cos [\{h - a) - 9° 13' 40"] = 0.0056570 + log cos \\{b + a) = 59° 59'] = 9.6991887 + log tan [i(B + A) = 83° 23' 39".3] = 10.93627 03 Rejecting 10 = 10.6411160 = log cot iC .-. C = 25° 44' 31".6. ar. CO. log cos [KB - A) = 39° 01' 29".3] = 0.1096506 + log cos [KB + A) = 83° 23' 39".3] = 9.0008369 + log tan [K^» + a) = 59° 59'] = 10.23826 89 Rejecting 10 = 9.4087564 = log tan ^c. .-. c = 28° 45' 05".2. 154. ScH. — "When Napier's Analogies are used for solving Pro&.27i(Z (159), tlie most expeditious and elegant method of resolving the ambiguity, is by means of the analogies themselves. Thus, in the above example, after having found that B = 57° 34' 51". 4, or 122° 25' 08". 6, or both, a simple inspection of the anal- ogy next used will determine the number of solutions. Napier's 1st may be written •• ,^ cos K^ + «) , ,,„ cot iC =: ^ r-^ tan KB + A). cos i{b — a) ^ ' Now iC < 90°, hence cot ^C is + . If, therefore, neither of the values of B renders cot iC —, there are two solutions. If one value renders cot^C +, and the other — , there is one solution and it corresponds to the value of B which makes cot ^C +. If both values of B render cot ^C — , there is no solu- tion. In the last example, we see that cos [K^ + «) = 59° 59], and cos [^(6 — «) = 9° 13' 40"] are both + . Also tan [KB + A) = 50° 58' 30". 7, or 83° 23' 39".3, or both] is + for both values of B. Therefore there are two solutions. 106 ' SPHEKICAL TRIGONOMETRY. 4. Given A = 95° 16', B = 80° 42' 10", and a = 57° 38', to solve the triangle. 1st. To find &, sin A : sin B : : sin a : sin b. ar. CO. log sin (A = 95' IG) = 0.001837 + log sitt (B = 80° 43' 10") = 9.994257 + log sin (a = 57° 38) = 9.926671 Rejecting 10 = 9^922765 == log sin h. : b = 56° 49' 57", or 123° 10' 03", or both. 2d. To find c, tan ic = ^^i|^-±-^ tan i(a + b). Now for b = 56° 49' 57", COS ^v** — ^/ tan |c is + ; but for b = 123' 10' 03 " tan ^c is — ; hence there is but one solu- tion, and that corresponds to the smaller value of b. ai-. CO. log cos [KA - B)= 7° 16' 55"] = 0.003517 + log cos [i(A + B) = 87° 59' 05"] = 8.546124 + log tan [K« + b)- 57° 13' 58"] = 10.191.352 Rejecting 10 = 8.740993 = log tan U. .-. c r= 6'° 18' 19". 3d. To find C, we may use (1) or (2) of Napiei-'s Analogies, or sin « : sin c : : sin A : sin C, the last of whicn is the most expeditious. ar. CO. log sin {a = 57° 38) = 0.073329 + log sin (c = 6' 18' 19") = 9.040705 + log sin (A = 95° 16') = 9.998163 Rejecting 10 = 9.112197 = log sin C .-. C = 7° 26' 22' This value is taken for C instead of its supplement, since C is opposite the smallest side c. 5. Given a = 70° 14' 20", h = 49° 24' 10", and c = 38° 46' 10"; to solve the triangle. COMPUTATIOK. a = 70° 14' 20" b = 49° 24 10" c = 38' 46' 10" 5 = 158° 24 40" is = 79° 12' 20" ar. co. log sin = 0.007753 ^s- a = 8° 58' 00" " " = 9.192734 is- b = 29° 48' 10" " " = 9.696370 U- c = 40° 26' 10" " " = 9.811977 2)18.708834 .-. log k = 9.354417 Napier's analogies. 107 log tan iA = \ogk- log sin {is - a) + 10= 10.161683. .-. A = 110" 51 16". log tan iB = log A; - log sin (is - 5) + 10 = 9.658047. .-. B = 48* 56' 04". log tan iC =\ogk - log sin {is — c) + 10 = 9.542440. .-. C = 38° 26' 48". 6. Given a = 109° 55' 42", B = 116° 38' 33", and c = 120°43' 37", to solve the triangle. COMPUTATION. A = 109° 55' 42" B = 116° 38' 33" C = 120° 43' dl" S = 347° ir 52" iS = 173° 38' 56" ar. co. log cos = 0.002683 iS - A = 63° 43' 14" " " = 9.646158 iS - B = 57° 00' 23" " " = 9.730035 iS - C = 52° 55' 19" " " - 9. 780247 2)1 9.165123 .-. log K = 9.582561 log cot ia = log K - log cos (iS ~ A) + 10 = 9.936403. .-. a - 98° 21' 38". log cot \h = log K - log cos (iS - B) + 10 r= 9.846526. .-. b = 109° 50' 20". log cot ie = log K - log cos (^S - C) + 10 = 9.802314. .\c = 115° 13' 28". ScH. 1.— The student can use the exercises in the preceding section to famil- iarize the metliods here given. In doing so, it will be well for him to seek the most expeditious soUitious. He will find that Examjiles iinder Prob. 1 require 11 logarithms by Napier's Analogies and {135), and 12 logarithms by Napier's Rules and {135). Examples under Pjrob. 2, when there is but one solution, require 10 loga- rithms by Napier's Analogies and {135), and 12 logarithms by Napier's Rules and {135). When there are two solutions, 15 logarithms are required by Napier's Analogies and {135), and only 14 by Napier's Rules alone, or by these rules and {135). Examples under Prob. 3 require but 7 logarithms by the method given in this section and 13 by the previous method. ScH. 2. — In cases in which the angles or sides are near the limits 0°, 90°, or J80°, so that the functions used in the particular solution change very rapidly in proportion to the arc, it is often possible to select one among the several methods which will give more accurate results than the others. There are also other methods which are better adapted to such cases than those here given. For these, as well as for much other interesting matter, and especially for the discussion of the General Spherical Triangle, American students have an excellent resource in the treatise of Professor Chauvenet of Washington Univer- sity, St. Louis. 108 SPHEEICAL TKIGONOMETRY. SECTION IV, AREA OF SPHERICAL TRIA^SGLES. 155, JProh, — Having the angles of a S2}lierical triangle given, to find the area. Solution. — [The solution is given in Pabt II. {613), and we simply re- produce the result in order to give completeness to this section.] The area is equal to the ratio of the spherical excess to 90°, or ^tt, into the trirectangular tri- angle. That is, letting the sum of the angles be S°, the area K, and the radius of the sphere 1, whence the area of the trkectangular triangle is ^tt, we have K = ^-^ X ^7t = S-n. In the latter expression S is the sum of the angles in terms of the radius, i. e.j CO CO ^ = 5r:29578' ^' ^PP^o^i^^tely, S = .-^ {9). EXERCISES. 1. "What is the area of a spherical triangle whose angles are 100°, 58°, and 62°, on a sphere whose diameter is 6 feet ? Solution. K = S — ;r = -^ — 3.14159 = .698, the area of a similar ti'i- angle on a sphere whose radius is 1. Hence, the area of the required triangle is .698 X 3* = 6.282. [The method given in Pakt II. (613) is more expedi- tious, but it is our purpose to illustrate the form here given.] 2. What is the area of a si3herical triangle whose angles are 170°, 135°, and 115°, on a sphere whose radius is 10 feet? Ans. 418.875 square feet. 3. What is the area of a spherical triangle whose angles are 150°, 110°, and G0°, on a sphere whose radius is 3 feet ? ISO. J^rob. — Having the sides of a sjyherical triangle given, to find the area. Solution. — The angles may be found by {14:8)^ and then the area by {15S). AREA OF SPHERICAL TRIANGLES. 109 But a more direct method is to find the spherical excess by means of LhuiU Iter's formula, which we will now produce. ^K = :l(A + B + C - TT) Whence tan ^K = tan ^[A + B + C - :nr] = tan [KA + B) - \(7t - C)] _ sin i( A + B) - sin \{Tt — C ) ~ cos KA~+ B) + cos \{Ti - C) ' (7, page 3t\ _ sin i(A + B) — cos ^C ~ cos i(A + B) + sin iC = [coaK^-6)-cos-k]cosiC_ ^^ ^^^ ^^^ 3^^^ [cos \{(i + 6) + cos \c\ sin iC _ cos \{a — b) — cos |c / sin js sin {U — c) (146 Id"^) ~~ cos ^a + b) + cos ic r sin (|s -- a) sin Us — b) ' sin i(a + c - &) sin i{b + c — a) / sin -^^ sin (-^g — c) „ p, ^, cos i(a + b + c) cos i(a + 6 — c) r sin (il-s — a) sin (U — b) ' ' sin=^ -Kis — 6) sin^ His — a) sin |8 sin (is — c) , _ a . \ cos' :i« cos'^ i(i.s — c) sin (^s — a) sin (^s — 5) ' "" sin*^ ii^s — d) sin^ i{U — a) sin is c os ^s s in jd^ — c) cos -Ki^ — <^) ,^^v cos'' is cos=^ i(is— c) sin i(is— a) cos i(is— «) sin i{is—b) cos i(i«— 6) .-. Tan iK =* .y/tan is tan i(|s - a) tan i(is - b) tan i(i« - c). (A) Having found K, the spherical excess, or what is the same thing, the area of a similar triangle on a sphere whose radius is 1, we have but to multiply K by the square of the radius in any given case. EXERCISES. 1. Given a = 98°, b = 110°, and c = 115°, to find the area of a spherical triangle, on a sphere whose radius is 4000 miles. COMPUTATION". log tan {\s = 80" 45') = 10.788185 + log tan [{\s-ia) = 31° 45'] = 9.791563 + log tan [{is-U) = 25° 45] = 9.683356 + log tan [{\s-ic) z= 23° 15] = 9.633098 2)39.896203 Rejecting 10 = 9.948101 = log tan ^K . .-. K =166° 20' 20" * The + is always taken ; otherwise, iK being > 90°, K would be i> 360' which is impoa- eible. (Part III., 256). 110 SPHERICAL TRIGONOMETRY. Whence area = i?^^— ' . irt (4000)' = ^^ . i^ (4000)» = 46,450,440, nearly. 2. Given a = 70° 14' 20", 5 = 49° 24' 10", and c = 38° 46' 10", to find the area of a spherical triangle on a sphere whose diameter is 8 feet. -^fis. 5.1, nearly. lo7. I*roh. — Having fico sides and their included angle given in a spherical tria^igle, to find the area. Solution.— Compute the other two angles by Napier's Analogies, and find rtM ^ , ^ , ^ cot ^ cot ^ + cos C . the area from the angles. [The formula cot ^K = = r^^ gives the spherical excess in terms of two sides and their included angle ; but it is of no practical value for finding the area, as it is not adapted to logarithmic compu- tation. For the manner of producing it and several other forms for K, see Todhunter's Spherical Trigonometiy, {103)]. PRACTICAL APPLICATIONS. [Note. — The three following problems are given merely to indicate to the student some departments of investigation in which Spherical Trigonometry is of essential service. The two sciences to which this branch of Pure Mathe- matics is indispensable, are Geodesy, or the mathematical measurement of the earth, and Astronomy.] ^roh, l.—To find the shortest distance on the earth^s surface be- tween two 2^oi7its whose latitudes and longi- tudes are knoiun. Sug's. — The shortest distance on the surface be- tween two points is the arc of a great circle joining the points. Hence, the Problem is : Given two sides (the co-latitudes) and the included angle (the differ- ence in longitude), to find the third side. Fig. 71. Ex. 1. Berlin is situated in Lat. 52° 31' 13" N., Lon. 13° 23' 52" R, and Alexandria, Egypt, in Lat. 31° 13' X., Lon. 29° 55' E. Wliat is the shortest distance in miles on the earth's surface between them, the earth being considered a sphere whose radius is 3962 miles ? Ans. 1G91.96 miles. PRACTICAL APPLICATIONS. Ill Ex. 2. A ship starts from Valparaiso, Chili, Lat.33° 02' S., Lon. 71° 43' W., and sails on the arc of a great, circle in a northwesterly direc- tion 3840 miles, when her longitude is found to be 120° W. What is her latitude ? Ajis. 52' 48" S. Ex. 3. A ship starts from Eio Janeiro, Brazil, Lat. 22° 54' S., Lon. 42° 45' W., and sails in a northeasterly direction on the arc of a great circle 624.44 miles, when her latitude is found to be 50° N. What is her longitude ? Ans. 2° 01' 14" W. JProb, 2. — To find the time of day frcmi the altitude of the sun, Sug's.— Let NESQ represent the projection of the concave of the heavens upon the plane of the meridian of observation. The equator of this concave sphere is simply the intersection of the plane of the earth's equator with this imaginary concave sphere, and its axis is the prolongation of the earth's axis, the poles being the points N and S where the axis pierces the imaginary concave. EQ is the projection of this celestial equator, NS the axis or the projection of a great circle perpendicular to the meridian of the observer (NESQ) and to the equator, HO the projec- tion of the horizon, and ZZ' the projection of the prime vertical (that is, a great circle of the heavens passing through the zenith of the observer and the east and west points in his horizon). Now let S' be the place of the sun at the time of observation. RS', the sun's declination, is known from the almanac ; LS', the sun's altitude, is measured with the sextant (or other instrument) ; and EZ is the latitude of the observer. Hence, in the spherical triangle ZNS' we know the three sides, viz., NS' = the co-declination of the sun, ZS' = the co-altitude of the sun, and ZN = the co- latitude of the observer. We may therefore compute the angle ZNS', which reduced to time gives the time before or after noon as the case may be. Ex. 1. On April 21st the master of a ship at sea in latitude SO"* OG' 20" N., observed the altitude of the sun's centre at a certain time in the forenoon and found it to be 30° 10' 40", and looking in the almanac found the sun's declination at that time to be 12° 03' 10" N. What was the time of day ? COMPUTATION. 90° - 30° 10' 40" = 90° - 12° 03' 10" = 90° — 39° 06' 20" = 59° 40' 20" 77° 56' oO" 50° 53' 40" 2 )188° 39' 50" 94° 19' 55 " a. c. log sin 94° 10' 55" ~ a. c. log sin 34° 30' 35" = log sin 16° 23' 05" = log sin 43° 26' 15" = 0.001242 0.246765 9.450381 9.837312 2) 19.535:0') log tan 30° 22' 03".3 = 9.707650 112 SPHERICAL TRIGONOMETRY. Therefore i the hour angle NNS' = 30' 2Z' 03 '.3, and the hour angle is 60° 44' 07". This reduced to time at 4 minutes to a degree, gives 4 h. 2 m. 56 s. be- fore noon, or 7 h. 57 m. 4 s. a. m. Ex. 2. In latitude 40° 21' X., when the declination of the sun is 3° 20' S., and its altitude 36° 12', what is the time of day ? Ans. 9 h. 42 m. 40 s. A. M. Ex. 3. In latitude 21° 02' S., when the sun's declination was 18° 32' X., and the altitude in the afternoon 40° 08', what was the time of day? Ans. 2 h. 3 m. 57" P. m. JProb. S,—To fijul the time of sunrising and sunsetting at any given place on a given day. Sug's. — The projection being the same as before, let M'RS'M represent the ap- parent diurnal path of the sun. Since S'M is described in 6 hours, the time taken to describe RS' is the time before 6 o'clock, at which the sun rises, i. e., passes the horizon HO. But the time requisite to describe RS', is the same part of 24 hours (360' angular measure) that the angle CNL (= arc CL) is of 360". Hence, the arc CL, in time, is the time before 6 o'clock at which the sun rises. In a similar manner, C/, in time, is seen to be the time after 6 o'clock when the sun is south of the equator. The solution of the prob- lem, therefore, consists in finding CL. Now, in Fig. 73. the triangle RLC, right angled at L, LR = the sun's declination at the time, and angle RCL = ECH = the co-latitude of the place.* From these data CL is readily found. Ex.1. Eequired the time of sunrise at latitude 42° 33' N., when the sun's declination is 13° 28' N. COMPUTATIOK". cot 47° 27' = 9.962813 tan 13° 28' = 9.379239 sin 12' 41' 52" = 9.342052 (12° 41' 52") X 4 gives the time before 6 o'clock as 50' 47". .'. The sun rises at 5 h. 09 m. 13 s. * This may be seen thus : Suppose a person to start from the equator at E and travel north. When he is at E. the south point of hia horizon (H) is at S ; and for every degree he goes north, the south pole (S) sinks a degree below his horizon. Hence, HCS = ^is latitude, and ECH = co-latitude. PRACTICAL APPLICATIONS. 113 Ex. 2. Required the time of sunrise at latitude 57° 02' 5-4" X., when the sun's declination is 23° 28' JST. Sun rises at 3 h. 11 m. 49 s. Ex. 3. How long is the sun above the horizon in latitude 58° 12' N., when the sun's dechnation is 18° 41' S., that is about Januaiy 25th ? Ans. 7 h. 35 m. 36 s. Ex. 4. What is the length of the longest day at Ann Arbor, Mich., Lat. 42° 16' 48".3, the sun's greatest declination being 23° 27' ? Ans. 15 h. 05 m. 50 s. [Note. — In such problems as the foregoing, several small corrections have to be made in order to entire accuracy, such, for example, as that for refraction in taking the altitude, and for the time required for the sun's disk to pass the hori zon. But they would be out of place here. ] THE EIs-D. INTRODUCTION TO THE TABLE OF LOGARITHMS [Note. — If the student understands the nature and use of logarithms so as to be able to use the common tables, it will not be necessary that he should read this introduction. Otherwise a careful study of it will be needful before reading Section 4 of the Plane Trigonometry.] i. A Lofjavith/in is the exponent by w^liich a fixed number is to be affected in order to produce any required number. The fixed number is called the Base of the System. III. Let the Base be 3 : then tho logarithm of 9 is 2 ; of 27, 3 ; of 81, 4 ; ol 19623, 9 ; for 3" = 9 ; 3=* = 27 ; 3* = 81 ; and 3' = 19083. Again, if 64 is the base, the logarithm of 8 is -J-, or .5, siuce 64", or 64' = 8 ; i. e., |, or .5 is the exponent by which 64, the base, is to be affected in order to produce the num« ber 8. So also, 64 being the base, ^, or .333 + , is the logarithm of 4, since 64 , or g^.33 3 + _ 4. ^ g^ 1^ Qj, 333 ^^ ig tij(j exponent by which 64, the base, is to be f affected in order to produce the number 4. Once more, since 64 , or 64''^^ -^ = 16, !, or .666 +,is the logarithm of 16, if the base is 64. Finally, 64~" or 64~"' = I, or .125 ; hence — |, or — .5, is the logarithm of \, or .125, when the base is 64. In like manuer, with the same base, — i, or — .333 +,is the loga- rithm of i, or .25. EXAMPLES. 1. If 2 is the base what is the logarithm of 4 ? of 8 ? of 32 ? of 128? of 1024? Solution. 7 is the logarithm of 128, if 2 is the base, since 7 is the exponent by which 2 is to be affected in order to produce the number 128. 2. If 5 is the base, what is the logarithm of 625? of 15625? of 125? of 25? a INTRODUCTION TO THE TABLE OF LOGARITHMS. 3. If 10 is the base, what is the logarithm of 100 ? of 1000 ? of 10,000? of 10,000,000 ? 4. If 2 is the base, what is the logarithm of J, or .25 ? of -J, or .125 ? of ^, or .03125 ? Ans. to the last, — 5. 5. If 8 is the base, of what number is |, or MQ + the logarithm? of what number is |-, or 1.333 +, the logarithm ? of what number is 2 the logarithm? of what number is 2\, or 2.333" + ? of what num- ber 3|, or 3.666 + ? Ans. to the last, 2048. ScH. Since any number with for its exponent is 1, the logarithm of 1 is in all systems. Thus 10* = 1, whence is the logarithm of 1, in a system in which the base is 10. 2. A System of LoffCirWuns is a scheme by which all num- bers can be represented, either exactly or approximately, by expo- nents by which a fixed number (the base) can be affected. 3. There are Ttvo Systems of Logarithms in common use, called, resi^ctively, the Briggean or Common System, and the Kajyierian or Hyperlolic System. The base of the former is 10, and of the latter 2.71828 -f. In speaking of logarithms of numbers, the com- mon logarithm is always signified, if no specification is made. 4. The logarithms of all numbers, except the exact powers of the base, indicate a power of a root, and are consequently fractional and usually only approximations. It is customary to write them in the form of decimal fractions. The integral part is called the Char- acteristic, and the fractional part the 3Iantissa. The charac- teristic can always be told by a simple inspection of the number itself; hence only the mantissa is commonly given in the table. 5» JProp. — TJie characteristic of the common logarithm of any mimber greater than icnity, is one less than the number of integral figures in the given number. III. The logarithm of 4685 is more than 3, because 10' = 1000, and less than 4, because 10* = 10,000; hence it is 3 + a fraction. The same method may be pursued to determine the characteristic of the logarithm of any other number greater than unity, and the truth of the proposition be observed. Thus the logarithm of 2o645.827 is 4, since the number lies between the 4th and 5th powers of the base, 10. 6*. I^rojy* — Tlie mantissa of a decimal fraction^ or of a mixed number, is the same as the mantissa of the number considered as integral. INTRODUCTION TO THE TABLE OF LOGARITHMS. 3 Dem. Suppose it is known log 2845672 = 6.454185. This means thai 10'*'*"* _ 2845672. Dividing by 10 successively, we have 1Q5.454185 ^ 284567.2, or log 284567.2* = 5.454185, 10* •*"»"» = 28456.72, or log 28456.72 = 4.454185, ^Qs.-,54i86 ^ 2845.672, or log 2845.672 = 3.454185, 10'-*"*" = 284.5672, or log 284.5672 = 2.454185, 10>-45«ie. ^ > 28.45672, or log 28.45672 = 1.4.54185, • io°-""«^ = 2.845672, or log 2.845672 = 0.454185. Now if we continue the operation of division, only writing 0.454185 — 1, 1.454185, meaning by this that the characteristic is negative and the mantissa positive, and the subtraction not performed, we have lQ-464 186 ^ 0.2845672, or log 0.2845672 =1.454185, lQ»:464ie6 ^ 0.02845672, or log 0.02845672 = 2^454185, lQ^464i85 ^ 0.002845672, or log 0.002845672 = 8.454185, etc., etc. Q. E. D. 7. Cor. T7ie characteristic of a numler consistiiig e^itirely of a decimal fraction, is negative, and one more than the numler of O's i7nmecliately following the decimal point, as appears from the last demonstration, or from the fact that 1"* = -jig- = .1 ; 10~' = ^hr = .01 ; 10-" = T^ = .001 ; etc., etc. 8, One of the most important uses of logarithms is to facilitate the multiplication, division, involution, and extraction of roots of large numbers. These processes are performed upon the following principles : 9» l?rop» 1. — The sum of the logarithms of two numbers is thb logarithm of their product. Dem. Let a be the base of the system. ' Let m and n be any two numbers whose logarithms are x and y respectively. Then by definition «■= = m^ and ay = n. Multiplying these equations together we have a^-^y = mn. Whence a; + y is the logarithm ofmn. q. e. d. 10. JProj)* 2. — The logarithm of the quotient of two numbers is the logarithm of the dividend minus the logarithm of the divisor. Dem. Let a be the base of the sj'^stem, and m and n any two numbers whose logarithms are, respectively, x and y. Then by definition we have rt^ = m, and a" = n. Dividing, we have a^-v = _. Whence x-y\^ the logarithm of ^-. Q. E. D. * This is the common abbreviation indicating the logarithm of a number, and should bo read "logarithm of 284567.2," not ''log 284567.2," which ie grossly inelegant. i INTKODUCTION TO THE TABLE OF LOGAEITH^diS. 11, I^i'Oj), 3. — TJie logarithm of a jjoiuer of a number is the logarithm of the number 7mdtiplied by the index of the jJOicer. Dem. Let a be tlie base, and x the logarithm of m. Then cF = m\ and raising both to any power, as the 2th, we have a^* = wr. Whence xz is the logarithm of the 2th power of m. q. e. d, 12, I^rop, 4, — The logarithm of any root of a number is the logarithm of the number divided by the number exjjressing the degree of the root. Dem. Let a be the base, and x the logarithm of m. Then a^ = m. Extract* - 2 — 2" 2/— ing the 2th root we have a""— \/m. Whence - is the logarithm ofVm. Q. e. d. TABLE OF LOGARITIDIS. IS. In order to apply the above principles practically, we need what is called a Table of Logarithms. That is, a table from which we can readily obtain the logarithm of any number, or the number corresponding to any logarithm. Such a table is given on pages 11 to 28. For methods of computing it, the student is referred to algebra. Its nature and manner of use will be learned from the two following problems : 14. Prob,— To fiid the logarithm of a number from the table. Solution. The logarithm of any number hehceen 1 and 100 is seen directly from the table on page 11. The column marked N contains the numbers, and the adjacent column to the right gives the logarithm of the corresponding number \o G places of fractious. Thus, log 7 = 0.845098 ; log 33 = 1.518514. The mantissa of the loganthm of any number exjJi'cssed with three figures is found in the column headed 0, on some one of the pages from 12 to 26 inclusive. The given number being found in the column marked N, the mantissa of its logarithm stands opposite. Tlie characteristic can be determined by (5), (6), (7). When but four figures are found opposite the number in the column, the two left-hand figures of the mantissa are the same as in the next mantissa above, which has six. Thus, log 443 = 2.646404. To find the logarithm of a number consisting of four figures. Let it be required to find the logarithm of 293G. Looking for 293 (the first three figures) in tlie column of numbers, and then passing to the right until reaching the column headed 6, the fourth figure, we find 7756, to wliich prefixing the figures 4f5, \7iiich belong to all the logarithms following them till some others are indicated, We have for the mantissa of the logarithm of 2936, .467756. But, as 3 is the INTRODUCTION TO THE TABLE OF LOGARITHMS. 5 logarithm of 1000, and 4 of 10,000, log 3930 is 3 and this decimal, or log 2936 = 3.467756. Tojind the logarithm of a number consisting of more than four figures. Let it be required to find the logarithm of 2815672. Finding the decimal part of logarithm of the first four figures 2845, as before, we find it to be .454082. Now the logarithm of 2846 is 153 (millionths, really) more than that of 2845, as found in the right-hand column, marked D. Hence, assuming that if an increase of the number by 1000 makes an increase in its logarithm of 153, an increase of 672 in the number will make an increase in the logarithm of tWtj, or .672 of 153, or 103, omitting lower orders, and adding this to .454082, we have .454185 as the mantissa of log 2845672. The integral part is 6, since 2845672 lies between the 6th and 7th powers of 10. Hence, log 2845672 = 6.454185. q. e. d. ScH. 1. If in seeking the logarithm of any number, any of the heavy dots noticed in the table are passed, their places are to be filled with O's, and the first two figures of the decimal of the logarithm taken from the column in the line below. Thus, log 3166 is 3.500511. This arrangement of the table is merely a convenient method of saving space. ScH. 2. The column marked D is called the column of Tabular Differences ; and any number in it is the difference between the mantissas found in columns 4 and 5, which is usually the same as between any two consecutive logarithms in the same horizontal line. The assumption made in using this difference, viz., that the logarithms increase in the same ratio as the numbers, is only approxi- mately true, but still is accurate enough for ordinary use. EXERCISES. 1. Find the logarithm of 2200. ....... Logarithm, 3.342423. 2. Find the logarithm of 24.3G Logarithm, 1.38G677. 3. Find the logarithm of 2.698 Logarithm., 0.431042. 4. Find the logarithm of 3585.9 Logarithm, 3.554598. 5. Find the logarithm of 42.6634 Logarithm, 1.630050.-1 6. Find the logarithm of 331.957 Logarithm, 2.521082.^.^ 7. Find the logarithm of 2519.38 Logarithm, 3.401294. -^ 8. Find the logarithm of .538329 Logarithm, 1.731047. , 9. Find the logarithm of .087346 Logarithm-, 2.941243. 10. Find the logarithm of .007389 Logarithm, 3.86858G. 15, ScH. 3. It will be observed that the tabular difference varies rapidly at the beginning of the table; hence, for numbers between 10000 and 11000 it is better to use the last two pages of the table. IG. JProh 2. — To find a number corresponding to a given hviarithm. 6 INTEODUCTION TO THE TABLE OF LOGARITHMS. Solution. Let it be required to find the number corresponding to the iofrarithm 5.5152G4 Looking in the table for the ?iext less mantissa, we find .515211, the number corresponding to which is 3275 (no account now being taken as to whether it is integral, fractional, or mixed ; as in any case the figm'es will be the same). Now from the tabular difference, in column D, we find that an increase of 133 (million ths, really) upon this logarithm (.515211), would make an increase of 1 in the number, making it 3276. But the given logarithm is only 53 greater than this, hence it is assumed (th ^ugh only approximately correct) that the increase of the number is ■f'h of 1, or 53 -r- 133 = .3984 + . This added (the figures annexed) to 3275, gives 32753984 + . The characteristic, being 5, indicates that the number lies between the 5th and 6th powei^s of 10, and hence has 6 integral places. .'. 5.515264 = log. 327539.84 + . Q. e. d. EXERCISES. 1. Find the number wliose logarithm is 1.240050. Number, 17.38. 2. Find the number whose logarithm is 2.431203. Number, 269.9. 3. Find the number whose logarithm is 3.503780. Number, 3189.91. 4. Find the number whose logarithm is 0.138934. Number, 1.377. 5. Find the number whose logarithm is 1.368730. _ Number, .233738. 6. Find the number whose logarithm is 2.448375. ^Number, .028078. 7. Find the number whose logarithm is 3.538630. Number, .003456. 8. Find the number whose logarithm is . 843970. _ Num.ber, 6.98184. 9. Find the number whose logarithm is 1.867372. iVl^mJer, .736837. 10. Find the number whose logarithm is .003985. _ Number, 1.00921. 11. Find the number whose logarithm is ¥. 723460. Number, .005290. APPLICATIOS 1. Find, by means of logarithms, the product of 57.98 by 18. SoLUTiojr. As the logarithm of the product equals the sum of the logarithms of the factors \9\ we find the logarithms of 57.98, and 18 from the table, and INTRODUCTION TO THE TABLE OF LOGARITHMS. 7 adding Ihern together, find the number corresponding to the sum. The latter number is the product required. Thus, log 57.98 = 1.763278 log 18 = 1.22o273 3.018551 = log 1043.64. 2. Multiply 23.14 by 5.062. 3. Multiply 0.00563 by 17. 4. Multiply 397.65 by 43.78. 5. Multiply 0.3.854 by 0.0576. 6. Find the continued product of 3.902, 597.16, and 0.0314728. 7. Multiply 832403 by 30243. 8. Multiply 9703407 by 90807. 9. Multiply 3.47 by 9.83. 10. Multiply 12.763 by 10.976. [Note. The examples in division below will offer additional exercise, if necessary.] Prod. 117.1347. Prod. 0.09571. Prod. 17409.117. Prod. 0.022199. M6, Prod. 73.3354. Prod. 25174363929.* Prod. 881137279449. Prod. 34.1101. Prod. 140.086688.* 1. Divide 24163 by 4567. Solution. Since the logaiithm of the quotient equals the logarithm of the dividend minus the logarithm of the divisor, we have the following operation t log 24163 = 4.383151 log 4567 - 3.659631 0.723520 = log 5.29078, which number is the quotient. 2. Divide 56.4 by 0.00015. Operation, log 56.4 = 1^751279 log 0.00015 = 4176091 Difference of log's = 5.575188 .*. The quotient is 376000. SuG. Observe that only the cJiaracteristic of the logaritlun is negative, and that in subtracting we are to regard the nature of the logarithmic numbers a« positive or negative. 3. Divide 461.02876 by 21.4. 4. Divide 25.49052 by 2.46. 5. Divide 17610.8248 by 37.6. 6. Divide .00144 by 1.2. 7. Divide .0000025 by .005. A71S. 21.5434. Alls. 10.362. Ans. 468.373. A71S. .0012. Ans. .0005. 8 • I^'TRODUCTION TO THE TABLE OF LOGARITHMS. 8. Divide 43.2 by .24. Ans. ISO 0. Divide 59.74514 by 1.36. Ans. 43.93025. 10. Divide .0001728 by 2.4. A)is. .000072. [Note. The examples in multiplication given above will afford additional exercise, if necessarj',] . 17. ^cu.— Arithmetical Comjyleinefit.—The arithmetical complement of a number is simply the remainder after subtracting the number from some particular fixed number. Thus, the a. c. of 5 with reference to 9 is 4 ; of 3. 6 ; of 7, 2; etc. The a. c. of 7 with reference to 10 is 3 ; of 4, 6 ; of 2, 8 ; etc. When required to subtract one number from another, we may, if we choose, add its a. c. and then subtract the number with reference to which the a. c. is taken. This process will give the true difference. Thus, if we are to subtract 6 from 9, we may add to 9 what 6 lacks of being 10 (10 — 6 = 4, the a. c. of 6 with reference to 10) and then subtract 10. 9 — 6 = 9 + 4 — 10. A few such questions as the following will render this simple process familiar. What number must I add to 57G, in order that I may subtract 100 from the sum, and get the same remainder as if I had subtracted 58 in the first instance ? Again, if I wish to take 87 from 160, what must I add to the latter, in order that I may subtract 40 from the result, and get the difference sought ? This principle is sometimes used in computing by means of logarithms. It is especially convenient when there are several multipliers and several divisors involved in the same computation. An example or two will make the process familiar. The complements of logarithms are usually taken with reference to 10. If the logarithm exceeds 10, 20 may be used, etc. 18. Eeqnired the result of the following combinations : 346 X 27.8 -^ 1156 X 3426 ~ 2.004 X 27 -^ 11.05. Operation. log 846 = 2.539076 log 27.8 = 1.444045 a. c. log 1156 = 6.937042 log 3426 = 8.534787 a. c. log 2.004 = 9.698102 log 27 = 1.431864 a. c. log 11.05 = 8.956638 84.541054 Rejecting 30.00000 as three complements are used. 4.541054 is the logaritlim of the re- quired result. .-. As 4.541054 = log 34757.92, the latter is the result sought [Note. The preceding examples can be used to familiarize this principle if thought desirable.] Suo. The a. c. of 2.468216 is 11.531784, since 2 is negative. An a. c. can bo written directly from the table with nearly the same ease as the logarithm itself, by writing from, left to right, and taking each figure from 9, except the INTRODUCTION TO THE TABLE OF LOGAKITHMS. 9 riglit-liand one, which is to be taken from 10. Thus, if the characteristic is 3, we write 6 • the next figure being 2, write 7 ; for 4, write 5, etc. 1. What is the cube of 32 ? Solution. Since the logarithm of the cube of a number is three times the logarithm of the number itself {IT), we have » log (32)' = 3 log 32 = 4.515450 = log 32767.9T, which number is the cube of 32, as accurately as the process gives it. (32)' by multiplication = 32768. 2. What is the cube root of 7896.34? Sua Log (7896.34)^ = ^ log 7896.34 - 1.299142 = log 19.913. .'. (7896.34)^ = 19.913. (Seei;^.) 3. What is the 20th power of 1.06 ? Ans, 3.2071. 4. What is the 5th poAver of 2.846 ? 5. What is the 5th root of 31152784.1 ? Ans. 31.52+. 6. What is the cube root of 30 ? Ans. 3.107 + . 7. What is the cube root of .03 ? SuG. Log .03 ="2.477121. Now to divide this by 3, we have to bear in mind that the characteristic alone is negative; i. e., 2.477121 = — 2 +_.477121, or — 1.522879. This divided by 3 gives - .507626, or - .507626 = 1.492374. But a more convenient metliod of effecting this division is to write for the — 2, -3 +_1, whence we have for "2.477121,-3 + 1.477121, which divided by 3 gives 1.492374, nearly. 8. Divide 3^261453 by 2, by 4, by 5. Last quotient, 1.4522906. 9. What is the 4th root of .00000081 ? Ans, .03. 10. What is the 7th root of 0.005846? Ans. .4707. 1. If 28.035 : 3.2781 : : 3114.27 : x, what logarithmic operations will fiud X ? 'SuG. The logarithm of the product of the means is the sum of their loga- rithms ; and the logarithm of the quotient of this product divided by the first extreme, is the logarithm of said product minus the logarithm of the other extreme. .-. log x = log 3.2781 + log 311427 - log 28.035 = 0.515622 + 3.493356 - 1.447700 = 2.561278. Hence, x = 364.1478 + . ^ 2. Given 72.34 : 2.519 : : 357.48 : x, to find x, by logarithms. X = 12.448. 3. Given 6853 : 489 : : 38750 : .r, to find x, by logarithms. a: = 2765. 10 INTRODUCTION TO THE TABLE OF LOGAKITHMS. SuG. The most elegant way to solve such propositions by logarithms is to take the sum of the logarithms of the means and the a. c. of the given extreme and reject 10, The result is log x. 4. Given 497 : 1891 : : 37G : x, to find re, using the a. c. log. Operation. log 1891 = 3.37G693 log 37G = 2.575188 a. c. log 497 = 7.303644 Sum, less 10 = 3.155524 = log 1430.62. .'. x =,1430.63. [Note. Solve the preceding in a similar manner, by using a. c. log.] Let the student give the reasons for the following : 1. Given (|)2 ^ {^Y = x, we have 2 log 2 = 0.602060 'i log 16 = 0.903090 a. c. 2 log 3 = 9.045758 a. c. Uog 5 = 9.475773 Sum, less 20 = 0.026681. .-. x = 1.0033+ . 2. Given V\ : x : : (3^)2 : ^/¥, to find x. Log x = 2 log 3 + i log 2 + § log 6 + a. c. 2 log 10 + a. c. i log 5 - 20 = .274039. .-. cc = 0.1879 + . 3. Given \/Tl5 X V^'^ : (0.0051)' :i x : j^^v Log a; = i log 115 + h log .016 + ^ log .32 + a. c. f log 1146 + a. c 2 log 0051 - 20 = 2.729701. .-. x = 536.66 +. TABLE I CONTAINING lOGAIlITHMS OF NUMBEES Feom 1 TO 11,000. N. Log. N. Log. N. Log. ' N. Log. 1 0-000000 26 1-414973 51 1-707570 j 76 1.880814 2 0'3oio3o 27 i-43i364 52 63 1-716003 77 1-886491 3 0-477I2I 28 1-447158 1-724276 1 78 1-892095 4 o.6o2o6o 1 29 1-462398 54 1-732394 79 1-897627 5 0-698970 30 I •47-7121 55 1-740363 80 1-908090 6 0-778151 0-845098 31 1-491362 56 1-748188 1 81 1-908485 7 32 1 -505150 57 1-755875 82 1-913814 8 0-903090 33 i-5i85i4 58 1-763428 83 1-919078 9 0-954243 34 i-53i479 59 1-770852 84 1-924279 10 I • 000000 35 1-544068 60 1.778151 85 1-929419 11 1-041393 36 1 -556303 61 1-785330 86 1-934498 12 1-079181 37 1-568202 62 1-792892 87 1 -989519 13 1-113943 38 1-579784 . 63 I -799341 88 1-944483 14 1-146128 39 1-591065 64 1-806180 89 1-949890 15 1-176091 40 1-602060 65 1-812913 90 1.954243 16 1-204120 : 41 1-612784 66 I -819544 91 1.959041 17 1-230449 1 42 1-623249 67 1-826075 92 1-968788 18 1-205273 43 1-633468 68 1-832509 1-3 1-968483 19 1-278754 44 1-643453 69 1-838849 94 1-978128 20 1 -301030 45 1-653213 70 1-845098 95 1-977724 21 1-322219 46 1-662758 71 I-85I258 96 I -982271 22 1-342423 47 1-672098 72 1-857333 97 1-986772 23 1-361728 48 I -681 241 73 1-863323 98 1-991226 24 1-3802II 49 1-690196 74 1-869282 , 99 1-995635 25 1-397940 50 1 -698970 75 1-875061 100 2 - 000000 Remark.— In the following Table, the first two figures, in the first column of Logarithms, are to be prefixed to each of the numbers, in the same horizontal line, in the next nine columns; but when a point (•) occurs, a is to be put in its place, and the two initial figures are to be taken from the next line below. 12 LOGARITHMS OF NUMBEKS. N. 1 ' 1 2 3 4 5 6 7 8 9 D. 100 101 102 103 104 105 106 107 108 109 1 4321 86oo 012837 7033 i 021 iSo ; 53o6 1 o384 03J424 7426 0434 4751 f.^ 7451 i6o3 57.5 0789 3826 7825 0868 5i8i 0451 368o 7868 2016 6125 •195 4227 8223 i3oi 5609 9876 4100 8284 2428 6533 •600 4628 8620 1734 6o38 •3oo 4521 8700 2841 6942 1004 5029 9017 2166 6466 •724 4940 9116 3252 7350 1408 5430 9414 2598 6894 1147 5360 9532 3664 7737 1812 5830 9811 ^029 7321 1D70 5779 9947 4073 8164 2216 623o •207 3461 7748 1993 '.It] 4486 8571 2619 6629 •602 3891 8174 24i5 6616 4896 8978 3021 7028 •998 432 428 424 f^6 412 408 404 400 396 110 111 112 113 lU 115 116 117 113 119 041393 5323 9218 053078 6905 060698 4438 8186 071882 5547 1787 5714 9606 3463 7286 1075 4832 8557 2250 5912 2182 oio5 3«46 7666 1452 5206 8928 2617 6276 2576 6495 •38o 423o 8046 1829 558o %n 6640 •766 46i3 8426 2206 5953 9668 3352 7004 3362 88o5 2582 6326 ••38 3718 7368 3755 7664 i538 5378 9185 2958 6699 •407 4o85 773i 4148 8o53 1924 5760 9563 3333 7071 •776 445 1 8094 4540 8442 2309 6142 9942 3709 7443 1 145 4816 8457 4932 883o 2694 6524 •320 4o83 7bi5 i5i4 5i82 8819 393 IP. 382 372 363 120 121 122 123 124 125 126 127 123 129 079 1 81 082785 636o 9905 093422 6910 100371 3804 7210 1 1 0390 9543 3i44 6716 •258 3772 7257 0715 4146 7549 0926 4122 7604 1059 4487 7888 1263 •266 3861 7426 •963 4471 7951 i4o3 4828 8227 1599 •626 4219 i3i5 4820 8298 1747 5169 8563 1934 4576 8i36 1667 5i6g 8644 2091 55io 8903 2270 i347 4934 8490 2018 55j8 8990 2434 585i 9241 26o5 1707 5291 8845 2370 5866 9335 2777 t»i9i 9579 2940 2067 5647 9198 2721- 62i5 9681 3ii9 653 1 2426 6004 9552 3071 6562 ••26 3462 6871 •253 3609 36o 357 355 35i 349 346 343 340 338 335 130 131 132 133 134 135 136 137 138 139 1 13943 7271 120D74 3852 TI05 i3o334 3539 6721 14301 5 7603 0903 4178 7429 o655 3858 7037 •194 3327 4611 7934 123l 45o4 7753 0977 7354 •5o8 3639 4944 8265 i56o 4830 8076 1298 4496 7671 •822 3951 5278 8595 1888 5i56 8399 1619 4814 7987 1136 4263 56ii 8926 2216 5481 8722 lilt 83o3 i45o 4574 5943 9256 2544 58o6 9045 2260 545i 8618 1763 4885 6276 9586 287I 6i3i 9368 258o 5769 8934 2076 5196 6608 Zl 6456 9690 2900 6086 9249 2389 5507 6940 •245 3525 6781 ••12 3219 64o3 9564 2702 58i8 333 33o 328 325 323 321 3i8 3i5 3i4 3u 140 141 142 143 144 145 146 147 148 149 146128 9219 152288 5336 83(2 i6i3b8 4353 73,7 170262 3i86 6438 9527 2594 5640 8664 1667 465o 7613 o555 3478 6748 9835 2900 5943 8965 1967 4947 7908 0848 3769 7o58 •142 32o5 6246 9266 2266 5244 8203 1141 4060 7367 •449 35io 6549 9567 2564 5541 8497 1434 435i 7676 •756 38i5 6852 9868 2863 5838 8792 1726 4641 7985 io63 4120 7154 •168 3i6i 6i34 9086 2019 4932 8:594 1370 4424 7457 •469 3460 643o 9380 23ll 5222 86o3 1676 4728 7759 •760 3758 6726 9674 2603 55i2 8911 1982 5o32 8061 1068 4o55 7022 tl 58o2 309 307 3o5 3o3 3oi 299 293 291 150 151 152 153 154 155 156 157 158 159 176091 8977 181844 4691 7521 190332 3i25 5900 8657 201397 638i 9264 2129 ^vl 0612 3403 6176 8932 1670 6670 9552 24i5 5259 8084 ^] 6453 9206 1943 6959 9b39 2700 5542 8366 1171 3959 6729 9481 2216 7248 •126 2985 5825 8647 i45i 4237 7oo5 9755 2488 7536 •41 3 3270 6108 8928 1730 45i4 7281 ..29 2761 7825 6391 9209 2010 4792 ,7556 •3o3 3o33 8ii3 •986 ih 6674 9490 228b 5069 7832 •577 31o5 8401 1272 4123 6956 9771 ^^67 5346 8107 •85o 3577 4407 7239 ••5i 2846 5623 8382 1124 3848 289 287 285 III -I It 272 N. 1 j 2 3 14 5 6 ! 7 8 9 D. LOGARITHMS OF NUMBERS. 13 N, ' 2 3 4 5 6 7 8 9 D. 160 204I20 4391 4663 4934 5204 5475 5746 6016 6286 6556 271 161 6826 7096 7365 7634 7904 8173 8441 ^V.9, 8979 9247 1 269 162 95i5 9783 ••5i •319 •586 •853 1121 i388 1654 1921 1 267 163 212188 2454 2720 2986 3252 35i8 3783 4049 i 4814 4379 266 164 4844 5i09 i 5373 i 5638 5902 6166 6430 j 6694 6957 9385 7221 264 165 7484 11^1 8010 8273 8d36 8798 9060 1 9823 9846 262 166 220108 0370 o63i 0892 ii53 1414 1675 , 1986 2196 2456 261 167 2716 2976 3236 3496 3755 4oi5 4274 i 4533 1 4792 5o5i lU 168 5309 5568 5826 6084 1 6342 ! 6600 6858 71 15 7372 7680 169 7887 8144 8400 8657 8913 9170 9426 9682 9988 •193 256 170 230449 0704 0960 12l5 1470 1724 1979 2284 2488 2742 254 171 2996 325o 35o4 3757 4011 4264 4317 4770 5028 5276 233 172 5528 5781 6o33 6285 6537 6789 7041 7292 7544 7795 252 173 ^y 8046 8297 8548^ 8799 9049 1D46 9299 9550 9800 ••30 •800 25o 174 240349 3o38 0799 1(?48 1297 1795 2044 2298 2541 2790 1$ 175 3286 3534 3782 4o3o 4277 4525 4772 5019 5266 176 55i3 5759 6006 6252 6499 6745 6991 7287 7482 7728 246 177 7973 821? 8464 8709 8954 1395 9198 9443 9687 tts •176 245 173 260420 0664 0908 3338 ii5i i638 1881 2125 2610 248 179 2853 3096 3580 3822 4064 4806 4548 4790 5o8i 242 180 255273 55i4 5755 5996 6237 6477 6718 6958 7198 7489 9888 241 181 7679 7918 8i58 8398 8637 8877 9116 9855 9594 11^ 182 260071 o3io 0548 0787 1025 1263 i5oi 1789 1976 4346 2214 183 245i 2688 2925 3i62 3399 3636 8878 4109 4582 287 184 4818 5o54 5290 5523 5761 5996 6282 6467 6702 6987 235 185- 7172 7406 7641 7875 8110 8344 8378 8812 9046 9279 284 136 95i3 9746 9980 •2l3 •446 •679 •912 1 144 1377 1609 283 187 271842 2074 23o6 2538 2770 3ooi 3233 8464 3696 3927 282 188 4i58 4389 4620 485o 5o8i 53ti 5542 5772 6002 6232 23o 189 6462 6692 6921 7i5i 7380 7609 7888 8067 8296 8323 229 ll'O 278754 8982 9211 9439 9667 9895 •123 •85i •578 •806 228 191 281033 1 261 1488 1715 1942 2169 2896 2622 2849 3075 227 192 33oi 3527 3753 3979 42o5 4431 4656 4882 6107 5882 226 193 5557 5782 6007 6232 6456 6681 6905 7180 7354 7578 225 194 7802 8026 8249 8473 8696 8920 9143 9866 ?^?? 9812 228 195 290035 0257 0480 0702 0925 1147 3363 1869 1591 2084 222 196 2256 2478 2699 2920 3i4i 3584 38o4 4025 4246 221 197 4466 4687 4907 5i27 5347 5367 5787 6007 6226 6446 220 198 6665 6884 7104 7323 7542 7761 7979 8.98 8416 8635 ^1^ 199 8853 9071 9289 9507 9725 9943 •161 •878 •595 •8i3 200 3oio3o 1247 1464 1681 1898 2114 2881 2547 2764 2980 217 201 3196 3412 3628 3844 4039 4275 4491 4706 4921 5i86 216 202 53DI 5566 5781 5996 6211 6425 6639 8778 6854 7068 72S2 2l5 203 7496 7710 7924 8i37 835i 8564 8991 9204 9417 218 204 9630 9843 ••56 •268 •481 •693 •906 1118 i33o 1542 212 205 311754 1966 2177 2389 2600 2812 3028 3234 3443 3656 211 206 3867 4078 4289 4499 4710 4920 5i3o 5840 5531 5760 210 207 5970 6180 6390 8481 6599 6809 7018 7227 7486 7646 7834 209 208 8o63 8272 8689 8898 9106 9314 9522 9780 9938 208 269 820146 o354 o562 0769 0977 1184 1891 1598 i8o5 2012 207 210 322219 2426 2633 2839 3046 3252 3458 3665 3871 4077 206 211 4282 4488 4694 4899 6950 5io5 53io 55i6 5721 5926 6181 203 £12 6336 6541 6745 7i55 53^^ 7563 7767 7972 8176 204 213 8380 8583 8787 8991 9194 9601 9805 •••8 •211 203 214 33o4i4 0617 0819 1022 1225 1427 i63o 1882 2084 2236 202 215 2438 2640 2842 3o44 3246 3447 3649 3850 4031 4253 202 216 4454 4655 4856 5o57 5257 5458 5658 5859 7838 6069 8038 6260 201 217 6460 6660 6860 7060 7260 7459 7659 8257 200 218 8456 8656 8855 9054 9253 9431 965o 9849 ••47 •246 198 219 340444 0642 0841 1089 1237 I4e5 I682 1880 2028 2225 N. 1 " 1 2 3 4 5 i ^ 7 8 9 D. K F 14 LOGARITHMS OF NU>rBERS. N. « i 1 2 3 .1 . 6 7 8 1 , D. 220 221 222 228 224 225 226 227 228 229 342423 4392 6353 83o5 350248 2i83 4108 6026 7935 9835 2620 4589 6549 85oo 0442 2375 43oi 6217 8125 ••25 2817 4783 6744 8694 o636 2568 4493 6408 83i6 •2l5 3oi4 4981 til 0829 2761 4685 6599 85o6 •404 3212 5178 7135 9083 1023 2934 4876 6790 8696 •593 3409 5374 7330 9278 1216 3 147 5o68 6981 8886 •783 36o6 5570 7523 9472 1410 3339 5260 7172 9076 •972 38oj 5766 7720 9666 i6o3 3532 5452 7363 9266 1161 3999 5962 7913 9860 1796 3724 5643 7554 9436 i35o 8110 ••54 5834 7744 9646 1539 \^ 195 1 194 193 193 192 191 \^ 230 231 232 233 284 235 236 237 238 239 361728 36i2 5488 7356 9216 371068 2912 4743 IVol 5675 7342 9401 1253 3096 4932 6759 858o 2io5 7729 9587 1437 3280 5ii5 6942 8761 2294 4176 6049 7915 9772 1622 3464 5298 7124 8943 2482 4363 6236 8101 ??o6 3647 5481 7306 9124 2671 455i 6423 8287 •143 IK 5664 7488 9306 2859 4739 6610 8473 •328 2173 4013 5846 7670 9487 3048 4926 6796 8639 •5i3 236o 4198 6029 7852 9668 3236 5ii3 6983 8845 •698 2344 4382 6212 8o34 9849 3424 53oi 7.69 9o3o •883 2728 4565 6394 8216 ••3o 188 188 187 186 i85 184 184 1 83 182 181 240 241 242 243 1 244 ; 245 ' 246 , 247 24S ; 249 : 380211 2017 38i5 56o6 I'Z 390935 2697 4432 6199 0392 2197 3995 5785 7568 9343 1112 2873 4627 6374 0573 2377 4174 5964 7746 9520 1288 3o48 4802 6548 0754 2557 4353 6142 %i 1464 3224 4977 6722 0934 2737 4533 6321 8101 9875 1641 3400 5i52 6896 III 5 2917 4712 6499 8279 ••5 1 3373 5326 7071 1296 3097 4891 6677 8456 •228 \% 55oi 7245 1476 3277 3070 6856 8634 •4o5 2169 3926 5676 7419 1 656 3456 5249 7034 881 1 •582 2345 4101 585o 7592 1837 3636 5428 vi 2521 4277 6025 7766 181 180 -I 178 177 176 176 175 174 250.' 251 252 253 254 '[ 255 ; 256 257 , 258 259 •397940 9674 40I40I 3l2I 4834 6540 8240 9933 41 1620 3300 8114 9847 1573 3292 5oo5 6710 8410 •102 1788 3467 8287 ••20 1745 3464 5176 688 1 8579 •271 8461 •192 1917 3633 5346 7o5i 8749 •440 2124 38o3 8634 •365 2089 3807 55,7 7221 8918 •609 2293 3970 8808 •538 2261 3978 5683 7391 9087 •777 2461 4i37 8981 •711 2433 4149 5858 756i 9257 •946 2629 4303 9154 •883 26o5 4320 6029 7731 9426 1114 2796 4472 9328 io56 2777 4492 6199 in 1283 2964 4639 9501 1223 294q' 4663 6370 8070 9764 I45i 3i32 4806 173 173 172 171 171 170 169 167 260 261 262 ' 268 , 264 ■. 265 : 265 267 268 269 414973 6641 83oi 9956 421604 3246 4882 65ii 8i35 9752 5i4o 6807 8467 •121 1768 3410 5045 6674 8297 9914 5307 6973 8633 •286 1933 3574 5208 6836 8459 ••75 5474 a •236 5641 73o6 8964 •616 2261 3901 5534 7161 8783 •398 58o8 7472 9129 •781 2426 4o65 5697 7324 8944 •559 5974 7638 9295 :^ 4228 5860 7486 9106 •750 6141 7804 9460 mo 2754 4392 6023 7648 9268 •881 63o8 7970 9625 1275 2918 4555 6186 7811 9^29 1042 6474 8i35 9791 1439 3082 47>8 6340 7973 9591 I203 167 166 1 65 1 65 164 164 1 63 162 162 161 270 1 271 £72 278 274 275 276 277 278 279 43i364 2969 4569 6i63 775i 9333 440909 2480 4045 56o4 i525 3i3o 4729 6322 7909 'X 2637 4201 5760 1685 3290 4888 6481 8067 9648 1224 2793 4357 5915 1846 345o 5048 6640 8226 9806 i38i 2950 43i3 6071 2007 36io 5207 3io6 4669 6226 2167 3770 5367 6957 8542 •122 1695 3263 4825 6382 2328 3930 5D26 7116 8701 •279 1 852 34T9 49^1 6537 2488 4090 5685 7275 8839 •437 2009 3576 5i37 6692 2649 4249 5844 7433 9017 •594 2166 3732 3293 6B48 2S09 4409 6004 7592 9175 •752 2323 3889 5449 7oo3 161 160 ii i58 \l^ 1 56 i55 N. 1 2 3 4 5 6 7 8 9 D. LOGARITHMS OF NU:MBEE3. 15 K 1 . 1 1 2 S 4 5 G 7 8 9 D. 280 281 282 283 284 285 286 287 288 289 447158 8706 450249 1786 33i8 4845 6366 7882 9392 460898 73i3 8861 o4o3 1940 3471 4997 65i8 8o33 9543 1048 7468 9015 0557 2093 3624 5i5o 6670 8184 9694 1198 7623 9170 07H 2247 3777 53o2 6821 8336 9B45 1348 7778 9324 o865 2400 3930 5454 6973 8487 9990 1499 7933 9478 101 3 2553 4082 56o6 7125 8638 •146 1649 8088 9633 1172 2706 4235 5753 7276 8789 •296 1799 8242 9787 i326 2859 4387 5910 7428 6940 •447 1948 8397 9941 1479 0012 4540 6062 7579 9091 2098 8552 ••95 i633 3i65 4692 6214 7731 9242 .743 2243 i55 1 54 1 54 1 53 1 53 l52 l52 i5i i5i i5o 290 291 292 293 294 295 296 297 298 299 462398 3893 5383 6868 8347 9822 471292 2706 4216 5671 2548 4042 5532 7016 8495 9969 1438 2903 4362 58i6 2697 4191 568o 7164 8643 •116 1 585 3049 45o3 5962 2847 4340 5829 73.2 8790 •263 1732 3195 46^3 6107 2997 4490 5977 7460 8933 •410 1873 3341 4799 6232 3i46 4639 6126 7608 Qo85 •557 2025 3487 4944 6397 3296 4788 6274 7756 9233 •704 2171 3633 5090 6542 3445 4936 6423 7904 9380 •85i 23i8 i^v- 5235 6687 6571 8o52 9527 •998 2464 3925 538i 6832 3744 5234 6719 8200 9675 1145 2610 4071 5526 6976 i5o 149 149 143 143 147 146 146 146 145 800 801 302 803 804 805 806 1 807 ' 808 1 309 477121 8566 480007 1443 2874 43oo 5721 7i38 855i 9958 7266 8711 oi5i 1 586 3oi6 4442 5863 7280 8692 ••99 8855 0294 1729 3i59 4585 6oo5 7421 8833 •239 7555 &999 0438 1872 33o2 4727 6147 7563 8974 •38o 7700 9143 o582 2016 3445 4869 6289 7704 9114 •520 7844 9287 0725 2159 3587 Sou 6430 7845 92DD •661 7989 943 1 0869 2302 3730 5id3 6572 7986' 9396 •801 8i33 9575 1012 2445 3872 5295 6714 8.27 9537 •941 8278 T.ll 2588 4oi5 5437 6855 8269 9677 io8i 8422 9863 1299 2731 4.57 5579 6997 8410 9818 1222 145 144 144 143 143 142 142 141 141 ■40 810 811 812- 813 814 815 816 817 818 819 491362 2760 4i55 5544 9687 D0I0D9 2427 3791 l502 2900 4294 5683 7068 8448 9«24 1196 2564 3927' 1642 3o4o 4433 5822 7206 8586 ??M 2700 4o63 1782 3i79 4572 5q6o 7344 8724 ••99 14-^ 2837 4199 1922 3319 47II 6099 7483 8862 •236 1607 2973 4335 2062 3458 485o 6238 7621 1744 3109 4471 2201 4989 6376 7759 9137 •5u 1880 3246 46©7 2341 3737 5i23 65i5 7897 9273 •643 2017 3382 4743 2481 3876 5267 6653 8o35 9412 •785 2ID4 35i8 4878 2621' 4oi5 5406 6791 8173 9550 •922 2291 3653 5oi4 140 139 139 139 i33 1 33 137 137 i36 1 36 820 821 822 823 824 825 826 327 823 ^ 829 5o5i5o 65o5 7856 9203 5io545 1 883 3218 4548 5874 7196 5286 6640 7991 9337 0679 2017 335i 4681 6006 7328 5421 6776 8126 9471 o8i3 2l5l 3484 48i3 6139 7460 5557 691 1 8260 9606 0947 2284 3617 4946 6271 7592 5693 7046 8395 9740 1081 24.8 3750 5o79 64o3 7724 5828 7181 853o 9874 12l5 255i 3883 52II 6535 7855 5964 7316 8664 •••9 1 349 2684 4016 5344 6668 7987 6099 7431 1482 2818 4149 5476 6800 8119 6234 7586 8934 •277 i6i6 2951 42S2 5609 6932 825i 6370 7721 9063 •411 1750 3oS4 44i5 5741 7064 8382 136 i35 i33 1 34 i34 i33 i33 i33 l32 l32 830 331 332 333 834 835 336 337 333 839 N. 5i85i4 9823 52II38 2444 3746 5o45 6339 763o 530200 8646 9959 1269 2575 3876 5i74 6469 7759 9043 0328 8777 ••90 1400 2705 4006 53o4 6598 7888 9174 0456 8909 •221 i53o 2835 4i36 5434 6727 8016 9302 o584 9040 •353 1661 2966 4266 5563 6856 8145 94 3 0712 9171 •484 1792 3096 4396 6985 8274 9559 0840 93o3 •6i5 1922 3226 4526 5822 8402 9434 •745 2053 3356 4656 5951 7243 853 1 9815 1096 9566 •376 2i83 3486 4785 6081 7372 8660 9697 1007 23 14 36i6 49'5 6210 75oi 8788 ••72 i35i i3i i3i i3i i3o i3o 129 129 III 128 1 2 3 4 5 6 ^ ' 9 D. 16 LOGARITHMS OF NUMBERS. N. 1 1 2 8 4 5 6 7 1 8- 9 1 D. 1 S40 531479 ! 1607 i 1734 1862 1990 j 2117 2245 1 2372 j 25oo 2627 128 841 2754 2882 i 3009 3i36 3264 1 3391 35i8 I 3645 ! 3772 3899 127 842 4026 41 53 ' 4280 4407 4534 4661 4787 U914 ! 5o4i 5167 III 848 5294 5421 : 5547 5674 58oo 5927 6o53 1 6180 1 63o6 6432 844 65o8 ; 6685 68 u 6937 7063 7189 8574 ' 8699 ^823 ^] 126 845 7819 7945 8071 8197 8322 8448 126 846 9076 , 9202 9327 9432 9578 9703 9829 9934 ••79 •204 125 847 540329 1 0455 od8o 0705 o83o 0955 io3o ; i2o5 i33o 1454 125 848 1679 1704 1 1829 1953 2078 2203 2327 2452 2576 2701 123 349 2823 2930 1 3074 3199 3323 3447 3571 ! 3696 3820 3944 124 850 • 544068 1 4192 43 16 4440 4564 46S8 4812 4936 1 5o6o 5i83 124 851 ! 5307 ! 543 1 5555 5078 58o2 5925 6049 6172 1 6296 6419 124 852 6543 6666 6789 6913 7o36 7.59 72S2 { 7403 3,1? 7652 123 858 : 7773 7S93 802 1 8144 8267 8389 85i2 1 8635 8881 123 854 9003 9126 9249 9371 9494 9616 9739 ! 9861 9984 •106 123 855 5502 28 o35i 047-i 0595 0717 0840 0962 1 10S4 1206 i328 122 856 1430 1572 1694 1816 1933 2060 2181 j 23o3 2423 2547 122 857 2663 2790 2911 3o33 3i55 3276 3398 1 3519 3640 3762 121 858 ' 3583 4004 4126 4247 4368 44&9 4610 ; 4731 4852 4973 121 S59 860 : 5094 52i5 5336 5457 5573 5699 5820 5940 6061 6I82 121 5563o3 6423 6544 6664 6785 6905 7026 i 7146 i 7267 7387 120 861 7507 7627 7748 8948 7868 7988 8108 8228 1 8349 i 8469 858^ 120 862 8709 8829 9068 9188 9308 9428 ' 9548 j 9667 9787 120 863 9907 ••26 •146 •265 •385 •5o4 •624 ! ^743 i ^863 •982 119 864 56IIOI 1221 1 340 1459 1578 1698 1817 ; 1936 2055 2174 119 365 2293 2412 253 1 265o 2769 2887 3oo6 i 3i25 3244 3362 119 SC6 3481 3600 1 3718 3837 3955 4074 4192 i 43ii 4429 4548 \;i 367 4666 4784 4903 502I 5i39 5257 5376 5494 5612 5730 868 5848 5966 6084 6202 6320 6437 6555 6673 6791 6909 118 869 7026 7144 7262 7379 7497 7614 7732 7849 7967 8084 118 870 568202 83i9 8436 8554 8671 8788 8905 9023 9140 9257 117. 871 .9374 949 » 9608 .9725 9842 9939 ••76 •193 •309 •426 117 872 570043 0660 0776 0893 lOIO 1126 1243 1359 1476 1392 "I 116 873 1709 1S25 1942 2058 2174 2291 2407 2323 2639 2755 374 2872 2988 3104 3220 3336 3432 3568 3684 3800 39.3 116 S75 4o3i 4147 4263 4379 5534 4494 4610 4726 4841 4957 5672 116 376 5.88 53o3 5419 5630 5765 588o 5996 6111 6226 ii5 877 6341 6457 6572 6687 6802 6917 7o32 ^47 8293 7262 7377 ii5 378 7492 7607 8754 mi 7836 7951 8066 8181 8410 8525 ii5 879 8639 8983 9097 9212 9326 9441 9555 9669 114 880 579784 9898 ••12 •126 •241 •355 •469 •583 •697 •811 114 881* 580925 1039 ii53 1267 i38i 1495 263 1 1608 1722 1836 1950 114 882 2063 2177 2291 2404 25i8 2745 2858 2972 3o85 114 883 3199 33i2 3426 3539 3652 3765 3879 3992 4103 4218 ii3 884 433 1 4444 4557 4670 4:83 4896 3009 5l22 5235 5348 ii3 885 5461 5574 5686 5799 5912 6024 6137 62 5o 6362 6475 ii3 886 887 6587 nil 6700 7823 6812 7935 6925 8047 7037 8160 7149 8272 7262 8384 c 7486 8608 7399 8720 112 112 888 8944 9o56 9167 9279 9391 95o3 90.5 9726 9838 112 889 9950 ••61 •173 •284 •396 •307 •619 .-,30 •842 •953 112 890 591065 1176 2288 1287 1399 i5io 1621 1732 1843 1935 2066 III 891 2177 2399 25lO 2621 2732 2843 ■ 2954 3o64 3175 4282 III 892 1 3286 3397 35o8 36i8 3729 3840 3950 ] 4061 4171 III 393 1 4393 45o3 4614 5827 4834 4945 5o55 I 5i65 5276 5386 no 894 ! 5496 56o6 5717 5937 6047 61 57 i 6267 6377 6487 no 395 , 6597 769D 6707 6817 6927 i:li 7146 7256 i 7366 i 7476 8681 no 306 t 7805 8900 7914 8024 8243 8353 1 8462 8572 no S97 ' 9009 9119 9228 9337 9446 1 9556 ! 966J 9774 109 393 9883 9992 600973 I 1082 •lOI •210 1 •319 •428 •537 '646 1 •755 •864 109 899 ■ 1191 1299 1 1408 i5i7 1625 1734 1 1843 ,951 109 N. 1 1 2 3 1 4 1 5 1 6 1 7 8 9 D. LOGARITHMS OF NUMBERS. 17 N. 1 I 2 3 4 5 6 7 8 9 D. 400 i 401 i 402 , 403 404 405 406 407 403 409 602060 3i44 4226 53o5 6331 7455 8526 ^ 9594 610660 1723 2169 325J 4334 54i3 6489 7562 8633 9701 0767 1829 2277 336i 4442 5521 6596 7669 8740 9808 0873 1936 2386 3469 455o 5628 6704 ^847 9914 0979 2042 2494 3577 4658 5736 6811 7884 8954 ••21 1086 2148 2603 3686 4766 5844 6919 7991 9061 •128 1192 2234 2711 3794 4874 5951 7026 8098 9167 •234 1298 2360 2819 3902 4982 6059 7.33 8203 2S i4o5 2466 2928 4010 5089 6166 7241 8312 9381 •447 i5ii 2572 3o36 4118 5197 6274 7348 8419 94b8 •554 1617 2678 108 lOb 108 108 107 107 107 107 106 106 410 411 412 413 414 415 416 417 41S 419 612784 3842 tVo 7000 8048 9093 620136 1176 2214 2890 llU 6o55 7105 8i53 9198 0240 1280 23i8 lit, 5io8 6160 7210 8257 9302 o344 1 384 2421 3l02 4139 52i3 6265 73i5 8362 9406 0448 1488 2525 32o7 4264 5319 6370 7420 8466 9311 o552 1592 2628 33 13 4370 5424 6476 7525 8571 9615 0656 1695 2732 3419 4473 5529 658i 9719 0760 1799 2835 3525 458 1 5634 6686 7734 8780 9824 0864 1903 2939 363o 4686 5740 6790 8884 9928 0968 2007 3o42 3736 4792 5»45 6895 7943 89«9 ••32 1072 2110 3146 106 106 I05 103 io5 io5 104 104 104 104 420 421 422 423 424 425 426 427 4£8 429 623249 4282 53i2 6340 7366 8389 9410 630428 1444 2457 3353 4385 541 5 6443 7468 8491 9512 o53o i545 2559 3456 4488 55i8 6546 llll 9613 o63i 1647 2660 3559 4591 5621 6648 7673 8695 9715 0733 1748 2761 3663 4695 5724 6731 7773 8797 ^^ 1849 2802 3766 4798 5827 6853 7878 8900 9919 0936 1931 2963 3869 4901 l$t 7980 9002 ••21 io38 2052 3o64 3973 5oo4 6o32 7o58 8082 9104 •123 1139 21 53 3i65 4076 5107 6i35 7161 8i85 9206 •224 1241 2255 3266 4179 52IO 6238 7263 8287 9308 •326 i342 2356 3367 io3 io3 io3 io3 102 102 102 102 lOI 101 430 431 432 433 4^4 435 436 4G7 438 439 633468 6488 IX 9486 640481 1474 2465 3569 437^ 5584 6588 9586 o58i 2563 3670 4679 5685 6688 7690 8689 9686 0680 1672 2662 3771 4779 3785 6789 7790 ^789 9783 0779 1771 2761 3872 4880 5886 6889 7890 8888 9885 0879 1871 2860 3973 4981 5986 6989 7990 8968 9984 0978 1970 2939 4074 5o8i 6087 7089 8090 9088 ••84 1077 2069 3o58 4175 5182 6187 7.89 8190 9188 •i83 1177 2168 3i56 4276 5283 6287 7290 8290 9287 •283 1276 2267 3255 4376 5J8J 6388 7390 8389 93^7 •382 1375 2366 3354 100 100 100 100 99 99 99 99 99 99 440 441 442 443 444 445 446 447 448 *449 643453 4439 5422 6404 7383 836o 9335 65o3o8 1278 2246 355i 4537 552i 65o2 7481 8458 9432 o4o5 1375 2343 365o 4636 56i9 6600 7579 85DD 9530 o5o2 1472 2440 3749 4734 5717 6698 7676 8653 9627 0599 1369 2536 3847 4832 58i5 6796, 7774 8750 9724 0696 1666 2633 3946 4931 59.3 6894 & 9821 0793 1762 2730 4044 5029 6011 6992 7969 8943 9919 0890 1839 2826 4143 5127 6110 7089 8067 9043 ••16 tl 2923 4242 5226 6208 v:ii 9140 •ii3 1084 2o53 30.9 4340 5324 63o6 7285 8262 9237 •210 11^1 2l50 3ii6 98 9.^ 9^ 98 98 97 97 97 97 97 450 451 452 453 454 455 456 467 458 459 653213 mi 6098 70D6 8011 8965 9916 660865 i8i3 3309 4273 5235 6194 7102 8107 9060 **II 0960 1907 34o5 4369 5331 6290 7247 8202 9155 •106 io55 2002 35o2 4465 5427 6386 7343 8298 9230 •201 ii5o 2096 3598 4562 5523 6482 7438 8393 9346 •296 1245 2191 3693 4658 5619 6577 7534 8488 9441 •391 1339 2286 3791 4734 5715 6673 til 9536 •486 1434 238o 3888 485o 58io 6769 7723 8679 9631 •58 1 i529 2473 3984 4946 5906 6864 7820 8774 1623 2569 40S0 5o42 6002 6960 7916 8870 9821 1718 2663 96 96 9b 96 96 9D 9? 9? 93 N. 1 ^ 1 2 3 * 5 « ' 8 9 1 D. 18 LOGARITHMS OF l^UMBERS. N. 1; 1 2 a 4 6 7 8 35i2 4454 lit 7266 8199 9i3i ..60 0988 1913 9 i D. 1 4 SO 461 462 463 464 465 466 467 463 469 662753 ! 3701 ■i 4642 1 55«i 1 65i8 1453 8386 1 670246 1 1173 2852 3795 4736 5675 6612 7546 8479 9410 0339 1265 I 2947 3»89 483o 7640 8J72 95o3 043 1 i358 3o4i 3983 4924 5ttd2 ^fs 8665 9596 0024 I45i 3i35 4078 5oi8 5956 6892 7826 8759 9689 0617 i543 323o 4172 5ll2 6o5o 6986 7920 8852 9782 0710 1636 3324 4266 5206 6143 7079 8oi3 8945 9875 0802 1728 3418 436o 7173 bio6 9o38 tl 1821 4548 5487 6424 8293 9224 .153 1080 2oo5 94 94 94 93 470 471 472 473 474 475 476 477 47S 479 480 451 432 453 484 455 486 487 433 4S9 1 672098 302I 3942 4861 5778 6694 n 1 9428 : 68o336 2190 3ii3 4o34 4953 5870 6785 7698 8609 9319 0426 2283 32o5 4126 5o45 5962 6876 7789 8700 9610 o5i7 2375 3297 4218 5i37 6o53 6968 7881 8791 9700 0607 2467 3390 43io 5228 6145 7059 7972 8»82 9791 0698 256o 3482 4402 5320 6236 7i5i 8o63 8973 9882 0789 2652 3574 4494 5412 6328 7242 8i54 0064 9973 0879 2744 3666 4586 55o3 6419 7333 8245 9155 ..63 0970 2836 3753 65ii 7424 8336 9246 •154 1060 UP, 6602 7516 8427 ii5i 92 92 92 92 92 91 91 91 91 91 681241 2145 1 ^°47 3947 4^45 5742 6636 7529 8420 9309 i332 2235 3i37 4037 4935 583i 6726 r, 9393 1422 232b 3227 4127 5o25 5921 681 5 mi 9486 i5i3 2416 3317 4217 5ii4 6010 6904 7796 8087 9375 i6o3 25o6 3407 4307 5204 6100 6994 7886 8776 9664 1693 2596 3497 4396 5294 6189 7083 7975 8865 9753 1784 2686 3587 4486 5383 6279 7172 8064 8953 9»4i 1874 2777 3677 4376 5473 6368 7261 8i53 9042 9930 1964 2867 3767 4666 5563 6458 735i 8242 9i3i ..19 2o55 2957 3857 4756 5652 6547 7440 8331 9220 •107 90 90 90 90 II 89 89 490 491 4y2 493 494 495 496 4y7 493 4y9 690196 1081 1965 2847 3727 46o5 5482 6356 7229 8101 0285 1 1 70 2o53 2935 3Si5 4693 5569 6444 7317 8i83 0373 1258 21-42 3o23 3903 4781 5657 653 1 7404 8275 0462 i347 223o 3iu 3991 4863 5744 6618 7491 8362 o55o 1435 23i8 3199 4078 4956 5&32 6706 7578 8449 0639 i524 2406 3287 4166 5o44 7665 8535 0728 1612 2494 3375 4234 5i3i 6007 6880 7732 8622 0816 1700 2353 3463 4342 5219 6094 6968 7839 8709 0905 1789 2671 355i 443o 5307 6182 7055 ^^ 0993 1877 nil 4517 5394 6269 8014 8883 89 88 88 83 83 88 87 ?7 87 87 500 501 502 503 504 505 506 507 503 509 700704 1 568 243 1 3291 5oo8 5864 6718 9057 9924 0790 i654 25i7 4236 5094 5949 68o3 9144 ••11 0877 1741 26o3 3463 4322 I'oll 68tt8 923i ••98 0963 1827 26S9 3549 4408 5265 6120 6974 9317 •184 io5o 19.3 2773 3635 4494 5350 6206 7059 9404 •271 ii36 1999 2861 3721 4579 5436 6291 7»44 9491 •358 1222 2086 2947 3807 4665 5522 6376 7229 9578 .444 i3o9 2172 3o33 ^??? 5607 6462 73i5 1395 2258 3119 3979 4837 5693 6547 7400 9751 .617 1482 2344 32o5 4o65 4922 5778 6632 7485 87 87 86 86 86 86 86 86 85 85 510 i! 707570 511 i: 8421 512 1^ 9270 512 1, 710117 514 0963 515 i' 1807 516 ! 265o 517 ■ 3491 513 ■ 433o 519 : 5167 7655 85o6 9355 0202 1043 IX 3575 4414 525i 7740 8591 9440 0287 Il32 1976 2H18 3659 4497 5335 7826 8676 9324 0371 1217 2060 2902 3742 458 1 5418 9609 0456 i3oi 2144 2986 3826 4665 55o2 7996 8081 8846 8931 9694 ; 9779 o54o 0620 i385 1470 2229 23i3 3070 3 1 54 3910 1 3994 4749 i 4833 5586 1 5669 8166 90i5 9S63 0710 1 554 2397 3238 4078 4916 5733 825i 9100 9948 0794 1639 2481 3323 4162 5ooo 5836 8336 ??]? 2566 3407 4246 5o84 5920 85 85 85 85 84 84 84 84 84 84 -X. 1 i 2 3 4 5 1 6 7 8 9 D. LOGARITBLMS OF NUMBERS. 19 N. i 1 2 3 4 5 6 7 8 i 9 I). 520 521 522 523 524 525 526 527 528 529 716003 6838 7671 85o2 9331 ; 720109 1 0986 i 181I j 2634 i 3456 6087 6921 lili 9414 0242 1068 1893 2716 3538 6170 7004 7837 8668 9497 o325 ii5i 1975 2798 3620 6254 70S8 7920 8751 9580 0407 1233 2o58 2881 3702 6337 8834 9663 0490 i3i6 2140 2963 3784 6421 7254 8086 8917 9745 0573 1398 2222 3o45 3866 65o4 7338 8169 9000 9828 0655 1481 23o5 3.27 3948 6588 7421 8253 9083 9911 :r^ 2387 3209 4o3o 6671 7504 8336 9165 9994 0821 1646 2469 3291 4112 6754 7587 8419 9248 ••77 0903 1728 2552 3374 4194 83 83 83 83 83 83 82 82 82 82 530 531 532 533 534 535 536 537 538 539 i 724276 ; 5095 591-2 6727 8354 9165 9974 4358 5176 5993 6809 7623 8435 9246 ••55 o863 1669 4440 5258 6075 6890 7704 85i6 9327 •i36 0944 1750 4522 5340 61 56 6972 7785 8597 9408 •217 1024^ i83o 46o>f 5422 6238 7053 7866 8678 iio5 191 1 4685 55o3 6320 7134 7948 8739 9370 •378 1186 1991 4767 5585 6401 7216 8029 8841 965 1 •459 1266 2072 4849 5667 6483 7297 8uo 8922 9732 •540 1347 2l52 4931 5748 6564 7379 8191 9003 9813 •621 1428 2233 5oi3 583o 6646 7460 8273 9084 9893 :s 23i3 82 82 82 81 81 81 81 81 81 81 540 541 542 543 .544 545 546 547 548 549 732394 3999 4800 ?;§^ 8781 9572 2474 3278 4079 4880 5679 6476 7272 8067 8860 965 1 2555 3358 4160 4960 5759 6556 7352 8146 8939 9731 2635 3438 4240 5o4o 5838 6635 7431 S225 9018 9S10 2715 35i8 4320 5l20 5918 6715 75ii 83o5 9097 9889 2796 3598 4400 5200 5998 6795 7590 8384 2876 3679 4480 5279 6078 6874 7670 8463 9256 ••47 2956 3759 456o 5359 6157 6934 7749 8543 9335 •126 3o37 3839 4640 5439 6237 7034 7829 8622 9414 •205 3.17 3919 4720 5319 63i7 7113 7908 8701 9493 •284 80 80 80 80 80 80 79 79 79 79 550 551 552 553 554 555 556 557 558 559 740363 Il52 Itt 35io 4293 5o73 5855 6634 7412 0442 I230 2018 2804 3588 4371 5i53 5933 6712 7489 052I 1 309 2096 2882 3667 4449 523i 601 1 6790 7567 0600 i388 2961 3745 4528 53o9 6868 7645 0678 1467 2254 3o39 3823 4606 5387 6167 6945 7722 0757 1 546 2332 3ii8 3902 4684 5465 6245 7023 7800 o836 1624 241 1 3196 3980 4762 5543 6323 7101 7878 0915 1703 2489 3275 4o58 4840 5021 6401 0994 111 3353 4i36 4919 5699 6479 7256 8o33 1073 i860 2647 343 1 4213 4997 5777 6556 7334 8110 79 79 79 ]t 78 7^ 78 78 78 560 561 562 563 564 565 566 567 568 ' 569 748188 8963 75^508 i?79 2048 2816 3583 •4348 5lI2 8266 9040 9814 o586 1 356 2125 2893 3660 4425 5189 8343 9118 9891 o663 1433 2202 2970 3736 4301 5265 8421 9195 9968 0740 i5io 2279 3o47 38i3 4578 5341 8498 9272 ••45 0817 1 587 2356 3i23 3889 4654 5417 8576 9350 •I23 0894 1664 2433 3200 3966 4730 5494 8653 9427 •200 0971 2 509 3277 4042 4807 5570 8731 9504 •277 1048 1818 2586 3353 4119 4883 5646 8808 9582 •354 II25 1895 2663 343o 4195 4960 5722 8885 9659 •43 1 1202 1972 2740 35o6 4272 5o36 5799 77 77 77 77 77 77 77 77 ^t 76 570 571 572 573 574 575 576 577 578 579 755875 6636 lit. 8912 9668 760422 1 1 76 1928 2679 595. 6712 7472 823o 8988 9743 0498 I25l 2003 2754 6027 6788 7548 83o6 9063 ??;? i326 2078 2829 6io3 6864 7624 8382 9139 9894 0649 1402 2i53 2904 6180 6940 7700 8458 9214 9970 0724 1477 2228 2978 6256 7016 7775 8533 9290 ••45 0799 i552 23o3 3o53 6332 7092 7831 8609 9366 •121 0875 1627 2378 3i:.8 6408 7168 7927 8685 9441 •196 0930 1702 2453 32o3 6484 7244 8oo3 8761 9517 •272 1023 1778 2329 3278 656o 7320 8079 8836 9592 •347 IIOI i853 2604 3353 76 76 76 76 76 ?5 75 75 75 N. 1 2 3 J . 5 6 7 8 9 1 D. 20 LOGARITHMS OF KUMBEES. N. \ 1 2 8 4 5 6 7 8 9 D. 580 ; 763428 35o3 3578 3653 3727 38o2 3877 3952 4027 4101 i 5S1 4176 425i 4326 4400 4475 455o 4624 1 4699 4774 4848 7? 582 i 4923 4998 5072 5i47 5221 5296 5370 5445 5520 5594 75 583 1 If^ 5743 58i8 5892 5966 6041 6ii5 6190 6264 6338 74 584 1 6487 6562 6636 6710 6785 6859 6933 7007 7°^^ 74 585 7i56 723o i3o4 8046 7379 J453 7527 7601 7675 8416 7749 7823 8564 74 586 lt& 7972 8120 8268 8342 8490 74 637 8712 87S6 8860 8934 9008 9082 91 56 9230 93o3 74 5S8 9^77 9451 9525 9599 9673 9746 9820 9894 9968 ••42 74 589 770u5 0189 0263 o336 0410 0484 0557 o63i 0705 0778 74 590 770802 0926 0999 1073 1 1 46* 1220 1293 1 367 1440 i5i4 74 591 i587 1661 1734 1808 1881 1955 2028 2102 2175 2248 73 592 2322 2395 2468 2542 26i5 2688 2762 2835 2908 2981 73 593 3o55 3128 3201 3274 3348 3421 3494 3567 4298 3640 3713 73 594 3-86 3860 3933 4006 4079 4132 4225 4371 4444 73 595 4317 4590 4663 4736 4809 5538 4882 4955 5028 5ioo 5173 73 596 5246 5319 5392 5465 56io 5683 5756 5829 5902 73 597 5974 6047 6120 6193 6265 6338 641 1 6483 6556 6629 *?? 598 6701 6774 6846 6919 6992 7064 7'j7 7209 7282 8006 7334 8079 73 599 7427 7499 7372 7644 7717 7789 7862 7934 72 600 778131 8224 8296 8368 8441 85i3 8585 8658 8730 8802 72 601 '8874 8947 9019 9091 9'^^ 9236 9308 9380 9452 9524 72 602 9596 9669 9741 98.3 9885 9957 ••29 •lOI •i"?^ •245 72 6»S 780317 0389 0461 o533 o6o5 0677 0749 0821 0893 0965 72 604 , 1037 1 109 1181 1253 i324 1396 1468 1 540 1612 1684 72 605 1755 1827 1899 1971 2042 2114 2186 2 258 2329 2401 72 606 2473 2544 2616 2688 2759 283 1 2902 2974 3o46 3ii7 72 607 3189 3260 3332 3403 3473 3546 36i8 3689 44o3 3761 3832 71 608 3904 3975 4046 4118 4189 4261 4332 4475 4546 71 609 4617 4689 4760 483 1 4902 4974 5o45 5ii6 5187 5259 V 610 785330 5401 5472 5543 56i5 5686 5757 5828 5899 5970 V 611 6041 6112 6i83 6254 6325 6396 6467 6538 6609 6680 71 612 6751 6822 6893 6964 7035 7106 7177 7248 7319 7390 71 618 7460 8168 7531 7602 7673 7744 7815 7885 7936 8027 8098 71 614 8239 83io 838 1 845i 8522 8593 8663 8734 8804 71 615 8875 8946 9016 9087 9157 9228 9299 9369 9440 9510 71 616 9381 9601 9722 9792 9863 9933 •••4 ••74 •144 •2l5 70 617 790285 0356 0426 0496 o567 ob37 0707 0778 0848 0918 70 618 0988 1039 1129 1 199 1269 1 340 1410 1480 i55o 1620 70 619 1691 1761 i83i 1901 1971 2041 2111 2181 2252 2322 70 620 792392 2462 2532 2602 2672 2742 2812 2882 2952 3022 70 621 3092 3i62 323i 33oi 3371 3441 35ii 358i 365i 3721 70 622 ' 3790 386o 3930 4000 4070 4.39 4209 4279 4349 4418 70 623 4488 4558 4627 4697 4767 4836 4906 4976 5045 5ii5 70 624 ; 5l85 5254 5324 5393 5463 5532 5602 5672 5741 58ii 70 625 1 5880 5949 6019 67.3 6088 6i58 6227 6297 6366 6436 65o5 69 626 1 6574 6644 6782 6852 692. 6990 7060 7129 7198 ^ 627 1 726a 7337 7406 7475 7545 7614 7683 7732 7821 & 69 6'i8 It, 8029 8098 8167 8858 8236 83o5 8374 8443 85i3 69 629 8720 8789 8927 8996 9065 9«34 9203 9272 69 630 i 799341 9409 9478 9547 9616 9685 9754 9823 9892 9961 69 631 800029 0098 0167 0236 o3o5 0373 0442 o5ii o58o 0648 69 632 1 0717 0786 o854 0923 0992 1061 1129 1 198 1266 i335 69 633 1404 1472 1 541 1609 1678 1747 I8l3 1884 1952 2021 69 634 2089 2i58 2226 2295 2363 2432 25oo 2568 2637 2705 69 635 2774 3457 2842 2910 2979 3o47 3ii6 3i84 3252 3321 3389 68 636 3525 3?94 3662 3730 3798 3867 3935 4oo3 4071 68 637 4139 4208 4276 4344 4412 4480 4548 4616 4685 4753 68 638 4821 4889 4957 5o25 5093 5i6i 5229 5297 5365 5433 68 639 55oi 5569 5^37 5705 5773 5841 5oo8 5976 6044 6112 68 K. 1 " 1 2 8 4 5 6 7' 8 9 D. LOGARITHMS OF NUMBERS. 21 N. 1 2 8 4 5 6 7 8 9 D. 640 806180 6248 63i6 6384 645 1 65i9 6587 6655 6723 6790 68 641 6858 6926 6994 7061 7129 7197 7264 7332 7400 l^^J 68 642 7535 8211 7603 7670 8346 7738 7806 7873 7941 8008 8076 8143 68 643 ll'S 8414 8481 8549 8616 8684 8751 8818 67 644 8886 9021 9088 9i56 9223 9290 9358 9425 9492 67 645 9560 9627 9694 9762 9829 9896 9964 ••3 1 ••98 •i65 67 646 810233 o3oo o367 0434 o5oi 0369 0636 0703 0770 0837 6n 647 0904 0971 1039 1106 11-3 1240 i3o7 1 374 1441 i5o8 67 648 1575 1642 1709 1776 1843 1910 1977 2044 2111 2178 67 649 2245 23j2 2379 2445 25l2 2379 2646 2713 2780 2847 67 650 812913 2980 3o47 3ii4 3i8i 3247 33i4 338i 3448 35i4 67 651 3581 3648 3714 3781 3848 3914 438i 3981 4048 4114 4181 67 652 4248 43i4 438i 4447 45i4 4647 4714 4780 4847 67 653 4913 4980 5046 5ii3 5i79 5246 53 1 2 5378 5445 55ii 65 654 5578 5644 5711 5777 5843 5910 5976 6042 6109 6175 66 655 6241 63o8 6374 6440 65o6 6573 6639 6705 6771 6838 66 £56 6904 6970 7o36 7102 7169 7235 7301 7367 7433 7499 8160 66 657 7565 763 1 U 7764 7830 7896 7962 8028 8094 66 658 8226 8292 8424 8490 8556 8622 8688 8754 882c 66 659 8885 8951 9017 9083 9149 9215 9281 9346 9412 9478 66 6G0 8t9544 9610 9676 9741 9807 9873 9939 •••4 ••70 •i36 66 661 820201 0267 0333 0399 0464 o53o 0095 0661 0727 0792 66 662 o858 0924 0989 io55 1120 1186 I25l i3i7 i3S2 1448 66 6G3 i5i4 i579 1645 1710 1775 1841 1906 1972 2037 2io3 65 664 2168 2233 2299 2952 2364 243o 2495 256o 2626 2691 2756 65 GG5 2822 2887 3oi8 3o83 3i48 32i3 3279 3344 3409 65 666 3474 3539 36o5 3670 3735 3Soo 3865 3930 3996 4061 65 667 4126 4191 4256 4321 43% 445i 45i6 458i 4646 471 1 65 663 4776 4841 4906 4971 5o36 5ioi 5i66 523i 5296 536i 65 669 5426 5491 5556 5621 5686 5751 58i5 588o 5945 6010 65 670 826075 6140 6204 6269 6334 6399 6464 6528 6593 6658 65 671 6723 6787 6852 6917 6981 7046 7111 7175 7240 73o5 65 672 7369 7434 8080 7499 7563 7628 7692 7757 7821 7886 795i 65 678 8oi5 8144 8209 8273 8338 8402 8467 853 1 8095 9239 64 674 8660 8724 8789 8S53 8918 9561 8982 9046 91U 9175 64 675 9304 9368 9432 9497 9625 9690 9754 9818 9882 64 676 9947 ••11 ••75 •i39 •204 •268 •332 •396 •460 •525 64 677 83o589 0653 0717 1358 0781 0845 0909 0973 1037 1102 1 166 64 678 I230 1294 1934 1422 i486 i55o 1614 1678 1742 1806 64 679 1870 1998 2062 2126 2189 2253 23i7 23Si 2445 64 680 832509 2573 2637 2700 2764 2828 2892 2956 3020 3o83 64 6S1 3i47 32II 3275 3338 3402 3466 353o 3593 3657 3721 64 682 3784 3848 3912 3975 4039 4io3 4166 423o 4294 4357 64 6S8 4421 4484 4548 461 1 4675 4739 4802 4866 4929 4993 64 684 5o56 5l20 5i83 5247 53io 5373 5437 55oo 5564 5627 63 685 5691 5754 5817 588 1 5944 6007 6071 6i34 6197 6261 63 686 6324 6387 645 1 65i4 6077 6641 6704 6767 6-830 6S94 63 687 6957 7020 70S3 7146 7210 7273 7336 7399 7462 7325 81 56 63 688 ■7388 7652 8282 77i5 7778 7841 7904 It] 8o3o 8093 63 689 8219 8345 8408 8471 8534 8660 8723 8786 63 690 838849 8912 8975 9o38 9101 9«64 9227 9289 9352 94i5 63 691 9478 9541 9604 9667 9729 9792 9855 9918 9981 ••43 63 692 840106 0169 0232 0294 0357 0420 04S2 o545 0608 0671 63 693 0733 0796 o859 0921 0984 1046 1 109 1172 1234 1297 63 694 1359 1422 1485 1 547 1610 1672 1735 1797 i860 1922 63 695 19S5 2047 2110 2172 2235 2297 2360 2422 2484 2047 62 696 2609 2672 2734 2796 2859 2921 2983 3046 3io8 3170 62 697 .3233 3295 3357 3420 348? 3544 36o6 .3669 3731 3793 62 698 3855 3918 3980 4042 4104 4166 4229 4291 4353 44i5 62 699 4477 4539 4601 4664 4726 4788 4b5o 4912 4974 5o36 62 N. 1 2 8 4 5 6 7 8 9 D. .-)9 LOGARITHMS OF XOIBERS. K. : 1 2 3 4 5 6 7 8 9 D. 700 845098 5i6o 5222 5284 5346 5408 5470 5532 j 5594 5656 62 701 5718 5780 5842 5904 5966 6028 6090 6i5i 1 62i3 6275 62 702 6337 6399 6461 6323 6585 6646 6708 6770 6832 6894 62 703 6955 7017 7079 7141 7202 7264 7326 7388 ^ 7511 8128 62 704 : 7373 7634 8189 8231 ; 880! 8866 7696 7758 7819 7881 7943 8004 62 705 83.2 8374 8433 8497 8559 8620 8682 8743 62 706 8928 8989 9604 905 1 9112 9174 9235 9297 9358 61 707 9419 9481 9542 9663 9726 9788 9849 9911 9972 61 70S 85oo33 0095 0156 0217 0279 o34o 0401 0462 0324 o585 6t 7oy i 0646 0707 0769 o83o 0S91 0952 1014 1075 1x36 1197 61 710 ; 851258 l320 i38i 1442 i5oJ 1 564 1625 1686 1747 1809 61 711 1870 1931 1992 2o53 2114 2175 2236 2297 2358 2419 61 712 2480 2341 2602 2663 2724 2785 2846 2907 2968 3029 61 71S 1 3090 3i5o 3211 3272 3333 3394 3455 35i6 3577 3637 61 714 3698 3759 3820 388i 3941 4002 4o63 4124 4i85 4245 61 715 i 43o6 4367 4428 4488 4549 4610 4670 4731 4792 4852 61 716 1 4913 4974 5o34 5095 5i56 5216 5277 5337 5398 5459 61 717 5519 5d8o 5640 5701 5761 5822 5882 5943 6oo3 6064 61 71S 6124 6i85 6245 63o6 6366 6427 6487 6348 6608 6668 60 719 ' 6729 6789 685o 6910 6970 703 1 7091 7i52 7212 7272 60 720 1 857332 7393 7453 75i3 8116 7574 7634 7694 7755 7815 7875 60 721 7935 8537 9i38 7995 8o56 8176 8236 8297 8357 8417 8477 60 722 8597 8657 8718 8778 8838 8898 8958 9359 •i58 9018 9078 60 723 9198 9^?^ 9318 9379 9978 9439 9499 ••98 9619 9679 60 724 selhl ti^ 9839 9918 ••33 •218 •278 60 725 0458 o5.8 0578 0637 0697 0757 0817 0877 60 726 1 0937 X 1036 1116 1176 1236 1295 i355 I4i5 1475 60 727 ; 1534 1 654 1714 1773 1 833 1893 2489 1952 2012 2072 60 728 i 2l3l 2I9I 2787 225l 23l0 2370 243o 2549 260S 2668 60 729 I 2728 2847 2906 2966 3o25 3o85 3i44 3204 3263 60 730 : 863323 3382 3442 35oi 356i 3620 368o 3739 3799 3858 59 731 i 3917 3977 4o36 4689 4155 4214 4274 4333 4392 4452 59 732 ! 45ii 4370 463o 4748 4808 4867 4926 4985 5045 59 733 ! 5io4 5i63 5222 5282 5341 5400 5459 55i9 5578 5637 59 734 i 6287 5755 58i4 5874 5933 5992 605T 6110 6169 6228 59 735 6346 6405 6465 6524 6583 6642 6701 6760 6819 59 736 1 6878 6937 6996 7055 7114 7173 7232 8468 735o 7409 59 737 7467 7526 7585 7644 7703 7762 7821 7939 8586 59 733 1 8o56 8ii5 8174 8233 8292 83 5o 8409 8527 59 739 , 8644 8703 8762 8821 8879 8938 8997 9o56 9114 9173 59 740 869232 9290 9349 9408 9466 9525 9584 9642 9701 9760 ^ 741 9818 9877 9933 9994 ••53 •ill •170 •228 •287 •345 ll 742 870404 0462 o52i o579 o638 0696 0755 o8i3 0872 n]t 743 0989 1047 1106 j II64 1223 1281 i339 1398 1456 58 744 ! 1573 1 2i56 i63i 1690 1748 1806 i865 1923 19S1 2040 Itf, 58 745 22l5 2273 233i 2389 2448 2306 2564 2622 58 746 ' 2739 2797 2855 2913 2972 3o3o 3o88 3i46 3204 3262 58 747 : 3321 3379 3437 3495 3353 36ii 3669 3727 1 3785 3844 58 743 1 3902 3960 4018 4076 4i34 4192 425o 43o8 4366 4424 58 749 4482 4540 4598 4656 4714 4772 4830 4888 4945 5oo3 58 750 875061 5119 5i77 5235 5293 535i 5409 5466 5524 5582 58 751 5640 5698 5756 58i3 5871 5929 5087 6045 6102 6160 58 752 6218 6276 6333 6391 6449 6564 6622 6680 6737 58 753 6795 6853 6910 6968 7026 7083 7141 7199 ! 7256 73.4 58 754 7371 7429 7487 7544 7602 7659 7717 7-774 7832 7889 58 755 1; 7947 8004 8062 81 19 8177 8234 8292 8349 8407 8464 57 756 ; 8522 8579 8637 8694 8752^ 8809 8866 8924 8981 9039 57 757 1 9096 9.53 9211 9268 9325^ 9383 9440 9497 9555 9612 57 753 •1 9669 9726 9-84 9>^4i i 9^98 9936 ••i3 ••70 •127 •i85 57 759 ,| 880242 0299 o356 o4i3 j 0471 o528 o585 0642 0699 0756 57 N. f 1 2 3 4 5 6 1 7 1 8 9 D. LOGARITHMS OF NUMBERS. 23 N. 1 1 2 3 4 5 6 7 8 9 1- 760 880814 0871 0928 0985 1042 1099 ii56 12l3 1271 i328 57 7G1 1385 1442 1499 1 556 i6i3 1670 1727 1784 1841 1898 57 762 1955 2012 2069 2126 2i83 2240 2297 2354 2411 2468 57 763 2525 258i 2638 2695 2752 2809 2866 2923 2980 3o37 57 764 3093 3i5o 3207 3264 3321 3377 3434 3491 3548 36o5 57 765 366i 3718 3775 3832 3888 3945 4002 4059 4n5 4172 57 766 4229 4285 4342 4399 4965 4455 4512 ^5^ 4625 4682 4739 57 767 4795 4852 4909 5022 5078 5192 5248 53o5 5i 76S 536i 5418 5474 553 1 5587 5644 5700 5757 5di3 58-0 57 769 5926 5983 6039 6096 6i52 6209 626D 6321 6378 6434 56 770 886491 6547 6604 6660 6716 6773 6829 6885 8067 6998 56 771 7054 71U 7167 7223 7280 7336 7392 79D3 7449 7561 56 772 7617 7674 8292 7786 7842 7898 8011 8123 56 773 8179 8236 8348 8404 8460 85i6 8573 8629 8685 56 774 8741 879T 8853 8909 8965 9021 9077 9134 9190 9246 56 775 9302 9358 9414 9470 9526 9582 9638 9694 9750 9806 56 776 9862 9918 9974 ••3o ••86 ••141 •197 •253 •309 •365 56 777 890421 0477 0333 0589 0645 0700 07 D6 0812 0868 0924 56 778 ^% io35 1091 1147 1203 1259 i3i4 1370 1426 1482 56 779 1593 1649 1705 1760 1816 1872 1928 1983 2039 56 780 892095 2l50 2206 2262 23i7 2373 2429 2484 2540 2595 56 781 265i 2707 2762 2818 2873 2929 298D 3o4o 3096 3i5i 56 782 8207 3262 33i8 3373 3429 3484 3D40 3595 3651 3706 56 783 3762 38i7 3873 3928 3984 4039 4094 4i5o 42o5 4261 55 .784 43i6 4371 4427 4482 4538 4593 4648 4704 4739 4814 55 785 4870 4925 4980 5o36 5091 5i46 5201 5257 53i2 5367 55 786 5423 5478 5533 5588 5644 5699 5754 5809 5864 5920 55 787 5975 6526 6o3o 6o85 6140 6195 6251 63o6 6361 6416 6471 55 788 658i 6636 6692 6747 6802 6857 6912 6967 7022 55 78y 7077 7132 7187 7242 7297 7352 7407 7462 7517 7572 55 790 897627 7682 7737 llf, 7847 7902 8451 79^7 8012 8067 8122 55 791 8176 823i 8286 8396 8D06 856i 861 5 8670 55 792 8725 8780 8835 8890 9437 8944 8999 9054 9109 9164 9218 55 793 9273 9328 9383 9492 9347 9602 9656 9711 9766 55 794 9821 9875 9930 9985 • •3g ••94 •149 •2o3 •258 •3l2 55 795 900367 0422 0476 o53i oj86 0640 0695 0749 0804 0859 55 796 0913 0968 1022 1077 ii3i 1186 1240 1295 1 349 1404 55 797 1458 i5i3 1567 1622 1676 1731 1785 1840 1894 1948 54 798 2003 2057 2112 2166 222Z 2275 2329 2384 2438 2492 54 799 2547 2601 2655 2710 2764 2818 2873 2927 2981 3o36 54 800 903090 3c44 3199 3253 3307 336i 3416 3470 3524 3578 54 801 3633 3687 3741 3795 3849 3904 3958 4012 4066 4120 54 802 4174 4229 4283 4337 4878 4391 4445 4499 4553 4607 466i 54 803 4716 4770 4824 4932 49S6 5040 5094 5 1 48 5202 54 804 5256 53io 5364 5418 5472 5526 558o 5634 5688 5742 54 805 5796 585o 5904 5958 6012 6066 6119 6658 6173 6227 6281 54 806 6335 6389 6443 6497 655i 6604 6712 6766 6820 54 807 6874 6927 6981 7035 7089 7143 7196 725o 7304 7358 54 808 741 1 7465 7519 8o56 7573 7626 7680 8217 7734 8270 7787 7841 8378 tt 54 809 7949 8002 8110 8i63 8324 54 810 908485 8539 8592 8646 8699 8753 8807 8860 8914 8967 54 811 9?L' 9074 9128 9181 9235 9289 9342 9396 9449 95o3 54 812 9556 9610 9663 9716 9770 9823 9»77 9930 9984 ••37 53 813 910091 0144 0107 0731 025l o3o4 o358 0411 0464 o5i8 0571 53 814 0624 0678 0784 o838 0891 0944 OQ98 io5i 1104 53 815 ii58 1211 1264 i3i7 1371 1424 1477 i53o i584 1637 53 816 i6go 1743 1797 i85o 1903 1956 2009 2063 2116 2169 53 817 2222 2275 2328 238i 2435 2488 2541 2594 2647 2700 53 818 2753 2806 2859 2913 2966 3019 3072 3i25 3178 3^31 53 819 3284 3337 3390 3443 3496 3549 36o2 3655 3708 3761 53 N. 1 2 3 4 5 6 7 8 9 D. 24 LOGAEITHMS OF NUMEEKS. N. 8'20 1 ^ 1 2 8 4 • v5} f6l ■ 7 - : 8 9 D. ■ 9i38i4 3867 3920 45o2 4026 4070 ~ 4608 4i32 4184 4237 4290 53 8:^1 4343 4396 4449 4555 4660 4713 4766 4819 53 8'J2 4872 4925 1 4977 5o3o 5o83 5i36 5189 5241 5294 5347 .53 82S i 5400 5453 DD05 5558 56ii ,5664 5716 5769 5822 5875 53 8'J-l 1 59-^7 5980 65o7 6o33 6o85 6i38 6191 6243 6296 6340 6875 6401 53 8'ir, 6454 6559 6612 6664 6717 7243 6770 6822 6927 53 82(1 6980 75o6 7033 7085 7.38 7190 7295 7348 7400 7433 53 827 7558 8o83 7611 7663 7716 7768 7820 7873 7925 SS 52 82S 8o3o 8i35 8188 8240 8293 8345 8397 845o 52 829 8555 8607 8659 8712 8764 88i6' 8869 89,21 8973 9026 52 830 ; 919078 9i3o 9183 9235 9287 9340 9392 9444 9496 9549 52 831 9601 9653 9706 9758 9810 9862 9914 9967 ••.9 ••71 52 832 ■ 920123 0176 0228 0280 o332 o384 0436 0489 o54i 0593 52 S33 i 0645 0697 0749 0801 0853 0906 0958 lOIO 1062 1114 52 834 ; 1 166 1218 1270 l322 1374 1426 1478 i53o 1 582 1 634 52 835 i 1686 1738 1790 1842 1894 1946 2??8 2o5o 2102 2.54 52 836 2206 2258 23lO 2362 2414 2466 2570 2622 2674 52 , 837 2725 2777 2829 3348 2881 2933 2985 3o37 3089 3i4o 3.92 52 52 838 3244 3296 3399 345i 35o3 3555 3607 3658 3710 830 3762 38i4 3865 3917 3969 4021 4072 4124 4176 4228 52 840 924279 4331 4383 4434 4486 4538 4589 4641 4693 4744 *52 841 4796 4848 4899 4951 5oo3 5o54 5io6 5i57 52og 5261 52 S42 53i2 5364 541 5 5467 55i8 5570 5621 5673 5725 5776 52 843 5828 5879 5931 5982 6o34 6o85 6137 6188 6240 6291 5i 844 6342 6394 6445 6497 6548 6600 665 1 6702 6754 68o5 5i 845 6857 6908 6959 701 1 7062 7114 7i65 7216 7268 7319 5i 846 7370 7422 7473 8037 7576 7627 7678 8191 7730 7781 8293 7832 8345 5i 847 i 7883 7935 7986 8088 8140 8242 5i 848 1 8396 8447 8498 8549 8601 8652 8703 8754 88o5 8857 9368 5i 849 1 8908 8959 9010 9061 9112 9163 9215 9266 9317 5i 850 929419 9470 9521 9572 9623 tii 9725 9776 9827 9579 5i 851 I 9930 9981 ••32 ••83 •i34 •236 •287 •338 •389 0898 5i 852 j 930440 0491 o542 0592 0643 0694 0745 0796 0847 5i 853 ' 0949 1000 io5i 1102 ii53 1204 1254 i3o5 i355 1407 5i 854 i 1458 1 509 i56o 1610 1661 1712 1763 1814 1 865 1915 5i 855 S 1966 2017 2068 2118 2169 2220 2271 2322 2372 2423 5i 856 2474 2524 2575 2626 2677 2727 2778 2S29 ll^ 2930 5. 857 29S1 3o3i 3082 3i33 3i83 3234 3285 3335 3437 5i 858 3487 3538 3589 3639 3690 3740 3791 3841 3892 3943 5i 859 3993 4044 4094 4145 4195 4246 4296 4347 4397 4448 5i 860 : 934498 4549 4599 465o 4^00 475i 4801 4852 4902 4953 5o 861 1 5oo3 5o54 5io4 5i54 5203 5255 53o6 5356 5406 5457 5o 862 ! 5507 5558 56o8 5658 5709 5759 5809 5860 5910 5960 5o 8'13 601 1 6061 6111 6162 6212 6262 63 1 3 6363 64.3 6463 5o 864 65i4 6564 6614 6665 6715 6765 68i5 6865 6916 6966 5o 865 7016 7066 7117 7167 7217 7718 '7267 73.7 7367 74.8 7468 5o 866 75i8 7568 7618 7668 7769 8269 7819 7869 7919 7969 5o 867 8019 8069 &119 8169 8219 8320 8370 8420 8470 5o 868 8520 8570 8620 8670 8720 8770 8820 8870 8920 8970 5o 869 9020 9070 9120 9170 9220 9270 9320 9369 9419 9469 5o 870 939519 9569 0068 9610 0118 9669 9719 0218 9769 9819 9869 99.8 9968 5o 871 940018 0168 0267 o3i7 0367 0417 0467 5o 872 o5i6 o566 c6i6 0666 1 0716 0765 081 5 0865 0915 0964 5o 873 1014 1064 1114 Ji63 12-. 3 1263 i3i3 i362 14.2 1462 5o 874 j5ii i56i 1611 1660 1710 1760 1809 1809 1909 .958 5o 875 2008 2o58 2107 2137 2207 2256 23o6 235D 240D 2455 5o 876 25o4 2554 2603 2653 2702 2752 2801 285 1 2901 2950 5o 877 3ooo 3o49 3099 3148 3.98 3247 3297 3346 3396 3445 49 878 3495 3544 3593 3643 36q2 3742 3791 3841 3890 3939 49 879 3989 4o38 4088 4i37 4186 4236 4285 4335 4384 •4433 49 N. 1 2 3 4 5 6 7 8 9 D. LOGARITHMS OF NUMBERS 25 N. ' 1 2 3 4 5 6 7 8 9 D. 880 1 944483 4532 458i 463 1 4680 4729 4779 4828 4877 1 4927 49 8S1 4976 5o25 5074 5i24 5173 5222 5272 5321 5370 ; 5419 49 882 5469 55i8 5567 56i6 5665 5715 5764 58 1 3 5862 5912 49 883 5961 6010 6059 6108 6i57 6207 6698 6256 63o5 6354 6403 49 884 6402 65oi 655i 6600 6649 6747 6796 6845 6894 49 885 6943 6992 7041 7090 7140 7189 7238 7287 7336 7385 49 886 7434 7453 p32 8022 ^07^ 7630 8119 7679 8168 7728 7777 8266 7826 83i5 7875 8364 49 887 7924 8413 ml 8217 49 883 85ii 8560 8609 8657 8706 8755 8804 8853 49 889 890 8902 8951 8999 9048 9097 9146 9195 9244 9292 9341 49 949390 9439 9488 9536 9585 9634 9683 9731 9780 9829 49 891 9878 9926 9975 ••24 ••73 •J21 •170 •219 •267 ^316 49 892 95o365 0414 0462 o5ii o563 0608 0657 n43 0706 0754 o8o3 49 893 o85i 0900 0949 0997 1046 1095 1192 1240 1289 49 894 i338 1386 1435 i4«3 1 532 i58o 1629 1677 1726 1773 S 895 1823 1872 1920 1969 2017 2066 2114 2i63 2211 2260 896 23o8 2356 24o5 2453 2502 255o 2599 2647 2696 2744 48 897 2792 2841 2889 2938 2986 3o34 3o83 3i3i 3i8o 3228 48 898 3276 3325 3373 3421 3470 35i8 3566 36i5 3663 3711 48 899 3760 38o8 3856 3905 3953 4001 4049 4098 4146 4194 48 900 954243 4291 4339 4387 4435 4484 4532 458o 4628 4677 48 901 4725 4773 4821 4S69 4918 4966 5oi4 5o62 5iio 5i58 48 902 5207 5688 5255 53o3 535i 5399 5447 5495 5543 5592 5640 48 903 5736 5784 5832 5880 5928 5976 6024 6072 6120 48 904 6168 6216 6265 63i3 636 1 6409 6457 65o5 6553 6601 48 •905 6649 6697 6745 6793 6840 6888 6936 6984 7032 7080 48 906 7128 7176 7324 7272 7320 7368 7416 7464 75x2 m 48 907 7607 7655 7703 775i 8229 7799 8277 7847 7894 7942 8421 7990 48 908 8086 8i34 8181 8325 8373 8468 85i6 48 i*09 8564 8612 8659 8707 8755 88o3 885o 8898 8946 8994 48 910 959041 ^^ 9137 9i85 9232 9280 9328 9375 9423 9471 48 911 9518 9614 9661 9709 9757 9804 9852 9900 9947 48 912 9995 ••42 ••90 •i38 •i85 •233 ^280 •328 •376 •423 48 913 960471 o5i8 o566 o6i3 0661 0709 0756 0804 o85i 0S99 48 914 0946 0994 1041 1089 ii36 1 184 I23l 1279 i326 1374 47 915 1421 1469 i5i6 i563 1611 1658 1706 1753 1801 1848 47 916 1895 1943 1990 2o38 2o85 2l32 2180 2227 2275 2322 47 917 2369 2843 2417 2404 25ll 2559 2606 2653 2701 2748 2795 47 918 2890 2937 2985 3o32 3079 3126 3174 3221 3268 47 919 33i6 3363 3410 3457 35o4 3552 3599 3646 3693 3741 47 920 963788 3835 3882 3929 3977 44.i8 4024 4071 4118 4i65 4212 47 9-21 4260 4307 4354 4401 4495 4542 4590 4637 4684 47 922 473 1 477« 4825 4-S72 4919 4966 5oi3 5o6i 5io8 5i55 47 923 5202 5249 5296 5343 5390 5437 5484 553 1 5578 5625 47 924 5672 5719 5766 58i3 586o 5907 5954 600 1 6048 6093 47 925 6142 6189 6658 6236 6283 6329 6376 6423 6470 65i7 6564 47 926 6611 6705 6752 6799 6845 6892 6939 6986 7033 47 927 7080 7127 7173 7220 7267 7314 7361 7408 7454 7501 47 928 929 .7548 8oi6 ui 7642 8109 7688 8i56 7735 8io3 7782 8249 7829 8296 llll 7922 8390 Itl 47 47 930 ' 968483 853o 8576 8623 86-0 8716 8763 8810 8856 8903 47 931 i 8950 8996 9043 9090 9136 9.83 9229 9276 9323 9369 47 932 9416 9463 9009 9556 9602 9649 9693 9742 9789 9S35 47 933 9882 9928 9973 ••21 ••68 •114 •161 •207 •254 •3oo 47 934 1 970347 0393 0440 04S6 o5>3 0579 0626 0672 0710 ii83 0765 46 935 0812 o858 0904 0951 09 V7 1044 1090 1137 1229 1693 46 936 1276 l322 1369 1415 14 I i5o8 1 554 i6ol 1647 46 937 1740 1786 1832 1879 1975 1971 2018 2064 2110 2137 46 938 2203 2249 2295 2342 23^8 2434 2481 2527 2573 2619 46 939 2666 2712 2758 2804 283i 2897 2943 2989 3o35 3082 46 N. 1 2 3 * 5 1 " 7 8 9 D. 26 LOGAEITHMS OF NUMBEES. N. 1 2 3 4 5 6 7 8 9 D. 940 973128 3i74 3220 3266 33i3 3359 34o5 345i 3497 3543 46 941 3590 4o5i 3636 3682 3728 3774 3820 3866 '4 4SU 3959 4oo5 46 942 4097 4143 4189 4235 4281 4327 4420 4466 46 943 45i2 4558 4604 465o 4606 4742 4788 4880 4926 46 944 4972 5oi8 5o64 5iio 5.D6 5202 5248 5294 5340 5386 46 945 5432 5478 5524 5570 56i6 5662 5707 5753 5799 5845 46 946 5891 6396 5983 6029 6oi5 6121 6167 6212 6258 63o4 46 947 63 5o 6442 6488 6533 6579 6625 6671 6717 6763 46 948 6808 6854 6900 6946 6992 7037 7083 7129 7175 7220 46 949 7266 7312 7358 7403 7449 7495 7541 7586 7632 7678 46 950 977724 8181 7769 8226 78.5 7861 83i7 7906 7952 7998 8043 8089 8 1 35 46 951 8272 8363 8409 8454 85oo 8546 8591 46 952 8637 8683 8728 8774 88i§ 8863 891 1 9366 8956 9002 9047 46 953 9093 9i38 9184 Q23o 9273 9321 9412 9457 95o3 46 954 9548 9594 9639 9685 9730 9776 9821 9867 X 9958 46 955 980003 0049 0094 0140 oi85 023l 0276 o322 0412 45 956 0458 o5o3 0049 0594 0640 0685 0730 0776 0821 0867 45 957 0912 0957 ioo3 1048 1093 1139 1184 1229 1275 l320 45 958 1366 I4U 1456 i5oi 1547 1592 1637 1 683 1728 1773 45 959 960 1819 1864 J 909 1954 2000 2045 2090 2i35 2181 2226 45 982271 23i6 2362 2407 2452 2491 2543 2588 2633 2678 45 961 2723 2769 2814 2859 2904 2949 2994 3o4o 3o85 3i3o 45 962 3175 3220 3265 33io 3356 3401 3446 3491 3536 358i 45 963 3626 3671 3716 3762 3807 3852 3897 3942 4392 3987 4o32 45 964 4077 4122 4167 4212 4257 43o2 4347 4437 4482 45 965 4527 4572 4617 4662 4707 4752 4797 4842 4867 4932 5382 45 96^ 4977 5022 5067 5112 5.57 5202 5247 5292 5337 45 967 5426 5471 55.6 556r 56o6 565 1 5696 5741 5786 583o 45 963 5875 6920 6369 5965 6010 6o55 6100 6144 6189 6234 6279 45 969 6324 64i3 6458 65o3 6548 6593 6637 6682 6727 45 970 986772 6817 6861 6906 6951 6996 7040 7085 7i3o 7175 45 971 7219 7264 7309 7353 7398 7443 7488 7532 8024 7622 45 972 7666 8ii3 7711 7756 8202 7800 7845 7890 Ifsi lilt 8068 ^l 973 8157 8247 8291 8336 8470 85i4 45 974 8559 &604 8648 8693 8737 8782 8826 8871 8916 8960 45 975 9000 9049 9094 9i38 9183 9227 9272 93i6 9361 94o5 45 , 9:6 9430 9494 95J9 9583 9628 9672 9717 9761 9806 9856 44 t)77 ,8o5 9903 3o 0783 9983 ••28 ••72 •.17 .•161 •206 •25o •294 0738 44 978 0428 0472 o5i6 o56i o6o5 o65o 0694 44 979 0827 0871 0916 0960 1004 1049 1093 1137 1182 44 980 991226 1270 i3i5 1359 l4o3 1448 1492 1536 i58o 1625 44 981 ^^669 1713 1758 1802 1846 itr, 1935 2377 1979 2023 2067 44 982 2111 2i56 2200 2244 2288 2421 2465 2509 44 983 2554 2598 2642 2686 2730 2774 2819 2863 2907 2951 44 984 2995 3o39 3o83 3127 3568 3172 3216 3260 33o4 3348 33?2 44 985 3436 3480 3524 36i3 3657 3701 3745 3789 3833 44 986 3877 3921 436i 3965 4009 4o53 4097 4141 4i85 4229 4273 44 987 43.7 44o5 4449 4493 4537 458i 4625 4669 5io8 4713 44 9SS 4737 4801 4845 4889 4933 4977 502I 5o65 5i52 44 989 5196 5240 5284 5328 5372 5416 5460 55o4 5547 5591 44 900 995635 5679 5723 5767 58ii 5854 5898 5942 5986 6o3o 44 901 6074 6.17 6555 6161 6205 6249 6293 6337* 6380 6424 6468 44 992 65i2 6599 6643 6687 6731 6774 6818 6862 6906 7343 44 993 6949 6993 7037 7080 7124 7168 7212 7255 7299 44 994 7386 7430 7474 7517 7561 7605 7648 8o85 7692 7736 8172 7779 8216 44 995 7823 8259 7867 7910 l%t 7998 8041 8129 44 996 83o3 8347 8434 8477 8521 8564 8608 8652 44 997 8695 8739 8782 8S26 8869 8913 9348 8956 9000 9043 9087 44 908 9.31 9174 9218 9261 93o5 9392 9435 9479 9522 44 999 9565 9609 9652 9696 9739 9783 9826 9870 9913 9957 43 K 1 2 1 3 4 5 6 7 8 9 D. LOGARITHMS OF NUMBERS. 27 N. 1 2 3 4 5 6 7 8 9 D. 1000 000000 0043 0087 oi3o 0174 0317 0260 o3o4 o347 0391 43 lOOI 0434 0477 052I o564 0608 o65i 0694 0738 i 0781 0824 43 1002 0868 "^l 0954 1 388 0998 io4i iob4 1128 1171 1 1214 1238 43 [oo3 i3oi I43i 1474 i5i7 I36l 1604 1647 1690 43 1004 1734 nil 1820 i863 IK 1950 1993 2o36 2080 2123 43 ioo5 002166 2209 2252 2296 2382 2420 2468, 25 1 2 2.05 43 ioo6 2598 2641 2684 2727 2771 2814 2857 2900 2943 2986 43 1007 3029 3073 3ii6 3i59 3202 3245 3288 333i 3374 3417 3848 43 1008 346 1 35o4 3547 3390 3633 3676 3719 3762 38o5 43 1009 , 3891 3934 3977 4020 4o63 4106 4149 4192 4235 4278 43 lOIO 004321 4364 4407 445o 4493 4536 4579 4622 4665 4708 43 lOII 4751 4794 4837 4880 4923 4966 5395 5009 5438 5o52 5095 5i38 43 I0I2 5i8i 5223 5266 5309 5352 5481 5524 5567 43 ioi3 5609 5652 5695 5738 5781 5824 5867 5909 6?8o 5995 43 IOI4 6o38 6081 6124 6166 6209 6252 6295 6338 6433 43 lOID 006466 65o9 6552 6594 6637 6680 6723 6765 6808 685 1 43 1016 6894 6936 7364 6979 7022 7065 7107 7i5o 7.93 7230 7278 43 IOI7 IO18 7321 7406 7449 7492 7534 7577 7620 7662 7703 43 7748 7790 7833 7876 l%l 7961 8004 8046 8i32 43 IOI9 8174 8217 8259 83o2 8387 8430 8472 8558 43 1020 008600 8643 8685 8728 8770 88i3 8856 8893 8941 9366 8983 43 1021 ' 9026 9068 9111 9153 9196 923s 9281 9323 9408 42 1022 9431 9493 9536 9578 9621 9663 9706 9748 9791. 9833 42 1023 9876 X 9961 •ooo3 •0043 .0088 •oi3o •0173 •0213 • 0258 42 - 1024 oio3oo 0385 0427 0470 05l2 o554 0597 0639 io63 0681 42 1020 010724 0766 0809 o85i 0893 0936 0978 1020 no5 42 1026 1147 1 190 1232 1274 i3i7 1359 1401 1444 i486 i528 42 1027 1570 i6i3 1655 1697 1740 1782 1824 1866 1909 1931 42 1028 1993 2o3o 2078 2 1 20 2162 2204 2247 2289 233 1 2373 42 1029 24i5 2458 2,500 2542 2584 2626 2669 2711 2753 2795 42 io3o 012837 2879 2922 2964 3385 3oo6 3048 3090 3i32 3174 3217 42 io3i 3259 33o. 3343 3427 3469 35ii 3553 3596 3638 42 I032 368o 3722 3764 3 806 3848 3890 3932 4353 IV4 4016 4o58 42 io33 4100 4142 4184 4226 4268 43io 4437 4479 4898 42 io34 4521 4563 46o5 4647 4689 4730 4772 4814 4856 42 io35 014940 5360 4982 5o24 5o66 5io8 5i5o 5192 5234 5276 53i8 42 io36 5402 5444 5485 5527 5569 56ii 5653 5695 5737 42 1037 5779 5821 5863 5904 5946 5988 6o3o 6072 6114 6i56 42 io38 6197 6239 6281 6323 6365 6407 6448 6490 6532 6574 42 1039 6616 6657 6699 6741 6783 6824 6866 6908 6950 6992 42 1040 017033 7075 7117 7159 7200 7242 7284 7326 7367 7409 42 io4i ^H 7492 7534 7576 7618 7659 7701 7743 7?84 7826 42 1042 7868 8284 7909 7951 7993 8o34 8076 8118 8139 8201 8243 42 1043 8326 8368 8409 845 1 8492 8534 857b 86.7 9033 8659 42 1044 8700 8742 8784 8825 8867 8908 9324 8950 9366 8992 9075 42 1045 019116 9i58 9199 9241 9282 9407 9449 9490 42 1046 9532 9573 9613 9656 9698 9739 9781 9822 9864 9903 42 1047 1048 9947 9988 • oo3o •0071 •oii3 • 0154 • 0193 .0237 i-0278 •0320 41 "o2o36i o4o3 0444 0486 0527 o568 0610 o65i 0693 0734 41 1049 0775 0817 0858 0900 0941 0982 1024 io65 1107 1 148 41 io5o 021 189 I23t 1272 13,3 i355 1396 1437 1479 l520 i56i 41 io5i i6o3 1644 i685 1727 1768 1809 l83l 1892 1933 2346 1974 41 1032 2016 2057 2098 2140 2181 2222 1 2263 23o5 2387 41 io53 2428 2470 25ll 2352 2593 2635, 2676 2717 2758 2199 41 io54 2841 2882 2923 2964 3oo5 1 3047 1 3o88 3417' 3458: 3499 3129 3170 3211 41 io55 023232 3294 3335 3376 3541 3582 3623 41 io56 3664 3705 3746 3787 3828: 3870 3911 4239 4280 ; 4i2i 3932 3993 4o34 41 io57 io58 4075 4116 4137 4198 4363 4404 4445 41 4486 4527 4568 4609 4630 4691 4732 4773 4814 4855 41 1059 4896 4937 4978 5oi9 5o6o 5ioi 5i42 5i83 5224 5265 41 N. 1 2 3 4 5 6 7 8 9 D. 28 LOGAEITHMS OF XU^IBERS. m 1 2 3 4 5 6 7 j 8 1 9 D. 1060 ! o253o6 5347 5388 5429 5838 6247 5470 55ii 5552 5593 ! 5634 i 5674 41 1061 1 1062 5715 6125 5756 6i65 5797 6206 5879 6288 5920 6329 5961 6370 6002 6411 6043 , 6084 6452 6492 41 41 1063 6533 6514 6982 7390 661 5 6656 6697 6737 6778 7186 6819 6860 6901 41 1064 6942 027350 7023 7064 7io5 7146 7227 7268 7300 7716 41 io65 7431 7472 75i3 7553 7594 7635 7676 8042 8o83 41 1066 7757 ml 7839 7«79 7920 & S002 8124 41 1067 1 8164 8246 82S7 8327 8409 8449 1 8490 853 1 41 1068 8571 8612 8653 8693 8734 8775 9181 88i5 8856 1 8896 8937 41 1069 8978 9018 9059 9100 9140 9221 9262 j 93o3 9343 41 1070 029384 9424 9465 9506 9546 9587 9627 9668 ! 9708 9749 41 1071 9789 9830 9871 T6 Itr, 9992 .oo33 •0073-0114 -oi54 41 1072 o3oi9D 0235 0276 0397 0438 04781 0319 o883 1 0023 1287 1328 0559 40 1073 1 0600 0640 0681 0721 0762 0802 0843 Xt 40 1074 1004 1045 io85 1126 1166 1206 1247 40 1073 o3i4o8 1449 1853 189? i53o 1570 1610 i65i 1691 1732 1772 40 1076 1812 1933 1974 2377 2014 2o54 2093 2i35 2173 40 1077 2216 2256 2296 2337 2417 2458 2498 2538 2578 2981 40 1078 2619 2609 2699 2740 2780 2820 2860 IZ 2941 40 1079 302I 3062 3l02 3i42 3182 3223 3263 3343 3384 40 1080 033424 3464 35o4 3544 3585 3625 3665 3705 1^43 3786 40 1081 3826 3866 3906 43o8 3946 3986 4388 4027 4067 4468 4107 4147 4187 40 1082 4227 4628 4267 4348 4428 45o8 4548 4588 40 1083 4669 4709 4749 4789 4829 4869 nziitPo it 40 1084 5029 5069 5109 5i49 5190 523o 5270 40 io85 o3543o 5470 55io 5550 5590 5630 5670 5710 i 5750 5790 40 1086 5830 5870 6?09 6?49 5990 6o3o 6070 6110 6i5o 6190 6580 7387 40 1087 6230 6269 6389 6429 6828 6469 6868 65o9 6008 .7307 6549 40 1088 6629 7028 6669 6709 7108 6749 7140 6789 6048 7347 40 1089 7068 7187 7227 7267 40 1090 037426 7466 7506 7546 7586 7626 7665 7705 8io3 8143 7785 8i83 40 1091 7825 1 8223 7865 IToi m 7984 8024 8064 40 1092 8262 8382 8421 8461 i 85oi 8541 858o 40 1093 8620 8660 8700 8739 8779 9i36 9176 8819 8859 8898 : 8o58 9295 ' 9335 8978 9374 40 1094 9017 9057 9097 9216 9253 40 1093 039414 9454 9493 9533 9573 9612 9652 9692 1 9731 9771 40 1096 9811 985o 9800 mi Itt • 0009 .0048 • 0088 ,-0127 1-0167 40 1097 040207 0246 0286 040D 0444 0484 1 o523 o563 40 1098 0602 0642 0681 0721 0761 0800 0840 0879 0019 1274 i3i4 0958 i353 40 1099 0998 1037 1077 1116 ii56 1195 1235 39 N. 1 2 3 4 5 6 7 8 9 D. TABLE II. CONTAINING NATURAL SI?^^ES AND COSHES, LOGAEITIIMIC SI:N^ES, COSINES, TAiNGENTS, AND COTANGENTS, EVERY DEGREE AND inXTTE OF THE QUADRANT. 30 TRIGOXCniETRICAL FUXCTIOXS.— 0°. Nat. Functions. Logarithmic F l-nctions + 10. N.sine.' N. cos. L. sine. D.1" L. COS. D.l" L. tang. D.l" L. cot 1 00000 1 Unit. 0-000000 10.000000 1 1 0-000000 ! Infinite. | 60 1 ooo2g Unit. 6.463726 50I7-I7 000000 00 6-463726 5oi7 17 13.5362741 59 2 00058 Unit. 764756 2934- 85 000000 00 1 764756 2934 83 235244 58 S 00087 Unit. 940847 2082 3i 000000 00: 940847 2082 3i 059153 57 4 001 16 Unit. 7.065786 i6i5 JJ 000000 00 ■ 7-06D786 i6i5 17 12.934214 56 837304 55 5 00145 1 Unit. 162696 i3i9 000000 00 162696 i3i9 69 6 001751 Unit. 241877 HID 75 9-999999 i oil 241878 1115 78 758122; 54 7 00204 Unit. 308824 966 53 999999 1 01 1 308825 t. 53 601 1 75 153 8 00233 Unit. 366816 852 54 999999 01 1 366817 54 633183152 9 00262 Unit. 417968 762 63 999999 999998 01! 417970 463727 762 63 582o3o'51 10 00291 i Unit. 463725 "689 88 _0I_ 689 88 536273; 50 11 oo32o j 99999 7T5051T8 629 81 9.999998 01 7'5o5i2o 629 81 12^494880 49 12 00349 99999 542906 579 36 999997 i 01 , 542909 IV, 33 457091 48 13 00378! 99999 577668 536 41 999997 01 577672 42 422328 47 14 00407 99999 609853 499 38 999996 i 01 609837 499 39 390143 46 15 00436 99999 639816 467 438 14 999996 , 01 639820 467 i5 36oi8o 45 16 00465 99999 667845 81 999995 1 01 667849 438 82 332i5i 44 17 00495 99999 694173 4i3 72 999995 1 01 694179 4i3 73 3o5S2i 43 18 oo524 99999 718997 391 35 999994 1 01 719003 391 36 280997 42 19 00553 99998 742477 371 \l 999993 : 01 ; 742484 371 28 237316 41- 20 oo582 99998 764754 353 999993 s±, 764761 35i 36 235239 ^"^ 21 006 1 1 99998 7-785943 336 72 9.999992; 01 7-785951 336 "73 12^214049 3y 22 00640 99998 806146 321 75 999991 1 01 1 8061 55 321 76 193845 33 23 00669 99998 825451 3o8 o5 999990 1 01 ' 825460 3o8 06 174540 37 24 00698 99998 843934 'M 47 999989 ; 999988 1 02 843944 295 49 l56o56 36 25 00727 99997 861662 88 02 861674 283 90 138326 85 26 00736 99997 878695 273 17 999988 1 02 i 878708 273 18 121292 34 27 00785 99997 895085 263 23 999987 1 02 895099 263 25 104901 33 28 00814 99997 910879 253 I^ 999986 1 02^ 910894 254 01 089106 32 29 00844 99996 926119 245 999985 1 •02 926134 245 40 073866 31 SO i 00873 99996 940842 7.955082 237 33 999983 ; •02 940858 237 35 059142 30 31 00902 00931 99996 229 80 9.9999821 .02 ■^ "9^500° 229 81 12-044900 2y 1 82 99996 96S870 222 73 999981 •02 \ ?s 222 75 o3iiii 23 £3 00960 99995 982233 216 08 999980 •02 1 216 10 017747 27 £4 00989 01018 99993 995108 80077S7 209 81 999979 •02 ' 995219 209 -83 004781 26 35 9999D 203 t 999977 •02 8-007809 203 & 11-992191 25 36 01047 99995 020021 198 999976 1 •02 020045 198 979933 24 £7 01076 99994 031919 193 188 02 999975 i •02 031945 043527 • o5 96S055 23 38 oiio5 99994 043501 01 999973 i •02 • o3 956473 22 39 on 34 99994 054781 i83 25 999972 1 •02 054809 iS3 •27 945191 21 40 01 164 99q93 065776 178 Jl 999971 1 •02 o658o6 178 •74 934194 20 "IT 01193 99993 8.076500 174 •41 -»^^ •02 8-076531 174 •44 11 -923460] 19 42 01222 99993 086965 170 • 3i •02 086997 170 •34 9i3oo3 18 43 oi25i 99992 097183 166 .39 999966 .02 097217 166 •42 002783 8g3o37 17 44 01280 99992 107167 162 .65 999964 : .03 107202 162 .68 16 45 oi3o9 99991 116926 159 • 08 999963 I •03 116963 159 10 15 46 oi338 99991 126471 i55 .66 999961 ! • o3 i265io l53 .68 873490 14 47 01367 99991 i358io l52 •38 99993Q ; 9999D8 j • o3 i3585i l52 •41 864149 13 43 01396 99990 144953 149 •24 •o3 144996 149 •27 855004 12 1 49 !1 014251 QQQQO 153907 146 • 22 999956 • o3 ID3932 146 '11 846048 ; 1 1 1 50 01454 i 999^9 162681 143 .33 999954 • o3 162727 143 • 36 85/273 10 51 oi483| 99989 oi5i3l 99989 8.171280 140 •54 9-999952 •03 8-171328 140 ^ 11-828672 9 52 179713 i37 86 999900 • 03 \llti i37 .90 820237 8 53 01 542 i 99988 187985 i35 29 999948 • o3 i35 .32 811964 7 54 01571 j 99988 196102 l32 80 999946 • o3 196156 l32 .84 8o3844 6 55 01600 1 99987 204070 i3o 41 999944 • o3 204126 i3o •44 795874 5 56 01629! 99987 211895 128 10 999942 • 04 211953 128 •14 788047 4 57 01653 99986 219581 125 87 999940 • 04 219641 125 .90 780359 3 58 01687 99vS6 227134 123 72 999938 ; • 04 227195 123 •7^ 772805 2 59 01716 999R5 234557 121 64 999936 • 04 234621 121 .68 765379 1 60 01745 999S5 241855 119.63 999934! -04 241921 119-67 758079 N. COS. :N. sine. L. COS. D.l" L. sine, j L.cot. D.l" L. tang. ' 89^ 1 TRIGOXOMETRTCAL FUXCTIOXS. — 1' 31 Nat. Functions, Logarithmic Functions + 10. 1 / N. sine. N. cos. L. sine. D.l" L. cos. JD.l", L. tang. 8-241921 119-67 L.cot. j 1 01745 99985 8-241855 119 63 9.999934 •04! 11-758079 60 1 01774 '999'^4 249033 117 68 999932 •04 1 249102 117 72 700898 59 21 oi8o3 99984 256094 ii5 80 999929 •041 256i65 ii5 84 743835 53 3' oi832 99983 263o42 n3 98 999927 -04 263ii5 114 02 736885 57 4 01862 99983 269881 112 21 999925 -04 269956 112 25 730044 66 5i 01 89 I 99982 276614 110 5o 999922 .04 276691 no 54 723309 65 6 01920 99982 283243 108 83 999920 -04 283323 108 ll 716677 54 7 01949 99981 01978,99980 289773 107 21 999918 .04 289856 107 710144 £3 8 296207 io5 65 9999 J 5 -04 296292 lOD 70 703708 62 9 02007 999S0 302546 104 i3 999913 -04 302634 104 18 697366 61 10 11 o2o36 99979 308794 102 66 999910 -04 308884 102 70 691116 50 02065 99979 02094 9997a 8-314954 lOI 22 9.999907 • 04 8-3i5o46 101 26 11-684954 4^ j 12 321027 99 82 999900 -04 321122 99 87 678878 48 13 02123 99977 327016 98 47 999902 -04 327II4 98 5i 672S86 47 14 02l52 99977 332924 97 14 999H99 -o5 333025 97 19 666975 46 15 O2181 99976 338753 95 86 999897 -o5 338856 95 90 661144 45 16 022II 99976 344504 94 60 999894 -o5 344610 94 65 655390 44 17 02240 99975 35oi8i 93 38 999891 -o5 350289 93 43 6497 'I 43 18 02260,99974 02298199974 355783 92 ;? 999888 • o5 355895 92 24 644105 42 19 36i3i5 % 999885 -o5 36i43o 91 08 638570 41 20 21 02327 1 99973 366777 90 999882 -o5 -o5 366895 8-372292 89 88 95 "85 633io5 40 02356 99972 8-372171 88 80 9.999879 11.627708 "39" 22 02385 99972 377499 87 72 999876 -o5 377622 87 •77 622378 38 23 02414 99971 382762 86 67 999873 -o5 382889 86 -72 617111 37 24 02443 99970 387962 85 64 999870 -o5 388092 393234 85 -70 61 1908 36 25 02472 99969 393101 84 64 909867 -o5 84 -70 606766 35 26 025oi 99969 398179 83 66 999864 -o5 3983 1 5 83 •71 6oi685 34 27 o253o 99968 4o3i99 82 71 999861 .o5 403338 82 t 596662 33 28 0256o 99967 408161 81 77 999858 -o5 4o83o4 81 591696 32 29 02589 02618 02647 99966 4i3o68 80 86 999854 -o5 4i32i3 80 •91 586787 31 30 31 99966 4i79'9 79 it 999851 -06 418068 80 -02 581932 30 99965 8-422717 79 09 9.999848 -06 8-422869 79 ~iT ii-577i3i 29 32 02676 99964 427462 78 23 999844 -06 427618 78 •3o 572382 28 33 02705 99963 432156 77 40 999841 -06 4323i5 77 •45 567685 27 84 02734 99963 436800 76 57 999838 -06 436962 76 -63 563o38 26 35 02763 99962 441394 75 77 999834 -06 44i56o 75 -83 558440 25 36 02792 99961 445941 74 99 999831 -06 446110 75 .05 553890 24 37 02821 99960 450440 74 22 999827 .06 45o6i3 74 .28 549387 23 88 o285o 99959 454893 73 46 999823 -06 455070 73 52 544930 22 39 02879 99939 99958 99957 459301 72 73 999820 -06 459481 72 •79 54o5i9 '21 40 41 02908 02938 463665 72 00 999816 -06 463849 72 06 536i5i 20 8-467985 71 29 9.999812 -06 8-468172 71 .35 11-531828 TIT 42 02967 99956 472263 70 60 999809 -06 472454 70 .66 527546 18 43 02996 j 99955 476498 69 91 999805 .06 476693 69 98 523307 IT 44 o3o25 ' 99954 480693 ^ 24 999801 -06 480892 69 .31 519108 16 45 o3o54 1 99953 484848 59 999797 -07 485o5o 68 -65 5i495o 15 46 o3o83 99952 488963 67 94 999793 •07 489170 68 -01 5io83o 14 47 o3ii2 199952 493040 ^ 3i 999790 -07 493250 67 -38 506750 IG 48 o3i4i 1 99951 497078 66 ^ 999786 .07 497293 66 -76 502707 12 49 o3 1 70- '99950 5oio8o 66 999782 -07 501298 66 -i5 498702 11 "50 51 03199-99949 03228 ' 99948 5o5o45 8.508974 65 64 48 "89 999778 9-999774 -07 ! 505267 65 .55 494733 10 9 8-509200 513098 64 96 11-490800 52 03257:99947 1 03286 : 99946 512867 64 3i 999769 •07 64 39 4%902 8 53 516726 63 75 999765 -07 516961 63 82 483o39 7 54 o33i6 99945 52o55i 63 •9 999761 • 07 520790 63 26 479210 6 55 j' o3345 ! 99944 524343 62 64 999757 -07 524586 62 72 475414 5 56 1 03374 99943 528102 62 11 999753 -07 528349 62 18 47i65i 4 57 1 o34o3 199942 531828 61 58 999748 -07 532080 61 65 • 467920 8 58 03432; 99941 535523 61 06 999744 -07 535779 61 i3 464221 2 59 1 1 03461 1 99940 539186 60 55 999740 -07 539447 60 62 460553 1 60 03490 ; 99939 542819 60-04 999735 • 07 543084 60-12 11-456916 |.N. COS. N. sine. L. COS. D.l" L.sine. 1 L.cot. D.l" L. tang. 1 ' 88° 32 TRIGOXOMETRICAL FUXCTIOXS. — 2" Nat. Fr^-cTioNS. LOGAKITHMIC FUNCTIONS + 10. ' [^N.sine. N. cos. L. sine. D. 1" L. COS. D.l" L. tang. 1 D. 1" L. cot. 1 o l\ 6 'si 9 10 03490 99939 o35i9 99938 0354S 99937 03577 99936 o36o6 99935 o3635 99934 03664 99933 o36q3 99932 03723 99931 03752 99930 03781 99929 8-542819 60 546422 1 59 557054 i 58 56o54ol 57 570836 ' 56 5742141 55 577566 I 55 04 55 06 58 11 65 •9 74 3o 87 44 9.999735 999731 999726 999722 : 999717 999713 999708 999704 999699 999&89 •07 •07; :SI .081 •081 .081 .081 -08! .08 .08 8-5430841 60 553817 1 58 557336 1 58 560828 j 57 564291 57 567727 56 571137 56 574520 55 577877 1 55 12 62 14 66 it 38 95 02 11-456916 453309 449732 446183 442664 439172 430709 432273 428863 425480 422123 60 69 58 57 66 55 54 53 52 % 11 12 13 14 15 16 17 18 19 20 o38io 99927 o3S39 99926 o3S68 99925 03897 99924 03926 99923 "03955 99922 03984 99921 04013 99919 04042 99918 04071 ,999 '7 8080892 584193 587469 590721 593948 597152 6oo332 603489 606623 609734 55 54 54 53 53 53 52 52 5i 5i 02 60 J9 11 00 61 23 86 49 9-999683 999680 999673 999670 999665 999660 999655 99965o 999645 999640 .08 •08: .081 .08 i .o8i -08 i .08: .081 .09 -09 8-581208 5845i4 587795 591051 594283 597492 600677 6o3839 60697a 610094 55 54 54 53 53 53 52 52 10 68 I] 47 08 It 11-418792 415486 4l2205 408949 399323 396161 393022 389906 tl 47 46 45 44 43 42 41 40 21 !; 04100 99916 22; 04129 99915 23 04159 99913 24 041SS 99912 -5 '04217 99911 26,; 04246 99910 27 jl 04275 99909 28 1104304 99907 29:04333 99906 30; 04362 99905 8-612823 5i 615891 1 5o 618937! 5o 621962 5o 624965 : 49 627948 i 49 630911 J 49 633854 ! 48 636776 ! 48 639680 '; 48 12 76 41 06 72 38 04 11 -06 9-999635 999629 999624 999619 999614 999608 999603 999597 999592 999586 -09 -09 • 09 -09 -09 -09 -09 -09 -09 -09 -09 -09 -09 -09 -10 - 10 -10 • 10 -10 -10 8.613189' 5i 616262 ; 5o 619313 1 5o 622343! 5o 625352 ! 49 628340 ; 49 63i3o8' 49 634256 48 637184 48 640093 48 21 85 5o i5 81 47 i3 80 48 16 11-386811 383738 380687 374648 371660 368692 365744 362816 359907 39 38 37 86 85 84 83 82 31 80 31 32 33 34 35 36 87 88 39 40 , 04391 , 99904 1 04420 99902 : 04449 99901 04478 99900 04507 99898 ,04536 99897 ; 04565 ; 99896 : 04594 99S94 ; 04623 99893 04653 99S92 8-642563 645428 648274 65iio2 65391 1 656702 659475 662230 664968 667689 47 47 46 46 46 45 45 45 45 12 82 •52 -22 21 35 06 9-999581 999575 999570 999564 999558 999553 999547 999541 999535 999529 8-642982 645853 648704 65i537 654352 657149 659928 662689 665433 668160 V 47 47 46 46 46 46 45 45 45 -84 53 22 t\ 3i 02 73 44 26 11.357018 29 354147 28 351296 27 348463 ; 26 345648 1 25 342851 ; 24 340072 , 23 337311! 22 334567 ■ 21 331840 20 41 42 43 44 45 46 47 4S 49 50 04682 ^ 99890 ,04711 i 99889 i 04740 ! 99888 04769 99886 0479S I 99S85 04827 1 99883 ; 04856 99882 04885 99881 ; 04914 '99879 1 04943 99878 675751 678405 681043 683665 686272 688863 691438 693998 44 44 44 43 43 43 43 42 42 42 :?? •24 97 -70 •44 •18 -92 -67 •42 9-999524 999518 999512 999006 999500 999493 999487 999481 999475 999469 • 10 -10 •10 -10 •10 -10 •10 - 10 • 10 -10 ' 8-670870 44 D73563 44 676239 44 678900 44 681544 43 684172 43 686784 ' 43 689381 i 43 691963 ' 42 694029 42 88 61 34 'bI 54 28 o3 11 n.329i3o,19 326437 1 18 323761 1 17 32II00 16 3i8456 15 3i5828 14 3i32i6 13 310619 12 3o8o37 ; 11 3o547i j 10 51 52 53 54 55 56 57 58 59 60 1 04972 99876 o5ooi 99875 o5o3o 99873 o5o59 99872 ;o5o88 99870 o5ii7 99869 o5i46 «99S67 ,o5i75 99866 o52o5 99S64 io5234 99863 8-696543 699073 701589 704090 706577 709049 7ii5o7 713902 716383 718800 42 41 41 41 41 40 40 40 40 40 17 9^ 68 44 21 97 74 -5i 29 -06 9-999463 909406 999450 999443 999437 999431 999424 999418 9994" 999404 -11 -11 •II . -11 •II -II -II -II , 8-697081 42 699617, 42 702139 41 704646 41 707140 1 41 709618; 41 712083 : 40 714534 40 716972 40 719396 40 28 o3 It 32 08 85 62 40 ■is 297861 295354 292860 290382 287917 285465 283028 280604 9 8 T 6 5 4 8 2 1 ■ N. COS. N. sine. L. COS. D. 1" 1 L. sine. 1 L.cot. i D.l" L. tang. / §7° 1 TRIGO:S"OMETRIC AL FUKCTIOXS. — 3°. 33 Nat. Functions. LooAKiTHMic Functions + 10. 1 1 2 S 41 I 7 8 ,1 11! 12 1 \l 15 16 17 18 19 20 N.sine.'N. COS. L. 6ine. D, 1" L. COS. ] D.l" L. tang. D. 1" 1 L. cot. 05234 05263 09292 05321 o535o 05379 05408 05437 05466 05495 o5524 99863 99861 99860 99858 99857 99855 99854 99892 99831 99849 99847 8.718800 721204 723595 725972 728337 730688 733027 735354 737667 739969 742259 40-06 39-84 39-62 39-41 39-19 38-98 38-77 38-57 38-36 38-i6 37-96 9.999404 999398 999391 999384 999378 999371 999364 999337 999330 999343 999336 •II •II •II •II •II -II -12 •12 •12 •12 •12 8^719396 721806 724204 726588 728959 73i3i7 733663 740626 742922 8^745207 747479 749740 751989 754227 736453 758668 760872 763o65 765246 40-17 39-95 im 39-30 3? -89 38-68 38-48 38-27 38-07 11-280604 278194 273796 273412 271041 2686S3 266337 264004 2616S3 259374 237078 60 59 6S 67 60 65 64 53 52 51 50 49 43 47 46 45 44 43 42 41 40 89 88 87 86 85 34 83 82 31 80 05553 o5582 o56ii o564o 05669 05698 Zd 05785 o58i4 99846 99844 99842 99841 99839 9983a 99836 99834 99833 99831 8-744536 746802 749055 751297 753528 755747 757955 760131 762337 764511 ^^56 37-37 37-17 36-98 36-79 36-6i 36-42 36-24 36 -06 9.999329 999322 9993 1 5 999308 999301 999204 999286 999279 999272 999265 •12 •12 •12 •12 •12 -12 -12 -12 -12 •12 37-87 37-68 37-49 37.29 37-10 36-92 36-73 36-55 36-36 36-i8 ii^254793 252521 230260 24801 1 245773 243547 24i332 239128 236935 234754 21 22 23 24 25 26 27 28 29 80 31 32 33 84 85 36 37 38 39 40 05844 05873 05902 05931 05960 05980 06018 06047 06076 o6io5 99829 99827 99826 99824 99822 99821 99819 99817 99815 99813 8-766675 768828 770970 773ioi 775223 777333 779434 781524 7836o5 785675 35-88 35-70 35.53 35-35 35.18 35-01 34-84 34-67 34-5i 34-3i 9.999237 999230 999242 999233 999227 999220 999212 999205 999197 999189 •12 •l3 -l3 -13 -13 • 13 • i3 • i3 • 13 • i3 8^767417 769578 771727 773866 773995 778114 780222 782320 784408 786486 36-00 35-83 35-65 35-48 35-31 35-14 34-07 34-80 34-64 34-47 11 •232583 230422 228273 226134 224005 221886 219718 217680 215392 2i35i4 061 34 06 1 63 06192 06221 o625o 06279 o63o8 1 06337 06366 06395 99812 99810 99808 99806 99804 99803 99801 99799 99797 99795 8.787736 789787 791828 793859 795881 797894 799897 801892 803876 8o5S52 34-i8 34-02 33-86 33-70 33-54 33-39 33-23 33.08 32.93 32-78 9.999181 999174 999166 999158 9991 5o 999142 999134 999126 999118 999 no • 13 • i3 •i3 • i3 • i3 -13 ■M • 13 • i3 8-788554 790613 792662 794701 796731 79S752 800763 802765 804758 806742 34-3i 34-i5 33-68 33-52 33-37 33-22 33-07 32-92 11-211446 2093S7 207333 2o52Q9 208209 201248 190242 193268 29 23 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 ~9" 8 7 6 5 4 8 2 1 41 42 43 44 45 46 47 48 49 '50 51 52 53 54 55 56 57 58 59 60 1 06424 06453 06482 o65ii 06540 06569 06598 06627 06656- 06685 99793 99792 99790 99788 99786 99784 99782 99780 99778 99776 8-807819 809777 811726 813667 815599 817522 819436 821343 823240 825i3o 32-63 32.49 32.34 32.19 32.05 3i .91 31.77 31.63 31.49 31.35 9.999102 999086 999077 999069 999061 999053 999044 999036 999027 •14 •14 •14 •14 •14 •14 •14 •14 8-808717 810683 812641 814589 816529 818461 820384 822298 824205 826103 32-78 32-62 32-48 32-33 32-19 32-o5 31-91 3i-77 31-63 3i-5o Il'l9i283 i8q3i7 1S7359 ,i3o4ii 1 3347 1 181539 179616 177702 173795 173897 .06714 06743 06773 06802 o683i 06860 06889 06918 06947 06976 99774 99772 99770 99768 99766 99764 99762 99760 99758 99756 8-827011 828884 830749 832607 834456 836297 838 i3o 839956 841774 843585 3l.22 3i.o8 30.95 30.82 30.69 30.56 30.43 3o.3o 30.17 3o.oo 9.999019 999010 998993 998984 998976 &^l 998930 998941 •14 •14 •14 •14 •14 •14 • i5 • 15 • i5 • i5 8-827992 i 829874 831748 ! 8336i3 1 835471 837321 839163 840998 842825 844644 3i-36 31-23 3i-io 30-96 3o-83 30-70 30-57 3o-45 30-32 30-19 ii^i720oS 170126 108202 166387 164529 162679 160S37 159002 107175 155356 ,N. COS. |N. sine. L. COS. D. 1" ' L. sine. | |; L. cot. D. 1" L. tang. 86° u TRIGONOMETRICAL FUXCTIOXS. — 4' Nat. Functions. Logarithmic Functions + 10. 1 N.8ine.|N.cos. L. sine. D.l" L. COS. D.l" . L. tang. Dl." L. cot 06976 99756 8-843585 3o-o5 9-998941 .i5 8.844644 30.19 11-155356 60 1 07005 1 99754 845387 29-92 998932 •10 846455 3o-07 153545 59 2 ' 07034 '99752 847183 29-80 99S923 • 15 848260 29-95 i5i74o 5S 8 ] 07063 9975o 848971 29-67 998914 .i5 85oo57 29-82 149943 57 4 : 07092 ! 99748 85o75i 29-55 998905 • i5 851846 29-70 148154 56 5 071 21 [99746 852525 29-43 998896 .i5 853628 29-58 146372 55 6 07 1 5o' 99744 854291 29-31 « .i5 855403 29-46 144597 54 7 07179 99742 07208,99740 856049 29-19 • 15 857171 29-35 142829 53 8 857801 29-07 998869 • 15 858932 29-23 141068 52 9 ! 07237 99738 859546 28.96 998860 • 15 860686 29. u j393i4 51 10 11 [07266! 99736 861283 2S-84 998851 • 15 862433 29-00 137567 50 107295 199734 8-863014 28-73 9.998841 • 15 8-864173 28-88 11 -135827 49 12 107324 99731 864738 28-61 998832 .i5 865906 28.77 134094 43 IS 07353 99129 866455 28 -50 998823 • 16 867632 28-66 132368 47 14 07382 99727 868160 28-39 998813 .16 869351 28-54 I 30649 46 15 0741 1 99725 869868 28-28 998804 .16 871064 28-43 128936 45 16 07440 99723 871565 28.17 998795 .16 872770 28-32 127230 125531 44 17 07469 07498 99721 873255 28-06 998785 .16 874469 28-21 43 IS 99719 874933 27.95 998776 .16 876162 28-11 123838 42 19 07027 99716 876615 27-86 998766 998757 .16 ■877849 28-00 122l5l 41 '20 07506199714 8782S5 27-73 .16 879029 27-89 120471 40 21 07085,99712 8.879949 27-63 9.998747 ^76 8-881202 27.79 11-118798 39 22 07614 99710 881607 27.52 998738 .16 882869 27.68 II7131 33 2S 07643 99708 883258 27.42 998728 .16 884530 27.58 1 1 5470 37 24 07672 99705 884903 27.31 998718 .16 8861 85 27-47 ii38i5 36 25 07701 99703 886542 27-21 998708 .16 887833 27.37 112167 85 26 07730 j 99701 888174 27-11 998689 .16 889476 27.27 110524 34 27 07759 99699 889801 27-00 .16 891112 27.17 108888 33 28 07788 99696 891421 26-90 998679 .16 892742 27-07 107258 32 29 07817 99694 893035 26-80 998669 •17 894366 26.97 105634 31 30 07846 99692 894643 26-70 998609 •17 895984 26-87 104016 30 'si 07875 99689 8-896246 26-60 9.998649 •17 8-897596 26.77 11-102404 "29" 32 07904 99687 897842 26-51 99S639 •17 899203 26.67 26.58 100797 28 33 07933 99685 899432 26-41 998629 •n 900803 099197 27 34 07962 ; 99683 901017 26-31 998619 •17 902398 26.48 097602 26 35 07991 j 99680 08020 1 99678 902596 26-22 998609 •17 903987 26.38 096013 25 36 904169 26-12 998599 •n 905570 26.29 094430 24 37 08049 99676 905736 26-03 998589 998078 •17 907147 26.20 092853 23 38 08078 99673 907297 908803 25-93 25-84 •17 908719 26-10 091281 089715 22 39 ,08107,99671 998568 •17 910285 26-01 21 40 41 ,o8i36' 99668 910404 25.75 998558 •17 911846 25-92 0881 54 20 :o8i65j 99666 8-911949 20.66 9.998548 •17 8-913401 25-83 11.086599 19 42 '08194 99664 913488 25-56 998037 •17 914951 25.74 o85o49 18 43 ; 08223 99661 9l5022 25-47 998527 •n 916495 25.65 o835o5 17 44 08252 199659 9i655o 25-38 998516 .18 918034 25.56 081966 16 45 , 08281 , 99657 918073 25-29 998506 .18 919568 25.47 080432 15 46 'o83io' 99654 919591 25-20 998495 998485 .18 921096 25-38 IfX 14 47 08339 : 99652 921103 25-12 .18 922619 25-30 13 48 08368 1 99649 922610 20-03 998474 .18 924136 25.21 075864 12 49 08397 99647- 924112 24-94 24-86 998464 .18 920649 25-12 074351 11 50 08426 ' 99644 925609 998453 .18 927156 25. o3 072844 10 TT 08455 i 99642 8-927100 24-77 9.998442 .18 8-928658 24.95 11-071342 9 52 08484 99639 928587 24-69 998431 .18 93oi55 24.86 069845 8 53 o85i3 99637 930068 24-60 998421 .18 931647 24-78 068353 7 54 08542 99635 931 544 24-52 998410 .18 933 1 34 24-70 066866 6 55 : 08571 99632 9330 1 5 24-43 l& .18 934616 24-61 065384 5 56 08600 , 99630 934481 24-35 .18 936093 24-53 063907 4 57 08629 99627 935942 24-27 998377 .18 , 937565 24-45 062435 3 58 oS658 99625 937398 24.19 998366 .18 ; 939032 24-37 060968 2 59 08687 99622 938850 24-11 998355 -18 I 940494 24 -30 059506 o58o48 1 60' 08716; 99619 940296 24-03 998344 .18 } 941952 24-21 ' N. COS. N. sine. L. COS. D.l" L. sine. L. cot. D.l" L. tang. ~^ 85° 1 TRIGONOMETRICAL FUXCTIOXS. 35 Nat. Functions. Logarithmic Functions +- 10. 1 j ' ! N.sine.' N. cos. L. sine. D. 1" L. COS. jD.l"; L. tang. D.1- L. cot ^\ 08716 ! 99619 8-940296 24-o3 9.998844 -19 '8.941952 24-21 ii^o58o48 60 1 08745 Qq6i7 941738 28-94 998333 : •19' 948404 24-13 056596 59 1 2 08774 99614 943174 28-87 998822 1 -19, 944852 24-05 o55i48;5S 1 3' o88o3 99612 944606 28-79 9988.1 1 •19: 946295 23.97 o587o5 57 4; o883i 99609 946034 28-71 998800 -19 947734 23-90 052266 66 5 08860 99607 947456 948874 23-68 998289 -19 949 J 68 28-82 o5o832 55 6 08889 99604 23-55 998277 : -19 930597 28.74 049403 54 7 08918 99602 950287 23-48 998266 •19 952021 28-66 047979 53 8 08947 99399 931696 28-40 998255 1 -,9 958441 28-60 046559 52 9 08976 99396 953100 28.82 998248 -19 954856 28-5i 045144 51 10 u 09005 99594 _954499 8-955894 28-25 998282 liil 956267 23-44 048788 50 090841 99591 28-17 9-998220 •I9i 8-957674 28.37 u-042326 ly" 12 09063 99388 957284 23-10 998209 .,9 959075 23.29 28.28 040925 4S 13 09092 99386 958670 28-02 .998 197 .191 960473 089527 47 14 091 21 1 99583 960052 22-95 998186 .19I 961866 23i4 088184 46 15 09150 99380 961429 22-88 993174 .19 968255 23-07 086745 45 16 Z'^ 99578 962801 22-80 998168 •19 964689 23-00 o3586i 44 17 99373 964170 22.73 998151 .19 966019 22-98 22-86 088981 43 18 09237 99572 965534 22-66 998189 .20 967394 968766 082606 42 19 09266 99370 966893 22-59 998128 .20 22-79 081234 41 20 09295 1 99367 968249 22-32 998116 .20! 970188 22-71 029S67 40 21 i 09324 99564 8-969600 22-44 9.998104 .20 8-971496 22-65 11^028504 oU 22 09353 99562 970947 22-38 998092 -20 972855 22.57 027145 SS 23 09382 99559 972289 22-31 998080 -20 974209 22.51 025791 37 "24' o'J4i I 99556 978628 22-24 998068 -20 975560 22.44 022440 36 25 09440 99553 974962 22.17 998056 -20 976906 22.37 022094 35 26 011469 99531 976293 22-10 998044 -20 978248 22.30 021752 34 27 09498 99548 977619 22.08 998082 .20 979586 22.23 020414 83 23 09527 99545 978941 21-97 998020 •20 980921 22-17 019079 32 29 09556 99542 980259 21-90 99S008 •20 982251 22-10 017749 01 80 81 09585 i 99540 981573 21.88 997996 •20 988577 22-04 016428 SO 09614 99537 3-982883 21-77 9.997984 -20 8-984899 21-97 ii^oiSioi 29 82 09642 99534 984189 21.70 997972 •20 986217 21.91 21.84 018788 2S 83 09671 99531 985491 21.68 997939 -20 987532 012468 27 84 I 09700 99528 $Vot'3 21.57 997947 -20 988842 21.78 011158 26 85 ij 09729 99526 21.50 997935 •21 990149 21-71 ooo85i 25 86 1 °9758 99523 989374 21.44 997922 -21 991401 21^65 008549 24 87 109787 99320 990600 21.88 997010 997897 997883 •21 992730 21^58 007250 23 88 j 09816 99317 991943 21.31 .21 994045 21^52 005955 22 89 1098451 99514 998222 21.25 .21 995337 21^46 004663 21 40 41 09874 995 n 09903 99508 994497 8-995768 21-19 997872 -21 996624 21^40 21^34 008876 20 21.12 9.997860 1 •21 ,8 -997908 11 -002092 TT 42 09932 99006 997086 21.06 997847 •21 1 999 '88 21.27 000812 IS 43 09961 995o3 998299 21-00 997835 •21 9-000465 21.21 10-999535 17 44 09990 99300 999560 20-94 997822 •21 001788 2I.l5 998262 16 45 I00I9 99497 9-000816 20-87 997809 -21 008007 21^03 996998 15 46 10048 ! 99494 002069 20-82 997797 -21 004272 995728 14 47 10077! 99491 oo33i8 20-76 997784 -21 005534 20.97 994466 13 48 ' 10106 99488 004568 20-70 997771 •21 006792 20-91 20-85 993208 12 49 ;ioi35 99485 oo58o5 20-64 997758 -21 008047 991933 11 50 51 IOI64 99482 007044 20-58 997745 •21 009298 20-80 ?9^7_o^ 10 IOI92 99479 9-008278 20-52 9-997782 •21 '9-010546 20-74 10 •989454 9 52 ! 10221 I 99476 009510 20-46 997719 -21 011790 018081 20-68 980210 3 53 10230 99473 010787 20-40 997706 -21 20-62 986969 7 54 10279 99470 011962 20-34 997698 •22 014268 20-56 983782 6 55 io3o8 99467 018182 20-29 997680 -22 oi55o2 20^5l 984498 5 56 10337 99464 014400 20-28 997667 •22 016782 20^45 988268 4 57 10366 99461 oi56i3 20-17 997654 -22 017959 20^40 982041 3 58 10895 99458 016824 20-12 997641 -22 019188 20^33 980817 2 59 10424 99455 oi8o3i 20-06 997628 •22 020408 20-28 ^??3S 1 60 10453 99452 019235 20-00 997614 •22 021620 20-28 |n.C09,|N. sine. L. COS. D. 1" L. sine. L. cot. D. 1" L. tang. 1 ' 84° 36 TRIGOKOMETRICAL FUNCTIONS.— 6^ Nat. Functions. Logarithmic Functions + 10. 1 1 2 3; t! l\ — i 11 121 13 14 15 17 13 19 20 21 22 23 24 25 26 27 23 29 30 81 82 8':> 84 35 36 87 83 89 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 5ti 57 53 59 60 N. sine. N. cos. Ksine. 1 D.l" L.C08. ,] D.l" L. tang. D.1'' L. cot 10453 99452 10482 99449 io5ii 99446 io54o 99443 10569 99440 10597 99437 10626 99434 io655 99431 10684 99428 10713 99424 10742,99421 9-019235 020435 02l632 022825 024016 025203 026386 027567 028744 029918 031089 20-00 19.95 -9-89 19-84 19.78 19.73 19.67 19-62 19-57 19-51 19-47 9-997614 997601 997588 997374 j 997361 997347 1 997334 j 997020 997007 997493 997480 -22 •22 .22 •22 .22 •22 .23 •23 .23 •231 .23 9-021620 022834 024044 02525l 026455 027655 028852 o3oo46 o3i237 032425 o336o9 20 20 20 20 20 19 19 ^9 19 19 19 23 17 II 06 00 95 90 »5 79 10.978380 972345 971148 969954 966391 60 59 58 57 56 55 54 53 52 51 50 10771 99418 10800 994 1 5 10820 99412 io858 99409 10887 .99406 10916 1 99402 10945 99399 10973,99396 11002 99393 iio3i i 99390 9-032257 033421 034582 035741 036896 o38o48 039197 040342 041485 042625 19-41 19-36 i9-3o 19-25 19-20 19.10 19-10 19.05 18.99 18-94 9-997466 997452 997439 997425 99741 1 997397 997383 997341 •23 -.11 -33, • 23 •23 9.034791 035969 037144 o383i6 039485 04065 1 04i8i3 042973 O44i3o 045284 19 19 19 19 19 19 19 J9 19 19 64 58 53 48 43 38 33 28 23 18 10-965209 96403 I 962856 961684 9605 1 5 %!!, 957027 955870 954716 49 43 47 46 45 44 43 42 41 40 "39" 38 37 36 35 34 83 32 31 30 1 1060 99386 11089 '99383 11 118! 99380 1 1147 99377 II 176 99374 ii2o5'99370 11 234 1 99367 11263 ! 99364 11291,99360 1 1 320 99357 9-043762 044895 046026 047154 048279 049400 o5o5i9 o5i63d 052749 053859 18.89 18.84 18-79 18-75 18.70 i8-65 18-60 18-55 18. 5o 18-45 9-997327 1 9973i3 997299 997285 997271 997257 997242 997228 997214 997199 •34 •24 •24 •24 •24 -24' •24' •24 i 9-046434 047382 048727 049869 001008 052144 053277 004407 o55o35 056659 19 19 19 18 18 18 18 18 18 18 i3 08 o3 98 93 89 84 79 74 70 10-953566 952418 901273 90oi3i 948992 947806 946723 945093 944465 943341 11340 '99354 1 13-8; 99351 11407:99347 1 1 1436 199344 11 1465 1 99341 11494,99337 1ID23 99334 11552 99331 iii58o! 99327 1 1609! 99324 Ti638!9932o 1 1667 99317 1 1696 99314 11725:99310 11754:99307 117831 993o3 11812 1 99300 11840199297 j 11869! 99293 j 11898199290 111927 99286 ' 11956.99283 1 1 1985 99279 : 12014 1 99276 12043 ' 99272 12071 1 99269 ! 12100' 99265 i 1 21 29' 99262 ji2i58 99258 12187 99255 9-054966 056071 057173 o5827i 059367 060460 o6i55i 062639 063724 064806 i8-4i 18-36 i8-3i 18-27 18.22 18.17 18. i3 18.08 18.04 17.99 9-997185 997170 997156 997141 997127 997112 997098 997083 997068 997053 •24 •24! •24' -24 •24' .25 .25 .25 9-057781 058900 060016 061 i3o 062240 063348 064453 965556 066655 067752 18 18 18 18 18 18 18 18 18 18 65 69 55 5i 46 42 37 33 28 24 iu-942219 941 100 939984 938S70 935547 934444 933345 932248 29 28 27 26 25 24 23 22 21 20 9-065885 066962 o68o36 069107 070176 071242 072306 073356 074424 075480 9-076533 077583 078631 079676 080719 081759 082797 083832 084864 080894 17.94 17-90 17.86 17.81 n-77 17-72 17-68 17-63 ^■M 17-50 17-46 17-42 17-38 17-33 17-29 17-20 17.21 9-997039 997024 997009 996994 996979 996964 996949 996934 996919 996904 .25 .25 .25 .25 .25 .25 9-U68846 069938 071027 072113 073197 074278 075306 076432 0775o5 078576 18 18 18 18 18 17 17 17 10 06 02 9'3 80 10.931154 930062 928973 927887 926803 925722 924644 923568 922495 921424 19 IS 17 16 15 14 13 12 11 10 9-996889 996874 996858 996843 906828 996812 996797 996782 996766 996751 •25 .20 .25 .20 .25 .26 .26 .26 .26 • 26 9.079644 080710 081773 082S33 1 083891 ; 084947 080000 j 087050 , 088098 1 089144 17 '7 17 17 17 17 17 17 17 76 V 'A ?! 01 .38 10-920356 919290 918227 917167 916109 9i5o53 914000 912950 91 1902 910856 9 8 7 6 5 4 3 2 1 N. COS. N. sine. L. COS. D.l" L. sine. L.COL D.l" L. tang. / I TRIGONOMETRICAL FUNCTIONS. — 7°. 37 Nat. Functions. Logarithmic Functions + 10. 1 1 2 8 4 5 ?! 81 91 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 80 N.sine. N. COS. L. sine. D.1" L. COS. ,] D.l" L. tang. D.l" 1 L. cot. 12187' 12216 12245! 12274 I2302 ' i233i 12360 12389' 12418 12447' 12476, 99235 99231 99248 99244 99240 99237 99233 99230 99226 99222 99219 9-085894 086922 087947 088970 089990 091008 092024 093037 094047 093056 '096062 . 17 17 17 16 16 16 16 16 16 16 i3 09 04 00 96 92 88 84 80 76 73 9.996731 996733 996720 996704 996688 996673 996657 996641 996625- 996610 > . 996594 .26 .26 .26 •26 •26 .26 •36 .26 .26 •26 •36 9.089144 090187 091228 092266 093302 094336 ^tl 097422 098446 099468 17- 17- 17 17 17 17 17 17 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 38 34 3o 27 22 \l 11 07 o3 99 95 84 80 76 72 ^^ 63 61 .58 •54 .50 .46 •43 it •32 •29 •23 io^9io856 909813 908772 907734 & 904633 9o36o5 902578 901554 900532 10^899513 898496 897481 896468 895458 894450 893444 892441 891440 890441 10-889444 888449 887457 886467 885479 884493 683509 8S2328 88 1 548 880571 60 59 58 57 56 55 54 53 52 51 50 49 48 47 48 45 44 43 42 41 40 39 88 37 30 35 34 33 32 31 30 i25o4, 12533 12562' 12591 12620 12649 12678 12706 12735 12764 12793 12822 12831 12880 12908 13937 12966 12995 i3o24 i3o53 99215 992 1 1 99208 99204 99200 99197 99193 99189 99186 99182 9.097065 098066 099065 100062 ioio56 102048 io3o37 104025 io5oio 105992 -16 16 16 16 16 j6 16 16 16 . 16 68 65 61 ll it 41 38 34 9-996578 996562 < 996546 996530 996514 996498 996482 996465 996449 996433 •27 •27 •27 •27 •27 •27 •27 •27 •27 •27 9-100487 ioi5o4 102519 103532 104542 io555o 106556 107559 108360 109559 99178 99175 99171 99167 99163 99160 99136 99152 99 J 48 99144 9-106973 107951 108927 109901 110873 111842 112809 113774 114737 113698 16 16 16 16 16 16 16 16 16 i5 3o 27 23 19 16 12 08 o5 01 97 9-996417 996400 996384 996368 9o635i 996335 996318 996302 996285 996269 •27 •27 •27 •27 •27 •27 •27 .28 .28 .28 9.110556 iii55i 112543 113533 114521 ii55o7 116491 117472 118452 119429 81 32 33 84 85 86 37 88 89 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 i3o8i i3iio i3i39 i3i68 i3i97 13226 13254 13283 i33i2 i334i 99141 99137 99133 99129 99125 99122 99118 99114 99110 99106 9-116656 117613 118067 119519 1 20469 121417 122362 i233o6 124248 125187 i5 i5 i5 i5 i5 i5 i5 i5 i5 i5 94 90 «7 83 80 ]t 69 66 •62 9-996252 996235 996219 996202 996185 996168 996131 996134 996117 996100 .28 •28 .28 •28 •28 .28 .28 .23 .28 .28 9.120404 121377 122348 123317 124284 125249 126211 127172 i28i3o 129087 16 16 16 16 16 16 16 i5 i5 i5 •22 •18 •i5 • u • 07 • 04 • 01 •97 .94 .91 10-879596 878623 877652 876683 875716 874751 873789 872828 871870 870913 29 28 27 26 25 24 23 22 21 20 13370 13399 13427 13436 13485 i35i4 13543 13572 i36oo 13629 13658 13687 13716 13744 13773 i38o2 1 3831 i386o 13889 13917 99102 99098 99094 99091 99087 990S3 99079 99073 99071 99067 9-126125 127060 127993 128925 129854 130781 131706 i3263o 133551 134470 i5 i5 i5 i5 i5 i5 i5 i5 i5 i5 .62 :8 •42 -.11 .32 •29 9.996083 996066 996049 996032 996015 993998 995980 993963 993946 993928 •29 •29 .29 •29 •29 •29 •29 •29 •29 •29 9-i3oo4i 130994 i3i944 132893 133839 134784 135726 136667 137605 138542 i5 i5 i5 i5 i5 i5 i5 i5 i5 i5 1^ • 84 .81 •77 •74 •71 •67 64 •61 58, 10-869939 869006 868o56 867107 866161 863216 804274 863333 862395 861438 19 18 17 16 15 14 13 12 11 10 ~9" 8 7 6 6 4 3 2 1 99063 99059 99055 99031 99047 99043 99039 99035 9903 1 99027 9.135387 i363o3 137216 i38i28 139037 139944 i4o85o 141754 142655 143555 i5 i5 i5 i5 i5 i5 i5 i5 i5 14 •25 -22 .19 •16 •12 ■2 -o3 -00 .96 9.995911 993894 995876 995859 993841 995823 993806 995788 995771 995753 •29 •29 •29 •29 •29 •29 •29 •29 •29 •29 9.139476 140409 141340 142269 143196 144121 145044 145966 146885 147803 i5 i5 i5 i5 i5 15 i5 i5 i5 i5 •55 •5i •48 •45 42 n 32 11 io-86o524 839391 858660 857731 856804 855879 854956 854034 853115 832197 N. COS. N.sine. L. COS. D.l" L. sine. 1 L. cot D.l" L. tang. ' 82° 1 38 TRIGONOMETRICAL FUNCTIONS. — 8" Nat. Functions. LoGARiTUMic Functions + 10. :N.sine. N. COS. L. sine. 1 D. 1" L. COS. D.l' !| L. tang. 1 D." L. cot |i39i7 99027 9-143555 i 14-96 9.995753 -3o 9.147803 15-26 10-852197 1 60 1 1 13946 99023 144453 1 14-93 995735 -3o 148718 15-23 851282 159 2 1 13975 99019 145349 i 14-90 995717 • 3o 149632 l5-20 85o368 58 8 14004 990 ID 146243 1 14-87 993699 •3o i5o544 15-17 849456 57 4 14033 9901 1 147136 j 14-84 995681 .30 i5i454 i5-i4 843546 56 5 14061 99006 148026 ! 14-81 995664 .30 152363 15-11 847637 55 6 \ 14090 99002 148915 i 14-78 995646 .30 153269 i5-o8 846731 54 7 14119 98998 149802 14-75 995628 •30 154174 i5-o5 845826 53 8 14143 98994 i5o686 14-72 995610 -30 I 55077 l5-02 844923 52 9 14177 98990 i5i569 14-69 995591 •3o 155978 14-99 844022 51 10 11 1 i42o5 ' 989% i5245i 14-66 _995^73_ • 3o 156877 14-96 843123 50 49 , I42J4I9S9S2 9-153330 14-63 9-995555 •3o 9.157775 14-93 10-842225 12 14263198978 154208 14-60 995537 -3o 158671 14-90 14-87 841329 48 13 14292 98973 i55o83 14-57 995319 • 30 159565 840435 47 U ! 14320 i 98969 155957 14-54 995501 -3i 160457 14-84 839543 46 15 14349:9^965 1 56830 14-51 993482 • 31 i6i347 14-81 838653 45 16 1437^:98961 157700 14-48 995464 -3i 162236 14-79 837764 44 17 14407 1 98957 158569 14-45 995446 -3i i63i23 14-76 836877 43 IS 14436 1 98953 159435 14-42 995427 -3i 164008 14^73 835992 42 19 14464 ! 9S948 i6o3oi 14-39 995409 -3i 164892 14-70 835 1 08 41 20 21 I44q3 ' 98944 161164 14-36 995390 -3i 165774 14-67 834226 40 14522 9S940 9-162025 14-33 9-995372 -3i 9.166654 14-64 10-833346 3D 22 14551 I9S936 162885 i4-3o 995353 -31 168409 i4-6i 832468 38 23 14580^98931 163743 14-27 995334 -3i 14^58 83i59i 37 24 14608 : 9S927 164600 14-24 995316 .3i 169284 14-55 830716 86 25 14637198923 165454 14-22 995297 • 3i 170157 14-53 829843 85 26' 14666 98919 1 66307 14-19 995278 -3i 171029 14-50 828971 84 27' 14695:98914 167159 168008 i4-i6 995260 -3i 171899 14-47 828101 83 2S 14723 1 98910 i4-i3 995241 •32 172767 14^44 827233 82 29 14732 9S906 168856 14-10 993222 •32 173634 14-42 826366 31 80 I478I I9S9O2 16970a 14-07 995203 -32 174499 14-39 825501 30 29 81, 14810 198897 9-170547 i4-o5 9-995184 -32 9-175362 14-36 10-824638 82 14833 i 98893 171389 14-02 995 1 65 -32 176224 14-33 823776 28 83 14867 : 98889 172230 i3-99 995146 -32 177084 i4-3i 822916 27 84 14896 !98S84 173070 13-96 993127 •32 177942 14-28 822058 26 85 14925 98880 173908 i3-94 995108 •32 178799 14-25 821201 25 86 14954 , 98876 174744 i3-9i 995089 -32, 179655 i8o5o8 14-23 820345 24 87 14932 98871 175578 13-88 995070 •32 14-20 819492 23 88 i5oii 98867 176411 i3-86 995o5i •32 i8i36o 14-17 818640 22 89 i5o4o' 98863 177242 13-83 995o32 .32 182211 14-15 817789 21 40 41 10069 '98858 178072 i3-8o 99501 3 •32 i83o59 14-12 816941 20 19 i5o97 198854 9.178900 13-77 9-994993 ^2" 9.183907 14-09 10-816093 42 i5i26| 98849 179726 13-74 994974 -32 1 184752 14-07 81524S 18 43 i5i55'9S845 i8o55i 13-72 994955 •32 1 185597 14-04 8i44o3 17 44 i5i84;9SS4i 181374 13-69 994935 •32 186439 14-02 8i356i 16 45 1 52 1 2 9S836 182196 13-66 9949 '6 .33 187280 188120 i3-99 812720 8iib8o 15 46 15241 98832 i83oi6 13-64 994896 -33 i3-?6 14 47 15270 98827 183834 i3-6i 994877 -33 188958 i3-93 811042 13 43 15299 98823 i8465i 13-59 994857 •33 i 189794 13-91 810206 12 49 15327 98818 185466 13-56 994838 -331 190629 13-89 8o8538 11 50 1 15356 98814 186280 i3-53 994818 •331 191462 13-86 10 61 15385 98809 9-187092 i3-5r 9-994798 -33 9-192294 i3-84 10-807706 "9" 52 i54i4 98805 187903 13-48 994779 •33; 193124 i3-8i 806876 8 53 15442 9S800 188712 13-46 994759 • 33: 193953 13-79 806047 7 54 1 547 1 9S796 lSg5ic^ 13-43 994739 -33 i 194780 13.76 803220 6 55; i55oo 98791 190323 i3-4i 994719 -33, 193606 13-74 804394 5 56 15529 98787 191130 i3-38 994100 -331 196430 13-71 803570 4 57 15557 98782 191933 i3-36 994680 •33 197253 13-69 802747 8 68 15586 98178 192734 i3-33 994660 -33' 198074 i3-66. 801926 2 69 15615198773 193534 i3.3o 994640 •33: 198894 13-64 801106 1 60 15643 j 98769 194332 13-28 994620 •33 199713 i3-6i 8002S7 ~\ N. COS., N. sine. L. COS. D.1" L. sine. L. cot. D.l" L. tang. ' 81° TRIGONOMETRICAL FUl^CTIOXS. — 9°. 39 Nat. Functions. Logarithmic Functions + 10. 1 N.sineJ N. cos. L. sine. D. 1" L. COS. D.l" 1 L. tang. D.l" L. oot. 10643 98769 9-194332 13-28 9.994620 .33 9.199713 13.61 10.800287 60 1 n672 98764 193129 13-26 994600 .33 1 200529 13-59 799471 798655 59 2|J 15701 I9S760 8^5730198755 195925 13-23 994 5So .33 1 201345 13-56 58 196719 l3-21 994560 •34 j 202159 13.54 797841 57 4 - - - 15738 98701 197511 i3-i8 994540 •34 1 202971 13.32 797029 56 5 15787 98746 198302 i3-i6 994519 •34 203782 13.49 796218 55 6 i58i6 98741 199091 i3.i3 994499 •34 204392 13.47 795408 54 7 !i5«43l9»7^7 199879 i3-ii 994479 •34 2o54oo 13.45 794600 53 8 I 1D873 92^32 200666 i3-o8 994439 •34 206207 13.42 793793 52 9 15902 98728 201431 i3-o6 994438 •34 207013 13.40 792987 51 10 11 1 593 1 15959 98723 202234 i3-o4 994418 •34 •34 1 207817 13.38 792183 50 49" 98718 9'2o3oi7 i3-oi 9.994397 9.208619 13.35 10.791381 12 15988 98714 203797 12-99 994377 •34 1 209420 13.33 79o58o 4S 13 16017198709 204577 12-96 994357 •34 I 210220 i3.3i 789780 47 14 1 16046 98704 205354 12-94 994336 •34 : 211018 13.28 788982 46 15 1 16074 ' 98700 2o6i3i 12-92 994316 •34 1 2ii8i5 13.26 788185 45 16 j i6io3 98695 Ii6i32| 98690 116160 98686 206906 12.89 994295 •34 : 212611 i3.24 787389 786595 44 17 207679 12.87 994274 •35 i 2i34o5 13^21 43 18 208432 12-85 994254 .35 214198 13^19 785802 42 11 16189 '98681 209222 12-82 994233 .35 214989 13^1? 785011 41 20 21 16218.98676 209992 12-80 994212 J5 213780 9.216568 i3^i5 784220 41) 39 16246 98671 9-210760 12-78 9-994191 .35 13.12 ro^8343T 22 16275 98667 2II326 12-75 994171 .35 217356 13.10 782644 38 23 i63o4 98662 21229I 12-73 994 i5o .35 218142 13.08 781858 37 24 16333, 9%57 2i3o55 12-71 994129 •35 218926 i3.o5 781074 86 25 i636i 9S652 2i38i8 12-68 994108 .35 219710 13. o3 780290 85 26 163901 98648 214579 12-66 994087 .35 220492 i3.oi 779508 1 34 27 16419 98643 215338 12-64 994066 •35 221272 12.99 778728 33 28 16447 9'^638 216097 I2-6l 994045 •35 222052 12^97 777948 32 29 16476 98633 216834 12-59 994024 •35 222830 12-94 777170 31 30 31 i65o5 ; 98629 16533" 9S624 217609 9-218363 12-57 994003 •35 2 236o6 12-92 776394 30 12-55 9.993981 •35"' 9.227382 12^90 10.775618 29 1 82 i6562 9S619 219116 12-53 993960 •35; 225i56 12^88 774844 28 83 16391 9S614 219868 I? -50 993939 993918 •35! 225929 I2^86 774071 27 34, 16620 98609 220618 12-48 •35 226700 12.84 773300 26 85: 16648 98604 221367 12.46 993896 •36 227471 I2-8l 772529 25 86 16677 9^600 222Il5 12-44 993875 993854 •36 228239 12.79 771761 24 37 16706 98595 222861 12-42 •36 229007 12^77 770993 : 23 S8 16734 98590 16763 98585 223606 12-39 993832 •36 229773 12^75 770227 122 39 224349 12-37 993811 •36 23o539 12-73 769461 ' 21 . 40' "if' 16792 98580 223092 12-35 993789 •36! 23l302 12.71 768698 : 20 16820; 98575 9-225833 12.33 9.993768 • 36 9.232065 12.69 10.767935 191 42; 16849 ' 98570 226573 12-31 993746 • 36' 232826 12.67 767174 18 43 16878 1 98565 227311 12-28 993725 •36 233586 12.65 766414 17 44 i 16906 1 98561 228048 12.26 * 993703 .36 234345 12.62 765655 16 f^ 16935 98556 228784 12-24 993681 • 36 235io3 12-60 764897 15 46 16964 98551 229318 12-22 993660 •36 235859 12.58 764141 14 47 16992 98546 230252 12-20 993638 • 361 236614 12.56 763386 13 48 17021 1 93541 230984 12.18 993616 •36 i 237368 12.54 762632 12 49 i7o5o- 98536 23i7i5 12.16 993594 •37: 238120 12-52 761880 11 r50 l7o^8:9853^ 232444 12.14 993572 •37! 238872 12. 5o 761128 10 17107198526 9-233172 12.12 9.993550 •37' 9.239622 12.48 10.760378 9 52 17136 ,98521 233899 12-09 993528 •37' 240371 12.46 759629 8 53 1 17164 1 98516 234625 12-07 993506 .37 i 241118 12.44 758882 7 "i 17193 I 9851 1 235349 236073 12-05 993484 •37 i 24i865 12.42 758i35 6 00 1 17222 98506 12-03 993462 •37 242610 12.40 737390 5 K 17250 98501 236193 12-01 993440 .37: 243354 12.38 756646 4 57 17279 98496 2375i5l 11-99 993418 •37 244097 244839 12.36 755903 8 58 17308 > 98491 238235 11-97 993396 •37^ 12-34 755161 2 59 17336:98486 238953 11.95 993374 •37 ■ 243379 12.32 754421 1 60 17365198481 239670 11.93 993351 •37 246319 12. 3o 753681 N. COS. N. sine. L. COS. D.l" 1 L. sine. ! L. cot. D.l" L. tang. 80° 1 40 TRIGONOMETRICAL FUNCTIONS. — 10°. Nat. Functions. Logarithmic Functions + 10. 1 1 N.sine.|N.co8. Lsine. D.l" L. COS. D.l" L.tang. D.l" L.cot 17365 1 98481 9-239670 11.93 9.993351 •37 9-246819 12.30 10-753681 60 1 17393 98476 240886 11-91 993329 •^7 247057 12-28 75.2943 59 2 17422 9S471 241101 11-89 993307 i'' 247794 12.26 702206 53 3 17451 198466 241814 11.87 998283 .37 248380 12-24 751470 57 4 17479198461 242526 11-85 998262 •37 249264 12.22 750736 56 5 17008 98455 17337 9S450 248287 11-83 998240 :^? 249998 12-20 730002 55 6 248947 ii-8i 998195 250780 12.18 749270 54 7 17565:98445 244655 11-79 .38 251461 12-17 743389 53 8 17594 93440 245363 11-77 11-75 998172 .38 252191 12.13 747809 52 91 17623:98435 246069 993149 • 38 252920 12.13 747080 51 10 ! 11 17651 9^430 246770 11.73 993127 .381 253648 12.11 746352 50 17680 9S425 9-24747« 11-71 9.998104 .38 9-254874 12.09 10-745626 49 12 1 17708 98420 248181 11-69 998081 -38 255100 12.07 744900 48 13 1 177^7 98414 248883 11.67 998059 -38 255824 I2-05 744176 47 14 { -17766 98409 249583 11.65 998086 -38 256547 12. o3 743453 46 15 17794 98404 250282 11.63 998018 .38 257269 12-01 742731 45 16 17823 98399 250980 11. 61 992990 • 38 257990 12.00 742010 44 17 17852 9S394 251677 n.59 992967 .38 258710 11.98 741290 43 18 17880 fM 252878 11-58 992944 .38 259429 11.96 740371 42 19! 17909 253067 11-56 992921 .38 260146 11.94 789834 41 20 : 17987 98878 253761 11-54 992898 .38 260868 11.92 739187 40 21' 17966 98873 9-254453 11.52 9.992875 .38 i 9.261578 11.90 10-788422 3i) 22 17993 98868 255x44 11. 5o 992852 • 38 262292 11.89 737708 33 23 18023 98862 255834 11.48 992829 992806 .39 263oo5 11.87 736993 786288 37 24 i8o52 98357 256523 11.46 .39 268717 11-85 86 25 1 808 1 93352 257211 11-44 992788 .39 264428 11-83 735572 35 26 18109 98847 257898 11.4s 992759 •39 265i38 ii-8i 734S62 34 27 i8i38 98341 2585b3 II. 41 992786 .39 265847 11-79 734153 33 28 18166 9S336 259268 11-89 992713 •39, 266555 11.78 788445 32 29 18195 98331 259951 11.37 992690 .39 267261 11.76 782739 31 30 31 18224 98825 260633 11.35 992666 .39 267967 11.74 782033 30 18252 98820 9-26i3i4 11.33 9-992643 •39. 9.268671 11-72 10-781829 29 82 18281 98815 261994 II. 3i 992619 •39, 269875 11-70 780625 23 33 i83o9 98310 262673 II. 3o 992596 •39, 270077 11.69 729928 27 34 18338 98304 263351 11-28 992372 .39, 270779 11.67 729221 26 35 18867 98299 264027 11.26 992349 .39 271479 11-63 728521 25 36 18895 98294 264703 11.24 992323 .39 272178 11.64 727822 24 37 18424 98268 265377 11.22 992301 .39 272876 11-62 727124 23 38 18452 98288 266o5i 11.20 992478 •40 273573 11.60 726427 22 39 1 848 1 98277 266723 11.19 992454 .40 274269 11.58 723781 21 40 41 i8509 98272 267895 II. 17 992480 .40 274964 11.57 725o36 20 i8538 1 98267 9-26do65 11. i5 9-992406 .40 9-275658 11-55 10.724342 ly 42 18567 98261 268734 iii3 992882 .40 276351 11-53 728649 18 43 18595 98256 269402 ii'ii 992839 -40 277043 n-5i 722957 17 44 18624 98250 270069 II. 10 992335 .40 277734 278424 11-50 722266 16 45 i8652 98245 270735 11-08 992811 • 40 11-48 721576 15 46 18681 {98240 271400 11-06 992287 -40 279113 11.47 720S87 14 47 18710 198234 272064 ii-o5 992268 .40 279801 280488 11.45 720199 13 48 18738:98229 272726 11-03 992289 -40 11.43 710012 718S26 12 49 18767:98228 278888 11. 01 992214 -40 281174 II. 41 11 50 51 18795 j 98218 274049 10.99 992190 -40 •40 28i858 11.40 718142 10 18824 98212 9-274708 10.98 9.992166 9.282542 11-38 10-717458 9 52 i8852 98207 275367 10.96 902142 •40 283^25 11-36 716775 8 53 18881 98201 276024 10.94 992II7 -41 288907 284388 11-35 716093 7 54 18910 98196 276681 10.92 992098 •41 11-33 715412 6 55 18988 98190 18967 98185 277337 10.91 10.89 992069 -41 285268 ii-3i 714732 5 56 277991 992044 -41 285947 11-80 714053 4 57 18995 98179 278644 10.87 992020 -41 286624 11.28 713376 8 58 19024 98174 279297 10.86 991996 •41 287801 11-26 713699 2 59 19052 98168 279948 10.84 99I97I -41 287977 11-25 712028 1 60 19081 98163 2S0599 10.82 991947 -41 288652 11-23 711848 N. COS. N. sine. L. COS. D.l" L. sine. L. cot D.l" L. tang. ' T9° 1 trigox(5metkical junctions. — 11= 41 42 TRIGONOMETRICAL FUXCTIOXS.— 12= Nat. Functions. Logarithmic Functions + 10. 1 ~0 N. sine J N. cos. L. sine. D. 1" L. COS. D.l" L. tang. 1 D. 1" L. cot. 20791 ' 97815 9-317879 ITs 9-990404 -45 9-327474! 10-35 10-672526 60 1 20820 97809 318473 990378 -45 328095! 10-33 671905 59 2 20848 ! 97803 319066 9-87 99o35i -45 328715 329334 10-32 671285 58 3 20877 ; 97797 319658 9-86 990324-45 io-3o 670666 57 4i 2090D 977QI 320249 9.84 990297! .45 329953- 10-29 330570 10-28 670047 56 5 209JJ 97784 320840 9-83 990270 •45 669430 6688. 3 55 6 20962 97778 32i43o 9-82 990243 •45 331187 10-26 54 7 20990 97772 322019 9.80 9902x5 -45 331803; 10-25 668197 53 8 21019! 97766 322607 9-79 990188 •45 332418' 10-24 667582 52 9 21047; 97760 323194 9-77 990161 •45 333o33' 10-23 666967 666354 51 10 11 21076: 97754 323780 9-76 990134 •45 333646; 10-21 50 49 21104 97748 'g^Uibb 9-75 9-990107 .46 9-334259] 10-20 10-665741 12 2Il32 97742 324950 9-73 990079 .46 334S71 1 10-19 665129 43 13 2II61 ! 97735 325534 9.72 990052 .46 335482' 10-17 664518 47 14 2II89! 97729 326117 990025 .46 3360931 10-16 663907 46 15 21218! 97723 326700 9-69 989997 ■46 336702 10- i5 663298 45 16 21246 1 97717 327281 9-68 9S9970 .46 337311I 10-13 662689 44 17 21273 1 97711 327862 9-66 989942 .46 337919 10-12 662081 43 18 2i3o3 1 97705 328442 9-65 9899.5 -46 338527 339133 lO-II 661473 42 19 2i33i 1 97698 329021 9-64 989887 .46 10- 10 660867 41 20 '2! . 2i36o 97692 329599 9-62 989860 •46 339739 io-o8 660261 40 '21388 976S6 9-330176 9-6i 9-989832 -46 9-340344 10-07 10-659656 39 2i. 21417 97680 330753 9 -60 9^9804 .46 340948 10-06 659o52 38 23 21445 97673 33i329 9-58 9^9777 .46 341552 10-04 658448 37 24 21474 97667 331903 9-57 9S9749 •47 342.56 io-o3 657845 36 25 2l5o2 9766. 332478 9-56 989721 •47 342757 10-02 657243 35 26 2i53o 97655 333o5i 9-54 989693 -47 1 343358 10-00 656642 34 27 2.559 97648 333624 9-53 9S9665 •47 343938 9-99 656042 33 2S 2087 97642 334195 9-52 9S9637 •47 344558 9-98 655442 32 29 2I6I6 97636 334766 9 -So 989609 •47 1 345 1 57 9-97 654843 81 SO 21644 97630 335337 9-49 989582 •47 1 343755 9-346353 9-96 9-94 654245 .0-653647 30 "29" 21672 97623 9-335906 9.48 9-989553 •47, 32 ! 21701 97617 336475 9.46 989525 ■47. 346949 9-93 653o5i 23 33 : 21729 97611 337043 9.45 989497 •47 347343 9-92 652455 27 34 '21758 97604 337610 9.44 9^9469 •47! 348141 9-91 65.859 26 35 21786! 97598 338176 9-43 98^441 -47 348735 9.90 661265 25 36 21814I 97392 338742 9-41 9S9413 •47 349329 9-88 650671 24 37 ! 21843 97585 339306 9.40 989384 •47 349922 33o5i4 9-87 650078 23 38 : 21871 97579 218991 97573 339871 9-39 9S9356 •47 9-86 649486 22 39 340434 9.37 989328 •47; "35iio6 9-85 648894 21 40 "if 42 43 ■ 219281 97566 340996 9-36 989300 •47 1 35.697 9-352287 9-83 6483o3 20 21956 97560 9-34i558 9.35 9-989271 •47; 9-82 10-647713 19 ^2.985 97553 342119 9-34 989243 •47 352876 9-8. 647124 18 22013 i 97547 342679 9.32 9892.4 •47 353465 9-80 646535 17 44 22041 ; 97541 343239 9-3i 989.86 •47 354053 9-79 645947 16 45 46 47 220701 97534 343797 9 -30 989.57 •47 354640 9-77 645360 15 22098 97528 344355 9-29 989128 .48 355227 9.76 644773 14 22126 97521 344912 9-27 989100 .48 3558.3 9.73 644187 13 48 221551 975i5 345469 9-26 989071 .48 356398 9-74 643602 12 49 22i83; 97508 346024 9-25 989042 •481 356982 357566 9.73 643oi8 11 50 "5]" 52 58 22212 97502 346579 9-24 9890.4 •481 9-71 642434 10 22240 97496 97489 9-347134 9-22 9-988985 •48: 9-358149 9-70 io-64i85i 9 22268 347687 9.21 988956 •48 1 358731 9-69 9-68 641269 8 22297 97483 348240 9-20 988927 -48 359313 640687 7 54 223251 97476 348792 9-19 988898 •48! 359893 9-67 640107 6 55 22353] 97470 349343 9-17 988869 -48! 360474 9-66 639526 5 56 22382 97463 349S93 9-i6 988840 •48: 36io53 9-65 638947 4 57 22410 97457 350443 9. ,5 98881 1 •49' 36i632 9-63 638368 3 5S 22438! 9745o 350Q92 9-14 988782 •49; 362210 9-62 637790 2 59 2246-: i 97444 33ID40 9-i3 988753 •49 1 362787! 9-6i 637213 1 60 , 22495 97437 352088 9. II 988724 -49 363364 9 -60 636636 n'coT: J?, sine. L. COS. D.l" 1 L. sine. 1 i L. cot. ! D. 1" i L. tang. ~ 77^^ 1 TRIGONOMETRICAL FUNCTIONS. — 13° 43 Nat. Functions. LoiiARiTHMic Functions ^- 10. 1 ■ N.sine. N. cos. L. sine. D. 1" L. COS. D.l" L. tang. Dl." j L. cot 22495 97437 9-352088 9-11 9-988724 •49 9 •363364 9-60 10. 636686 60 1 22523 97430 352635 j 9-10 988695 .49 363940 9.59 686060 59 2j! 22552 i 97424 353i8i 9.09 988666 .49 36451 5 9-58 635485 53 8 2258o 97417 353726 9-08 988636 988607 •49 365090 9-57 684910 57 4!' 22608 9741 1 354271 9-07 •49 365664 9-55 634336 56 5 22637 97404 354^:115 9-o5 988578 •49 366237 9-54 688763 55 6 22665 19-398 3553581 9.04 988548 •49 366810 9-58 688190 54 7 8 9 22693 97391 355901 1 9 -03 988519 .49 367382 9-52 682618 53 22722 97384 356443 9-02 988489 •49 367953 9-51 682047 52 22750 97378 356934 1 9-01 938460 •49 368524 9-5o 681476 51 10 ! ^778 97371 3070241 8-99 988430 •49 369094 9-49 680906 50 iM 22807 97365 9-353064 8-98 9-988401 •49 9^369663 9-48 io.63o337 49 12 22835^97358 3586o3 8-97 988371 •49 1 870232 9-46 629768 43 131 22863! 97351 359141 8-96 988342 •49; 370799 9-45 629201 47 ui 22892:97345 359678 8-95 988312 -So! 371367 9-44 628633 46 15 i 22920197338 36021 5 8-93 988282 -So! 371933 9-43 628067 45 16 22948 97331 360752 8-92 988252 -50 1 372499 9-42 627001 44 17 22977 97325 23oo5 97318 361287 8-91 988223 -5o! 373064 9-41 626986 43 13 361822 8.90 988193 .50 373629 374193 9-40 6263TI 42 19 23o33|973ii 362356 8-89 8-88 988163 -So lit 623807 41 2u 21 ! 22 23o62 97304 362889 988133 -5o 374756 623244 40 23090 97298 9-3634?2 8-87 8-85 9-988103 -So, 9^3753i9 9-37 10-624681 89 23ii8 97291 363934 988073 -5o 375881 9.35 624119 33 2:} 23 146 97284 364485 8-84 988043 -So 376442 9-34 623558 37 21 28175 97278 365oi6 8-83 988013 -So 377003 9.33 622997 36 25 23203 97271 365546 8-82 987983 -So 377563 9-32 622487 85 20 2323l 97264 366075 8-81 987953 -So 878122 9-3i 621878 34 27 23260 97257 366604 8-80 lt]tll -5o 878681 9-80 621819 33 2S 23288 97251 367i3i 8-79 -50 879289 9-20 9-28 620761 82 29 233i6 97244 367639 8-77 987862 .50 379797 620208 31 80 81 23345 97237 368i85 8-76 987832 -51 38o354 9-880910 9.27 9-26 61-9646 10-619090 618534 80 29 23373 97230 9-368711 8-75 9-987801 .5i 32 23401 97223 369236 8-74 987771 -5i 881466 9-25 23 88 23429 97217 369761 8-73 987740 -5i 882020 9-24 617980 27 84 23458 97210 370285 8-72 987710 .5i 882575 9-23 617425 26 85 23486 97203 370808 8-71 987679 .5i 888129 9-22 616871 25 8(5 235i4 97196 371330 8.70 987649 -5i 383682 9^21 616818 24 87 23542 97189 371852 8-69 987618 -51 884284 9^20 615766 23 83 23571 97182 372373 8-67 8-66 98^588 .5x 884786 9.19 6i52i4 22 89 23599 97176 372894 987557 -5i 885337 9-ia 614668 21 40 41 23627 97169 373414 8-65 987526 .5. 385888 9-17 614112 20 23656 97162 9-373933 8-64 9-987496 -5i 9.886488 9^i5 io^6i3l>b2 TiT 42 23684 97155 374452 8-63 987465 -5i 886987 9-14 618018 13 43 23712 97148 374970 8-62 987434 -51 887536 9-i3 612464 17 44 23740 97141 373487 8-61 987403 -52 888084 9-12 611916 16 45 23769 97134 376003 8-60 987372 -52 888631 9-u 611869 15 46 23797 97127 376519 8-59 8-58 987341 •52 889178 9-10 610822 14 47 23825 97120 377033 987310 -52 889724 9-00 9-08 610276 13 43 23853 97113 377549 8-57 987279 •52 890270 609780 12 4y 23882 97106 378063 8-56 987248 •52 890815 9-07 609185 11 •0 51 23910 97100 378577 8-54 987217 •52 891860 9 -06 608640 10 23938 97093 9-379089 8-53 9-987186 •52 9-891908 9 -05 10-608097 "9" 52 23966 97086 3-9601 8-52 987155 •52 892447 9.04 607533 8 53 1 23995 97079 3Soii3 8-51 987124 -52 892989 9-08 607011 7 54 24023 97072 330624 8-5o 987092 •52 893531 9-02 606469 6 55 24o5i 97063 38ii34 8-49 987061 •52 894078 9-01 603927 6o5386 5 56 24079 970D3 3Si643 8-48 987030 -52 894614 9-00 h 4 57 24108 97o5i 382152 8-47 986998 -52 395154 604846 3 58 24i36 97044 332661 8-46 986967 -52 395694 604306 2 59 124164:97037 333168 8-45 986936 -52 896233 8-97 608767 1 60 ;! 24192 |97o3o 333675 8-44 986904 •52 1 396771 8-96 608229 IjN. COS. N. sine. j L. COS. I D.l" L. sine. 1 L. cot. D.l" L. tang. ' 76<^ 1 44 TRIGOXOMETRIC AL F CXCTIOXS. — 14". Nat. Functions. Logarithmic Fuxctioxs + 10. 1 N.sine. X. cos. L. sine. D.l" L. COS. D.l L. tang. D. 1" L.cot 1 24192 97o3o 9.353675 8.44 9.986904 .52 9-396771 8.96 io-6o3229 60 1 24220 9-023 3S4182 8-43 986873 .53 , 397309 896 602691 59 1 2 24249 QTOl5 384687 8.42 986841 •53 397846 8.95 602154 53 1 8 24277 9700S 385i92 8-41 986809 .53 . 398333 8-94 601617 57 1 4 243o5 Q'ooi 385697 8.40 9S6778 .53 398919 8-93 601081 56 / 5 24333 96994 386201 8. "3^ 9S6746 .53 399455 8-92 600545 55 6 24362 96987 3S6704 95^714 .53 399990 400524 8-91 600010 54 7 24390 969S0 3S7207 ^^7 986683 •53 8-90 599476 53 8 24418 96973 387-09 8-36 986631 •53 401 o53 8-89 598942 52 9 24446 96966 38^210 8.35 986619 •53 401591 8-88 598409 51 10 24474 96959 388711 8.34 986587 •53 402124 9-402656 8.87 597876 50 11 245o3 q6q52 9-389211 8-33 Q. 986555 •53 8-86 10.597344149 12 2j53i 96945 3S9711 8.32 986523 .53 4o3i87 40371a 8.85 5o68i3 43 13 24559 96937 390210 8-31 986491 .53 8.84 596282 47 14 245S7 96930 24615 96923 390708 8-30 986459 .53 404249 ^ 404778 8.83 595751 46 15 391206 8.28 986427 •53 8-82 595222 45 16 24644 96916 391703 8.27 986395 •53 4o53o8 8-81 594692 1 44 1^, 24672 ' 96909 '& 8.26 986363 •54 405836 8-80 594 1 64 1 45 18; 24700 96902 8-25 986331 .54 406364 8.70 8.78 593636 1 42 19' 24728 96894 393191 8.24 986299 .54 406892 593108 41 20; 21; 24756 ' 96SS7 393685 8-23 986266 •54 407419 9.407945 8.77 592581 40 24784 ■ 96880 9.394179 394673 8.22 9.986234 .54 8.76 10-592055! 89 221 24813.96S73 8.21 986202 •54 , 408471 8.75 591529133 591003 37 23 i 24841 1 96866 395166 8-20 986169 .54 408997 409321 8.74 24 24869 1 96S58 395658 8-19 8.18 8.17 986137 •54 8.74 590479 36 25 26 24897 1 9685 1 24925 1 96844 396100 396641 986104 986072 •54 •54 410045 410569 8.73 8.72 589955 589431 35 34 27 24954 1 96837 397132 8.17 986039 •54 41 1092 8.71 588908 33 28 24982196829 25oio 196822 25o38| 96815 39-621 8.16 986007 •54 4ii6i5 8.70 588385 32 29 398111 8.i5 985974 •54 412137 8.69 8.68 587864 31 30 31, 39^600 8.14 985942 •54 .55 412658 9.413179 587342 30 25o66 : 96807 "9^99088 8-13 ''$^?6 8-67 10.586821 29 S21 25094 1 96800 399575 8.12 .55 413699 8-66 586301 23 S3' 25l22 96793 400062 8.11 985843 .55 414219 8-65 535781 27 84; 25i5i 96-S6 400549 8-10 985811 .55 414735 8-64 585262 26 85' 25179:96778 4oio35 8.09 985778 .55 415257 8.64 584743 25 86' 25207 07"! 401320 8-08 985745 • 55 415775 8.63 584225 24 87 25235 96-64 402005 8-07 985712 • 55 416293 8.62 583707 23 8S 25263 96706 402489 8.06 985679 .55 416810 8.61 583190 22 39 25291 96749 402972 8-05 985646 .55 417326 8-6o 582674 21 40 41 25320 96-42 403455 8-04 98561 3 • 55 417842 8.59 582158 20 25348 96734 9.403933 8-o3 9.985580 1 .55 9.418353 8-58 10.581642 19 42 25376 96-27 404420 8-02 985547 1 .55 j 418873 8-57 581127 13 43 25404 96719 404901 8-01 985514 .55 419337 8-56 58o6i3 17 44 25432 96712 4o5382 8.00 985480 .55 419901 8-55 p?? 16 45 25460 ' 96705 4o5S62 ]-ll 985447 .55 42o4i5 8-55 15 46' 25488 , 96697 406341 985414 • 56 420927 8-54 l]r£ 14 47 2 55 1 6 96690 25545 : 96682 406820 7-97 985380 • 56 421440 8-53 13 43 407299 7.96 985347 • 56 421952 8-52 578048 12 49 25573 ! 96675 407777 7.95 985314 • 56 422463 8-5i 577537 11 50 51 25601 '96667 408254 7-94 985280 • 56 422974 8-5o 577026 10 25629 ' 96660 9.408731 7-94 9.985247 i • 56 9.423484 8-40 8-48 10.576316 9 52 25657 , 96653 409207 7.93 985213 1 • 56 423993 4245o3 576007 8 53 25685 96645 409682 7.92 985 180 • 56 8-48 f^in 7 54 25713 96633 410157 7.91 985146 • 56 425oii 8-47 6 55 25741 96630 410632 985,13 • 56 425519 8-46 574481 5 56 25760 ■ 06623 411106 985079 1 • 56 426027 8-45 573973 4 57 257qS 066 1 5 41 1 579 985045 ; • 561: 426534 8-44 573466 8 58 25826 96608 4l2o52 7-87 9S5011 • 56 1: 427041 8-43 572939 2 59 25854 96600 412524 7.86 984978 •56 |i 427547 8.43 572453 1 60 25882 96593 412996 7-85 984944 •561 428052 8.42 571948 N. COS. N. sine. L. COS. 1 D.l" L. sine. 1 1 L. cot. D.l" L. tang. » 75° 1 TRIGONOMETRICAL FUNCTIONS. — 15°. 45 Nat. Functions. LOGARiTHiiic Functions + 10. 1 M N. sine. N. cos. L. sine. D.l" 1 L, cos. |D.l":j L. tang. D.l" L.cot. oi 25832 '06393 9-412996 7-85 9.984944 .57 19.428052 8.42 10-571948 60 1; 25910 :965a5 413467 7-84 984910 •57' 428007 8.41 571443 59 2 25988 96578 418938 7-83 984876 •57 429062 8-40 670988 5S 8 25966 96370 414408 7-83 984842 •57 429566 8.39 570434 57 4 25994 96062 414878 7-82 984808 •57 480070 8.38 569980 1 5(5 5 26022 96555 415347 7-8i 984774 •57 480573 8.38 569427 j 55 568920 j 54 6 26o5o 96D47 4i58i5 7-80 984740 •57 481075 8.37 7 26079 96040 416283 7-79 984706 •57 43 1 577 8.36 568423 53 8 26107 26135 96032 416751 7-78 984672 •57 482079 8.35 567921 52 9 96524 417217 7-77 984687 •57 432580 8.34 • 567420 1 51 10 11 26163 26191 965n_ 417684 7-76 984603 •57 433o8o 8.33 566920 50 96509 9-4i8i5o 7-75 9.984569 •57 9.433580 8.32 10-566420 4 'J 12 26219 96502 4i86i5 7-74 984535 •57 434080 8.32 565920 43 13 26247 i 96494 419079 7-73 984000 •57 434579 8.3i 565421 47 14 26275 I 96486 419544 7-73 984466 •57 435078 8.3o 564922 46 15 263o3 i 96479 420007 7-72 984432 .58 435576 8.29 564424 45 16 26331 1 96471 420470 7.71 984397 .58 486073 8.28 568927 44 17 26359 ; 96463 420933 7-70 984868 .58 436570 8.28 563430 i 4:5 1 18 263S7 1 96456 421390 ]:ll 984828 .58 Sltl 8.27 562983 42 19 26415:96448 421857 4223i8 984294 984269 .58 8-26 562487 41 20 21 26443 I 96440 7-67 .58 488059 8-25 561941 40 26471 196433 9.422778 7.67 9.984224 .58 9.438554 8.24 10-561446 39 22 26500:96425 423238 7.66 984190 .58 489048 8.23 560952 33 23 26528 96417 428697 7-65 984155 .58 489543 8-23 660457 37 24 265561 96410 424 1 56 7-64 984120 .58 440086 8.22 559964 3G 25 26584 96402 424615 7-63 984085 .58 440529 8.21 609471 35 26 26612 96394 425073 7-62 984050 .58 441022 8.20 608978 34 27 26640 96386 425530 7-6i 984015 .58 44i5i4 8.19 558486 33 28 26668 96379 425987 7 -60 988981 • 58 442006 8.19 607994 32 29 26696 96371 426443 7-6o 988946 .58 442497 442988 8.18 557508 31 80 26724 96363 426S99 7-59 988911 .58 8.17 667012 30 81 26752 96355 9-427354 7-58 9.988875 .58 9.443479 8.16 IO-65602I 2ir 82 26780 96347 427809 428263 7-57 988840 .59 1 443968 8.16 606082 28 83 26808 96340 7-56 988800 .59 1 444458 8.i5 500042 27 84 26836 96332 428717 7-55 988770 .59 444947 8-14 55oo53 26 85 26S64 96324 429170 7-54 988735 .59 445435 8.i3 654565 25 86 26892 96316 429623 7-53 988700 .59 445923 8.12 664077 553589 24 87 26920 96308 480075 7-52 988664 .59 44641 1 8.12 23 88 26948 96801 480527 7-02 988629 .59 446898 8. II 658102 22 89 26976 96293 480978 7-5i 9S3594 .59 447384 8.10 662616 21 40 "if 27004 96285 481429 7-5o 983558 .59 447870 8-09 552 1 3o 20 27032 96277 9.431879 7-49 9.983523 .59 9-448356 8-09 10.551644 |iy 42 27060 96269 482829 432778 ]■■% 988487 .59 448841 8.08 551169 13 43 27088 96261 983452 .59 449326 8-07 600674 ' 17 44 127116196253 488226 I'^l 983416 .59 449810 8-06 600190 ! l() 45 127144 96246 488675 7-46 988881 .59 450294 8.06 6^9706 ! 15 46 27172 96238 434122 7-45 988845 .59 450777 8.o5 649228 u 47 27200 96230 434569 7-44 & •So 451260 8-04 648740 13 48 27228 j 96222 485oi6 7-44 .60 451743 8-o3 648267 12 49 1272561 96214 435462 7-43 988233 .60 452225 8.02 647775 u 51 j 272-84 ' 96206 480908 7-42 988202 .60 402706 8-02 647294 i 10 ; 27312196198 9.486853 7-41 9.988166 .60 9.453187 458668 8-01 10-546813 9 52 27340196190 486798 7-40 988130 .60 8-00 646882 8 58 27368 96182 437242 7-40 988094 .60 454148 7-99 546862 7 54 27396 96174 437686 ]-M 983o58 .60 454628 7-99 645372 6 55 27424 96166 488129 988022 .60 455107 7.98 644898 5 56 27452 96158 488572 1-^1 982986 -60 455586 7-97 644414 4 67 274^0 96i5o 439014 7.36 982950 -60 406064 7.96 643986 3 58 27008 96142 439456 7-36 982914 .60 456542 7.96 643458 2 59 27536 96134 439897 440338 7.35 982878 -60 457019 7.95 542981 1 60; 27564 96126 7-34 982842 .60 457496 7-94 642604 ~| N. COS. N. sine. { L. cog. D.l" L. sine. L.cot. D.l" L. tang, j ' 1 I 46 TRIGONOMETRICAL FUNCTIONS.— 16' Nat. FiTKCTiONs. LOGABITHMIC FUNCTIONS + 10. ' N. sine. N. cos. Ksine. j D.l" L. COS. D.l" L. tang. D.l" L. coL Oi 27364 96126 9-440338; 7-34 9.982842 1 -60 9.437496 7-94 lo- 042504 60 1 } 27092 961 18 440778 7.33 982800 .60 437973 7-93 542027 59 2, 27620 96110 441218: 7.32 982769 .61 438449 7-93 54i55i 53 3 2764S 9'Jio2 441658 J 7-3. 982733 •61 458925 7-92 541075 57 4 27676 9O094 442096 ; 7-3i 982696 •61 439400 7.91 040600 06 5 27704 96086 442535 ; 7-30 982660 •61 459873 7-90 540125 55 6 j 27731 96078 442973 ' 7-29 982624 .61 460349 7-90 539651 54 7 j' 27759 96070 443410 7-28 982087 •61 460823 7.89 539177 53 8 127787 96062 443847 • 7.27 982031 .61 461297 7^88 538703 52 9' 278i5 q6o54 444284 7.27 982514 .6. 461770 7.88 538230 51 10' 27S43 96046 444720 7-26 982477 • 61 1 462242 7-87 537753 50 49 11 27871 96037 9.443155 7-20 9.982441 .61 ^62714 7.86 10-537286 12 27899 96029 445590 7-24 982404 •61 463186 7-?^ 536814 43 13 27927 96021 446025 7-23 982367 .61 463658 7-80 536342 47 14 27955 96013 446439 446893 7-23 982331 .61 464129 7-84 535871 46 15 27983 96005 7.22 982294 .61 464599 ^1^ 535401 45 16 28011 95997 447326 7.21 982237 .61 465069 7-83 534931 44 17 28039 9^939 447739 7.20 982220 .62 465539 7-82 534461 43 13 28067 93981 443191 7.20 982183 .62 466008 7.81 533992 42 19 28oq5 q5972 443623 7.19 982146 .62 466476 7.80 533524 41 20 28123 93964 449034 7-18 982109 .62 1 466945 7.80 533o55 40 21 . 2Si5i 95950 9.449435 7.16 9.932072 .62 0.467413 7-79 10-532587 39 22 28178 9594S 449910 982035 .62 1 467880 7.78 532 1 20 33 23 28206 93940 45o345 7.16 981998 .621 468347 7.78 53i653 37 24 28234 93931 450770 7.10 981961 .62 468814 7-77 531186 30 25 28262 95923 451204 7-U 981924 981886 .62 469280 7.76 530720 35 26 ' 28290 95913 45i632 7-13 .62 469746 7.75 530254 34 27 ; 283 18 95907 452060 7-i3 981849 •02 470211 7.75 029789 33 28 ; 28346- 93S98 452488 7-12 981812 .62 470676 7-74 529324 32 291' 28374 93890 452910 7.11 981774 .62 471141 7-73 523839 31 30 28402 95S82 453342 7.10 981737 .62 471605 7-73 528395 30 SI 28420 o5^-4 9-453768 7.10 9.981699 •63 9.472068 7.72 10-527932 29 32. 2S457 9^^^-'^ 454194 7.09 981662 •63 472332 7'7i 527468 23 33 i 28435 95357 454619 7.08 981620 .63 472993 7.71 527005 27 34i:2S5i3 95849 455044 7-07 981587 • 63 473437 7.70 526543 26 351 2S54I 93841 450469 7.07 981049 • 63 473919 7.69 526081 25 86! 28569 9^332 455S93 7.06 981512 •63 474381 7.69 525619 24 37 i 28597 93824 4563i6 7.00 931474 .63 474842 7.68 525i58 23 38 1 28625 95816 39 1 28652 95807 406739 7-04 981436 •63 475303 7.67 524697 22 407162 7-04 981399 • 63 473763 7.67 524237 21 40 28680 95-99 457584 7 -03 981361 .63 476223 7.66 523777 20 19 41 28708 95-91 9.453006 7.02 9-981323 •63 ,'9.476683 7.60 10-523317 42 28-36 95-82 453427 7.01 981280 .63 477142 7.60 522858 13 43' 28764 95774 453348 7.01 931247 • 63 477601 7-64 522399 17 44 2S792 95766 459268 7.00 981209 .63 478059 7-63 521941 16 451 28820 95757 45o653 6-99 981171 •63 478517 7-63 521483 15 46 28847 93^49 460108 6-98 981133 • 64 478975 7.62 521025 14 47 ,28875 95-40 460527 6.98 981095 .64 479432 7.61 520563 ] 13 48 28903 95732 460946 6-97 981057 .64 479889 7-6i 5201 1 1 12 49 28931 95724 461 364 6-96 981019 •64 480345 7-60 519655 111 50 28959 95^i5 461782 6-95 980981 .64 480801 7-59 519199 10 51 j 28987 93707 9.462199 6.95 9-980942 •64 :9-48i257 ]'U 10.518743 9 52 1 29010 95698 462616 6-94 980904 .64 481712 518288 8 53 ' 29042 q56qo 463o32 6-93 980^66 .64 482167 ^1^ 517833 7 54 29070 95681 463448 6-93 980827 .64 482621 7-57 517379 6 55 29098 95673 463864 6-92 980789 .64 483075 7-56 516920 5 56 ,29126 95664 464279 6-91 980750 •64 483529 7-30 516471 4 57 129154 95'j56 464694 6-90 9807 1 2 •64 4839S2 7-55 5 1 601 8 3 58 29182 95647 465 1 08 6.Q0 980673 .64 484435 7-54 5i5565 2 59 29209 9563g 465522 980635 .64 484887 7-53 5i5ii3 1 60 '29237 95530 463935 6-88 980596 .64 485339 7.53 5i466i N. COS. X. sine. L. COS. D.l" j L.sine. j KcoL 1 D.l" L. tang. 73^ 1 TRIGONOMETRICAL FUNCTIONS. — 17°. 47 Nat. Functions. Logarithmic Functions + 10. 1 _1 N.sine. N. cos. L. sine. D.1" L. COS. JD.l" L. tang. D. 1" L. cot 1 29237 95630 9.460935 6-88 9-980596 -64 9-485339 7-55 io-5i466i!60 1 29265 95622 466348 6-88 980508 1 .64 485791 7.5a 514209 [59 2 29293 j 90613 466761 6-87 980519 ■ 65 486242 7.5, 5i3758;5S 3 29321 956o5 467173 6-86 980480 •60 486693 7-5i 513307; 57 4 29348 95596 467585 6-85 980442 •65 487143 7-5o 512857! 56 5 29376 95588 467996 6-85 980403 -65 487593 7-49 512407 55 6 29404 95579 468407 6-84 98o364 • 65 488043 7-49 5,1957 54 7 29432 95571 468817 6-83 980325 • 65 4S8492 7-48 5iioo8 53 8 29460 95562 469227 6-83 980286 • 65 488941 7-47 5iio59J52 9 29487 95554 469637 6-82 980247 .65 489390 7-47 5io6,u 51 10 11 295i5| 95545 470046 6-8i 980208 .65 • 489838 .7.46 510162 150 29543 95536 9-470455 6-80 9-980169 .65 9.490286 7-46 10-5097,4 49 12 29571 95528 470863 6-80 980130 • 65 490733 7-45 509267 43 13 29399 95519 471271 '6-79 980091 • 65 491180 7-44 308820 47 14 29626 955,1 471679 6-78 980052 .65 491627 492073 7-44 508373 46 15 29604 905o2 472086 6-78 980012 • 65! 7-43 507927 45 16 29682 95493 472492 6-77 979973 .65 492519 7-43 50748, 44 17 29710 95485 472898 6-76 979934 .66' 492960 7-42 507035 43 18 29737 95476 473304 6-76 979895 .66 493410 7-41 506590 42 19 29765 95467 473710 6-75 979855 .66 493854 7-40 506146 41 20 21 29793 1 95459 4741 1 5 6-74 979816 .66' . 494299 7-40 5o57o, 40 29821 95450 9-474519 6.74 9-979776 .66 9-494743 7-40 ,o-5o5257 89 22 29849 95441 474923 6-73 979737 .66 495186 7-39 504814 83 23 29876 95433 475327 6.72 979697 .66 495630 7-38 504370 37 24 29904] 95424 473730 6-72 919658 .66' 496073 7-37 503927 86 25 29932 I 95415 476i33 6-71 979618 • 661 49651 5 7-37 5o3485 35 26 29960; 95407 476536 6-70 979079 .66 496957 7-36 5o3o43 34 27 29987 95398 476938 6-69 979539 .66 497399 7-36 5o26oi 33 28 3ooi5! 95389 477340 6-69 979499 .66 497841 7-35 5o2i59 32 29 30043 i 95380 477741 6-68 979409 .66 498282 7-34 501718 31 30 81 30071 30098^ 95372 478142 6-67 979420 .66 498722 9.499163 7-34 7-33 50,278 io-5oo837 80 2y 95363 9-478542 6-67 9-979380 .66 32 30126 I 95354 478942 6-66 979340 .66 499603 7-33 5oo397 28 33 3oi54i 95345 479342 6-65 979300 .67 500042 7.32 499908 27 34 30182; 95337 479741 6-65 979260 •67 500481 7-3i 499319 26 35 30209; 95328 480140 6-64 • 979220 -67 000920 7-3. 499080 25 36 30237 \ 95319 480539 6-63 979180 •67' 5oi359 7.30 498641 24 87 3o265; 95310 480937 6-63 979140 •67 501797 7.30 498203 .23 38 30292 1 95301 481334 6-62 979100 .67 502235 7.29 497765 ! 22 1 89 3o32oi 95293 481731 6-6i 979059 .67 502672 7.28 497328 21 40 3o348| 95284 482128 6.61 979019 •67 5o3i09 7.28 496891 10-496454 20 ^^1 30376 i 90270 9-482525 6-60 9-978979 .67; 9-5o3546 7-27 42 3o4o3i 95266 482921 6.59 978939 .67 003982 7-27 496018 18 431 3o43i i 90257 4833 16 6.59 978898 -67 5o44 1 8 7-26 495582 ]7 44 1 30459! 90248 483712 6-58 978858 .67' 504854 7-25 490146 16 45! 30486; 95240 484107 6.57 978817 •671 505289 7-25 4947" 15 46 1 3o5i4 9523i 484501 6-57 978777 •67: 505724 7-24 494276 14 47 30042 95222 4S4895 485289 6-56 978736 .67 1 5061O9 7-24 493841 13 48 30570 95213 6-55 978696 •68, 506593 7-23 493407 12 49 3o597 95204 485682 6-55 978655 -68 507027 7-22 492973 11 «0 3o625 95195 486070 6-54 978615 •68! 507460 7-22_ 492040 10-492107 10 9 3o653} 95186 1 9-486467 6-53 9-978574 • 68 9-507893 7-21 52 3o68o| 95177 486860 6-53 978533 .68, 5o8326 7-21 491674 8 53 30708! 95168 487201 6-52 978493 -68* 508759 7-20 491241 7 54 30736 90159 487643 6.5i 978402 -68 i 509191 7-19 49080Q 490378 6 55 30763 95i5o 488034 6-5i 978411 -68! 509622 ?:;? 5 56 30791 95142 488424 6-5o 978370 -68 5 10004 489946 4 57 30819 95i33 488814 6-5o 978329 -68! 5 1 0485 7-,8 489015 3 58 30846 95124 489204 6-49 6-48 978288 -68 5 ! 09 1 6 7-'7 489084 2 59 1 30874 90115 489093 978247 -68 5ii346 7.16 488604 1 601 30902 95106 489982 6-48 978206 .68 511776 7.16 488824 N. COS. iN. sine.] L. COS. D. 1" L. sine. L. cot. D. 1" L. tang. ' 72^ 1 48 TRIGONOMETRICAL FUNCTIOl^S. — 18" Nat. Functions. Logarithmic Functions + 10. 1 ' •N.sineJ N. cos. L. sine. D. 1' L. COS. D.l" L. tang. D.l" L. cot 30902 93106 9.489982 6.48 9-978206 • 68 9.511776 7.16 10.488224 60 1 30929 1 90097 490311 6 48 978165 .68 5l2206 7.16 487794 59 2 30907 I 90088 490759 6 47 ^''I'l^ .68 512635 7-15 487365 58 S 30980 1 90079 491147 6 46 978083 .69 5i3o64 7-14 486936 57 4 31012190070 491535 6 46 978042 .69 513493 7-U 486507 5(5 6 3 1 040 95061 491922 6 45 978001 .69 513921 7-13 486079 55 6 3 1 068 95o52 492308 6 44 977959 .69 514349 7. .3 485651 54 7 31095 95043 492695 6 44 977918 .69 514777 7.12 485223 53 8 31123 90033 493081 6 43 911^11 .69 5i02o4 7.12 484796 52 9 3II0I 95024 493466 6 42 977835 .69 5i563i 7.11 484369 51 10 11 31178 95oi5 493851 6 42 977794 .69 5i6o57 7.10 483943 50 49 3 1 206 95006 9.494236 6 41 9.977702 .69 9.516484 7.10 io.4835i6 12 3.233 94097 494621 6 41 977711 .69 516910 7-09 483090 48 13 3 1 26 1 949><8 495oo5 6 40 977669 .69 517335 U 482665 47 U 31289,94979 495388 6 39 977628 .69 517761 482239 46 15 3i3i6 94970 495772 6 39 977586 .69 518180 7.08 481815 45 16 3i344 94961 496154 6 38 977544 .70 518610 7-07 481390 44 17 3i372 94952 496037 6 37 9775o3 .70 519034 7-o6 480966 43 18 3i399 94943 496919 6 37 977461 .70 519458 7.06 480042 42 19 31427 94933 497301 6 36 977419 .70 519882 7.05 480118 41 20 31404 94924 497682 6 36 977377 .70 •70 52o3o5 7.05 ^47969! 40 39 21 31482 94915 9.498064 ~b 35 9.977335 9.520728 7-04 10.479272 "22 3i5io 94906 498444 6 34 977293 .70 521101 7-03 478849 38 23 31037 94^97 498825 6 34 977201 •70 521573 7.03 478427 37 24 31065 94888 499204 6 33 977209 •70 521995 7.03 478005 36 .25 31D93 94878 499584 6 32 977167 .70 522417 522838 7.02 477583 35 26 3i62o 94869 499963 6 32 977125 .70 7.02 477«62 34 27 31648 94860 5oo342 6 3i 977083 .70 523259 7.01 476741 S3 28 31670 9485 1 500721 6 3i 977041 .70 523680- 7.01 476320 32 29 3 1 703 94842 501099 6 3o 976999 .70 524100 7.00 475900 31 30 31730 94832 501476 6 29 976907 .70 524520 6.99 475480 SO 31 31708 94823 9.001804 6 29 9.976914 •70 1 9.524939 6.99 10-475061 29 32 31786 94814 50223l 6 28 976872 •71 525359 6.98 474641 28 33 3i8i3 94805 502607 6 28 976830 •71 525778 6.98 474222 27 34. 31841 94795 94786 502984 6 27 976787 •71 526197 6.97 4738o3 26 35 3 1 868 5o336o 6 26 976745 •71 5266 1 5 6-97 473385 25 36 31896 94768 ■ 5o3730 6 26 976702 •71 527033 6.96 472967 24 37 31923 5o4tio 6 25 976660 •71 527451 6.96 472049 23 38 31951 94758 504485 6 25 976617 •71 527868 6.95 472132 22 39 31979 94749 504860 6 24 976574 •71 528285 6.95 471713 21 40 41 32006 94740 5o5234 6 23 976532 •71 528702 6.94 471298 20 32034 94730 9.5o56o8 "T 23 9.976489 •7» 9.529119 6.93 10-470881 19" 42 32061 94721 005981 6 22 976446 •71 529530 6.93 470465 18 43 32089 94712 5o6354 6 22 976404 .71 529950 6.93 470o5o j 17 44 321 16 j 94702 ^ 506727 6 21 976361 •71 53o366 6.92 469634 16 45 32144 94693 507099 6 20 976318 •71 530781 6.91 469219 468804 15 46 32171 1 94684 507471 6 20 976275 •71 531196 6-91 14 47 32199 1 94674 507843 6 «9 976232 •72 531611 6-90 468380 467975 467561 13 48 32227 94665 508214 6 »9 976189 .72 532025 6.90 6.89 12 49 32204 94656 5o8585 6 18 976146 .72 532430 532853 11 50 32282 94646 508956 6 18 976103 .72 6.89 467147 10 51 32309:94637 9.509326 6 n 9.976060 •72 9.533266 6.88 10.466734 9 52 32337 94627 509696 6 16 976017 .72 533679 6.88 466321 8 53 32364 94618 5 10065 6 16 975974 .72 534092 5.87 465908 7 54 32392 04609 510434 6 i5 975930 .72 534504 •; 6.87 465496 465o84 6 55^32419^94599 5io8o3 6 i5 975887 .72 534916 '6.86 5 56 I: 32447 I q45qo 511172 6 14 973844 .72 535328 • 9.86 464672 4 57 32474 94080 5ii54o 6 i3 975800 .72 535739 6.85 464261 3 58 32002 94071 511907 512275 6 i3 970757 .72 536 1 5o 6-85 463850 2 59 32529 94561 6 12 970714 .72 536561 6.84 463439 463028 1 60 32557 j 94552 512642 6-12 975670 • .72 1 536972 6-84 |N. COS. N. sine. L. COS. D.l" L. sine. 1 L. cot D.l' L. tang. 1 •71° 1 TRIGONOMETRICAL F TXCTIOXS. — 19". 49 Nat. Functions. Logarithmic Functions -t- 10. 1 ' 1 1 N.sine.!N. cos.| L. sine. D. 1" L. COS. |] D.l" ; L. tang. D 1." L. cot. 1 o! 1' 2; 3 4 \ 7 8 9 10 11 12 13 14 15 16 17 18 io 20 "2r 23 23 24 25 26 27 23 29 30 32557 1 32584 32612 33639 32667 32694 32722 32749 32777 32804 32832 32859 32887 32914 32942 32969 132997 33024 :33o5i 1 33079 33io6 94552 94542 94533 94523 94514 94504 94495 94485 94476 94466 94457 9-512642 5 1 3009 5x3375 5i374i 514107 514472 514837 5l5202 5 1 5566 5i593o 516294 6-12 6.11 6.11 6.10 6-09 6.09 6-o8 6-08 6-07 6.07 6-06 9.973670 973539 973496 975432 973408 975365 975321 975277 975233 •73 •73 1 9-536972 537382 537791 538201 5386x1 539020 539429 539837 540245 540653 541061 6-84 6-83 6.83 6-82 6-82 6-81 6-8i 6-80 6-80 6-79 6-79 10-463028 60 462618 59 402208 1 58 461798 1 57 461389156 460980 1 55 460371 1 54 460163 153 439.755 j 52 459347 1 51 435939 50 94447 94438 94428 94418 94409 94399 94390 94380 94370 94361 9.516657 517020 517382 517743 518107 518468 518829 519190 519551 519911 6-o5 6-05 6.04 6.04 6.o3 6-02 6.02 6-01 6-01 6.00 9-975189 973143 975101 97^037 973013 974969 974923 974880 974836 974792 1 •74 •74 •74 •74 •74 •74 •74 •74 •74 •74 •74 •74 •74 •74 •75 9.541468 541875 542281 542688 543094 543499 543905 544310 544715 545119 6.78 6-78 6.77 6.77 6.76 6.76 6.75 6.73 6-74 6-74 10-438532 458123 437719 437312 436906 456301 456095 433690 455285 43488 X 49 48 47 46 45 44 43 42 41 40 33i34 33i6i '83189 j 33216 33244 1 33271 1 33298 : 33326 ! 33353 |3338i 94351 94342 94332 94322 943 1 3 943o3 94293 94284 94274 94264 9.520271 52063 1 520990 521349 521707 522066 522424 522781 523i38 523495 6-00 5-99 5.98 5.?8 5-97 5.96 5.96 5.93 5-93 9.974748 974703 974659 974614 974570 974523 974481 974436 974391 974347 9-545524 545928 546331 546735 547138 547540 547943 548345 548747 549149 6-73 6.73 6.72 6.72 6.71 6.71 6-70 6-70 6-69 6-69 10.454476 434072 453669 453265 432862 432400 452037 431655 431233 43o85i 39 33 37 36 35 34 33 82 31 30 29 28 27 26 25 24 r3 22 21 20 31 32 S3 34 35 86 37 38 39 40 ' 33408" 1 33436 ' 33463 ' 33490 i335i8 33545 1 33573 '■■ 33600 133627 1 33655 94254 94245 94233 94225 94215 94206 94196 94186 94176 94167 9.523«52 524208 524564 524920 525275 525630 525984 526339 526693 527046 5-94 5.94 5.93 5.93 5.92 5.91 3.91 5.90 5-90 5-89 9-974302 974257 974212 974167 974122 974077 974032 973987 973i)42 973897 •75 •75 •75 •75 3 I 9-549530 549951 55o352 550752 55ii52 55x552 551952 552351 552750 553x49 6-68 6-68 6-67 6-67 6-66 6-66 6-65 6-65 6-65 6-64 10-430450 430049 449648 449248 445843 44S448 448048 447649 447250 440&3I 41 42 43 44 45 46 47 43 49 50 i 33682 133710 133737 ; 33764 33792 33819 33846 33874 |! 33901 433929 94137 94147 94137 94127 94118 94108 94098 94088 94078 94068 9.527400 527753 528io5 528458 528810 529161 529513 529864 53021 5 53o565 5-89 5-88 ^ 5.88 5-87 5.87 5.86 5.86 5.85 5.85 5-84 9-973852 973807 973761 973716 973671 973623 973530 973535 973489 973444 •75 .76 .76 •76 .76 -76 -76 -76 9-553548 553946 554344 554741 555x39 555536 555933 556329 556725 1 55712X 6.64 6.63 6.63 6-62 6.62 6-6x 6-6x 6-60 6.60 6-59 10-446452 446054 443036 443239 444861 444i64 444067 443071 443275 442579 19 18 17 16 15 i^ 12 11 10 51|!339d6 52 i| 33983 53 |340M 54 34038 55 !i 34065 50 ; 34093 57,; 34120 58 I 34147 591:34175 60 ; 34202 94o58 94049 94039 94029 94019 94009 93999 93989 93979 93969 9-53o9i5 531265 53i6i4 531963 532312 532661 533009 533357 533704 534052 5-84 5-83 5.82 5-82 5-81 5.81 5-80 5-80 5-79 5-78 9.973398 973332 973307 973261 973215 973169 973124 973078 973o32 973986 -76 .76 -76 -76 •76 -76 -76 -76 •77 •77 9-557517 557913 5583o8 55S702 559097 539491 559885 56 1 066 6.59 6.59 6-58 6-58 6.57 6.57 6.56 6-56 6-55 6.55 10-442483 442087 44I6Q2 441298 440903 440309 4401x5 439721 439327 435934 9 8 7 6 5 4 3 2 1 N. COS. N.6ine.| L. COS. D.l" L. sine. : L. cot. D.l" L. tang. 1 ' 1 70° 1 50 TRIGONOMETRICAL FUXCTIOXS. — 20° Nat. Fvnctions. LooARiTUMic Functions + 10. 1 T 3 4 5 6 7 S 9 10 11 12 13 14 15 Id 17 13 19 20 N.sme.|N.cos. L.sine. 'd 1" j L. COS. D.l" L. tang. D.l" L. cot 34202 34229 34207 34284 343 1 1 34339 34366 34393 34421 34448 34475 93969 93909 93949 93939 93929 93919 93909 93899 93889 93879 93869 9.534052 534399 534745 530092 535438 535783 536129 536474 5368i8 537163 537507 5 5 5 5 5 5 5 5 5 5 5 78 77 77 77 76 76 75 74 II 9.972986 972940 972894 972848 972802 972755 972709 972663 972617 972570 972524 •77 •77 •77 •77 •77 •77 •77 •77 •77 •77 •77 9-56io66 561409 66i85i 562244 562636 563028 563419 563811 564202 564592 564983 6.55 6.54 6.54 6.53 6.53 6.53 6.52 6-52 6-51 6.5i 6.5o 10.438934 438541 433149 437756 437364 436189 435798 435408 435017 60 59 53 57 56 55 54 53 52 51 50 345o3 ' 34530 34507 1 34084 34612 34639 ' 34666 34694 34721 ,34748 93809 93849 93839 93329 93319 93809 93799 93789 93779 93769 9-537851 538194 538538 538880 • 539223 539565 539907 540249 540590 540931 5 5 5 5 5 5 5 5 5 5 72 72 71 71 70 70 68 9-972478 972431 972385 972338 972291 972245 972198 972i5i 972105 972058 .78 -78 9.560373 565763 5661 53 566542 566932 567320 563486 568873 6.5o 6.49 6.49 6-49 6-48 6.48 6-47 6-47 6.46 6-46 10-434627 434237 433847 433458 433068 432680 432291 431902 431014 431127 49 4S 47 46 45 44 43 42 41 40 •2i 22 23 24 25 26 27 25 29 30 ol 32 33 34 35 36 37 33 39 40 34775 ■ 34803 34830 34857 34884 3491^ 34939 34966 34993 35o2i 93759 9374S 9373b 93708 93698 93688 93677 93667 9-541272 541613 541953 542293 542632 542971 543310 543649 543987 544325 5 5 5 5 5 5 5 5 5 5 66 65 65 64 64 63 63 9-972011 971964 971917 971870 971823 971776 971729 9716S2 971635 971588 t :?^ •79 •79 •79 •79 9.569261 069648 570035 570422 570809 571,95 571581 571967 572352 572733 6.40 6.40 6.45 6-44 6-44 6.43 6.43 6.42 6.42 6.42 10-430739 43o3o2 429965 429578 429191 428800 428419 428033 427648 427262 39 38 37 36 35 34 33 32 31 30 35048 35075 35102 ,35i3o 3oi57 35i84 352II 35239 35266 35293 93657 93647 93637 93626 93616 93606 93596 93585 93070 93565 9-544663 545000 545338 545674 54601 1 546347 546683 547019 547354 547689 5 5 5 5 5 5 5 5 5 5 62 62 6i 61 60 60 % 58 58 9.971540 971493 971446 971398 97i3oi 97i3o3 971256 971208 971161 971113 •79 •79 •79 •79 •79 •79 •79 •79 •79 •79 9.573123 573507 573892 574276 574660 575044 575427 570810 576193 576576 6.41 6.41 6.40 6-40 6.39 6.39 6-39 6-33 6-38 6.37 10-426877 426493 426108 425724 425340 424956 424373 424190 423807 423424 29 28 27 26 25 24 23 22 21 20 41 42 43 44 45 46 47 4S 49 50 ; 35320 1 35347 ; 35375 ! 35402 35429 35456 35484 ;355ii 135533 135565 93555 93544 93534 93524 93514 93oo3 93493 93483 93472 93462 9.548024 548359 548693 549027 549360 549693 550026 55o309 550692 55io24 5 5 5 5 5 5 5 5 5 5 56 56 55 55 54 54 53 53 9-971066 971018 970970 970922 970874 970827 970779 970731 970683 970635 .80 .80 .80 -80 •80 .80 .80 .80 •80 -80 9.576908 577341 577723 578104 578486 578867 579248 %tll 58o389 6.37 6.36 6-36 6.36 6.35 6.35 6.34 6.34 6/34 6.33 10.423041 422659 422277 421896 42i5i4 421133 420752 420371 419991 419611 19 IS 17 16 15 14 13 12 11 10 51 52 53 54 55 56 57 58 59 60 ,35592 ! 35619 i 35647 135674 j 35701 35728 35755 30782 35810 35837 93452 93441 93431 93420 93410 93400 93389 93858 9.501306 551687 552018 552349 552680 553010 555341 553670 554000 554329 5 5 5 5 5 5 5 5 5 5 52 52 52 5i 5i 5o 5o 49 49 48 9-970586 970538 970490 970442 970394 970345 970297 970249 970200 970152 .80 .80 .80 .80 .80 .81 -81 .81 .81 .81 9.580769 581149 58i528 582286 582665 533043 583422 583300 584177 6.33 6-32 6-32 6-32 6-31 6-3i 6-30 6-3o 6-29 6-29 10.419231 4i83oi 418472 418093 417714 417335 416957 416573 416200 41 5823 9 8 7 6 5 4 8 2 1 t N. COS. X.sin*. L. COS. D. 1" L.8ine. i 1 L. cot. D.l" L. tang. 69 1 TRIGONOMETRICAL FUNCTIONS.— 21< 51 Nat. Functions. Logarithmic Functions + 10. ' |!N.8lne.|N.co8 L. sine. D. 1" L. COS. D.l" 1 I* tang. D.l" L. cot. ! 35837 j 93358 9-554329 5.48 9.970152 .81 9-584177 6-29 10 ^41 5823 1 60 1 35864 ! Q3348 554658 5-48 970103 -81 584555 6-29 410445 : 5y 2 35891 93337 554987 5-47 970055 -81 584932 6-28 4i5o68 ' 58 3 35918 93327 55531 5 5-47 970006 .81 585309 6-28 414691 57 4 35945 93316 555643 5-46 969957 •81 585686 6-27 4i43i4 56 5 35973 93306 555971 5.46 969909 •81 586062 6^27 413938 55 6 36000 93295 556299 5.45 969860 .81 586439 6.27 4i356i 54 7 36027 93285 556626 5-45 9698 1 1 .81 58681 5 6^26 4i3i85 53 8 36o54 93274 5^6953 5-44 969762 -81 581190 i 6^26 412810 52 9 36o8i 93264 557280 5-44 Q69714 .81 587566 6^25 412434 51 10 36 1 08 93253 557606 5.43 969665 •81 587941 6-25 412059 50 11 36i35 93243 0-557932 5-43 9.969616 -82 9-588316 6^25 10-411684 ; 4j 12 36162 93232 558258 5-43 969567 .82 588691 6-24 4ii3o9 1 48 13 36190 93222 558583 5-42 969018 -82 589066 6^24 410934 47 14 36217 93211 558909 5-42 969469 .82 589440 6^23 4io56o 46 15 36244 93201 559234 5.4. 969420 -82 589814 6-23 410186 45 16 36271 93190 559558 5-41 969370 .82 590188 6^23 409812 44 17 36298 93180 559883 5-40 969321 .82 590562 6^22 409438 43 18 36325 93169 560207 5.40 969272 .82 590935 6^22 409065 42 19 36352 93i5g 56o53i 5.39 969223 -82 591308 6-22 408692 41 20 36379 93148 56o855 5.39 969.73 .82 591681 6^21 408319 4437 607322 4-75 961123 -93 646199 5-69 353801 7 54 40514 91425 607607 4-75 961067 .93 646340 5-68 353460 6 55 4o54i 91414 607892 4-74 9610'! -93 646881 5-68 353119 352778 5 56 40567 91402 608177 4-74 960935 -93 647222 5-68 4 57 40594191390 608461 4-74 960899 960843 -93 647562 5.67 332433 3 58 40621191378 608745 4-73 -94 647903 5-67 352097 2 59 40647 9 '366 609029 4-73 960786 .94 648243 5-67 351757 1 60 40674 91355 609313 4-73 960730 •94 648583 5-66 351417 IN. COS. jN. sine L. COS. 1 D. 1" L. sine. L. cot. D.l" L. tang. * 66° 1 54: TRIGONOMETRICAL FL'NCTIOXS.— 24°. Nat. Functions. Logarithmic Functions + 10. ' N. sine. 'n. COS.! Ksine. 1 D. 1" L. COS. D.I" L. tang. 1 Dl." L. cot ! 40674! 91355 1 : 40700 1 91843 2 j40727!9iJ3i 8 140753191819 4 l| 40780 1 91807 5 140806191295 6 40833 j 91 283 7 : 40860 91272 8 ;, 40886 191 260 9 1140913191248 10 1 40939 91286 9.609813 609J57 609S30 610164 610447 610729 611012 611294 611576 6ii858 612140 4-73 4-72 4-72 4-72 4-71 4-7' 4-70 4-70 4-70 4.69 4.69 9-960780 960674 • 960618 960561 960303 960448 960892 960835 960279 960222 960165 •94 .94 .94 •94 •94 .94 •94 .94 •94 .94 •94 9.643588 648928 649263 649602 649942 65o28i 65o62o 630959 65 1 297 65i686 631974 5.66 5.66 5.66 5.66 5.65 5-65 5.65 5.64 5-64 5-64 5-63 io.85i4i7 351077 350787 350398 350033 349719 349380 349041 348708 348364 348026 60 59 58 57 56 55 54 53 52 51 50 49 43 47 46 45 44 43 42 41 40 11 1 12 13 14 15 16 17 13 19 20 40966,91224 40992 91212 41019 91200 41043 91188 41072I91176 41098 1 91 164 4II25 91152 4ii5i 91140 41178 91128 41204 91116 9-612421 612702 612983 618264 613545 618825 6i4io5 614335 614665 614944 4-69 4-68 4-68 4-67 4-67 4-67 4-66 4-66 4-66 4-65 9-960109 960052 l^l 059882 959835 959768 959711 959634 959596 •9? .93 -93 .93 .93 .95 .93 .93 9.652812 652650 652988 658826 653663 654000 654337 654674 65501 1 655843 5.63 5.63 5.63 5.62 5.62 5.62 5.61 5-61 5-61 5-61 10.347688 347350 347012 346674 346887 346000 345663 345326 344989 344652 21 1 22 23 24 25 26 27 23 29 SO 41281 91104 41257 91092 41284 91080 41810 91068 4i387 9io56 ■41863 91044 41890; 91082 41416 91020 '41443 91008 41469^90996 9.615228 6i55o2 615781 616060 616888 616616 616894 617172 617430 617727 4-65 4-65 4.64 4-64 4.64 4-68 4-63 4-62 4-62 4-62 9.959539 959482 959425 959868 959810 959253 95908. 939028 .93 -93 -95 .96 .96 .96 .96 .96 .96 9.655684 656020 656856 656692 657028 657864 65i699 658o84 658369 658704 5.60 5.60 5-60 5-59 5-58 5-58 10.344816 343980 343644 343308 342972 342636 842801 341966 841681 341296 89 88 87 86 85 84 83 32 31 30 31 32 33 34 35 36 87 33 8'J 40 14149^ -90984 4i522 90972 41549 '90960 41575 90948 ; 41602 90986 141628 90924 4i655 90911 ,41681 90899 41707:90887 '41734 90875 9-6i8oo4 618281 6i8558 618884 619110 619886 619662 619988 620218 620488 4-6i 4-6i 4-6i 4-6o 4-6o 4-6o 4-59 4-59 4-59 4-58 9-958965 958^08 958850 958792 938734 958677 958619 ■ 938561 958303 ?58445 .96 -96 .96 .96 .96 -96 •97 •97 9.659089 1 659878 659708 i 660042 660876 660710 661043 661877 661710 662043 5-58 5.57 ni 5.56 5.56 5.55 5.55 10-840961 340627 340292 389958 339624 XT, 388628 388290 887957 29 28 27 26 25 24 23 22 21 20 41 42 43 44 45 46 47 48 49 50 '; 41760 90868 i 41787 '90851 41818:90889 41840 90826 41866 90814 41892 190802 41919 90790 41943:90778 41972 '90766 ,41998190753 9.620763 621088 621818 621587 621861 622185 622409 622682 622956 628229 1 4-58 4-57 14.37 1 4-57 4-56 4-56 4-56 4-55 4-55 1 4-55 9.958887 958829 958271 958213 958 1 54 958038 957979 957921 937868 •97 •97 •97 •97 •97 •97 •97 •97 •97 •97 j 9.662876 662709 668042 663375 668707 664089 664871 664703 i 665o35 i 665366 5-55 5.54 5.54 5.54 5-54 5-58 5-53 5-53 5.53 5-52 10-337624 337291 336958 336625 386293 335961 335629 334684 19 18 17 ]6 15 14 13 12 11 10 51 52 53 54 55 56 57 5S 59 60 142024 90741 I42051 190729 j 42077 j 907 17 142104190704 1 42180 90692 142156:90680 142188:90668 j 4^209 ! 90655 142285 '90648 142262:90681 g-6285o2 628774 624047 624819 624591 624863 625i35 625406 625677 625948 1 4-54 1 4-54 I.4-54 4-53 4-53 ! 4-53 4.52 4-32 4-52 4.51 9.957804 957746 957628 I 937570 9575.1 937432 937898 937335 937276 t .98 .98 -93 .98 .98 .98 1 9.665697 J 666029 666860 666691 ! 667021 1 667852 1 667682 ! 668013 1 668343 668672 5-52 5-52 5.5i 5-51 5.51 5-5i 5-5o 5-50 5.5o 5.50 10-884808 388971 333640 338809 332979 382648 882818 381987 381657 381828 9 8 7 6 5 4 8 2 1 JN. COS. N. sine L. COS. 1 D. 1" J L. sine. 1 L. cot. D.l" L. tang. 1 ' 65° 1 TRIGOl^OMETRICAL FUXCTIOXS. — 25= 55 Nat. Functions. Logarithmic Functions + 10. N.sine. N. COS. L. sine. D. 1" L. COS. DA" L. tang. D 1" L. cot. 42262 906:31 9-625948 4-5i 9.957276 .98 9.668673 5.5o io.33i327 60 1 42288 90618 626219 4-5i 937217 .98 669002 5-49 330998 69 2 423i5 90606 626490 4-5i 957158 .98 669332 5-49 33o668 53 8 42341 90594 626760 4-5o 937099 .98 669661 5-49 330339 57 4 42867 9o582 627030 4-30 937040 .98 669991 5-48 330009 56 5 42394 90569 627300 4-5o 956981 .98 670320 5.48 329680 55 6 42420 90557 627570 4.49 936921 •99 670649 5-48 329351 54 7 42446 90545 627840 4-49 936862 •99 670977 5.48 329023 53 8 42473 ! 90532 628109 4-49 956803 •99 671306 5-47 328694 52 9 42499 90520 628378 4-48 956744 •99 671634 5-47 328366 51 10 42323 9o5o7 628647 1 4-48 956684 •99 671963 5.47 328037 50 11 42332 90493 9.628916 4-47 9-956625 •99 9.672291 5-47 10-327709 49 12 42378 90483 629185 4-47 956566 •99 672619 5.46 327381 43 13 42604 90470 629453 4-47 9565o6 •99 672947 5-46 327053 47 14 4263 1 90458 629721 4.46 956447 •99 673274 5-46 326726 46 15 42637 90446 6299S9 4-46 956387 •99 673602 5-46 326398 45 16 42683 90433 630237 4.46 956327 •99 673929 5.45 326071 44 17 42709 90421 63o524 4.46 956268 •99 674257 5-43 325743 43 13 42736 90408 630792 4-45 956208 1. 00 674584 5.43 325416 42 19 42762 ] 90396 63 1039 4-43 956148 1-00 674910 5-44 325090 41 20 42788 1 90383 63i326 4-45 956089 I -00 675237 5-44 324-63 40 21 42813190371 9.631593 4-44 0-956029 I -00 9-675564 5-44 10.324436 3y 22 42841 90338 631859 4.44 955969 I -00 675890 5.44 324110 33 23 42867 1 90346 632123 4-44 953909 1. 00 676216 5-43 323784 37 24 42894 90334 632392 4-43 955849 1. 00 676543 5.43 323457 36 25 42920 9032 1 6326j8 4-43 955789 1-00 676809 5.43 323i3i 35 26 42946 9o3oq 632923 4-43 955729 I- 00 677194 5-43 322806 34 27 42972 90296 633189 4-42 955669 i-oo! 677320 5.42 322480 33 28 42999 90284 633454 4-42 935609 1-00 677846 678171 5-42 322154 32 29 43o23 90271 633719 4-42 955548 1. 00 5.42 321829 31 30 43o5i 1 90259 633984 4-41 955488 1. 00 678496 5.42 32i5o4 30 31 43077 90246 9. 634249 4-41 9.955428 1.01 9.678821 5-41 10.321179 29 32 43 104 90233 634314 4.40 955368 I-OI 679146 5-41 320854 23 33 43i3o 90221 634778 4.40 955307 I -01 679471 5.41 320520 320205 27 34 43 1 56 90208 635042 4.40 955247 I -01 5.41 26 35 43182 : 90196 6353o6 4-39 955186 I -01 680120 5.40 319880 25 36 43209 90183 635570 4-39 955126 I -01 680444 5-40 319556 24 37 43235 90171 635834 4-39 955o65 I-OI 680768 5.40 i3i9232 23 33 43261 |90i58 636097 4-38 955oo5 I-Ol 681092 5.40 318908 22 39 43287 I 90146 636360 4-38 954944 I .01 681416 5-39 3 18584 21 40 41 43313! 90133 636623 4-38 954883 I-Ol 681740 5.3^ 318260 20 43340 1 90120 9.636886 4-37 9.954823 1.01 g. 682063 5-39 10-317937 ly 42 43366 90108 637148 4-37 954762 I-Ol 682387 5-39 317613 18 43 43392 90095 63741 1 4-37 954701 l-Ol 682710 5-38 317290 17 44 43418 900S2 637673 4-37 4-36 954640 I-Ol 683o33 5-33 316967 16 45 43445 90070 637935 954379 £-01 683356 5-38 3 16644 15 46 43471 90037 638197 4-36 954518 1.02 683679 5-38 3i632i 14 47 43497 90045 638458 4-36 954457 1.02 684001 5.37 3 15999 13 48 43323 90032 638720 4-35 954396 1-02 684324 5.37 3 15676 12 49 f4 90019 638981 4-35 954335 1-02 684646 5-37 3 15354 11 ^50 90007 639242 4-35 954274 1-02 684968 5.37 3i5o32 10 51 43602 ; 89994 9.639503 4-34 9.954213 iT^ 9-685290 5-36 io-3i47'o 9 52 436281899^1 639764 4-34 954152 1-02 685612 5-36 314388 8 53 43634 89968 640024 4.34 954090 1-02 685934 5-36 3 1 4066 7 54 43680 89956 640284 4-33 954029 1-02 686255. 5-36 3i3745 6 65 43706 89943 640544 4-33 953968 1.02 686577 5-35 3i3423 5 56 43733 89930 640804 4-33 953906 1-02 686898 5-35 3i3io2 4 57 43759189918 641064 4-32 953845 1.02 687219 5-35 812781 8 58 43783 ; 89905 641324 4-32 953783 I.02I 687540 5.35 3 1 2460 2 69 438 n 1 89892 64 1 583 4-32 953722 i.o3 687861 5.34 3i2i39 3ii8i8 1 60 43837 1 89879 641842 4-3i 953660 I -03 688182 5.34 l! N. COS. N. sine. L. COS. D.l" L. sine. 1 L.cot D.l" L.tang. ' 1 64° 1 56 TRIGONOMETRICAL FUXCTIOXS.— 26°. Nat. FuKcnoxs. Logarithmic Functions •+• 10. ' ,X.sme.;N. COS. L. sine. D. 1" L. cos. D.1-; L. tang. D.l" L. cot 43837 89879 9.641842 4 •31 9.953660 i.o3 9.688182 5.34 io-3ii8i8 60 1 43S63 89^7 642101 4 3i 953599 i.o3 688502 5.34 311498 59 2 43839 89S54 642360 4 3i 953537 i.o3, 688823 5.34 311177 58 8 43916 8984. 642618 4 3o 953475 103 689143 5-33 310857 57 4 43942 8ob28 642877 4 3o 953413 i-o3 689463 5.33 3 10537 56 5 43968 89S I 6 643 1 35 4 3o 953352 1-03 689783 5-33 310217 55 6 43994 S9803 643393 4 3o 953290 i.o3 690103 5.33 309897 54 7 ; 44020 89790 6436DO 4 29 9D3228 i-o3 690423 5-33 309377 53 8 44046 89777 643908 4 29 953166 1-03 690742 5-32 309238 52 9 44072 189764 644165 4 29 953104 1-03 691062 5.32 308938 51 10 44098189752 644423 4 28 953042 i.o3 691381 5-32 308619 50 11 44124 189739 9.644680 4 28 9-952980 1-04 9-691700 5.31 io-3o83oo 49 12 44i5i 1S9726 644936 4 23 952918 1.04 692019 5-31 307981 4S 13 '44177 8o7l3 645193 4 27 932803 104 692338 5.3i 307662 47 U 1 442o3 ' 89700 645450 4 27 952793 1.04 692656 5.31 307344 46 15 ! 44229 S^bSi 645706 4 27 952731 1-04 692975 5.31 307025 45 16 4425! 89674 645962 4 26 952669 1-04 693293 5-3o , 306707 3o63S8 44 17 44281 89662 646218 4 26 952606 1-04 693612 5.30 43 13 44307 44333 89649 646474 4 26 952344 1.04 693930 5-3o 306070 42 19 89530 646729 4 25 952481 1.04 694248 5-3o 3o5752 41 20 1 44359 89623 646984 4 25 952419 1.04 694366 5.29 3o5434 40 21 i 44385 S9610 9-647240 4 25 9.952356 1.04 9.694883 5.29 io-3o5ii7 8y 22 4441 1 89597 647494 4 24 952294 1.04 693201 5.29 304799 3S 23 44437 5Q5ti4 647749 4 24 952231 1.04 695518 5.29 304482 37 24 ; 44464 89571 648004 4 24 952168 1-05, 695836 5.29 304164 36 25 1 44490 89558 64S258 4 24 952106 1.03 696153 5-23 3o3847 35 26 44516 S9545 6485i2 4 23 932043 i.o5' 696470 696787 5.23 3o353o 34 27 ;; 44542 89532 64S766 4 23 951980 i.o5 5-23 3o32i3 33 28 ;! 44568 89519 649020 4 23 931917 i.o5 697103 5-23 302897 32 29 1 44594 89506 649274 4 22 931834 1.03 697420 5.27 3o258o 31 30 ; 44620 89493 649327 4 22 931791 i.o5 697736 5-27 302264 80 31 44646 894:50 9-649781 4 22 9.951728 i.o5 9-698053 5.27 10.301947 2y 32 44672 S3 '44698 89467 65oo34 4 22 95 I 665 1.03 69S369 5-'2 3oi63i 23 89454 65o2b7 4 21 951602 i.o5 698685 5.26 3oi3i5 27 34 1; 44724 89441 65o539 4 21 93.539 I 05. 699001 5.26 300999 26 85 j 44750 89428 650792 4 21 931476 i-od! 699316 5.26 3oo684 25 36 447i6 89415 65 1044 4 20 931412 1.03, 699632 5-26 3oo368 24 37 44802 89402 651297 4 20 951349 1.06' 699947 5.26 3ooo53 23 38 44828 8o3S9 661549 4 20 951286 1.06 700263 5.23 299737 22 39 44854 89376 1 65iSoo 4 19 951222 1-06 700578 5.23 299422 21 40 448S0 89363 652052 4 •9 951159 1.06 700893 5.25 299107 20 19 41 , 44906 89350 9-6523o4 4' 19 9.951096 1.06 9.701208 5.24 10.298792 42 1 44932 89337 65z555 4 18 931032 1.06 7oi523 5.24 298477 18 43 1 44958 89324 652806 4 18 950968 1.06 701837 5-24 298163 17 44 ,44984 89311 653o57 4 1 3 950905 1.06 702152 5.24 297848 16 45 1 45oio 8929S 6533oS 4 18 95c84i i.o6i 702466 5-24 297534 15 46 1 45o36 89285 653558 4 17 950778 1.06 702780 5.23 297220 14 47 , 45062 89272 6538o8 4 n 950714 |.o6 703093 ^'3 296905 13 4S ! 45o88 89239 654o59 4 17 95o65o i-o6 ?o1^^ 5-23 296591 12 49 i 45i 14 89245 654309 4 16 95o5.% 1.06 5.23 296277 11 50 '4Pi4o 89232 654558 4 16 95o522 1.07 704036 5.22 295964 10 51 '45166,89219 9-654808 4 16 "9^950438" 1-07 9.704350 5.22 10-295650 9 52 '45192 '89206 655o53 4 16 950394 1-07 704663 5-22 295337 293023 8 53 45218 89193 6553o7 4 i5 95o33o 1-07 704977 5.22 7 54 45243 89180 655556 4 i5 950266 1.07 705290 5-22 294710 6 55 '45269 89167 6558o5 4 i5 950202 1.07, 7o56o3 5-21 294397 5 56 i452o5 8qi53 656o54 4 14 9501 38 1-07 705916 5-21 294084 4 57 ,45321 89140 656302 4 14 950074 1.07 706228 5.21 293772 3 53 145347 89127 656551 4 14 950010 1-07 706541 5.21 293459 2 59 1 453-3 8qii4 656799 4 i3 949945 1.07. 706854 5.21 293146 1 60 ;| 45399 S9101 657047 4-i3 949881 1.07 707166 5-20 292834 jl K COS. X. sine. L. COS. D. 1" L.sme. 1 Kcot D.l" L.tang. ~^ 63^ TRIGONOMETRICAL rUNCTI0KS.~27°. 57 Nat. Functions. Logarithmic Functions + 10 1 ' N.skie.'N.cos. L. sine. D. 1" L. COS. D.V'\ L. tang. D 1" L-cot. '45399 '89101 0.607047 4-i3 9.949881 1.07! 9.707I66 5 20 10.292834 60 1 43425 89087 657295 4-i3 949816 1-07 707478 5 20 292522 59 2 : 45451 89074 607042 4-12 949752 1-07; 707790 20 292210 5S 3 ' 45477 89061 657790 4-12 949688 708102 5 20 291898 57 4 i 455o3 89048 608037 4-12 949623 1.08 708414 5 19 291606 56 5 45529 89035 658284 4-12 949558 1.08 708726 5 •9 291274 55 6 45554 89021 658531 4-11 949494 I- 08! 709037 5 19 290963 54 7 45580 8900S 658778 4-11 949429 1.08' 709349 5 19 290661 53 8 45606 88995 609020 4-11 949364. 1.08 709660 5 19 290340 62 9 45632 889:^1 659271 4-10 949300 1.08 70997 1 5 18 290029 51 10 11 45658 45684 88968 659517 4-10 949235 1. 08 710282 5 18 289718 50 88955 9.609763 4-10 9-949170 1.08 9.710693 5 18 10.289407 4y 12 45710 88942 660009 4-09 949100 1-08 710904 5 18 289096 48 13 45736 88928 660255 4-09 949040 1.08 711215 5 18 288780 47 14 45762 88915 66o5oi 4.09 948975 1.08 711625 5 17 288475 46 15 45787 S8902 660746 4.09 948910 1.08 711836 5 17 288164 45 16 4581 3 8^888 660991 4-o8 948845 i-o8 712146 5 17 287854 44 17 45839 88875 661236 4-o8 948780 1.09 712456 17 287544 43 18 45865 8S062 661481 4-o8 948715 1.09 712766 5 16 287234 42 19 ; 45891 8S848 661726 4-07 948650 1-09 1.09 713076 5 16 286924 41 20 21 : 45917 88835 661970 4-07 948584 713386 5 16 286614 40 45942 88822 9.662214 4-07 9.948519 1.09 9-713696 0" 10 10-286304 oy 22 45968 8S808 662459 4-07 948454 1.09 714005 5 16 285990 83 23 45994 88795 662703 4 -06 948388 1-09 714314 5 i5 286086 87 24 46020 88782 662946 4 -06 948323 1-09 1.09 714624 5 10 '285376 286067 284758 86 25 : 46046 88768 663190 4-o6 948207 714933 5 10 85 26 ' 46072 88755 663433 4-o5 948192 1-09 710242 5 i5 34 27 46097 88741 663677 4-o5 948126 1.09 7io55i 5 14 284449 33 28 46123 88728 663920 4-o5 948060 1.09 7i586o 5 14 284140 32 29 46149 88715 664163 4-o5 947995 1.10 716168 5 14 283832 81 80 ,46175 88701 664406 4.04 9479'^9_ I-IO 1. 10 716477 5 14 283523 80 31 46201 88688 9.664648 4.04 9.947863 9.716785 T 14 10.28J210 2'J 32 46226 88674 664891 4-04 947797 1. 10 717093 5 i3 282907 23 33 46252 88661 665i33 4-o3 947731 I'lO 717401 5 i3 2b2099 27 34 146278 88647 665370 4-o3 947665 I. 10 717709 5 i3 2^2291 26 35 1 463o4 88634 665617 4-o3 947600 l.IO 718017 i3 2 -j 1983 25 36 i 46330 88620 665859 4-02 947533 I-IO 718325 5 .i3 2^1676 24 37 ' 46355 88607 666100 4-02 947467 l.IO 718633 5 12 281367 23 38 46381 88093 666342 4-02 947401 I. 10 718940 5 .12 281060 22 39 46407 88580 666583 4-02 947335 1.10 719248 5 •12 280762 21 40 41 46433 88566 666824 4-01 947269 1.10 719555 5 12 280445 20 j 46458 '8S5"53" 9.667065 4-01 9-947203 l-lO 9.719862 5 12 io.28oi38 "iy~ 42 146484 88539 667305 4-01 947 1 36 I-Il 720169 5 11 279831 18 43 j465io 83526 667546 4-01 947070 I. II 720476 5 11 279624 17 44 ' 46536 88512 667786 4-00 947004 I-U 720783 5 II 279217 16 45 i 46561 88499 668027 4-00 946937 1. 11 721089 5 11 278911 15 40 1 46587 88485 668267 4-00 946871 I.I 1 721396 5 11 278604 14 47 46613 88472 88458 6685o6 3.99 946804 1 .11 721702 5 10 278298 13 48 , 46639 668746 3.99 946738 1 .11 722009 5 10 277991 12 .49 , 46664 88445 668986 3-99 94667 1 1 .11 722310 5 10 277680 11 50 ! 46690 88431 669225 3-99 946604 1 .11 l.Il 722621 5 10 277379 10 51 46716 88417 9.669464 3-98 9.946538 9.722927 5 10 10.277073 y 52 46742 88404 669703 3.98 946471 1. 11 ^ 723232 5 09 276768 8 53 ' 46767 88390 669942 3-98 946404 1. 11 723538 5 09 276462 7 54 46793 88377 670181 3-97 946337 111 723844 5 09 276166 6 55 46819 88363 670419 3-97 946270 1.12 724149 5 09 275851 5 56 46844 88349 670658 3-97 946203 1.12 724454 5 09 275646 4 57 46870 88336 670S96 3-97 946 1 36 1.12 724759 5 08 276241 3 58 ' 46896 8832 2 671 i34 3-96 946069 1.12 725o65 5 08 274935 2 59 46921 883o8 671372 3.96 946002 1-12 720369 5 08 274631 1 60 46947 8S295 671609 3-96_ 945935 1.12 725674 5 08 274326 N. 008.;^. sine. L. COS. D.l" L.8ine. L. cot ^ 1" L. tang. ' 62° 1 58 TRIGOXOMETRICAL FUNCTIOXS. — 28" Nat. Functions. Logarithmic Functions + 10. 1 ' N.sine. N.cos. L. sine. D.l" L. COS. D.l" L. tang. D.l" L.cot : 46947 882^5 9-671609 3-96 9-945935 I -12 9-725674 5-08 10-274326 60 1 46973 88281 671847 3-95 945^68 1-12 725979 5-08 274021 59 2 '46999 88267 672084 3-95 945800 1-12 726284 5-07 273716 58 8 47024 88234 672321 3-95 945733 I -12 726588 5-07 273412 57 4 47o5o 88240 672558 3-95 945666 I -12 726892 5.07 273108 56 5 47076 8S226 672795 3-94 945598 I -12 727197 5-07 272803 55 6 47101 8S213 673032 3-94 945531 I -12 727501 5.07 272499 54 7 '47127 88199 673268 3-94 945464 ;i-i3 727805 5.06 272195 53 8 i 47153 88i85 673505 3-94 945396 ;i.i3 728109 5-06 271891 52 9 47173 88172 673741 3-93 945328 i.i3 728412 5-06 27i5b8 51 10 47204 881 58 673977 3-93 945261 |i-i3 728716 5-06 271284 50 11 47229 88144 9-674213 3-93 9-945193 '1-13 9-729020 5-06 10-270980 \ 49 1 12 47255 88i3o 674448 3-92 943125 |i-i3 729323 5-05 270677 48 13 47281 88117 674684 3-92 945o58 ii-i3 729626 5-o5 270374 47 14 47306 88! o3 674919 3-92 944990 i»-'3 729929 5-o5 270071 46 15 47332 &80S9 675155 3.92 944922 |i-i3 730233 5-o5 269767 45 16 47358 88075 675390 3-91 944854 ii-i3 73o535 5-o5 269465 44 17 47383 88062 675624 3.91 944786 i-i3 73o838 5-04 ' 269162 43 18 '47409 8S048 675859 3-91 944718 {I -13 731141 5-04 268859 42 19 47434 88o34 676094 3-91 944650 ]i-i3 731444 5-04 268556 41 20 47460 88020 676328 3-90 944582 I -14 731746 5-04 268254 40 21 47486 88006 9-676062 3-90 9-944514 |i-i4; 9.732048 5-04. 10-267952 sy 22 4751 1 87993 676796 3-90 944446 I -14, 732351 5-o3 267649 38 23 47337 ; 87979 677030 3-90 944377 '•14; 732653 5-o3 267347 87 24 47362 87965 677264 3-89 944309 1-14, 732955 5-03 267045 36 25 47588. 87931 677498 3-89 944241 1-14, 733257 5-o3 266743 35 26 47614 87937 677731 3-89 944172 I -14: 733558 5-03 266442 34 27 47639 87923 677964 3-88 944104 ,1-14 733860 5-02 266140 33 2S 47665 87909 678197 3.88 944o36 1-14 734162 5-02 265838 32 29 47690 87896 678430 3-88 943967 ji-14 734463 5-02 265537 31 SO 47716 87882 '87C.68' 67S663 3-88 943899 1-14; 734764 5-02 265236 30 81 47741 9-678895 3-87 9 -943830 i-i4i 9-735066 5-02 10-264934 29 32 477-i5, 736871 5-01 263129 23 33 47920 1 87770 68o5i9 3-85 943348 I -15: 737171 5.00 262829 22 39 47946 i 87756 680750 3-85 943279 I -15, 737471 5-00 262529 21 40 47971 '87743 680982 3-85 943210 i-i5; 737771 5-00 262229 20 41 ,47997 87729 9-68i2i3 3-85 9-943141 I -151 9-738071 5-00 10-261929 19 42 .48022 87715 681443 3-84 943072 i-i5: 738371 5-00 261629 18 43 48048 87701 681674 3-84 943oo3 i-i5, 738671 4-99 261329 17 44 148073 87687 6S1905 3-84 942034 I -15; 942864 i-i5j 738971 4-99 261029 16 45 ! 48099 87673 682135 3-84 739271 4-99 260729 15 46 ; 48124 87659 682365 3-83 942795 I -16, 739370 4-99 260430 14 47 ;48i5o 87645 682595 3-83 942726 I -16; 739870 4.99 26oi3o 13 48 48175 87631 682825 3-83 942656 I -16 740169 4-99 259831 12 49 ; 48201 ,87617 683o55 3-83 942587 I -16 740468 4-98 259532 u 50 '48226 87603 683284 3-82 942517 1-16I 740767 4-98 259233 10 51 ,48252 87589 9-683514 3-82 9-942448 1-16; 9-741066 4-98 10-258934 "T 52 : 48277 8757D 683743 3.82 942378 I -16, ^ 741365 4-98 258635 8 53 483o3 87561 683972 3-82 942308 1-16 741664 4-98 258336 7 54 '48328 87546 684201 3-81 942239 I -16 741962 4-97 258o38 6 55 48354 87532 684430 3-81 942169 I -16 742261 4-97 257739 5 56 .48379:87518 684658 3-81 942099 I -16, 742559 j 4-97 742858 ! 4-97 257441 4 57 48405 j 87504 684887 685ii5 3-8o 942029 i-i6| 257142 3 58 48430 87490 3-80 941959 i-i6 743 1 56 j 4-97 256844 2 59 ' 48456 87476 685343 3-80 941889 I -17 743454 4-97 256546 1 60 48481 187462 685571 3-80 941819 '-17 743752 ! 4-96 256248 i N. COS. N. sine. L. COS. D. 1" L, sine. L. cot. { D. 1" L. tang. ' 61° 1 TRIGOXOMETRICAL FL']S^CTIONS. — 29\ 59 Nat. Functions. Logarithmic Functions + 10. / N.sme.N. cob. L. sine. D. 1'' L. COS. D.l" L. tang. D.l" L. cot. 48481 87462 9-685571 3 .80 9-941819 1-17 9-743752 4.96 10-256248 60 1 485o6 87448 685799 3 •79 941749 1-17 744o5o 4-96 255950 59 2 48532 87434 686027 3 •79 941679 1-17 744348 4.96 255652 58 S 48557 87420 686254 3 •79 941609 1-17 744645 4-96 255355 57 4 1 48583 87406 686482 3 79 941539 1-17 744943 4-96 255o57 56 5 4S608 87391 686709 3 .78 941469 1-17 745240 4-96 254760 55 6 48634 87377 686936 3 78 941398 1-17 745538 4-95 254462 54 7 48659 87363 687163 3 78 941328 1-17 745835 4-93 254165 53 8 48684 87349 48710 87335 687389 3 78 941258 1-17 746132 4-93 203868 52 9 687616 3 77 941187 1-17 746429 4-95 253571 51 10 48735 87321 687843 3 77 941117 1-17 746726 4-95 253274 50 11 48761 87306 9-688069 T 77 9-941046 1-18 9-747023 4.94 10-252977 49 12 148786 87292 688295 3 77 940975 1-18 747319 4.94 252081 48 13 488ri '87278 6S852I 3 76 940900 1-18 747616 4.94 252384 47 14 48837 87264 688747 3 76 940834 1-18 747913 4-94 252087 46 15 48862 87200 688972 3 76 940763 1..8 748209 4.94 251791 45 16 48888 87235 689198 3 76 940693 1-18 7485o5 4-93 251495 44 17 48913 87221 689423 3 75 940622 ;:;? 748801 4-93 251199 43 18 48938 87207 689648 3 75 94o5oi 749097 749393 4-93 250903 42 19 48964 87193 689873 3 75 940480 i-i8 4-93 250607 41 20 48989 87178 690098 3 75 940409 1-18 749689 4-93 25o3ii 40 21 49014 87164 9-690323 3" 74 9-940338 1-18 1.18 9-749985 4-93 io-25ooi5 39 22 49040 87150 690548 3 74 940267 750281 4-92 249719 38 23 49065 87136 690772 3 74 940196 1-18 750076 4-92 249424 37 24 49090 8712J 690996 3 ^ 940125 1-19 750872 4-92 249128 36 25 49116 87107 691220 3 73 940054 1-19 751167 4-92 248833 35 26 49141 87093 691444 3 73 939982 1-19 751462 4-92 248538 34 27 49166 87079 691668 3 73 93991 1 .-19 751757 4-92 248243 33 28 49192 87064 691892 3 72 939840 1-19 752o52 4-91 247948 32 29 49217 87050 692115 3 72 939768 1-19 752347 4-91 247653 31 80 49242 87036 692339 3 72 939697 1-19 752642 4-91 247358 30 31 49266 87021 9-692002 3 72 9-939620 I -19 9-752937 4-91 10-247063 29 82 49293 87007 692785 3 71 939554 1-19 753231 4-91 246769 28 38 49318 86993 693008 3 71 939482 1-19 753526 4-91 246474 27 84 49344 86978 693231 3 71 939410 1.19 753820 4-90 246180 26 85 49369 86964 693453 3 71 939339 I -19 754115 4-90 245885 25 86 49394 86949 693676 3 70 939267 1-20 754409 754703 4.90 245591 24 37 49419 86935 693898 3 70 939195 1-20 4-90 245297 23 38 49445 86921 694120 3 70 939123 1-20 754997 4-90 245oo3 22 89 49470 86906 694342 3 70 939052 1-20 755291 4.90 244709 24441 5 21 40 49495 86892 694064 3 69 938980 1-20 755585 4-89 20 41 49521 86878 9-694786 3 69 9-938908 1-20 9-755878 4-89 10-244122 19 42 49546 86863 695007 3 69 938836 1-20' 756172 4-89 243828 18 43 49571 86849 695229 3 69 938763 1-20, 756465 4-89 243535 17 44 49596 86834 695450 3 68 938691 I -20! 756759 4-89 243241 16 45 49622 86820 695671 3 68 93S619 1-20 757052 4-89 242948 15 46 49647 8680 5 695892 3 68 938547 1-20 757345 4-88 242655 14 47 49672 86791 696113 3 68 938475 1-20! 757638 4-88 242362 13 48 49697 86777 696334 3 67 938402 I-2I 757931 4-88 242069 12 49 49723 86762 696554 3 67 938330 1-21 758224 4-88 241776 11 50 49748 86748 696775 3- 67 938258 1.2I| 758517 4-88 241483 10 51 49773 86733 9-696995 3. 67 9-938180 I-2IJ 9-758810 ' 4-88 10-241190 9 52 4979B 86719 697210 3- 66 938n3 I-2Ii 759102 4-87 240898 8 53 49824 86704 697435 3 66 938040 1-21 759395 4-87 240605 7 54 49849 86690 697654 3 66 937967 I-2I 759687 4-87 24o3i3 6 55 49874 86675 697874 3 66 937895 I-2I 759979 4-87 240021 5 56 49899 86661 698094 3- 65 937822 I-2I 760272 4-87 239728 4 57 49924 86646 698313 3- 65 937749 I-2I 76o564 4-87 239436 3 58 499^0 86632 698532 3- 65 937676 I-2I 760856 4-86 239144 2 59 49975 86617 698751 3 65 937604 I-2l| 761 148 4-86 238852 1 60 5oooo 866o3 698970 3-64 937531 I-2Ij 761439 4-86 238561 N. COS. |n. sine. L. COS. D. 1" L. sine. L.cot. D.l' L.tang. ' 60^ 1 60 TRIGONOMETRICAL FUXCTIOXS. — 30^ Nat. FUKCTION3. Logarithmic Functions + 10. 1 |N.8liie.N.co8. L. sine. D. 1" L. COS. D.I" 1 Ltang. Dl." L. cot. 1 50000 866o3 9-698970 3.64 9.937531 1.21 1 9-761439 4-86 !io.23856i 60 1 ; 50025 86588 699189 3-64 937458 122 761731 4-86 238269 59 2 'j5oo5o 865i3 699407 3.64 937385 1-22 762023 4-86 237977 58 s .50076 86559 699626 3-64 937312 1-22 762314 4-86 237686 57 4 i5oioi 85544 699844 3-63 93723s 1-22 762606 4-85 237394 56 5 ;5oi26i 86530 700062 3.63 937165 1-22 762897 4-85 237103 55 6 !5oi5i [86515 700280 3-63 937092 1.22 763 1 88 4-85 236812 54 7 50176 865oi 70049S 3-63 937019 1-22 763479 4-85 236521 53 8 '5o2oi 86486 700716 3-63 936946 936872 1-22 763770 4-85 236230 52 9 1 50227 B6471 700933 3-62 1-22 764061 4-85 235930 235648 51 10 i 5o252 ' 86457 7oii5i 3.62 936799 1.22 764352 4.84 50 11 1 50277 1 86442 9-701363 3.62 9-936725 1.22 9-764643 4-84 10-235357 49 12 i 5o3o2 I 86427 701585 3-62 936652 1.23 764933 4-84 235067 43 13 1 50327 !864i3 701802 3-61 936578 1-23 765224 4-84 234776 47 14 1 5o352 1 86398 702019 3-61 9365o5 1.23 765514 4-84 234486 46 15 !5o377 86384 702236 3-61 936431 1-23 7658o5 4-84 234195 45 16 30403 86369 702452 3.61 936357 1-23 766095 4-84 233905 44 17 ! 50428 86354 702669 3-60 936284 1-23 766385 4-83 233613 43 18 50453 86340 702885 3-60 936210 1-23 766675 4-83 233325 42 19 50478 86325 7o3ioi 3-6o 936i36 1-23, 766965 4-83 233o35 41 '20 00303 863 10 703317 3-60 936062 1.23| 767255 4-83 232745 40 8y 21 i5o528:862o5 9.703533 3.59 9.935988 1-23. 9-767545 4-83 10-232455 22 5o553 862S1 703749 3.59 935914 1-23 767834 4-83 232166 88 23 50578 86266 703964 3.59 935840 1-23! 768124 4-82 231876 37 24 5o6o3 86251 704179 3.59 935766 1 -.24 768413 4-82 23 1 587 86 25 50628 86237 704395 it 935692 1-24' 768703 4-82 231297 85 26 5o654 86222 704610 935618 I-24j 768992 4-82 23 1008 84 27 50679 86207 704825 3.58 935543 I -24! 769281 ■ 4-82 230719 83 28 50704 86192 705040 3.58 935469 I -241 769570 4-82 23o43o 82 29 50729 86178 705254 3.58 935395 1-24; 769860 4-81 23oi4o 31 80 .50754 86 1 63 705469 3.57 935320 1-24: 770148 4-8i 229852 80 31 50779 86148 9-705683 3-57 9-935246 1-24 9.770437 4-81 10 -229563 29 82 30804 86 1 33 705898 3.57 935171 1-24! 770726 4-81 229274 228985 28 33 50829 861 ig 706112 3.57 935097 1-24 771015 4-81 27 34 5o854 86104 706326 3.56 935022 1-24 77i3o3 4-8i 22S697 228408 26 35 50879 86089 706539 706753 3-56 934948 1-24 771592 4-81 25 36 50904 86074 3.56 934873 1-24 77i8ao 4-80 228120 2-1 37 50929 86059 706967 3-56 934798 1-25 772168 4-8o 227832 23 38 50934 86043 707180 3.55 934723 1-25 772457 4-8o 227543 22 39 50979 86o3o 707393 3.55 934649 1.25' 772745 4-80 227255 21 40 5 1 004 8601 5 707606 3.55 934574 1.25; 773o33 4-8o 226967 20 41 51029 86000 9.707819 3-55 9-934499 I.25| 9.773321 4-8o 10.226679 "19" 42 5io54 859S5 708032 3-54 934424 1.25 773608 4-79 226392 IS 43 51079 85970 708245 3.54 934349 1-25 773806 774184 4-79 226104 17 44 5no4 85956 708458 3.34 934274 1-25 4-79 2258i6 16 45 51129 85941 70S670 3.54 934199 934123 1.25 774471 4-79 225529 15 46 5ii54 83926 708882 3-53 1.25 774759 4-79 225241 14 47 51179 85911 709094 3.53 934048 1.25 775046 4-79 224954 13 48 5 1 204 85896 709306 3.53 933973 1-25 775333 4-79 4-7» 224667 12 49 31229 85881 709518 3-53 933898 1-26 775621 224379 11 50 ! 51 5i254 85866 709730 3.53 933822 1.26J 775908 4-78 224092 10 31279 85851 9-709941 3.52 9.933747 1-26: 9.776105 7764S2 4-78 io.2238o5 9 52 5i3o4 85836 710153 3-52 933671 1.26' 4-78 2235i8 8 53 5i329 85821 710364 3.52 933596 .-26 776769 4-78 223231 7 54 5 1 354 858o6 710375 3-52 933520 1-26 777055 4-78 222945 6 55 5i379 85792 710786 3.51 933445 -26; 777342 4-78 222658 5 56 5 1 404 85777 710997 3-51 933369 93329! -26' 777628 4-77 222372 4 57 51429 85762 7 1 1 208 3.5i -26 777915 4-77 222085 8 58 51454 185747 711419 3.5i 933217 .26 778201 4-77 221799 2 59 51479 85732 711629 3.50 933141 • 26 778487 4-77 22l5l2 1 60 1 1 5i5o4 85717 711839 3.50 933066 .26, 778774 4-77 221226 N. COS. N. sine. L. COS. D.l" L.8ine. I Lcot. D.l" L. tang. ' 1 59° 1 TRIGONOMETRICAL FUNCTIONS.— 31°. 61 Nat. Functions. LoGARiTHsiic Functions + 10. 1 N.8ine.|N. COS. L. sine. D. 1" L. COS. |l w' L. tins. 1 D. 1 " L.cot. 5i5o4 135717 9.711839 3 5o 9.933066 'i •26 9-778774 4-77 10-221226 60 1 5i529 185702 7i2o5o 3 5o 932990 !i •27 779060 4 77 220940 59 2 5 1 554 85681 712260 3 5o 932914 ji •27 779346 4 76 220654 58 3 5i579 85672 712469 3 49 932838 I •27 779632 4 76 2 20368 57 4 5i6o4 85657 712679 3 49 932762 1 •27 779918 4 76 220082 56 5 51628 185642 712889 3 49 932685 I •27 780203 4 76 219797 55 6 5 1 653 85627 713098 3 49 932609 I •27 780489 4 76 2I95II 54 7 51678 85612 7i33o8 3 49 932533 I •27 780773 ! 4 76 219225 53 8 5 1 703 85597 713517 3 48 932457 I •27 781060 4 76 2 1 8940 52 9 51728 85582 713726 3 48 932380 I •27 781346 4 75 218654 51 10 51753 85567 713935 3 48 932304 I :11 781631 1 4 jL 218369 60 11 5.778 8555i 9-714144 3 48 9-932228 I •27 9.781916 1 4 75 10.218084 j 49 12 5i8o3 85536 714352 3 47 932i5i ji •27 782201 4 75 217799 1 48 13 51828 85521 714061 3 47 932075 I -28 782486 4 75 217514 [ 47 14 5i852 855o6 714769 3 47 931998 |i -28 782771 4 75 217229 1 46 15 51877 85491 714978 3 47 931921 |i -28 783o56 4 73 216944 45 16 51902 85476 7i5i86 3 47 931845 I .28 783341 4 75 • 216659 44 17 01927 85461 715394 3 46 931768 I -28 783626 4 74 . 216374 43 18 51952 85446 7i56o2 3 46 931691 I -28 783910 4 74 216090 42 19 51977 85431 715809 3 46 931614 I .28 784195 4 74 2i58o5 41 20 52002 85416 716017 3 46 931537 I .28 784479 4 74 21 552 1 40 21 52026 85401 9-716224 3 45 9-931460 II "^ 9.784164 4 74 10.215236 89 22 52o5i 85385 716432 3 45 ^31383 I -28 785048 4 74 214932 38 23 52076 85370 716639 3 45 93i3o6 li • 28; 785332 4 73 214668 37 24 52I0I 85355 716846 3 45 931229 ;I • 29 7856x6 4 73 214384 36 25 52126 85340 717053 3 45 93ii52 li •29 785900 4 73 214100 35 26 52i5i 85325 717259 3 44 931075 I •29 786184 4 73 2i38i6 1 34 1 27 52175 85310 717466 3 44 930998 1 • 29 786468 4 73 213532 33 28 52200 85294 717673 3 44 930921 ii •29; 786752 4 73 213248 32 29 52225 85279 717879 3 44 930843 1 •29 787036 4 73 2 1 2964 31 30 52250 85264 718085 3 3' 43 930766 I •29 787319 4 72 212681 30 2y 52275 85249 9-718291 43 9-930688 I •29: 9-787603 4 72 10.212397 82 52299 85234 718497 3 43 930611 I •29 787886 788170 4 72 212114 23 33 52324 852i8 718703 3 43 93o533 I •29' 4 72 2ii83o 27 84 52349 85203 718909 3 43 930456 I •29; 788453 4 72 2ii547 26 35 52374 l85iS8 719H4 3 42 930378 I •29 788736 4 72 211264 25 86 52399! 85173 719320 3 42 93o3oo I -3o' 789019 4 72 210981 1 24 37 52423 |85i57 719525 3 42 930223 I .30' 789302 4 71 210698 ! 23 88 52448 85i42 719730 3 42 930145 I •3o' 789585 1 4 71 2 1041 5 22 89 52473 85137 719935 3 41 930067 1 • 3o 789868 1 4 71 2IOl32 21 40 52498I85II2 720140 3 41 ^92_9?89_ I -3o 790i5i 9-790433 4 71 209849 20 41 52522 ' 85096 9-720345 3" 41 9-929911 I T3^ 4 71 10-209567 ! 19 1 42 52547 i85o8i 720549 3 41 929833 I -30 790716 1 4 71 209284 18 43 52572 ! 85o66 720754 3 40 929755 1 .30 790999 j 4 71 209001 17 44 52597 85o5i 720958 3 40 929677 I .30 791281 4 71 208719 16 45 52621 85o35 721162 3 40 929599 I .30 791563 4 70 208437 15 46 52646 85o2o 721366 3 40 929521 I .30 791846 4 70 2081 54 14 47 52671 85oo5 721570 3 40 929442 I .3o| 792128 j 4 70 207872 13 48 52696 84989 721774 3 39 929364 I .3/ 792410 4 70 207590 12 49 52720 84974 721978 3 39 929286 I •31 792692 4 70 207308 11 50 .52745 84959 722181 3 39 929207 I -31 792974 4 70 207026 10 51 52770 i 84943 9-722385 T 39" 9-929129 I .31 9-793206 i 4 70 10-206744 9 52 52794 ! 84928 722588 3 j§ 92oo5o I .31 793538 4 69 206462 8 53 52819:84913 52844 , 84897 722791 3 -3. 793819 4 69 206181 7 54 722994 3 38 .31 794101 4 69 205899 6 55 52869 i 84882 723197 3 38 928815 I .31 794383 4 69 2o56i7 5 56 52893 , 84866 723400 3 38 928736 I • 31 794664 4 69 205336 4 57 52918! 84851 7236o3 3 37 928657 I .31 794945 4 69 2o5o55 3 58 52943 ; 84836 7238o5 3 37 928578 1 -3.; 795227 4 60 204773 2 59 52967 84820 724007 3 37 928499 I -31 795508 4 6^ 204492 1 60 52992 1 84805 724210 3-37 928420 1 -3i 795789 4-68 204211 N. COS. N, sine. L. COS. D.l" L. sine. L. cot D.l" L. tang. ' 5§= 1 62 TRIGONOMETRICAL FUXCTIOXS. — 32^ Nat. Functions. Logarithmic Functioxs + 10. / |N.sine.|N.co3. L. sine. | D. 1" L. COS. D.l" L. tang. D.l" L.cot ; 52992 84805 9-724210 ' 3 •37 9-928420 1-32 9-795789 4-68 10. 2042 II 60 1 53017 j 84789 724412 ! 3 •37 928342 1-32 796070 4-68 203930 59 2 53o4i 84774 724614 ' 3 •36 928263 1-32 796351 4-68 203649 203368 53 3 53o66 84759 724816 : 3 •36 928183 1-32 796632 4.68 57 4 53091 84743 725017 ; 3 • 36 928104 !i-32 796913 4-68 203087 56 5 |53n5 84728 725219 ' 3 -36 928025 |1.32 797194 4-68 202806 55 6 53140 84712 725420 ; 3 .35 927946 I1.32 797475 4-68 202325 54 7 53164.84697 723622 ; 3 .35 927^67 I1.32 797755 4-68 202245 53 8 53189:84681 725823 ■ 3 .35^ 927787 1-32 798036 4-67 201964 1 52 9 153214 84666 726024 i 3 .35 927708 1-32 798316 4-67 2016S4 51 10 53238 ' 8465o 726225 i 3 35 ^ 927629 [1.32 798596 4-67 201404 50 11 53263 84635 9.726426 1 1 34 9.927^49 :i-.'^2, 9-798877 4-67 10-201123 49 12 '53288 84619 •726626 ' 3 34 927470 1.33' 799137 4-67 200843 4S 13 '53312 84604 726827 3 34 927390 1.33! 799437 4-67 200563 47 U 53337 84588 727027 3 34 927310. 1.33' 799717 4-67 200283 46 15 ' 53361 84573 727228 3 34 927231 1.33, 799997 4-66 2oooo3 45 16 '53386 84557 727428 3 33 927151 1-33 800277 4.66 199723 44 17 ,5341 1' 84542 727628 3 33 927071 ii-33; 800557 4-66 199443 43 18 1 53435 84526 727828 3 33 926991 11-33 8oo836 4-66 199164 42 19 ,53460 84011 728027 3 33 926011 I1.33 926831 !i.33, 801 1 16 4-66 198884 41 20 53484 S4495 728227. 3 33 801396 4-66 19S604 40 21 53509 84480 9 -.728427 3 32 9-926751 1-33 9-801675 4-66 10. 198325 39 22 i: 53534 84464 72S626 3 32 926671 1.33 801955 4-66 198045 S3 23 ; 53558 S4448 728825 3 32 926591 1.33' 802234 4-65 197766 87 24 ! 53583 84433 729024 3 32 9265ii 1-34: 8025 1 3 4-65 197487 86 25 53607 84417 729223 3 31 926431 1-34' 802792 4-65 197208 85 26 ;' 53632 84402 729422 3 3i 926351 1-34 803072 4-65 196928 84 27 ! 53656 84386 729621 3 3i 926270 1.34; 8o335i 4-65 196649 83 23 i:5368i 843TO 729820 3 3i 926190 1.34 8o363o 4-65 196370 32 29 1] 53705 , 84355 730018 3 3o 9261 10 1.34 803908 4-65 196092 31 80 !i 53730 , 84339 730216 3 3o 926029 !i.34 804187 4-65 195813 30 31 153754 84324 9-73o4i5 1 3 3o '■untti'di 9-804466 4.64 10-195534 29 82 I 53779 ' 84308 73o6i3 3 3o 804745 4.64 195255 28 S3 53804 '84292 730811 3 3o 9257S8 1 1. 34' 8o5o23 4.64 194977 27 U 53828 84277 731009 1 3 29 925707 ii.34: 8o53o2 4-64 194698 26 85 53853 84261 731206 3 29 925626 ;i.34' 8o558o 4-64 194420 25 36 53877 84245 •731404 f3 1602 3 29 925545 I1-35 8o5859 4.64 194141 24 G7 53902 84230 3 29 925465 ji.35 806137 4.64 193863 23 38 ,53926 84214 731799 3 It 925384 1.35: 806415 4-63 193585 22 39 ,53951 84.98 731996 3 9253o3 I1.35' 806693 4-63 193307 21 411 i 53975 84182 732193 3 28 925222 I1.35, 806971 4-63 193029 20 41 '54000 84167 9.732390 3 28 9-925141 .;1.35 9-807249 4-63 10-192751 19 42 54024 84i5i 732587 3 28 925060 1.35 807527 4-63 192473 18 43 54040 84135 54073 84120 732784 3 28 924079 1-35 807805 4-63 192195 17 44 732980 3 27 924897 1.35, 808083 4-63 191917 16 45 54097184104 733177 3 27 924816 1.35: 8o836i 4-63 191639 15 46 54122:84088 733373 3 27 924735 1.36; 8o8638 4-62 191362 14 i7 54146 ! 84072 733569 3. 27 924654 1.36' 808916 4-62 191084 18 43 54171 84057 733765 3- 27 924572 1-36; 809193 4-62 190807 12 49 54195 84041 733961 3. 26 924491 1-36 809471 4-62 190329 11 50 54220 ' 84025 734157 3. 26 924409 'i.36 809748 4-62 190252 10 51 54244 84009 9-734353 "3^ W 9-924328 17736 924246 ii.36 9.810025 4-62 10.189975 9 52 54260 83994 5429! 83978 734549 3. 26 8io3o2 4-62 189698 8 53 73^744 3. 25 924164 ;i.36 8io58o 4-62 189420 7 54 54317 83962 734939 3. 25 9240S3 1.36 810857 462 188866 6 55 54342 83946 73513! 3- 25 924001 ji.36 811134 4-61 5 56 54366 8393o 735330 3- 25 923919 i.36j 811410 4-61 188590 4 57 54391 83015 54415 83899 54440 83883 735525 3. 25 923837 1.36 811687 4-61 1883 1 3 3 58 735719 3. 24 923755 1.37 811964 4.61 i88o36 2 59 735914 3. 24 923673 '1.37 812241 4-61 187759 1 60 54464 83867 736109 3.24 923591 1.37 812517 4.61 187483 N. COS. N. sine. L. COS. D. 1 " L. sine. ji L.cot. 1 D. 1"| L.tang. / 57- 1 TRIGONOMETRICAL FUNCTIONS. — 33". 63 Nat. Functions. 1 Logarithmic Functions + 10. "' N.eine.l N. COS. L. sine. D. 1" L.COS. D.l"|i L. tang. 1 D. 1" L.cot. 54464 1 83867 9.736109 7363o3 3-24 9-923591 •37! 9-812517 4-6i 0-187488 60 1 54488 1 83851 3-24 923509 •37 812794 4-6i 187206 59 2 54513] 83835 786498 8-24 928427 ■ •87, 818070 4-6i 186930 53 8 54537 1 83819 786692 8-23 928845 ••371 818847 4-60 186653 57 4 54561 838o4 736886 3-23 928268 1-37 818628 4-6o 186877 56 5 54586 83788 787080 3-23 928181 1.37, 818899 814173 4-6o 186101 55 6 54610 83772 787274 3.28 928098 1-37' 4-60 185825 54 7 54635 1 83756 787467 8-23 928016 1. 871 814452 4-60 185548 53 8 54659 ; 83740 54683 .83724 787661 3-22 922933 1-37I 814728 4-6o 185272 52 9 ! 787855 3-22 92285i 1.87! ■8i5oo4 4-6o 184996 51 10! 54708 1 83708 788048 3-22 922768 1-88! i.38| 815279 4-6o 184721 50 11 1 54732 , 83692 9-738241 3-22 9-922686 9-815555 4-59 10-184445 49 12 1 54756 ! 83676 788434 3-22 922608 1-881 81 583 1 4-59 184169 48. 13 54781 1 83660 788627 3-21 922520 1-8^1 816107 4-59 188898 47 14! 54805 ! 83645 788820 8-21 922488 i-88i 816882 4-59 188618 46 15 1 54829 183629 789013 3-21 922855 1-381 8i6658 4-59 188342 45 16 ! 54854 836i3 789206 3-21 922272 1-38; 816933 4-59 188067 44 17 54878 83597 789898 3-21 922189 1-381 817209 4.59 182791 43 18 1 54902 83581 789090 3-20 922106 I •38! 817484 4-59 182516 42 19 54927 83565 789788 3-20 922028 1-38' 81^23? ^12 182241 41 20 21 54951 83549 789975 3-20 921940 1-881 1-39 4-58 181965 40 54975 83533 9-740167 3-20 9-921857 9-8i83io 4-58 10-18x690 89 22 54000 83517 740359 3-20 921774 1-89 8x8585 4-58 i8i4i5 33 23 55024 83501 74o55o 8-19 921691 818860 4-58 181140 37 24 55048 83485 740742 8-19 921607 1.89 819185 ^•^^ i8o865 36 25 55072 83469 83453 740934 3-. 9 921524 1.89 819410 4-58 180590 85 26 55097 741125 8-19 921441 1 -39! 819684 4-58 180816 34 27 •55I2I 83437 741816 3-19 921857 1.39I 819959 4-58 180041 33 23 ,55145 83421 74i5o8 8-i8 921274 1-391 820284 4-58 179766 82 29 ' 55169 83405 741699 3-i8 921190 1-89; 82o5o8 4.57 179492 31 30 55194 83389 741889 3-18 921107 1-39, 820-83 4-57 179217 80 81 ,155218 83373 9-742080 3-18 9-921028 1-89 9-82'io57 4-57 10-178948 29 32 ; 55242 83356 742271 3-18 XA 1-40, 82x882 4-57 178668 28 33 ! 55266 83340 742462 3-17 1-40 821606 4-57 178894 27 34 : 55291 83324 742652 8-17 920772 I -40 82x880 4.57 178120 26 35 !553i5 833o8 742842 8-17 920688 1.40, 822x54 4-57 177846 25 36 i 55339 83292 743o33 8-17 920604 1-40 822429 4-57 177571 24 37 . 55363 83276 748228 3.17 920520 ..40 822708 ^1z 177297 23 3S , 55388 83260 748418 3-i6 920486 1-40 822977 4-56 177028 22 89 55412 83244 748602 8-i6 920352 I -40! 828250 4-56 X767DO 21 40 1 55436 83228 748792 8-i6 920268 1-40 823524 4-56 176476 20 '~i\ 1 55460 83212 9.748982 8-i6 9-920184 1.40 9-82-8798 4-56 10-176202 19 42 1 55484 83i95 744171 3.)6 920099 1-40 824072 4-56 175Q28 18 43 55509 83x79 744861 3-i5 920013 1-40 824345 4-56 1756D5 17 44 55533 83i63 744550 3-15 919981 1-4., 824619 4-56 175381 16 45 55557 83 1 47 744789 744928 8-i5 919846 1-41 8248q3 4-56 175107 15 46 55581 83i3i 8-i5 919762 1-41 825x66 4-56 174884 14 47 556o5 83ii5 745117 3-i5 919677 1-41 825439 4-55 174561 13 48 55630 83098 745806 8-14 919598 !i-4i 825718 4-55 174287 12 49 55654 83o82 745494 3.14 919508. 1-41 825986 4-55 I74bi4 11 90 55678 83o66 745688 3-14 919424 1-41 826259 4-55 178741 10 '51 jl 55702 83o5o 9-745871 3-14 9-919339 |i-4i 9-826532 4^55 10-178468 9' 52 55726 83o34 746039 746248 3.14 919254 |i-4i 826805 4-55 178195 8 &3 55750 83oi7 3-i3 919169 1-41 827078 4-55 172922 7 64 55775 83001 746486 3-13 919085 I-4I 827351 4-55 172649 6 55 f^ 1 82985 746624 3-i3 919000 I-4I 827624 4-55 172876 5 56 82969 746812 8-i3 918915 1-42 827897 4-54 172103 4 57 !j 55847 82953 746999 747187 3.i8 918880 1-42 828170 4-54 171880 8 58 55871 82936 3-12 918745 1-42 828442 4-54 171558 2 59 l| 55895 82920 747374 3-12 918659 1-42 828715 4-54 171285 1 60 :| 55919 82904 747562 3-12 918574 ;i-42 828987 4-54 171018 |i N. COS. 'n. sine L. COS. 1 D. \" L. sine. 1 L.cot , D. 1" L. tang. ' 56° 1 64 TRIGONOMETRICAL FCXCTIONS. — 34° Nat. Functions. Logarithmic Functions + 10. . 1 N.sine.^N. COS. L. sine. D. 1" L. COS. D.l" L. tang. D.l" L. cot 55919 82904 9-747562 3 12 9-918574 I 42 9-828987 4 54 10-171013 60 1 55943 82887 747749 3 12 918489 I 42 829260 4 54 170740 59 2 55968 82871 747936 3 12 918404 1 42 829532 4 54 170468 58 3 55992 82855 748123 3 II 9i83i8 I 42 829805 4 54 170195 57 4 |56oi6 82839 748310 3 II 918233 1 42 830077 4 54 169923 56 5 56040 82822 7484Q7 3 II 918147 I 42 83o349 4 53 169651 55 6 56o64 82806 7486S3 3 II 918062 I 42 83o62i 4 53 169379 54 7 1 56o88 82790 748870 3 II 917076 1 917891 I 43 830893 4 53 169107 53 8 56II2 82773 749056 3 10 43 83ii65 4 53 168835 52 9 56 1 36 82757 749243 3 lO 917805 I 43 8I1437 4 53 168563 51 10 11 56 160 82741 749429 3 10 917719 I 43 831709 4 53 168291 50 56 1 84 82724" 9-74o6i5 3 10 9-917634 I T3 9-83i98i 4 53 iO'i68oi9 49 12 56?o8 8270S 749801 3 10 917548 I 43 832253 4 53 167747 48 13 56232 82692 749987 3 09 917462 I 43 832525 4 53 167475 47 14 56256 82675 750172 3 09 917376 I 43 832796 4 53 167204 46 15 56280 82659 82643 75o358 3 09 917290 I 43' 833o68 4 52 166932 45 16 563o5 750543 3 09 917204 I 43 833339 4 52 16666 I 44 17 56329 82626 750729 3 S 917118 I 44 83361 1 4 52 166389 166118 43 18 56353 82610 750914 3 917032 I 44 833882 4 52 42 19 56377 82593 751099 3 08 916946 I 44 834154 4 52 165846 41 20 56401 82577 751284 3 08 916859 I 44 44 834425 4 52 165575 40 21 56425 82561 9-751469 3 08 9-916773 I 9.834696 4 "52 io.i653o4 3y 22 56449 82544 75i654 3 08 916687 . 44 834967 4 52 i65o33 38 23 56473 82528 75i839 3 08 916600 1 44 835238 4 52 164762 37 24 56497 825ii 752023 3 07 9i65i4 I 44 835509 4 52 164491 36 25 56521 82495 752208 3 07 916427 I 44 335780 4 5i 164220 35 26 56545 82478 752392 3 07 916341 I 44 836o5i 4 5i 163949 163678 34 27 56569 82462 ■ 752576 3 07 916254 I 44 836322 4 5i 33 28 56593 82446 752760 3 07 916167 I 45 836593 4 5i 163407 32 29 56617 82429 752944 3 06 916081 I 45 836864 4 5i i63i36 31 30 31 56641 82413 753128 3 06 915994 I 45 837134 4 5i 162866 30 56665 82396 9-753312 3 06 9-915907 I % 9-837405 4 5i 10-162595 29 32 56689 82380 753495 3 06 915820 I 837675 4 5i 162325 28 83 56713 82363 753679 3 06 915733 1 45 837946 4 5i 162054 27 34 56736 82347 753862 3 o5 915646 I 45 838216 4 5i 161784 26 35 56760 82330 754046 3 o5 915559 I 45 838487 4 5o i6i5i3 25 36 56784 82314 75ir229 3 o5 915472 I 45 838757 4 5o 161243 24 87 568o8 82297 754412 3 o5 915385 I 45 839027 4 5o 160973 23 38 56832 82281 754595 3 o5 915297 1 45 839297 4 5o 160703 22 89 56856 82264 754778 3 04 9i52io I 45 839568 '4 5o 160432 21 40 5688o 82248 754960 3 04 9i5i23 1 46 839838 4 5o 160162 20 41 56904 8223/ 9-755143 3 04 9 -915035 1 46 9-840108 4 5o 10-159892 19 42 56928 82214 755326 3 04 914948 I 46 840378 4 5o i5q622 IS 43 56952 82198 82181 755508 3 04 914860 I 46 840647 4 5o 159353 17 44 56976 755690 3 04 914773 I 46 840917 4 49 159083 16 45 57000 82165 755872 3 o3 914685 1 46 841187 4 49 I6y8i3 15 46 57024 82148 756o54 3 o3 914598 I 46 841457 4 49 158543 14 i7 57047 82132 756236 3 o3 914510 I 46 841726 4 49 158274 13 43 57071 82115 756418 3 o3 914422 I 46 841996 4 49 1 58004 12 49 57.595 82098 756600 3 o3 •914334 I 46 842266 4 49 157734 11 50 071 19 82082 756782 3 02 914246 I il 842535 4 49 157465 10 ' T 51 157143 82065 9-756963 3" 02 9-914158 I 47 9.842805 4' 49 10-157195 52 [5ti67 82048 757144 3 02 914070 I 47 843074 4 49 156926 8 53 157191 82032 757326 3 02 913982 1 47 843343 4 49 156657 156388 7 54 : 572.5 82015 757507 3 02 913894 I 47 843612 4 49 6 55 i 57238 81999 757688 3 01 9i38o6 1 47 843882 4 48 156118 5 56 57262 81982 757869 3 01 913718 I 47 8441 5i 4 48 155849 4 57 57286 81965 758o5o 3 01 9i363o I 47 844420 4 48 1 55580 8 58 :573jo 81949 758230 3 01 913541 I 47 844689 4 48 i553ii 2 59 57334 81932 758411 3 01 913453 I 47 844958 4 48 155042 1 60 57358 81915 758591 3 01 913365 ii 47 845227 4.48 154773 11. COS. N. sine. L. COS. D. 1" L. sine. | L. cot. ! D. 1" L. tang. ' 55° TRIGONOMETRICAL FUNCTIONS. — 55°. 65 Nat. Functions. Logarithmic Functions + 10. j 1 •N.sine. N. COS. L. sine. | D. 1" L. COS. D.l" L. tang. D.l" L. cot. 1 1 57358 81915 9.758591 3-01 9.913355 1-47 i 9-845227 4.48 10.154773 60 1 57381 81899 758772 3.00 913276 913187 |i-47 845496 4-48 1 54504 59 2 574o5 81882 758952 3-00 1-48 845764 4-48 154236 53 3 ^%l 8 1 865 759132 3-00 913099 1-48 846033 1 4-48 153967 57 4 81848 759312 3-00 9i3oio '1.48 846302 i 4-48 153698 56 5 ^lAll 8i832 759492 3-00 912922 1-48 846570 1 4-47 1 53430 55 6 57501 8i8i5 759672 2.99 912833 1.48 846839 4-47 i53i6i 54 7 57524 81798 759852 2-99 912744 1.48 847107 4-47 152893 53 8 57548 81782 760031 2-99 912655 1.48 847376 4-47 152624 52 9 57572 81765 7602 I I 2-99 912566 1.48 847644 4-47 152356 51 10 575q6 81748 760390 2-99 912477 1.48 8479 '3 4-47 152087 50 11 57619 81731 9.760569 2.98 9.912388 i~48 9.848181 4-47 io-i5i8i9 4'J 12 57643 81714 760748 2.98 912299 1-49 848449 4-47 i5i53i 4S 13 57667 81698 81681 760927 2-98 912210 1-4Q 848717 4-47 i5i283 47 14 57691 761106 2.98 ■912121 1-49 848986 4-47 i5ioi4 46 15 57715 81664 761285 2-98 9i2o3x 1-49 849254 4-47 1 50746 45 16 57738 81647 761464 2-98 9IIQ42 911853 1.49 849522 4-47 i5o4-S 44 17 57762 8i63i 761642 2.97 1.49 849790 4-46 l502I0 40 18 57786 81614 76:821 2-97 911763 1.49 85oo58 4.46 149942 42 19 57810 81597 761999 2-97 91 1674 1-49 85o325 4-46 149673 41 20 57833 8i58o 762177 2-97 9II584 1-49 85o593 4.46 149407 4" 21 57857 8 1 563 9.762356 2-97 9-911495 1-49 9.830861 4-46 10.149139 3y 22 57881 81546 762534 2.96 911405 1-49 851129 4-46 I48S71 38 23 57904 8i53o 762712 2.96 9ii3i5 I -So 85i396 4-46 148604 37 24 57928 8i5i3 762889 2-96 911226 1-50 85 1 664 4-46 148336 36 25 57952 81496 763067 2.96 9ili36 i-5o 85i93i 4-46 14S069 35 26 57976 81479 763245 2.96 911046 i-5o 852199 4-46 147801 34 27 57999 81462 763422 2.96 910956 I -50 852466 4.46 147534 S3 28 58023 81445 763600 2-95 910866 i-5o 852733 4-43 147267 32 29 58047 81428 763777 2.95 910776 910686 1-50 853001 4-45 146999 31 30 31 58070 81412 763954 2.95 1-50 853268 4-45 146732 So 58094 81395 9.764131 2.95 9-910596 1-50 9-853535 4-45 10.146465 2y 32 58ii8 81378 764308 2.95 9io5o6 i-5o, 853802 4-45 146198 2S 33 58i4i 8i36i 764485 2-94 910415 1-50 854069 4-45 145931 27 34 58i65 81344 764662 2.94 910825 i.5i 854336 4-45 145664 26 35 58189 8i327 764838 2.94 910235 i-5i 8546o3 4-45 145397 i45i3o 25 36 58212 8i3io 765oi5 2-94 910144 i.5i 854870 4-45 24 37 58236 81293 765191 2.94 910004 i-5i: 855137 4-45 144S63 23 38 58260 81276 765367 2-94 909963 i.5ii 855404 4-43 144396 22 39 58283 81259 765544 2.93 909873 909782 1-51 855671 4-44 144329 21 40 58307 81242 765720 2.93 i.5i! 855933 4-44 144062 20 41 58330 81225 9.765896 2.93 9-909691 1-51 9-856204 4.41 10.143796 ly 42 58354 81208 766072 2.93 909601 i.5i 856471 4-44 143329 18 43 58378 81191 766247 2.93 909510 i-5i. 856737 4.44 143263 17 44 58401 81174 766423 2.93 909410 909328 i.5i 837004 4.44 142996 16 45 58423 81I57 766598 2-92 1-52 857270 4-44 142730 15 46 58449 81140 766774 2.02 909237 1.52 857537 4.44 142463 14 47 58472 81123 766949 2.92 909146 1-52 857803 4.44 142197 13 48 58496 81106 767124 2.92 909055 908964 1-52 858069 4.44 141931 12 49 58519 58543 81089 767300 2-92 1.52i 858336 4.44 141664 11 r50 81072 767475 2-91 908S73 1-52, 858602 4.43 141398 10 51 58567 8io55 9.767649 2.91 9 •908781 1.52! 9-858868 4-43 io.i4ii32 9 52 58590 8io38 767824 2.91 908690 1-52 859134 4.43 140866 8 53 586i4 81021 ?a??? 2.91 908599 ..52| 859400 4.43 140600 7 54 58637 81004 2-91 908507 1.52' 839666 4-43 140334 6 55 58661 80987 768348 2-90 908416 1.53 859932 4.43 140068 5 56 58684 80970 768522 2-90 908324 1-53: 860198 4-43 139802 4 57 58708 80953 768697 2.90 908233 1-53; 860464 4.43 189536 8 53 58731 80936 768871 2.90 908141 1-53, 860730 4-43 189270 2 59 58755 80919 769045 2-90 908049 1.53; 860995 4.43 189005 188789 1 60 58779 80902 769219 2.90 907958 1.5:5 861261 4-43 (t N. COS. N.sine. L. COS. D.l" L. sine. L.cot D.l" L. tang. 54^ 66 TRIGOXOMETIilCAL TUXCTIOXS. — 36°. Nat. Functions. Logarithmic Functions + 10. 1 N.SlTlft N. COS. L. sine. D. 1" L. COS. ^! L. tang. Dl." L.cot 58779 i 80902 9.769219 2 §9 9.907958 .53 g. 861261 4-43 10.138739 60 1 58802 ; 80885 769393 2 907866 • 53' 861527 •43 188473 59 2 58826 1 80867 769566 2 89 907774 .53 861792 42 i382o3 68 3 58849 5S873 8o85o 769740 2 89 907682 •53: 862038 42 13794a 57 4 8o833 769913 2 89 907590 .53 862323 42 137677 56 5 58896 80816 770087 2 1 907498 •53i 862589 42 137411 55 6 58920 80799 770260 2 907406 .53: 862854 42 137146 54 7 58943 80782 770433 2 88 907314 -54 863 1 19 42 i3688i 53 8 58967 80765 770606 2 88 907222 •54 863385 42 i365i5 52 9 58990 8s7.^8 770779 2 88 907129 .54 863650 42 i3635o 51 10 59014 80730 770952 2 88 907037 -54 863915 42 i36o85 50 -K 11 59037 j 80713 9.771123 2 "SS' 9.906945 •54 9^864180 1 4 42 10.135820 49 12 59061^80696 771298 2 87 906852 .54. 864445 4 42 135555 43 13 59084 80679 771470 2 87 906769 .54: 864710 4 42 135290 47 14 59108,80662 771643 2 87 906667 -54, 864975 4 41 135020 46 15 59131 80644 771815 2 87 906575 1 •54 865240 4 41 134760 45 16 59154180627 771987 2 87 906482 -54i 8655o5 4 41 134495 44 17 59178 80610 772159 2 87 906389 1 .55, 865770 4 41 l3423o 43 X IS 59201 80593 772331 2 86 906296 •55 866035 4 41 133965 42 19 59225 80576 7725o3 2 86 906204 1 .55i 866300 4 41 133700 41 20 21 59248 8o558 772675 2 86 906111 1 • 55: 866564 4 41 133436 40 59272 j8o54i 9.772847 2 W 9.906018 I "T55i 9.866829 4 41 10.133171 3 J 22 59295 1 8o52< 773018 2 86 905925 1 • 551 867094 4 41 132906 33 23 59318 8o5o7 773190 2 86 9o5S32 1 •55 867358 4 41 132642 37 24 59342 80489 773361 2 85 900739 I .55 867623 4 41 132377 36 / 25 5o365 80472 773533 2 85 905645 I • 55! 867887 4 41 i32ii3 35 26 59389 80455 773704 2 85 905552 I • 55, 868i52 4 40 i3i848 34 27 59412 8o438 773875 2 85 905459 I • 55 868416 4 40 i3i584 33 23 59436 80420 774046 2- 85 9o5366 I • 56: 868680 4 40 l3i32o 32 29 59459 80403 774217 2- 85 905272 I • 56i 868945 4 40 i3io55 31 SO 5?482 8o386 774388 2 84 905 179 I • 56| 869209 4 40 130791 30 31 { 59506 8o368 9-774558 2- 84 9.900085 I 1^' 9-869473 4 40 io.i3o527 '^y 32 59529 8o35i 774729 2- 84 904992 I • 56> 869737 4 40 i3o263 23 33 59552 80334 774899 2 84 904898 I -56; 870001 4 40 129909 27 34 59576 8o3i6 775070 2 84 904804 I .56 870265 4 40 129735 26 35 59599 80299 775240 2 84 9047 1 1 I .56 870529 4 40 129471 25 36 59622 80282 775410 2 83 904617 I .56 870793 4 40 129207 21 37 59646 80264 775580 2 83 904523 I .56 871057 4 40 128943 23 38 59669 39 59693 80247 775750 2- 83 904429 I •57 871321 4 40 128679 22 8o23o 775920 2 83 904335 I •57 871585 4 40 128415 21 40 1 59716 {80212 776090 2 83 904241 I •57; 871849 4 39 I28i5i 20 41 1 59739 1 80195 9.776259 2 83 9-904147 I -57' 9.872112 4 39 10.1278SS 19 42 ' 59763 80178 776429 2 82 904053 I -57' 872376 4 39 127624 13 43 i 59786 80160 776598 2 82 903959 1 •57: 872640 4 39 127360 17 44 ; 59809 '80143 776768 2 82 903864 • •57 872903 4 39 127097 16 45 t 59832 80125 776937 2 82 903770 I •57 873T67 4 39 126833 15 46 59856; 80108 777106 2 82 903676 I -57 873430 4 39 126570 14 47 1 59879 j 80091 777275 2 81 9o358i 1 •57 873694 4 39 i263o6 13 4S , 59902 : 80073 777444 2 81 903487 I •57 873957 4 39 126043 12 49 1 59926 ; 8oo56 777613 2 81 903392 1 .58 874220 4 39 123780 11 50 51 59949 ' 8oo38 777781 2 81 903 29S I •58 874484 4 39 I255i6 1 10 1 59972 80021 9.777930 2 JT 9-9o32o3 I • 08 9-874747 4' '39 10. 125203 9 52 599q5 1 8ooo3 7781 19 2 81 903 1 o3 I • 58 875010 4 39 124990 8 53 ,60019! 799^^ 778287 2 80 9o3oi4 I • 58 875273 4 38 124727 7 54 60042 ! 79968 778455 2 80 902919 I • 58 875536 4 38 124464 6 55 6oo65 7995 1 778624 2 80 902824 1 • 58 875800 4 38 124200 5 56 i 60089 79934 778792 2 80 902729 I •58 876063 4 38 123937 4 57 1 60112 i 79916 778960 2 80 992634 I • 58 876326 4 38 123674 3 53 60135:79899 59 601 58 79881 779128 2 80 902539 I .59 876589 4 38 123411 2 779295 2 79 902444 I .39 876851 4 38 I23i49 1 60 ^ 60182 j 79864 779463 2-79 902349 • • 59, 877"4 4-38 122886 N. COS. N. sine. L. COS. D. 1" L. sine. ' L. cot. 1 D. 1" L. tang. ' 1 53^ 1 trigono:metrical functions. — 37^. e^l Nat. Functions. Logarithmic Functions + V). j ' JN.sine. N. COS. L. sine. D. 1" L. COS. E M" L. tang. D. 1" L. cot. ''60182 ^867 9-779463 2-79 9-902349 1 i'> 9-877114 4.38 10-122886 60 1 li 6o2o5 79846 779631 2.79 902253 I -59 877377 4-38 122623 oil 2 60228 79829 779798 2-79 9021 58 I •^9 877640 4-38 122360 58 S 6o25i 79811 779966 2-79 902063 I •59, 877903 4-38 122097 57 4 60274 79793 780133 2.79 901967 1 901872 I 19, 878165 4-38 12i83d 5*^ 5 60298 79776 780300 2-78 .59' 878428 4.38 121572 oo 6 ,6o32i 79758 780467 2.78 901776 I .59 87S691 4-38 i2i3o9 54 7 6o344 79741 780634 2.78 901681 |i .59 878953 4-37 121047 53 8 60367 79723 780801 2-78 901685 |i 19. 879216 4-37 120784 52 9 60390 79706 780968 2.78 901490 |i .59' 879478 4-37 120322 51 10 60414 79688 781134 2.78 901394 |i .60, 879741 4-37 120259 .=^.(1 1 11 , 60437 79671 9.781301 2-77 9.90B298 ji .60 9.880003 4-37 10-119997 12 ' 60460 79553 781468 2-77 901202 I .60: 880265 4-37 II9735 13 ; 60483 79635 781634 2.77 901106 I .60 88o528 4-37 119472 14 i 6o3o6 79618 781800 2-77 90101.0 ji .60' 880790 4-37 II921O 15 i6o529 79600 781966 2.77 900914 I -60' 88io52 4-37 16 :6o553 79583 782132 2.77 9008 1 8 I •60' 88i3i4 4-37 1 1 8686 44 17 60676 79565 782298 2.76 900722 I •60 881576 4-37 118424 4:3 18 : 60599 79547 782464 2.76 900626 I .60 881839 4-37 118161 4-2 19 ; 60622 79530 782630 2.76 000529 I •60 882101 4-37 117899 41 117637 40 20 : 60645 79512 782796 2.76 900433 I •6ij ~6\ 882363 4-36 21 i 60668 7Q4Q4 9-782961 2-76 9.900337 I 9-882625 4-36 10-117375 31" 22 1 60691 79477 783127 2.76 900240 |i -61 882887 883148 4-36 II7113 8S 23 60714 79459 783292 2.75 900144 [i •61 4-36 116852 87 24 '60738 79441 783458 2.75 T°% I -61 883410 4-36 I 16590 86 25 60761 79424 783623 2.75 .61, 883672 4-36 II6328 85 26 ' 60784 79406 783788 2.75 899854 I .61: 883934 4-36 116066 34 27 60807 79388 783953 2.75 899757 I '^A 884196 4-36 ii58o4 S3 28 1 6o83o 79371 784118 2.75 899660 I -611 884457 4-36 115543 82 29 , 6o853 79353 784282 2-74 899564 I .61 884719 4-36 ii528i 81 30 31 ! 60876 79335 784447 2-74 899467 I .62 884980 4-36 ll5020 80 60899 79318 9-784612 2-74 9-899370 I .62 9-885242 4-36 10-114758 21. 32 160922 79300 784776 2-74 899273 I .62 8855o3 4-36 114497 2S 83 60945 79282 784941 2-74 899176 I .62 885765 4-36 114235 27 34 60968 79264 785io5 2-74 .62 886026 4-36 113974 26 85 60991 79247 785269 2.73 .62| 886288 4-36 113712 25 86 ■6ioi5 79229 785433 2.73 898884 I -62! 886549 4-35 ii345i 24 87 |6io38 79211 785597 2.73 898787 I .62i 886810 4-35 1 13190 1 23 88 j6io6i 79193 785761 2-73 898689 I .62 887072 4-35 112928 j 22 89 1 61084 79176 785925 2.73 898592 I -62 887333 4-35 112667 1 21 40 '161107 I 79i58 786089 2.73 898494 I -63 887594 4-35 112406 ' 20 41 6ii3o 79140 9-786252 2.72 9-898397 I -63 9-887855 4.35 10-112145 VJ 42 J6ii53 79122 786416 2.72 898299 1 •63) 888116 4.35 111S84 ' IS 43 61176 79105 786579 2-72 898202 I .63' ^^■77 4.35 111623 17 44 1161199 79087 786742 2-72 89S104 I .63 888639 4.35 iii36i 1<) 45 61222 79069 786906 2.72 898006 |i • 63 888900 4-35 1 1 1 1 00 15 46 ^61245 7905 1 787069 2-72 897908 I 897810 I .63 889160 4.35 110840 14 47 1 61268 79033 787232 2.71 .63, 889421 4.35 110579 iio3i8 13 48 ,61291 79016 787395 2.71 897712 I .63 889682 4-35 12 .49 ;6i3i4 78998 787557 2.71 897614 ji .63 889943 4-35 110057 11 50 161337 78980 787720 2-71 897516 I -63 890204 4-34 109-96 10 51 i!6i36o 178962 9.787883 2.71 9.897418 I -64 9.890465 4.34 10-109J35 y 52 6i383 78944 788045 2.71 897320 I -64 890725 4-34 109275 s 53 61406 78026 788208 2.71 897222 I .64 890986 4-34 109014 10S753 7 54 61429 78908 788370 2.70 897123 II .64 891247 4-34 6 55 6i45i 78891 788532 2-70 897025 ;i .64 891507 4-34 108493 5 56 [61474 78873 788694 2-70 896926 ,1 .64 891768 4-34 108232 4 57 161497 78855 788856 2.70 896828 I •64 892028 4-34 107972 3 58 6i52o 78837 789018 2-70 896729 I -64 892289 4-34 107711 2 59 6i543 78819 789180 2-70 896631 'i •64 892549 4-34 107431 1 60 61 566 78801 789342 2.69 896532 [i .64 892810 4-34 107190 jN. COS. N. sine L. COS. D. 1" L. sine. } L. cot. D. 1" L. tang. ' 52° - 1 68 TEIGOXOMETKICAL FUXCTIOXS. — 38°. Nat. Functions. Logarithmic Functions + 10, 1 N.sine.! N. cos. L. sine. D.l" L. COS. D.l" L. tang. D.l" Loot 6i566i 78801 9-789342 2.69 9-896532 1-64 9-892810 4-34 10-107190 GO 1 61689 1 78783 789504 2.69 896433 1-65 893070 4-34 106930 59 2 61612 1 78765 789665 2.69 896335 1-65 i 893331 4-34 106669 5S 3 ,61635178747 789827 2.69 896236 1-65 893391 4-34 1 06409 57 4 61658178729 789988 2-69 896137 1-65 893831 4-34 106149 56 5 161681 78711 790 '49 2-69 896038 1-65 8941 1 1 4-34 105889 55 6 61704 78694 7903 10 2-68 895939 1-65 1 894371 4-34 105629 54 7 61726178676 790471 2-68 895840 1-65 1 894632 4-33 105368 63 8 61749 1 78658 790632 2-68 895741 1-65 1 894892 4-33 io5io8 52 9 1.61772 78640 790793 2-68 895641 1.65 895152 4-33 104848 51 10 61795 78622 790954 2-68 895542 1-65 1-66 I 895412 <-33 104588 50 11 61818 78604 9.791115 2-68 9-895443 : 9-895672 4-33 10.104328 49 12 6 I 84 I 178586 7912-5 2-67 895343 1-66 895932 4-33 1 04068 48 13 61864 S 78568 791436 2-67 893244 1-66 896192 4-33 io38o3 47 14 61887178550 791596 2-67 895145 1-66 896452 4-33 103548 46 15 61909 78532 791757 2-67 895045 1-66 896712 4-33 103288 45 16 61932 78514 791917 2.67 894945 1-66 895971 4-33 io3o29 44 17 61933 78496 792077 2.67 894846 1-66 897231 4-33 102769 43 IS 61978 78478 792237 2-66 894746 1-66 897491 4-33 102509 42 I'J 62001 78460 792397 2-66 894646 1-66 897751 4-33 102249 41 20 62024 78442 792557 2-66 894546 1-66 898010 4-33 101990 40 ::l 62046" 78424 9-792716 2-66 9-894446 1.67 1 9-89S270 4-33 10- ion3o 39 li:: 62069 78405 792876 2-66 894346 1.67 898530 4-33 101470 3S 23 62092 78387 793o35 2-66 894246 1.67 89S789 4-33 101211 37 24 62115 78369 793195 2-65 894146 1-67 II& 4-32 100951 36 25 62138 1 78351 1 793354 2-65 894046 1-67 4-32 100692 100432 35 26 62160 78333 7935i4 2-65 893946 1-67 899568 4-32 34 27 62183 7831 5 793673 2-65 893846 1-67 899827 4-32 100173 33 2S 62206 78297 793832 2-65 893745 1-67 900086 4-32 099914 32 2& 62229 78279 793991 2-65 893645 1-67 900346 4-32 099654 31 oO 6225l 78261 794 1 5o 2-64 893544 1-67 900605 4.32 099395 30 ol 62274178243 9 -794308 2-64 9-893444 1-68 9-900864 4-32 10-099136 29 32 62297 78225 794467 2-64 893343 1-68 901124 4-32 098876 23 33 62320 1 78206 794626 2-64 893243 1-68 901383 4-32 098617 27 34 62342 1 78188 794784 2-64 893142 1-68 901642 4-32 098358 26 35 62365 '78170 794942 2-64 893041 1-68 901901 4-32 098099 25 36 62388 78152 795101 2-64 892940 1-68 902160 4-32 097840 24 37 62411 78134 795259 2-63 • 892839 1-68 902419 4-32 097581 23 3S 62433 781 16 795417 2-63 892739 1-68 902679 902938 4-32 097321 22 39 62456 i 78098 795575 2-63 892638 1-68 4.32 097062 21 40 62479 78079 795733 2-63 892536 1-68 903197 4-3i 096803 20 41 62502 178061 9-795891 2-63 9-892435 1-69; 9 -903455 • 4-3i 10-096545 19 42 62524 i 78043 796049 2-63 892334 1-69 903714 4.31 096286 IS 43 62547 '< 78025 796206 2-63 892233 1-69 903973 4-3i 096027 17 44 62570 1 78007 796364 2-62 892132 1-69 904232 4-3i 095768 16 45 62592 J7798S 796321 2-62 892030 1-69 904491 4-31 095509 15 46 62615 , 77970 796679 2-62 891929 1-69I 904730 4-31 095250 14 47 62638 77952 796836 2-62 891827 1.691 1-69 903008 4-31 094992 13 48 62660 1 77934 796993 2-62 891726 905267 4-3i 094733 12 49 62683 1 77916 62706 i 77897 797 i5o 2-61 891624 1-69! 905526 4-3i 094474 11 50 797307 2-6l 891523 1-70, 905784 4-3i 094216 10 IT" 62728 77879 9-797464 2-61 9-891421 I -70! 9.906043 4-3i 1 10-093937 9 52 62731 77861 797621 2-61 891319 1-70, 9o63o2 4-3i 093698 8 53 62774 77843 797777 2-61 891217 1.70 906360 4-3i 093440 7 54 62796 77824 797934 2.61 891115 1-70, 906819 4-3i 093181 6 5d 62819 77806 798091 2-61 891013 i-7o| 907077 4-31 092923 5 56 62842 77788 798247 2-61 890911 1-70 907336 4.31 092664 4 57 62864 1 77769 798403 2-60 890809 1.70 907594 4-31 092406 3 58 62887 77751 798560 2 -60 890707 1-70, 907852 4.31 092148 2 59 62909 77733 798716 2.60 890605 1-70 908111 4-3o 091889 1 60 62932 77715 798872 2-60 89o5o3 1.70 908369 4.30 091631 N. co8.,N.8ineJ L. COS. D.l" L. sine. ! L. cot. D.l" L. tang. ' 51° 1 TRIGONOMETRICAL F CNCTIONS. — 39=" 69 Nat. Functions. Logarithmic Functions + 10. 1 ' N.sine.|N. COS. L. sine. D. 1" L. COS. I ).l" L. tang. Dl." L. cot. 60 ■ 62932 77715 9.798872 2-60 9-890503 I .70 9-908369 4-3o 10-091631 1 62955 77696 799028 2 -60 890400 1 •71 908628 4 3o 091372 59 2 62977 77678 799184 2 -60 890298 I •7' 908886 4 3o 091 1 14 58 3 63ooo 77660 799339 2.59 890195 I •71 909144 4 3o .090856 57 4 63022 77641 799495 2.59 890093 I •7' 909402 4 3o 090598 56 5 63045 77623 799631 2-59 '&i ; •71 909660 4 3o 090340 55 6 63o68 77605 799806 2.59 •7' 909918 4 3o 090082 54 7 63090 77586 799962 2.59 889785 I •71 910177 4 3o 089823 53 8 63ii3 77568 800117 2.59 889682 I •71 910435 4 3o 089365 52 9 63i35 77550 800272 2-58 889579 1 •7" 9106^3 910931 4 3o 089307 51 10 63 1 58 7753i_ 800427 2-58 889477 ' •71 4 3o 089049 50 4y 11 63 1 80 775i3 9.800582 2-58 9-889374 I ••72 9-911209 4 3o 10-088791 12 63203 77494 800737 2.58 889271 I •72 911467 4 3o 088533 48 13 63225 77476 800892 2.58 889168 I •72 911724 4 3o 088276. 47 14 1 632^8 77458 801047 2-58 889064 I •72 911982 4 3o 088018 46 15 i 63271 77439 801201 2.58 888961 I •72 912240 4 3o 087760 45 16 1 63293 77421 8oi356 2-57 888858 1 •72 912498 4 3o 087502 44 17 1 633 16 77402 8oi5ii 2.57 888755 I .72 912756 4 3o 087244 43 18 63338 77384 8oi665 2.57 888651 1 .72 9i3oi4 4 29 086986 42 19 63361 77366 801819 801973 2.57 888548 1 .72 913271 4 29 086729 41 '20 63383 77347 2.57 888444 I •73 •73 913529 4 29 086471 40 "3'7 21 63406 77329 9.802128 2.57 9- 88834 1 I 9-913787 4 29 io-o862i3 22 63428 77310 802282 2-56 888237 « •73 914044 4 29 085956 33 23 63451 77292 802436 2.56 888134 I •73 914302 4 29 085698 37 24 63473 77273 802089 2-56 888o3o I •73 914560 4 29 085440 36 25 634Q6 77255 802743 2-56 887926 I •73 914817 4 29 b85i83 35 26 635i8 77236 802897 2-56 887822 I •73 915075 4 29 084925 34 27 63540 77218 8o3o3o 2-56 887718 1 .73 913332 4 29 084068 33 28 63563 77199 8o32o4 2-56 887614 I •73 915390 4 29 084410 32 29 63585 77181 803357 2.55 887510 I •73 • 915847 4 29 084153 31 30 31 636o8 77162 8o35ii 2.55 887406 1 •74 916104 4 29 083896 80 29 63630 1 77144 9.803664 2-55 9-887302 9-916362 4 29 10-083638 32 63653 77125 8o38i7 2.55 887198 1 •74 916619 4 29 o8338i 28 33 63675 77'07 803970 2-55 887093 1 •74 916877 4 29 083.23 27 34 63698 770S8 804123 2-55 886989 •74 917134 4 29 082866 26 35 63720 77070 804276 2.54 886885 I •74 917391 4 29 082609 25 36 63742 77o5i 804428 2.54 886780 1 •74 917648 4 29 082352 24 37 63765 77033 804581 2-54 886676 •74 917905 4 29 082095 23 38 63787 77014 804734 2.54 886571 •74 918163 4 28 081837 22 39 63810 76996 804886 2.54 886466 •74 1 918420 4 28 o3i58o 21 40 41 63832 76977^ 8o5o39 2.54 886362 •75 •75 9 '8677 4 28 o8i323 20 63854 76939 9-8o5i9i 2-54 9-856257 ! 9-918934 4 28 10-081066 19 42 63877 76940 805343 2-53 836i52 •75 919191 4 28 080809 18 43 ,63899 76921 805495 2-53 886047 •■'^ 919448 4 28 o8o552 17 44 63922 76903 8o5647 2.53 883942 885837 •75 919705 4 28 080295 o8oo38 16 45 63944 76884 til?: 2.53 •75 919962 4 28 15 46 63966 76866 2.53 885732 ■■'^ 920219 4 28 079781 14 47 • 63989 76847 806 I o3 2-53 885627 •75 920476 4 28 079524 13 48 (64011 76828 806254 2.53 885522 •75 920733 4 28 079267 12 49 i 64033 76S10 806406 2.32 885416 •75 920990 4 28 079010 11 r50 ! 64056 76791 806557 2-52 8853 1 1 -76 1 921247 4 28 078753 10 51 1 64078 76772 9-806700 ~J^5T 9-885205 -76 1 9 -921503 4 28 10-078497 9 52 164100 76754 806860 2-52 885ioo -76 921760 4 28 078240 8 53 64123 76735 807011 2-52 884994 •76 922017 4 28 077983 7 54 64145 76717 807163 2.52 884889 884783 •76 922274 4 28 077726 55 64167 76698 807314 2-52 1-76 922530 4 28 077470 5 56 64190 76679 807465 2-51 884677 -76 922787 4 28 077213 4 57 64212 76661 807615 2.5l 884572 -76 923044 4 28 076956 8 58 64234 76642 807766 2-51 884466 -76 923300 4 28 076700 2 59 64256 i 76623 807917 2-51 884360 -76 923557 4 27 076443 1 60 64279 76604 808067 2.5l 884254 '•77 9238i3 4-27 076187 ■ |N. COS. N. sine L. COS. D. 1" L. sine. !| L.cot. D.l" L. tang. 50° 1 70 TRIGONOMETRICAL FUXCTIOXS. — 40°. Nat. FrxcTioNs. LOGAKITHailC FlTNCTIOKS + 10. ' iN.sme.'N. cos.| L. sine. D. 1" L. COS. JD.l' L. tang. D.l" 1 L. cot 0l| 64279 '76604 9.808067 808218 2-5l 9-884254 11.77 9.923813 4-27 110.076187 60 1 1 64301 I 76586 2-5l 884148 ;i.77 924070 4-27 1 075930 59 2 ! 64323 1 76367 8o8368 2-5l 884042 1-77 924327 4-27 075673 53 s ! 64346 1 76548 8o85i9 2-5o 883936 '1.77 924583 4-27 075417 57 4 ; 64368.76530 808669 2-5o 883829 1-77 883723 1-7? 924840 4-27 075160 56 5 64390^76511 808819 2-5o 925096 4.27 074904 55 6 ; 64412 ,76492 808969 2-5o 883617 ..77 925352 4-27 07464S 54 7 64433 76473 8091 19 2-5o 883510 1-77 925609 925865 4-27 074391 074135 53 8 64457 76455 809269 2-5o 883404 ;i-77 4-27 52 9 64479176436 809419 2-49 883297 :i-78 926122 4-27 073878 51 10 1 64301 ,76417 809569 2-49 883I9I J -78 926378 4-27 073622 50 11 1 64524 i 76398 12 j 64546 7638o 9-809718 2-49 1 9-883o84 I .78: 9-926634 4-27 10.073366 49 809868 2-49 882977 1-78 882871 1.78 926890 4.27 073110 48 13 i 645681 76361 810017 2-49 927147 4-27 072853 47 14 i 64590 < 76342 810167 2-49 882764 !i.78 927403 4.27 072597 46 15 ,64612:76323 8io3i6 2-48 882657 ;i-78 927659 927913 4.27 072341 45 16 ,64635 176304 810465 2-43 882550 '1-78 4-27 072085 44 17 i: 64657 76286 810614 2-43 S82443 ,1-78 928171 4-27 071829 43 18 ; 64679 76267 810763 2-43 8ftj336 1-79 928427 4-27 071573 42 19 1 64701 76248 810912 2-43 882229 ,^'79 028683 4-27 071317 41 20 64723 76229 811061 2-43 8S2121 1-79! 928940 4-27 071060 40 oi ' i'^n^o 96210 9-811210 2-43 9-882014 1-79 9.929196 881907 ,1.79; 929452 4-27 10-070804 89 11 ! Af ^^ 76192 8ii358 2-47 4-27 070548 88 23 1 64790 76173 8n5o7 2-47 881799 1.79! Q29708 4-27 070292 070036 87 24 i 64812 76154 8ii655 2-47 881692 1-79 929964 4-26 36 25 648341 76135 81 1804 2-47 88i584 1.79 930220 4-26 069780 85 26 j 64856 '761 16 811932 2-47 881477 ,1-79 930475 4.26 069525 84 27 i 64878 76097 812100 2-47 88,369 1.79 930731 4-26 069269 33 28 i 64901 76078 812248 2-47 881 261 1-80 930987 4-26 069013 82 29 64923 76059 812396 2-46 88ii53 1-80 931243 4-26 068757 81 30 64945 76041 1 812544 2-46 881046 ;i-8o 931499 4-26 o685oi SO 81 ,,64967 76022 9-812692 2-46 9.88093s 1.80 9.931755 4-26 10-068245 29 32 || 64989 76003 812840 2-46 88o83o '1-80 932010 4.26 067990 28 33 1; 65oi I 75984 812988 2-46 880722 1-80 932266 4-26 067734 27 34 : 65o33 75963 8i3i35 2-46 880613 1-80 932522 4.26 067478 26 35 65o55 1 75946 8i3283 2-46 88o5o5 1-80 932778 4-26 067222 25 36 ,63077175927 8i343o 2-43 880397 |i-8o 933o33 4-26 066967 24 37 1 63100 73908 813578 2-45 880289 i-8i 933289 933543 4.26 066711 23 33 ; 65i22| 75889 813725 2-45 880180 1-81 4-26 066455 22 39 65i44l 753-0 813872 2-43 880072 1-81 933800 4-26 066200 21 40 65i66.75S5i 814019 2-45 879963 1-81 934056 4-26 063944 20 41 65i88 175832 9-814166 2-45 9-879855 I -81 9.934311 4-26 10-0656S9 065433 19 42 65210 |758i3 8i43i3 2-43 879746 :i-8r 934567 4.26 18 43 ; 65232 75794 814460 2-44 879637 1-81 934823 4.26 063177 17 44 1! 65254 ' " 73773 814607 2-44 879529 I -81 935078 426 064922 16 45 1 65276 75756 814753 2-44 879420 I -81 935333 4.26 064667 15 46 65298 73738 814900 2-4< 879311 I -81 935589 4.26 0644 II 14 47 1 65320 j 75719 816046 2-44 879202 1-82 935844 4.26 0641 56 13 43 , 65342 1 73700 8i5i93 2-44 879093 1.82 936100 4-26 063900 12 49 65364 j 7568o 815339 2-44 878984 1-82 878875 '1.82 936355 4.26 063645 11 50 ,65386 175661 81 5485 2-43 936610 4-26 063390 10 51 ,65408175642 9-8i563i 2-43 9-878766 J -82; 9-936866 4-25 io-o63i34 9 52 6543o 1 75623 815778 2-43 : 878656 1.82 1 937121 4-2^ 4-23 062879 8 53 65452 ! 75604 815924 2-43 878547 1-82 937376 062624 7 54 '65474175585 816069 2-43 878438 1-82 937632 4-25 062368 6 55 163496 75566 8i63i5 2-43 878328 1-82 937887 4.25 0621 i3 5 56 ' 655i8 75547 8i636i 2-43 878219 i-83| 938142 4-25 06 I 858 4 57 65540 75528 8i65o7 2-42 878109 1-83, 938398 1 4.25 061602 8 58 65562 1 75509 8i6652 2-42 877999 I -831 938653 4-25 061347 ■2 59 65584 ■ 75490 816798 2-42 877890 I -83! 877780 I -831 938908 4-25 061092 1 60 656o6l 75471 816943 2.42 939163 1 4-25 060837 N. COS. N. sine. L. COS. B. 1" ' L. sine. 1 '1 L. cot. D. 1" | L. tang. 49° 1 TRIGONOMETRICAL FUXCTIOXS. — 41' 71 Na^. Functions. Logarithmic Functions + 10. 1 ' N.sine., N. cos L. sine. D.l" L. COS. D.l" 1 L. tang. D.l" 1 L. cot l656o6| 75471 9.816943 2.42 9.877780 1.83 ' 9.939163 4-25 10-060837 60 1 656281 75452 817088 2.42 877670 1-83 1 939418 4-25 o6o5S2 59 2 • 65650 75433 817233 2-42 877560 1-83 939673 4-25 060327 58 3 65672 75414 817379 2-42 877450 1-83 939928 4-25 060072 57 4 ' 65694 75395 817524 2-41 877340 1-83 940183 4-25 059817 56 5 65716 75375 817668 2-41 877230 1-84 940438 4-25 059562 55 6 ; 65738 75356 817813 2-41 877120 1-84 940694 4-25 059306 54 7 65759 75337 817958 2-41 877010 1-84 940949 4-23 05903 1 53 8 1 65781 75318 818103 2-41 876899 1.84 941204 4-25 058796 52 9 ! 658o3 75299 818247 2-41 876789 1.84 941458 4-25 058542 51 10 165825 75280 818392 2-41 876678 1-84 941714 4-25 058286 60 ir 65847 75261 9-8i8536 2-40 9.876568 1.84 ■ 9.941968 4-25 io.o58o32 49 12 165869 75241 818681 2-40 876457 1-84 942223 4-25 057777 48 13 J 65891 75222 818825 2-40 876347 1.84 942478 4-25 057522 47 14 65913 75203 818969 2 -40 876236 1-85 942733 4-25 057267 46 15 65935 75184 819113 2-40 876125 1-85 942988 4-25 057012 45 U 65956 75i65 819257 2-40 876014 1-85 943243 4-25 056757 44 17 65978 75146 819401 2-40 875904 1.85 943498 4-25 o565o2 43 18 66000 75126 819545 2.39 875793 1-85 9437D2 4-25 056248 42 19 66022 75107 819689 2.39 875682 1-85 944007 4-25 055993 41 20 66044 75088 819832 2.39 875571 1-85 944262 4-25 055738 40 21 66066 75069 9-819976 2-39 9-875459 1-85 1 9-944517 4-25 10.055483 39 22 66088 75o5o 820120 2.39 875348 1-85 944771 4-24 055229 3S 23 66109 75o3o 820263 2.39 875237 1-85 945026 4-24 054974 87 24 66i3. 75011 820406 2.39 2-38 875126' 1.86 945281 4-24 054719 36 25 66 1 53 74992 82o55o 875014 1-86 945535 4-24 o5446d 35 26 66175 74973 820693 2-38 874903 1-86 945790 4-24 054210 34 27 66197 74953 820836 2-38 874791 874680 1.86 946045 4-24 053955 33 28 66218 74934 820979 2-38 1.86 946299 4-24 053701 82 29 66240 74915 821122 2-38 874568 1.86 9465*4 4-24 053446 31 30 66262 74896 821265 2-38 874456 1.86 946808 4-24 053192 30 29 31 66284 1 74876 9-821407 2-38 9-874-344 1.86 9.947063 4-24 10.052937 32 663o6 74857 82i55o 2-38 874232 1.87 947318 4-24 052682 28 83 66327 74838 821693 2.37 874121 1.87 947572 4-24 052428 27 84 66349 74S18 821835 2-37 874009 1.87 947826 4-24 052174 26 85 66371 74799 821977 2-37 873896 ..87 948081 4-24 o5i9ig 25 36 66393 74780 822120 2.37 873784 1-87 948336 4-24 o5i664 24 87 66ii4 74760 822262 2.37 873672 1.87 948590 4-24 o5i4io 23 38 664361 74741 822404 2.37 873560 ..87 948844 4-24 o5ii56 22 39 66458 1 74722 822546 2.37 873448 1.87 It^ 4-24 o5o90i 21 40 66480 1 74703 822688 2-36 873335 1-87 4-24 o5o647 20 41 665oi 174683 "9^82283^ 2-36 9-873223 \ii 9.949607 4-24 io-o5o393 19 42 66523 74664 822972 2-36 873110 949862 4-24 o5oi38 IS 43 66545 ! 74644 823ii4 2-36 t^llt 1-88 950116 4-24 049884 17 44 66566 74625 823255 2-36 1-88 950370 4-24 049630 IG 45 66588 74606 823397 2-36 1]IUI 1-88 950625 4-24 049375 15 46 66610 74586 823539 2-36 1-88 950879 4-24 049121 14 17 66632 74567 823680 2.35 872547 1.88 95ii33 4-24 04^^867 13 4S 66653 74548 823821 2-35 872434 1.88 95x383 4-24 04S612 12 49 66675 74528 823963 2-35 872321 1.88 951642 4-24 048358 11 5t) 66697 74509 824104 2.35 872208 1-88 951896 4-24 048104 10 51 66718 74489 9-824245 2-35 9.872095 87I98I 1-89 9.952150 4-24 10-047850 ~~9~ 52 66740 74470 824386 2.35, 1.89 952405 4-24 047595 8 53 , ^A'^i'. 7445i 824527 824668 2.35 871868 1-89! 952659 4-24 047341 7 54 iii^i 7443 1 2-34 871755 ..89J 952913 4-24 047087 6 55 66805 74412 824808 2-34 871641 ,.89 953167 4-23 046833 5 56 66827 66848 74392 824949 2-34 871528 1.89 953421 4-23 046579 04632D 4 57 74373 825090 825230 2-34 87I4I4 1.89 953675 4-23 8 58 66870 74353 2-34 871301 1.89 953929 4-23 046071 2 59 66891 74334 825371 2-34 871187 1-89 954183 4-23 045817 1 eo 66913 74314 8255u 2-34 871073 1.90 954437 4-23 045563 N. COS. N. sine.' L. COS. D. 1" L. sine. L. cot. D.l" L. tang. f 48° '1 72 TEIGOXOMETRICAL FUXCTIOXS. — 42^^ Nat, FxrNCTioNs. LOGAKITHMIC FUNCTIOXS + 10 1 N. sine.! N. COS. L. sine. | D. 1" L.COS. 'r 1 L.tang. D . 1" Kcot 1 66913I74314 9-825511 2 •34 9-871073 1.90 ; 9-954437 7 23 IO-045563 , 60 1 66935 74295 825651 ] 2 -33 870960 870846 1-90 1 954691 4 23 045309 59 2 66956 j 74276 825-91 1 2 -33 1.90 954945 4 23 o45o53 ■ 5^. 3 66978 74256 825931 , 2 •33 870732 1-90 955200 4 23 044800 57 4 66999 74237 826071 1 2 .33 870618 1.90 955454 j 4 23 044546 56 £ 67021 74217 826211 2 33 870504 1.90 955707 4 23 044293 044039 55 6 67043 74198 826351 2 33 870390 1-90 955961 4 23 54 7 blobi 74178 826491 i 2 33 870276 1.90 956213 ! 4 23 043783 53 8 67086 74159 826631 ! 2 33 870161 1-90 956469 956723 4 23 04353 1 52 9 67107 74139 826770 j 2 32 870047 ^•9. 4 23 043277 51 10 67129 74120 826910 1 2 32 869933 1^1 936977 4 23 O43023 50 11 67151 74100 9-82-049 2 32 9-869818 1-91 9-957231 4 "2r 10-042769 4y 12 67172 '74080 82-189 2 32 869704 1-91 957485 4 23 o425i5 43 13 67194 74061 82-328 I 2 32 869589 1-91 957739 4 23 042261 47 ]^ 67215; 74041 827467 2 32 86Q474 i-9i| 957993 4 23 042007 46 15 67237 74022 827606 2 32 869360 i-9i| 1-91 958246 4 23 041754 45 16 67258 74002 827745 2 32 869245 958500 4 23 041 5oo 44 17 67280 73983 827884 2 3i 869130 1-91 958754 4 23 041246 43 18 67301 73963 828023 2 3i 869015 1-92 959008 4 23 040992 42 19 67323 ; 73q44 828162 2 3i 868900 1-92 959262 4 23 040738 41 20 6i344 ! 73924 8283oi 2 3i 868783 1-92, 959516 4 23 040484 40 21 67306 73904 9.828439 828578 2 3i 9-868670 1-92; 9-959769 960023 4 23 10-040231 39 22 67387 ! 73885 2 3i 868555 1-92; 4 23 039977 £3 2S I 6T4CQ 173863 828716 2 3i 868440 1-92 960277 4 23 039723 37 24 167430 73846 828855 2 3o 868324 1-92 00531 4 23 039469 * 039216 36 25 1 67452 -73826 828993 2- 3o 868209 1-92: 960784 4 23 35 26 67473 ii38o6 829131 2 3o 868093 1-92 961038 4 23 038962 34 27 67495173787 829269 2 3o 867978 867S62 1-93 961291 4 23 038709 1 33 28 67516173767 829407 2 3o 1.93 961545 4 23 038453 1 32 29 '67538 73747 829545 2- 3o 867747 1.93 961799 4 23 03S20I 1 31 30 I; 67559 73728 829683 2- 3o 867631 .•93 962002 4 23 037948 1 30 31 67580 1 73708 9-829821 2 29 9867515 1-93 9-962306 4 23 10-037694 i 2y 32 ' 67602 i 73688 829959 1 2 29 867309 867283 1.93 962560 4 23 037440 28 33 67^23; -366q 830097 2 29 ..93 962813 4 23 037187 I 27 34 67645. 7364q[ 830234 1 2- 29 867167 1-93 963067 4 23 036933 1 26 35 67666 73529 83o372 2 29 867051 1.93 963320 4 23 o3668o 25 86 67688:73610 83o5o9 2 29 866^19 866703 1-94 963574 4 23 036426 ■ 24 37 ' 67709 1 73590 830646 2- 29 1-94 963827 4 23 036173 1 23 83 67730; 73570 830784 2- 29 i^§4 964081 4 23 035919 22 39 67702 73551 830921 2- 28 866586 I •94; 964335 ! 4 23 035665 21 40 : 67773 73531 1 83io58 2- 28 866470 1:941 964588 i 4 22 035412 20 41 67793,73511 9-831195 2 28 9-866353 1-94 9-964842 I 4 22 io-o35i58 19 42 67816173491 831332 2 23 866237 1.94: 960095 i 4 22 o349o5 18 43 6783-' 734-3 831469 2- 28 866120 1.94: 965349 i 4 22 o3465i 17 44 67859 '73452 83 1606 2- 28 866004 1-95; 965602 4 22 034398 16 4? 67880,73432 831742 2- 28 865887 1.95 965855 4- 22 034I45 15 4f 67901 i 73413 83.1879 832015 2- 28 865770 1.95; 966109 4 22 033891 033638 14 47 67923 73393 2 27 865653 ..95 1 .95, 966362 4- 22 13- 4S 67944 1 73373 832152 2- 27 865536 966616 4- 22 033384 12 49 67965*73353 832288 2- 27 865419 1.95: 966869 4- 22 o33i3i 11 50 67987 73333 832425 2- 27 865302 1.95 967123 4- 22 032877 10 51 68008 '73314 9-832561 ; 2 27 9 -865 1 85 1.95 9-967376 4- ~ri 10-032624 9' 52 6S029 i 73294 832697 2 27 86D068 i-95i Zslt 4- 22 032371 8 53 68g5i i 73274 832833 2- 27 864950 1-95; 4- 22 032117 7 54 '68072173254 832969 2 26 864833 1 -961 968136 4 22 o3i864 6 55 68093:73234 8331 03 2 26 864716 1-961 968389 4 22 o3i6ii 5 56 '68ii5i732i5 833241 2 26 864598 864481 ,.96! 968643 4 22 o3i357 4 57 ,68i36 173195 833377 2 26 1-96; 968896 4 22 o3iio4 8 53 .i68i57 73175 . 833512 2 26 864363 1-96; 969149 969403 4 22 o3o85i 2 59 ,68179 73 1 55 833648 2 26 864245 ..96 4 22 o3o597 1 60 68200,73135 833783 2-26 864127 1.96! 969656 4-22 o3o344 N. COS. N.sine. L. COS. ' D. 1" L. sine. ^ L. cot 1 D. 1" L. tang. ~^ 1 TRIGONOMETRICAL FUXCTIOXS. — 43^ 73 Nat. Functions. LOGAKITHMIC FUNCTIONS + 10. 1 ' N.sine.i N. cos. 1 L. sine. 1 D. 1" L. COS. |l ).i" ' L. tang. D. 1" L- cot. 68200 73,35 9-833783 2 26 9-864127 I .96 : 9-969656 4-22 io-o3o344 60 1 68221 73116 8339,9 2 25 864010 I -96 1 969909 4-22 030091 59 2 6S242 73096 834054 2 25 863892 I •97 1 970162 4-22 029838 58 8 , 68264 73076 834189 2 25 863774 1 •97 1 970416 4-22 029584 57 4 i 68285 73o56 834325 2 25 863656 J ■97 970669 4-22 029331 56 5 i 683o6 73o36 834460 2 25 863538 1 •97 970922 4-22 029078 55 6 ] 68327 73016 834595 2 25 863419 •97 971175 4-22 028825 54 7 68349 72996 834730 2 25 863301 •97 971429 4-22 028571 53 8 68370 72976 834865 2 25 863 1 83 •97 97,682 4-22 0283,8 52 9 6839. 72957 834999 2 •24 863o64 •97 97,935 4-22 028065 51 10 68412 72937 833,34 2 24 862946 -98 -98 972,88 4-22 027812 50 11 68434 72917 9-835269 ~2 24 9-862827 1 9^972441 4-22 10-027559 1 49 1 12 ] 68455 72897 835403 1 24 862709 .98 972694 4-22 027306 4S 13 68476 72877 835538 2 24 862590 -98 972948 4-22 027052 47 U 68497 72857 835672 2 -24 862471 -98 973201 4-22 026799 46 15 ;685i8 72837 835807 2 24 862353 -98 973454 4-22 - 026546 45 16 68539 728,7 83594, 2 24 862234 -98 9-3707 4-22 026293 44 17 68561 72797 836075 2 23 862, ,5 -98 973960 4-22 026040 43 18 68582 7 -'77 7 836209 2 23 86,996 -98 9742,3 4-22 025787 42 11> 68603 72757 836343 2 23 86,877 -98 974466 4-22 025534 41 2<;» 21 68624 68645 72737 836477 2 -23 86,758 _199 .99 974719 4-22 025281 40 72717 9-8366,1 2 23 9-86,638 9-974973 4-22 10-02D027 39 22 68666 72697 836745 2 23 86,5,9 •99 975226 4-22 024774 3S 23 68688 72677 836878 2 23 86,400 , •99 975479 4-22 024521 37 24 68709 72657 8370,2 2 22 86,280 •99 975732 4-22 024268 36 25 68730 72637 837,46 2 22 86,161 , •99 975985 4-22 0240,5 35 26 68751 726,7 837279 2 22 861041 1 •99 976238 4-22 023762 34 27 68772 72597 837412 2 22 860922 , 860802 I •99 976491 4-22 023509 33 28 68793 72577 837546 2 22 -99 976744 4-22 023256 82 2y 68814 72557 837679 2 22 860682 2 -00 976997 4-22 o23oo3 31 30 31 68835 72537 837812 2 22 86o562 2 -00 -00 977250 4-22 022750 80 68857 72517 9-837945 2 22 9-860442 2 9 -977503 4-22 10-022497 29 32 68878 72497 838078 2 2, 86o322 2 -00 977756 4-22 022244 28 38 68899 72477 8382,1 2 21 860202 2 •00 978009 4-22 02,99, 02,738 27 34 68920 72457 838344 2 2, 860082 2 •00 978262 4-22 26 35 68941 72437 838477 2 2, 859962 2 859842 2 -00 9785,5 4-22 02,485 25 86 68962 72417 8386,0 2 2, • 00 978768 4-22 02,232 24 37 68983 72397 838742 2 2, 859721 2 -01 979021 4-22 020979 23 38 69004 72377 838875 2 2, 839601 2 -01 979274 4-22 020726 ! 22 1 39 69025 72357 839007 2 2, 859480 2 •0, 979527 4-22 020473 21 40 69046 72337 839140 2 20 859360 2 -0, 979780 4-22 020220 20 19 41 69067 72317 9-839272 2 20 9-859239 2 -01 9-980033 4-22 10-0,9967 42 69088 72297 839404 2 20 859,19 2 -01 980286 4-22 0197,4 18 43 69109 72277 839536 2 20 858998 2 -0, 980538 4-22 0,9462 17 1 44 6qi3o 72257 839668 2 20 858877 2 -0, 980791 4-21 0,9209 16 45 69151 72236 839800 2 20 858756 2 -02 98,044 4-21 0,8956 15 46 69172 72216 . 839932 2 20 858635 2 -02 981297 4-21 0,8703 14 47 69193 72,96 840064 2 '9 8585,4 2 •02 98,550 4-21 o,845o 18 48 69214 72176 840196 2 '9 858393 2 •02 98,803 4-2, 0,8,97 12 49 69235 72156 840328 2 '9 858272 2 -02 982056 4-21 017944 11 5<> 69256 72,36 840459 2 19 858i5i 2 -02 982309 4-21 017691 10 51 69277 721,6 9-840591 2 19 9-858029 2 •02 Q-982562 4-21 10-017438 9 52 69298 72095 840722 2 19 857908 2 -02 9828,4 4-21 0,7,86 8 53 69319 72075 840854 2 •9 857786 2 •02 983067 4-2, 0,6933 7 54 69340 72055 840985 2 \% 857665 '2 -o3 983320 4-2, 016680 6 55 69361 72035 841116 3 857543 2 .o3 983573 4-21 016427 5 56 69382 720,5 841247 2 ,8 857422 2 .o3 983826 4-21 016174 4 57 69403 7,995 841378 2 ,8 857300 2 -o3 984079 4-21 015921 8 58 69424 71974 841509 2 ,8 857178 2 -o3 984331 4-21 0,5669 2 59 69445 7,954 841640 2 ,8 857056 '2 -o3 984584 4-21 oi54,6 1 60 69466 71934 841771 2 18 856934 2 ^3 984837 4-21 0,5,63 N. COS. N.sine. L. COS. "d! \" L. sine. | L. cot. D. 1" L. tang. ' 46° 1 74 TRIGOXOMETRICAL FL'XCTIOXS. — 44' NiT. Functions. Logarithmic Fcnctioxs + 10. 1 X.sine.'N. cos. L. sine. D. 1" L. COS. 1 L. tang. Dl." L. cot. 69466 71934 9-841771 2.18 9-856034 '2 856812 '2 03; 9.984837 4.21 io.oi5i63 60 1 69487 71914 841902 2-l8 o3 985090 4-21 014910 59 2 69508 71894 842033 2.18 856690 '.2 04 985343 4-21 014657 58 S 69329 71873 842163 2-17 856568 2 04 985596 4-21 014404 57 4 69549 71853 842294 2.17 856446 2 04 985848 4-21 0i4i52 56 5 69570 71833 842424 2-17 856323 2 04 986101 4-21 013899 55 6 69591 I71813 842555 2-17 856201 2 04 986354 4-21 013646 54 7 69612 71792 842685 2-17 856078 2 04 986607 4-21 013393 53 8 69633171772 842815 2-17 855956 2 855833 2 04 986860 4-21 oi3i4o 52 9 69654 '71752 842946 2- 17 04 9871 12 4-21 012888 51 10 11 69675 71732 843076 2-17 855711 2 o5, 987365 4-21 01 2635 50 69696 7171 I 9.843206 2-16 9.855558 2 o5 9.987618 4-21 10012382 49 12 69717,71691 843336 2-l6 855465 2 o5 987871 988123 4-21 012129 43 13 69737 '7 167 1 843466 2-16 855342 2 o5 4-21 01 1877 47 U 69753,71650 843595 2-16 855219 2 o5 988376 4-21 011624 46 15 69779*71630 843725 2-16 855096 '2 o5 988629 4-21 011371 45 16 69800:71610 843855 2.16 854973 2 o5 988882 4-21 OI1I18 44 17 69821 71390 843984 2-16 854S50 2 o5 9S9134 4-21 010866 43 18 69842! 71569 8441 U 215 854727 2 06 989387 4-21 oio6i3 42 19 69862 1 71549 844243 2-l5 834603 I2 06 989640 4-21 oio36o 41 20 69883 1 71529 844372 2.l5 &544S0 2 06 989893 4-21 010107 40 ~2F 69904 7i5oS 9-844502 2-15 9.854356 2 06 Q. 990145 4-21 10-009855 39 22 69925 '71488 844631 2-l5 854233 2 06 990398 4-21 009602 38 23 69946 71468 844760 2-l5 854109 2 06 99065 I 4-21 009349 87 24 69966 1 71447 844889 215 853986 ,2 06 990903 4-21 SS 36 25 69987^71427 845018 2-15 853862 12 06 991 1 56 4-21 35 26 70008 71407 845147 2-l5 853738 2 06 991409 4-21 008591 34 27 70029 '71386 845276 2-14 853614 2 07 991662 4-21 O08338 33 23 70049 7 1 366 845405 2-14 853490 \7 07 991914 4-21 608086 82 29 70070; 71345 845533 2.14 853366 12 07 992167 4-21 007833 31 30 70091 1 7i325 845662 2.14 853242 2 _o7 992420 4-21 007580 30 31 70112 7i3oj 9-845790 1 2-14 9-853ii8 2 •07 9.992672 4-21 10-007328 29 32 70132:71284 845919 2-14 852994 2 852869 2 07 992925 4-21 007075 28 83 70153 71264 846047 2-14 07 993178 4-2! 006822 27 34 70174 71243 846175 2-14 852745 \2 07 993430 4-21 006570 26 35 70195171223 846304 2-14 852620 2 07 993683 4-21 oo63i7 25 36 70215 71203 846432 2l3 852496 \2 08 993936 4-2! 006064 24 37 70236 71 182 846560 2.l3 852371 2 08 994189 1 4-21 oo58ii 23 83 1 70237! 71 162 846688 213 852247 2 08 994441 j 4-21 005559 22 89 70277:71141 846816 j 2-l3 852122 '2 08 994694 4-21 oo53o6 21 40 70298 71121 846944 2-l3 851997 2 08 994947 1 4-21 oo5o53 1 20 1 41 70319 71100 9-847071 i 2-l3 9.851872 \2 "^ 9.995199 1 4-21 993452 4-21 10-004801 19 42 70339 71080 847199 2.l3 851747 12 08 004548 18 43 7o36o: 71039 847327 2.l3 85i622 '2 .08 995705 '4-21 004295 17 44 7o38i 1 71039 847454 212 85i497 2 09 995937 : 4-21 004043 16 45 70401 71019 847582 2-12 85i372 2 09 996210 ; 4-21 003790 15 46 70422 ! 70998 ESS 2-12 85 1246 2 09 996463 4-21 003537 14 47 70443 i 70978 212 85ii2i 2 09 996715 4-21 003285 13 48 70463 i 70957 847964 84?ogi 2-12 850996 2 09 996968 4-21 oo3o32 12 49 70484 ' 70937 2-12 350870 2 09 997221 4-2! 002779 11 50 7o5o5 70916 848218 212 850745 2 oq 997473 4-21 002527 10 51 70523 70896 9.848345 212 9-85o6i9 2 09 9-997726 1 4-21 10-002274 ! 9 1 52 70546 70875 848472 211 850493 2 10 4-21 002021 8 53 70567 70855 848599 2.|I 830368 2 10 998231 4-21 001769 7 -54 70587 . 70834 848726 211 85o242 2 10 998484 4-21 ooi5i6 6 55 70608; 70813 848852 2-11 85oii6 2 10 998737 4-21 001263 6 56 70628 , 70793 848979 211 849990 2 10 998989 4-21 OOIOII 4 57 70649 70772 849106 2-11 849864 2 10 999242 4-21 000738 3 58 70670 70752 849232 211 849738 2 .10 999495 4-21 000305 2 59 70690 70731 849359 849483 2-II 8496 IT 2 .10 999747 4-21 0002 53 1 60 70711170711 2-II 849485 2-10 lO-OOOOOO i 4'2I 10'«00000 1 N, COS. N. sine. L. COS. ' D. 1" L. sine. ' L. cot. , D. 1" L. tang. 45° 1 TABLE III. PEEOISE OALCULATIO^f FUNCTIONS NEAR THEIR LDIITS. 76 SIXES OF SMALL ANGLES. log . sin. x"= 4.685575 + log. X - diff. rOR THE SIXES OF S3IALL AXGLES. Anarles. Second?. Diff. Angles. Seconds Diff. 1 Angles. 1 Seconds. Diff. o" 1° 29' 5o" 5390 2^ 7'3o'' 765o ,?-5o'' 540 3o' 5o" 545o ^r 1 8' 10" 7690 100 900 2 3 3i'4o" 55oo 5; 53 54 55 8'45" 7725 lOl 102 io3 20' 20" 23' 5o" 1220 i43o 32' 3o" 33' 3o" 555o 56io 9' 20" 10' 7760 7800 V' 1620 4 5 34' 20" 566o 10' 40" 7840 104 io5 29', 5°!! 1790 6 35' 10" 57.0 56 II' i5" 7875 106 33' 3o" 1900 36' 5760 58io ll 11' 5o" 7910 35' 2100 7 8 36' 5o" 12' 3o" 7950 a 37' 2C" 2240 37' 40' 5860 i3' 5" 7985 39' 3o ' 2370 9 10 38'3o" 5910 icJ'40' 8020 log no 41' 3o" 2490 II 12 i3 39' 3o" 5970 61 14' 20" 8060 III 112 ii3 114 ii5 43' 20" 45' 10" 2600 2710 40' 20" 41' 10" 6020 6070 62 63 • 64 65 i5' i5'35" 8100 81 35 47' 2820 41' 5o" 6110 16' 10" 8170 48' 40" 2920 14 i5 42' 40" 6160 16-45" 8203 5o' 20" 3020 16 43' 3o" 6210 66 tl 69 70 17' 20" 8240 116 52' 3l20 44' 10" 6200 17' 55" 8275 53' 3o' 3210 \l 45' 63oo i8'3o" 83io VA 55' 33oo 45' 5o" 63 5o 19' 5" 8345 56' 3o' 3390 19 20 46' 3o" 6390 19' 40" 838o 119 120 58' 34S0 47' 20" 6440 71 74 75 20' 1 5" 8415 59' 20" i<>oo'4o" 3560 21 48' 6480 20' 5o" 845o 121 122 123 124 125 3640 22 23 24 25 48' 5o" 6530 21' 25" 8485 2' 3720 49' 3o" 6570 22' 8320 3' 20- 38oo 5o' 20" 6620 22' 35" 8555 4' 40' 388o 26 5i- 6660 76 23' 10" 8590 126 5' 5o" 3950 5i'5o" 6710 23' 45" 8625 8' 10" 4020 27 28 29 3o 52' 3o" 6750 ^l 24 20" 8660 4090 53' 10" 6790 24' 55" 8695 8730 9 20" ■4160 54' 6840 <9 80 20' 3o" 10' 3o" 4280 3i 32 33 54' 40" 6880 81 82 S3 26' 8760 i3i l32 i33 i34 i35 II' 40" 43oo 55' 20" 6920 26' 35" 8795 8825 12' 5o" 4370 56' 10" 6970 27' 5" 14' i5' 4440 45oo 34 35 56' 5o" 57' 3o" 7010 7o5o 84 85 27' 40" 28 10" 8S60 8890 16' 10" 17' 10- iTo 36 58' ro" 58' 50* ?T„ 86 ll 28' 45" 29' i5" 8925 8955 1 36 1 37 1 38 139 140 18' 10' 4690 59' 3o'- 7170 29' 5o" 8990 19' 20" 20' 20" 4760 4820 39 40 20 00' 10" 5o" 7210 725o 89 90 3o' 20" 3o' 55" 9020 9055 21' 20" 4880 41 I '40" 7300 91 3i'25" oo85 141 22' 20" 4940 2' 20" 7340 o2' Q120 23' 20" 24' 20" 25' 10 " 5ooo 5o6o DIIO 42 45 3' 3' 35" 4' 10" 7380 7415 7450 92 93 94 93 32' 3o" 33' 5" 33' 35" 9i5o 9185 9215 142 143 144 145 26' 10 ' 27 10" 28' 10" 5170 523o 5290 46 47 48 49 4' 5o" 5' 3o;' 6' 10" 7490 753o 7570 96 ll 34' 5" 34' 40" 35' 10" 9245 9280 9310 146 2q 5340 6' 5o" 7610 35' 40" 9340 2q' 5o" 5390 7'3o" 765o 99 36' i5" 9375 149 TAXGEXTS AK"D C0TA:N^GEXTS OF SMALL AXGLES. n Jog . tan. x"^ 4.685575 + log. X + diff. log . cot. x"= 15.314-125 - log. X - diff. FOR TANGEi^TS AND COTAi^GENTS OF SMALL ANGLES. Ane^les. Seconds. Diff. Angles. Seconds. Diff. Angles. Seconds. Diff. o" 1° 3'3o" 38io 5o 5i 52 53 lo3o'io" 5410 i lo" 43o 4' 10" 38D0 3c' 3o" 543o 100 ii' lo" 670 I 4' do" 3890 3i' 5460 lOI 14' 10" Bdo 2 3 5'3o" 3930 3i'3o" 5490 102 io3 17' 1020 6' 3960 32' 5520 19' 1440 4 5 6' 40" 4000 D4 55 32' 20" 5540 104 lOD 21' 1260 6 7' 20" 4040 56 59 60 32' 5o" 5570 106 23' i38o 7' 5o" 4070 33' 10" 5590 24' 5o" 1490 7 8 8'3o" 4110 33' 40" 5620 107 108 26' 3o" 1 590 9' 4140 34' 10" 565o 27' 5o" i6io 9 10 9' 40" 4180 34' 3o" 5670 log no 29' 20" 1760 T T 10' 20" 4220 61 62 63 35' 5700 3o' 40" 32' 1840 1920 12 i3 10' do" ir3o" 42DO 4290 35' 20" 3d' 5o" 5720 5750 1 1 1 112 ii3 33' 10" 1990 12' 4320 64 65 36' 10" 5770 34' 20" • 2060 i4 i5 12' 3o" 43do 36' 40" 5800 114 Ii5 35' 3o" 36' 40" 2l30 2200 16 i3'io" i3' 40" 4390 4420 66 67 68 69 70 37' 10" 37' 3o" 583o 585o 116 37' 5o" 2270 ;^ 14' 10" 44DO 3s' 588o ]\l 38' 5o" 233o 14' do" 4490 38' 20" 5900 39' 5o" 2390 '9 20 i5' 20" 4D20 38' do" 5930 119 120 40' 5o" 245o i5'5o" 455o 39' 10" 5950 41' do" 2DIO 21 16' 20" 4d8o 71 39' 3o" 5970 121 42' do" 2D70 263o 22 23 24 25 17' 4620 72 73 74 75 40' 6000 122 123 43' do" 17' 3o" 465o 40' 20" 6020 44' 40" 2680 18' 4680 40' 5o" 6o5o 124 125 45' 40" 46' 3o" 2740 2790 26 18' 3o" '9' 4710 4740 76 4i' 10" 41' 40" 6070 6100 126 47' 20" 2840 27 28 19' 3o" 4770 77 78 42' 6120 127 128 48' 10" 2890 20' 4800 42' 3o" 6i5o 49' 2940 29 3o 20' 3o" 483o 11 42' 5o" 6170 129 i3o 40' 5n" 2990 3i 32 33 34 35 21' 4860 81 82 83 84 85 43' 10" 6190 i3i l32 i33 1 34 i35 5o'4o" 5i'3o" 3040 3090 2.'3o" 22' 4890 4920 43' 40" 44' 6220 6240 52' 20" 3 1 40 22' 3o" 49D0 44' 3o" 6270 53' 3i8o 23' 49«o 44' do" 6290 53' 5o" 323o 36 37 38 39 23' 3o" 5oio 86 89 45' 20" 6320 1 36 III 54' 40" 3280 24' 5o4o 45' 40" 6340 » 55' 26" 56' 3320 3360 24' 3o" 25' 5070 5ioo 46' 40' 20" 6360 6380 56' 5o" 3410 2D' 3o" 5i3o 46' 40" 6400 139 40 90 140 57' 3o" 58' 10" 34DO 41 26' 5i6o 47' 10" 6430 3490 26' 3o" 5190 91 47' 3o" 6450 141 58' do" 3530 42 43 44 45 26' 5o" 52IO 92 93 48' 6480 142 143 59' 3o" 3570 27' 20" 5240 48' 20" 65oo 1» 0'20" 3620 27' Do" 5270 94 9D 48' 40". 6520 144 ■45 r 366o 46 28' 20" 53oo 96 49' 6540 146 \' Ao" 3700 28' 40" 5320 49' 20" 6560 2' 10" 3730 47 48 49 29' 10" 535o 97 98 49' 40" 9580 ',% 2' 5o" 3770 29' 40" 5380 5o' 10" 6610 3' 3o" 38 10 3o' 10" 5410 99 5o'3o" 663o u, 78 r TAXGEXTS AND COTAXGEXTS OF SMALL AXGLES. lo^. tan. a"= 4.GS5575 + log. A+ cliff. I02:. cot. A"= 15.314425 - loor. a- diff. FOR TAXGEXTS AXD COTAXGEXTS OF SMALL AXGLES. Anirles. ^Jfe do' 3o' 5o' 5o" 5i' 10" 5i'3o" D2' 52' 20" 5a' 40" 53' 53' 20" 53' do" 54' 10" 54' 3o" 54' 5o" 55' 10" 55' 3o" 55' 5o" 56' 10" 56' 3o" 56' do" 57' 10" 57' 40" 58' 58' 20" 58' 40" 59' 59' 20" 59' 40" 2° 00' 00" 20" 40" r r 20" r4o" 2' 2' 20" 2 40" 3' 3' 20" 3' 40" 4' 4' 20" 4' 40" 5' 5' 20" 5' 40" 6' 6' 20" 6' 40" 7; 7 20" 7 40' Seconds. Diff. 663o 66D0 6670 669c 672c 674c 6760 67S0 6800 683o 685o 6870 6890 69 10 6930 69D0 6970 6990 7010 7o3o 7060 7080 7100 7120 7140 7160 7180 7200 7220 7240 7260 7280 7300 7320 7340 7360 7380 ^400 7420 7440 7460 7480 7000 7D20 7340 7D60 7580 7600 7620 7640 7660 i5o IDI 1D2 i53 1D4 IDD 1 56 i57 IDS 159 160 161 162 i63 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 18, 182 1 83 184 i8d 186 187 188 189 190 191 192 193 194 19D 196 197 198 199 Angles. 7 40' 8' 8'i5' 8' 3o' 8 5o' 9 lo' 9' 3o' 9' 5o' 10' 10' 10' 20' 10' 40' 10' 55' u' i5' 11' 35' 11' 55' 12' i5' 12' 35' 12' 55' i3' i5' i3'35' i3'5o' 14' 10' 14' 3o' 14' 45' i5' 5' i5' 20' i5'4o' i5' 55' 16' i5' 16' 3o' 16' 5o' 17' 5' 17' 25' 17' 40' 18' 18' i5' 18' 35' 18' 55' 19' i5' 19' 3o' 19' 45' 20' 5' 20 20' 20' /iO' 20' 55" 21' i5" 21' 3o" 2i'45" 22' 5" 22' 20" 22' 35" Seconds. 7660 7680 7695 7710 773o 77D0 7770 7790 7810 7820 7840 7855 7875 7895 791D 793d 7955 7973 7995 SoiD 8o3o 8o5o 8070 8o85 8io5 8120 8i4o 8i55 8175 8190 8210 8225 8245 8260 8280 8295 83i5 8335 8355 8370 8385 8405 8420 8440 8455 8475 8490 &DOD 8D25 8540 8555 Diff. 200 201 202 203 204 205 206 207 208 209 210 211 212 2l3 214 2l5 216 217 218 219 220 221 222 223 224 2 25 226 227 228 229 23o 23l 232 233 234 235 236 237 238 239 240 241 242 243 244 245 246 247 248 249 Angles, 2° 22' 35' 22 55' 23' 10' 23' 3c' 23' 45' 24' 24' 20" 24' 35' 24' 55' 25' 10' 2D' 25" 25' 45" 26' 26' 20" 26' 35" 26' 5o" 27' 10" 27' 25" 27' 45" 28' 28' i5" 28' 35" 28' 5o" 29' 10" 29' 25" 29' 40" 3o' 3o'i5" 3o' 3o" 3o' do" 3i' 5" 3r2o' 3r35" 3i' 55' 32' 10" 32' 2D" 32' 40" 32' 55" 23' i5" 33' 3o" 33' 45" 34' 34' i5" 34' 3o" 34' 45" 35' 35' 20" 3d' 35" 35' 5o" 36' 5" 36' 20" Seconds. 8555 8575 8590 8610 8625 8640 8660 8675 869.N 8710 8725 8745 8760 8780 8795 8810 883o 8845 8865 8880 8895 8915 8930 89D0 8965 8980 9000 9015 9o3o 90D0 9065 9080 909D 9115 9i3o 9145 9160 9175 919D 921C 9225 9240 9255 9270 9285 9300 9320 9335 9350 9365 9380 TABLE IV. CONTAININQ THE NATURAL TANGENTS AND COTANGENTS E^^RY DEGREE AND MINUTE OF THE QUADRANT. 80 NATURAL TAXGEJsTS. ] l i 1° 1 r 3° 1 4= 5° 6° 7° 8° 9° n , oooooo 017455 034921! o524o8 069927 087489 io5io4 122785 i4o54i 158384 60 291 746; 52.2 2699; 070219 7782 5398 3980 0837 8683 59 2 582 018037 3281 55o3 ODII 8075 8368 5692 3375 n34 8981 58 3 873 5795 0804 5987 3670 3966 i43i 9279 57 4 001 164 619' 6086 3575I 1096 8661 6281 1728 9577 56 5 454 910 019201 6377 3866 1389I 8954 6575 4261 2024 9876 55 6 745 6668 41 58 1681 9248 7163 4557 2321 160174 54 I 002036 6960 445o 1973 9541 4852 2618 0472 53 327 618 725, 4742 2266 9832 7458 l\% 2915 0771 52 9 020074 7542 5o33 2558 090127 -&% 3212 1069 5i |368' 5o 10 909 365 7834 8125 5325 285i 0421 5738 35o8 003200 656 5617 3i43 0714 8340 7o34 38o5 1666 49 1965 48 12 491 782 021238' 8416 5909 3435 1007 8635 6329 6625 4102 i3 8707 6200 3728 i3oo 8920 9223 4399 4696 2263 47 i4 004072 529 8999 6492 6784 4020 :ii^ 6920 2562 46 i5 363 820 9290 43i3 9518 7216 4993 2860 45 i6 654' 0221 1 1 958i 7076 46o5 2180 9812 75i2 5290 5587 3i5q 44 3458 43 17 945 402 9873 7368 4898 2474 110107 7807 8io3 i8 005236 6o3 527 984 040164 7660 5190 5483 2767 0401 5884 3756 42 19 0456 Ife 3o6, 0695 8399 6181 4o55 41 20 818 023275 0747 5775 3354 0990 1284 8694 6478 4354 40 21 006109! 566 io38 8535 6068 3647 ^ X6776 4652 ii 22 400 857 024148 i33o 8827 636i 3941 :i?? 7073 4951 23 t 1621 9119 6653 4234 9582 7370 525o 37 24 439 1912 941 1 . 6946 4528 2168 9877 7667 5540 36 23 007272! 730 2204 9703 7238 4821 2463 130173 7964 4848 35 26 563! 025022 2493 060287 7531 5ii5 2757 0469 0765 8262 6147 34 27 854 3i3 2787 7824 8116 5408 3o52 8559 8856 6446 33 28 008145 604 3^78 0579 5702 3346 1061 6745 32 29 436 895 3370 0871 8409 5995 3641 1357 9154 7044 3 1 3o 727 026186 366 1 ii63 8702 6289 6583 3936 i652 9451 7343 3o 3i 009018.1 477 3952 1455 8994 9287 423o . 1948 9748 7622 29 32 ^309 768 4244 1747 6876 4525 2244 1 50046 7941 28 8240 27 33 600 027059 4535 2039 9580 7170 4820 2540 o343 34 80. 010181 ^35o 4827 233i 9873 7464 5ii4 2836 0641 8539 26 8838 25 35 641 5ii8 2623 o8oi65 V^. 5409 3x32 0938 1236 36 472! 933 541a 2915 0458 5704 3428 9137,24 37 763' 028224 5701 3207 0751 8345 5999 3725 1533 9437 23 38 oiio54J 5i5 ^\ 3499 1044 8.638 6294 4021 i83i 9736 22 39 345' 806 '^oVz i336 8932 6588 4317 2.29 170035 21 40 ^^^i °^'515 6576 1629 9226 6883 46i3 2426 o334 20 41 6867 4375 1922 9519 7178 'A2 2724 o634 19 0933 18 42 012218: 679 7159 4667 22l5 9813 7473 3022 43 509 970 745o 4939 25o8 J00107 Vf. 55o2 3319 1233; 17 44 800I 030262 7742 8o33 525i 2B01 0401 8o63 5798 3617 i532: 16 45 013091 1 553 5543 3094 o6q5 0989 8358 6094 3915 i83ii i5 46 382{ 844 8325 5836 3386 8653 6390 42i3 2i3i; i4 47 673! o3ii35 8617 6128 3679 1282 8948 6687 6983 45io 243o i3 48 964J 426 8908 6420 3972 1576 9243 4808 2730 1 2 49 JI4254 717 9200 6712 4265 1870 9538 7279 7576 5io6 3o3o u 5o 545 032009 1% 7004 4558 2164 9833 5404 3329 lO 5i 836i 3oo 7296 485 1 2458 120128 7872 5702 3629 9 3920 8 4228 7 52 '% It 050075 7589 5i44 2752 0423 8169 8465 6000 53 0366 7881 ^.73 5437 3o46 0718 6298 54 709' o33i73 0658 5730 ! 3340 ioi3 8761 6596 4528 6 55 016000! 465 1 0949 8465 6023 1 3634 ^^^ 9o58 6894 4828; 5 56 291! 756 1 1241 8738 63i6 ! 3928 1 i6o4 9354 7192 5l27|4 57 582 1 o34o47 I i533 9o5o 6609 4222 1 1899 965 1 i£ 5427' 3 58 873 338 017164 63o 1824 9342 6902 ' 45i6 ' 2194 2489 9948 5427: 2 ^ 2116 9635 7196 4810 , 140244 6027 I 89° 88° 87° 86° 85° 84° 83° 82° 1 81° 80^ Na tural C o-tangf mts. P. to j:-, 4-CJ 4-85 I 4-86 4-87 4-88 1 4-89 4-91 4.93 I4.96 1 4-98 NATTJKAL TANGENTS. 81 is 10° 11° 12° 13° 14° 15° 16° 17° 18° 19° "1 176327 194380 212557 23o868 249328 267949 286745 3o573i 324920 344328' 60 6627 4682 2861 1175 9637 8261 7060 6049 5241 4653, 59 4978: 58 6927 4984 3i65 i48i 9946 8573 7375 6367 5563 7227 5286 3469 1788 250255 8885 & 6685 5885 53o4i 57 7527 5588 3773 2094 o564 9197 7003 6207 563o 56 7827 8127 5890 4077 2401 0873 9309 8320 7322 6528 5955 55 6 6192 4381 2707 Ii83 9821 8635 7640 685o 6281 54 I 8427 6494 4686 3oi4 1492 270133 8950 7959 8277 7172 6607! 53 8727 6796 4990 3321 1801 0445 9266 7494 6933 52 9 9028 7099 5294 3627 2111 0737 9581 8396 7817 328139 7259 5i 10 179328 I 9740 I 215599 233934 252420 271069 289896 308914 347583 5o II 9628 77o3 5903 4241 2729 i382 290211 9233 8461 79" 49 12 9928 8oo5 6208 4548 3039 3348 1694 0327 9552 8783 8237' 48 i3 180229 83o8 65i2 4855 2006 0842 9871 310189 9106 8563' 47 14 0329 8610 6817 4162 3658 2319 ii58 9428 8889' 46 i5 0829 8912 7121 5469 3968 263 1 1473 o5o8 975i 92i6|43 i6 ii3o 92i5 7426 5776 4277 2944 1789 0827 330073 9542 44 \l i43o 9517 7731 5o83 4587 3256 2103 1 146 0396 9868 43 I73i 9820 8o35 6390 4897 3569 2420 1465 0718 3501951 42 19 2o3i 200122 8340 6697 5207 2555i6 3882 2736 1784 1 041 0522 41 20 182332 200425 218645 237004 274194 293o52 312104 33i364 350848: 40 21 2632 0727 8950 73.2 5826 4307 3368 2423 1687 1175 39 22 2933 io3o 9254 7619 6i36 4820I 3684 2742 2010 i5o2'38 23 3234 1333 9559 IVm 6446 5i33 4000 3062 2333 1829' 37 24 3534 1635 9864 6756 5446 43i6 338i 2656 2156; 36 25 3835 1938 220169 8541 7066 5759 4632 3700 2679 24831 35 26 4i36 2241 0474 8^8 7377 6072 4948 4020 3302 2810' 34 27 4437 2544 0779 1084 9156 7687 6385 5265 4340 3623 3i37;33 28 4737 2847 9464 7997 6698 558i 4659 3949 3464' 02 29 5o38 3149 1 389 9771 83o7 701 1 5897 4979 4272 379i|3i 3o 185339 203432 221693 240079 2586i8 277325 296213 3i5299 334393 354119: 3o 3i 5640 3755 2000 o386 8928 7638 653o 5619 4919 4446| 29 32 5941 4o58 23o5 0694 9238 795 1 6846 5939 5242 4773 28 33 6242 436i 2610 1002 9549 8265 7163 6258 5566 5ioi 27 34 6543 4664 2916 i3io 9859 8578 7480 6578 5890 5429' 26 35 6844 4967 3221 1618 260170 8891 7796 6899 62i3 5756I 25 36 7145 5271 3526 1925 0480 9205 8ii3 7219 6537 6084 24 ll 7446 5574 3832 2233 0791 9519 843o 7539 6861 6412 23 8048 lUl 4i37 2341 1102 9832 8747 9063 7839 8179 7i85 6740 22 39 4443 2849 i4i3 280146 75og 7068 21 357396 20 40 188349 206483 224748 243137 261723 280460 299380 3i85oo 337833 81 57 41 865 1 6787 5o54 3466 2o34 0773 9697 8820 7724! 10 8o32i 18 42 8952 7090 5360 3774 2345 1087 300014 9141 8481 43 9253 7393 5665 4082 2656 1401 o33i 9461 8806 838o' 17 44 9555 til 5971 4390 2967 3278 1715 0649 9782 9i3o 8708, 16 45 9856 6277 4698 2029 0966 320103 9434 9037! i5 46 190157 83o4 6583 5007 53i5 3589 2343 1283 0423 9779 9365j 14 S 0459 8607 6889 3900 2657 1600 0744 340103 9694I i3 0760 891 1 7194 5624 4211 2971 1918 1063 0428 36oo22i 12 49 1062 9214 7300 5932 4523 3286 2235 i386 0752 o35i} 11 5o 19x363 209518 227806 246241 264834 283600 302353 321707 341077 360679I 10 5i 1665 9822 8112 6549 6858 5i45 3914 2870 2028 1402 1008 9 52 . 1966 210126 8418 5457 'M 3i88 2349 1727 1337! 8 •53 2268 0429 0733 8724 7166 5768 35o6 2670 2032 16661 7 54 2570 9o3i 7475 7784 8092 6079 4857 3823 IZ 2377 1995I 6 55 2871 1037 9337 6391 5172 4141 2702 23241 5 56 3173 i34i 9643 6702 5487 4439 3634 3027 26531 4 J"' 3475 1645 9949 8401 7014 58oi 5095 3955 3332 2982 33i2 3 58 3777 4078 1940 2233 230253 8710 7326 6116 4277 3677 2 59 o562 9019 7637 643 1 541 3 4598 4002 3641 _^ 79' 78° 77° 7G° 75° 74° 73° 72° 71° 70° n Natural Co-tangeuts. to] ^; 5.01 5-05 5.09 5-13 5.17 5.22 5.27 .33 5-39 5.46 82 NATURAL TA:N-GE.NTS. 20= 21= 22° 23° 24° 25° 26° 27° 28° 1 29° 1! ^1?^ 383864 404026 424475 445229 4663o8 487733 509525J 531700 8093, 98921 2083 554309 60 ^J?? 4365! 4818 5577J 6662 4689 5o 5070, 58 4629 47o3 5 1 62 5926 7016 8433, 510258 2456 3 nu 4866 5o42 55o5 6275 7371 88i3| 0625 2829 32o3 545o 57 4 5200 538o 5849 bbi^i 7725 6973 8080 9174 X 583 1 56 5 56i8 5534 5719 6192 9534 3577 6212 55 6 5948 5868 6o58 6536 7322 8434 9895 1726 3950 6593; 54 I 6278 6202 6397 6880 6736] 7224 7671 8789 8020 9144 490236 2093 4324 6974: 53 7333 32 6608 6536 0617 0978 2460 4698 9 6938 6871 7075J 7568 8369 9499 2828 5072 7736 5i 10 367268 387205 407414 427912 448719 469834 491339 5i3i95 535446 5581 18 5o II 7598 7540 7753 8206 9068 470209 1700 5821 88??, 4? 13 Itt 7874 6209 1% 8601 9418 o564 2061 3930 6195 i3 8945 9768 0920 2422 4298 6370 9263 47 14 8589 8544 8771 9289 450117 1275 2784 4666 6945 7319 96451 46 i5 8919 8879 9111 9634 0467 i63i 3i45 5o34 56002 7 1 45 i6 [I 9250 958 1 9214 9549 9450 9790 4ioi3o ^m 0817 1167 2342 35o7 3869 5402 0770 0409 44 0791143 991 1 9884 0668 ml 2698 4231 6i38 8443 1174 42 '9 370242 3902191 0470 ioi3 3034 4593 6307 8820 1556 4. 20 370573 390554 4 I 08 10 43i358 452218 473410 53 1 7 516875 539.95 561939' 40 2J22 39 21 0804 0889 ii5o 1703 2568, 3766 ' '^^H 9571 22 1235 122D 1490 i83o 2048 2919: 4122 5679 7613 u^it 27o5i 38 23 1566 i56o 23o3 2739 3269 4478 6042 7982 b35i 3o83 37 24 1897 1806 223l 2170 3620 4835 6404 0698 3471 36 25 2228 25ll 3o84 3971 4322 5i,i 6767 8720 \Z 3854 35 26 2559 2567 2903 285i 343o 5548 7i3o 9089 4238| 34 27 2890 3192 3532 3775 4673 5905 7492 9458 1826 4621! 33 28 3222 3239 4121 5o24 6262 7835 6218 9828 2203 5oo5' 32 29 3553 3574 3873 4467 5375 6619 520197 2579 5389' 3i 3o 373885 393910 414214 434812 455726 ''% 498582 52o567 542956 3332 563773 6i37 3o 3i 4216 ss 4554 5i58 6078 8045 93o8 0937 i3o7 29 32 4548 48o5 5236 5504 6429 7690 'Se 6541 2« 33 4880 49' 9 5850 6781 8047 9672 1677 6925 7310 27 34 52II 5253 5577 ?197 6543 7i32 84o5 5ooo35 2047 4463 26 35 5543 5592 5919 7484 8762 0399 0763 2417 4840 4694 8079 25 36 5875 5928 6260 6889 7836 8188 9120 2787 5218 24 3? 6207 6265 6601 7236 9477 1127 3i58 5595 8464 23 38 6539 6601 6943 7582 8540 9835 1491 3528 635o 8849 22 39 6872 6938 7284 7929 8892 480193 i855 J899 9234 21 40 377204 397273 417626 438276 459244 48o55i 502219 2583 524270 546728: 569619 20 41 7536 7611 Itl 8622 9596 0909 4641 7106 570004 '9 42 7869 8201 It ,t? 9949 1267 2948 33i2 50I2 7484 0390 18 43 865 1 46o3oi 1626 5383 7862 8240 0776; 17 44 8534 8622 tnl 9663 o654 2343 3677 5754 1161 16 45 8866 8960 44001 1 1006 4041 6125 8619 1547 i5 46 l^. 9297 9634 9677 o358 1359 2701 4406 t^l ^l io33 14 u 420019 0795 1712 3o6o 4771 2319 i3 2703! 12 9864 9971 o36i 1033 2065 3419 5i36 7240 9755 49 380197 38o53o 400309 0704 1400 2418 3778 55o2 76.2 55oi34 3092' 11 5o 400646 421046 441748 462771 484.37 5o5867 ''Ite 55o5i3 5734781 10 5i o863 00841 1389 2095 3i24 4496 6232 0892 3865 9 4252! 8 52 1 196 l322 1731 2443 3478 383 1 4833 6598 8728 i6?o 53 i53o 1660 2074 2791 5214 7329 9100 4638 7 54 i863 m 2417 3i?9 4i85 5574 9473 2o3o 5o26 6 55 2196 253o 2759 3487 4538 5933 2695 8061 9845 2409! 54.3| 5 56 2673 3l02 3835 4892 6293 530218 27891 5800; 4 u 2863 3ou 3445 41 83 5246 6633 8427 0591 3169 6187 3 l^l 335o 3788 4532 5600 70i3 It ■^d 3549 6575 2 59 3688 4132 4880 5954 7373 3929 6962 I GD" 68° 67° 66° 65° 64° 63= 62= 61° 60= Natural Co-tangents. for. 5-53 560 5 68 5 76 5 S5 5 95 6 05 6 16 6.28 1 6 40 NATURAL TANGENTS. 83 d 5 30° 31° 32° 33** 34° 35° 36° 37° 38° a.j 577350 600861 624869 649408 674509 700208 726343 753554 781286 809784 60 , I 1738 I25l 5274 ' 9821! 4q32| 0641 650235 5355! 1075 6987! 4010 1754 810266 59 2 8126 i653 5679 7432 4467 4Q23 538o 2223 0748, 58 3 85i4 204(5 6o83 i 0649' 5779' 1 509' 7877 2692 123o!5t 4 8903 2445 6488 io63 62o3| 1943 8322 3i6i 1712 56 5 IX 2842 6894 '477 6627 2377 8767 92i3 5837 363 1 2195 55 6 323? 7299 1892 7031 2812 6294 6731 4x00 2678 54 7 580068 3633 7704 8110 1 23o6 7475 3246 9658 4570 3i6i 53 8 0457 4o32 2721 7900I 368i 730x04 7209 5o4o 3644 52 9 0846 4429 85i6 3x36 8324 41 16! " 035o 7667 ' 758125 ! 55io 4x28 5i 10 581235 604827 628921 • 9327 1 653551 678749 704551 730996 785q8i! 814612 5o II 1625 5224 3966 4382 9n4 4987 1443 85831 6431 55?ol 4? 12 2014 5622 9734 9399 5422 1889 9041 XI i3 24o3 6019 63oi4o 4797 52i3 680023 5858 2336 9300 6o65! 47 14 lltl 6417 0546 0430 6294 6730 2783; 9Q3Q iSb5\ 6549! 46 83361 7034; 45 i5 68i5 0953 i36o 5629 0876 3230 7604x8 i6 3573 7213 6043 l302 7166 3678 0877 88081 75iQ 44 \l Itl 761 1 1767 6461 1728 7603 ^039 4x25 i336 9280 8oo5| 43 8010 2174 6877 2x54 4573 1796I 9752 849X 42 8076 41 '9 4743 8408 258i 7294 258o 8476 502I 2236 700223 20 585i34 608807 632988 3396 657710 683007 3433 708013 9350 735469 762716, 790697] 8I9463I40I 21 5524 9205 8127 5qi7 6J66 3176 1170 qq4q'3o 22 5qi5 63o6 9604 38o4 8544 3860 9788 3636 1643 8204351 38 23 6iooo3 4211 8961 4287 710225 68i5 ss 2117 09221 37 24 ^l o4o3 4619 9379 4714 0663 7264 2390 1409 36 23 0802 5o27 9796 5i42 IIOI 7713 5oio 3o64 1897 35 20 7479 1201 54361 660214 5569 1539 8162 5480 3538 2384 34 27 2»70 8262 1601 5844 o63i 5997 '977 8611 5941 4012 2872 33 28 2001 6253 1049 6425 2416 9061 6403 4486 3360 32 29 8653 2401 6661 1467 6853 2854 9311 6865 496 X 3848 3, 3o 589045 6I280I 637070 66x886 687281! 713293 739961 767327 795436 824336 3o 3i 9437 3201 7479 23o4 7709' 3732 7404x1 7789 5911 6386 4825| 29 32 9829 36oi 7888 8298 2723 8x38; 4171 0862 8252 53x4: 28 33 590221 4002 3i4i 8567 46x1 i3x2 8714 6862 58o3| 27 34 o6i3 4402 8707 3560 8995 5o5o 1763 9177 8290 6292; 20 67821 25 35 1006 4803 9117 ]%t 9425 5490 5930 6370 2214 9640 36 l3q8 5204 9527 9854 2666 770x04 7272 24 37 1791 56o5 9937 48x8 690283 3117 o567 8706 •7762 8252 23 38 2184 6006 640347 5237 07x3 6810 3569 io3i 9242 22 39 2077 64o8 0757 5657 1 143 7250 4020 1495 9719 8743 21 40 592970 616809 641 167 666077 69x572 717691 744472 771959 242J 800 1 96 829234 20 41 3363 7211 1578 6497 2oo3 8i32 4925 0674 9725 \l 42 3757 7613 8oi5 1989 6917 2433 8573 5277 2888 n5i 83o2i6 43 4i5o 2399 7337 2863 9014 583o 3353 1629 0707 17 44 4544 84x7 2810 2758 8179 3294 9455 6282 38x8 2107 1199 16 45 iti 8819 3222 3725 720339 6735 4283 2585 'X i5 46 9221 3633 8599 4x56 7189 4748 3o63 14 % 5725 9624 4044 9020 4387 5oi8 0781 7642 8096 5214 3542 2676 i3 6120 620026 4456| 9442 1223 568o 4021 3169 12 49 65i4 0429 4868, 9863 545o 1665 8549 6146 4300 3662! Ill 5o 596908 620832 645280 670284 695881 722108 749003 776612 804979' 834155! 10 1 5i 73o3 1235 5692 0706 63x3 255o 9458 7078 5438 4648 qI 52 8488 i638 6104 1128 6745 If, 75o366 7545 5938 5i42 8 53 2042 65i6 i55o 7177 8012 6418 5636 I 54 2445 6929 1972 7610 387q 0821 8479 6898 6i3o 55 8883 2849 7342I 2394 8042 4323 1276 8946 7379 6624 5 56 9278 3253 7755 2817 8475 4766 1731; 94x4 8340 7119 4 57 967^ 3657 8168 3240 8908 9^41 5210 2187 9881 7614 8109 3 58 600069 0465 4061 858 1 3662 5654 2642 780349 8821 2 59 4465 8994 4085 9774 6098 3098 0817 93o3 8604 ' 59° 58° 57° 5G° 55° 54° 53° 52° 51° 50° a Natural Co-tangeuts. 1 ^^^6 53 fc.67 6.82 6.97 7.14 7-3i 7-5o 7.70 7.92 8.24 1 81 NATUKAL TAI^GENTS. 5 40° } 41° 42° 43° 44° 45° 46' 47° 48° 49° 60 839I00I 869287 900404 9325 1 5; 9656S9 I- 00000 1-03553 1-07237 1.11061 i-i5o37 I 9595; 9-98 0931 3o59 6231 oo58 36i3 7299 1126 5io4 It 2 840092: 870309J 1458 o588 0820! iq85 36.3. 6814 0116 3674 7362 \IU 5172 3 4148 7377 0175 3734 7425 5240 57 4 1084 i332 25i3 46o3 '^r, 0233 3794 7487 1321 53o8 56 5 i58i 1843 3o4i 5238 0291 3855 7550 1387 5375 55 6 2078 2356' 3569 5783 9067 o35o 39i5 7613 1452 5443 54 7 2575 2868 409S 6329 9632 o4o8 3976 7676 I5i7 55ii 53 8 3073 338i 4627 6873! 970196 0467 4036 7738 7801 1582 5579I 52 1 9 3571 3894 5 1 56 7422I 0761 o525 4097 1648 5647 31 10 844069 874407, 905685 937968 971326 i-oo583ji.o4>38 1.07864 1.117,3 1-15715 5o II 4567 4920 6215 83i5 1892 0642 42l» 7927 1778 til It 12 5o66 5434 6745 9o63 2458 0701 4279 7990 1844 i3 5564 5948 7275 9610 3o24 0759 4340 8033 1909 1973 5919 47 i4 6o63 6462! 7805 940 1 58 3590 0818 4401 8116 5987 46 ID 6562 6976^ 8336 0706 4157 0876 4461 8179 2041 6o56 45 i6 7062 nil 8867 1255 4724 0935 4522 8243 2106 6124 44 \l 7562 8062 9398 9930 i8o3 5291 0994 4583 83o6 2172 6192 43 8521 2352 5839 1053 4644 8369 2238 6261 42 '9 8562 9037 910462 2902 6427 1112 47o5 8432 23o3 6329 1.16398 41 20 849062 879553 ■910994 943451 976996 1-01170 1-04766 1-08496 1.12369 40 5 ^ 9563 880060 i526 4001 7364 8i33 1229 1288 4827 8539 2435 6466 59 22 850064 0585 2o59 4552 4888 8622 25oi 6535 38 23 o565 1102 2592 5l02 8703 i347 4949 8686 2567 66o3 i'^ 24 1067 1619 3i25 5653 9272 1406 5010 8749 2633 6672 36 23 1 568 2i36 3659 4193 5204 9842 1465 5072 5i33 881 3 2699 6741 35 26 2070 2653 6756 980413 i524 8876 2765 283 1 6809 6878 34 27 2573 3i7i 4727 7307 0983 1 583 5194 8940 33 28 3075 3689 5261 7859 i554 1642 5255 9003 2897 6947 32 29 3578 4207 5796 8412 2126 1702 53,7 9067 2963 7016 3i 30 854081 884725 9i633i 948965 982697 1-01761 1.05378 1.09131 I- 13029 1-17085 3o 3i 4584 5244 6866 9318 3269 1820 5439 9195 3096 7154 11 32 5087 5763 7402 950071 3842 1879 55oi 9238 3x62 7223 33 5591 6282 7938 0624 4414 2037 5562 9322 3228 7292 27 34 6095 6802 8474 1178 4987 5624 9386 3295 7361 26 35 6599 7321 9010 1733 5560 fei34 5685 945o 336i 743oj 23 36 7104 7842 8362 9547 2287 2117 5747 5^09 9514 3428 75oo 24 37 7608 920084 2842 6708 2176 9578 3494 756Q| 23 38 81 13 8882 0621 3397 7282 2236 5870 9642 356i 7638 22 39 8619! 9403 ii59 3953 7857 988432 2295 5932 9706 3627 7708 21 40 859124; 889924 921697 954308 1-02355 1-05994 1-09770 1.13694 1-17777 20 41 963o 890446 2235 5o64 9007 24i4 6036 9834 3761 7846 \t 42 8601 36 0967 2773 5621 9582 lili 6117 9899 3828 7916 43 0642 1489 3312 6177 6734 990158 6179 9963 3894 7986 17 44 1148 2012 385i 0735 2593 6241 1-10027 3961 8o55 16 45 1655 2534 4390 49^0 7292 i3ii 2653 63o3 0091 01 56 4028 8i25 i5 46 2162 3o57 7849 1888 2713 6365 4095 8194 14 47 2669 358o 5470 8407 2465 im 6427 0220 4162 8264 i3 48 Ull 4io3 6010 8966 9524 3o43 6489 0285 4229 8334 12 49 4627 655i 3621 2892 6551 0349 4296 8404' 11 1 5o 864193 895i5i 8174 960083 994199 1-02932 1.06613 1.10414 I- I 4363 1-18474, 10 1 5i 4701 5675 0642 4778 3012 6676 0478 443o 8544 t 52 5209 5718 6199 1202 5357 3072 6738 o543 4498 4565 8614 53 6724 8715 1761 5936 3i32 6800 0607 8684 I 54 6227 7249 9257 2322 65i5 3192 6862 0672 4632 8754 55 6736 7774 9800 2882 7095 3232 6925 0737 4699 8824 5 56 7246 8299; 93o342 3443 7676 33l2 6987 0802 tltl 8894 4 u 2756 8266 8823 o885 4004 8256 3372 3433 7049 0867 8964 3 9351 1428 4565 8837 9418 7112 0931 4902 9035 2 59 8776 9877 1971 5127 3493 7174 0996 4969 9105 1 49° 48° 47' 46° 45° 44° 43° 42° 41° 40° Natural C 0- tangents. ro?^«-38 8-64 8-92 9-21 9-53 O-OQ 1. 02 1.06 I>10 ,.,5 ] NATURAL TANGENTS. 85 ifi. 50° 51° 52° 53° 54° 55° 56° 57° 58° 59° n 1-19175 1.23490 1-27994 1-32704 1-37638 i.428i5'i-48256 1-53986 i-6oo33 1-66428 60 I 9246 3563 8071 2785 7722 2903 8349 4o85 0137 6538 59 6647; 58 2 9316 3637 8148 2865 7807 2992 8442 4i83 0241 3 9387 3710 8225 2946 7891 3o8o 8536 4281 0345 6757 57 4 nil 3784 3858 83o2 3026 ^ 3169 3258 8629 4379 4478 0449 o553 6867 56 6978 55 7088 54 5 lilt 3io7 8722 8816 6 9599 3931 3187 8145 3347 4576 0657 I 9669 4oo5 8533 3268 8229 3436 8909 90o3 4675 0761 7198,53 9740 iVi 8610 3349 83i4 3525 4774 o865 7309 52 9 9811 8687 3430 i.384?4 36i4 9097 4873 0970 7419 5i 10 I. 19882 1-24227 1.28764 I-335I1 1-437031.49190 1.54972 1.61074 1-67530 5o 9953 43oi 8842 3592 8658 3792 9284 507. 'AM 7641149 7752, 48 12 1-20024 4375 8919 3673 8653 388i 9378 §170 i3 0095 444Q 4523 8997 3754 8738 3970 9472 5269 5368 1388 7863' 47 14 0166 9074 9i52 3835 8824 4060 9566 1493 79741 46 8o85 45 i5 0237 4597 3916 8909 4i49 9661 5467 1598 i6 o3o8 4672 9229 3998 8994 5239 9755 9849 5567 I2o8 8196 44 \l 0379 4746 9307 4079 mt 432Q 4418 5666 83o8! 43 045 1 4820 9385 4160 9944 5766 5866 1914 8419 42 '9 0522 4895 9463 4242 9230 45o8 i.5oo38 2019 8531I41 20 1-20593 1-24969 1-29541 1-34323 1-39336 1-44598 4688 i.5oi33 1.55966 1.62123 1-68643; 40 21 o665 5044 9618 44o5 9421 02 2B 6o65 223o 8754 39 8866; 38 22 0736 0808 5ii8 9696 4487 9507 4778 486S 0322 6i65 2336 23 5193 9775 4568 9593 0417 6265 2442 8979' 37 24 0879 5268 9853 4650 9679 4958 05l2 6366 2548 9091136 25 0951 5343 9931 4732 9764 5o49 0607 6466 2654 9203 35 26 1023 5417 1-30009 4814 985o 5i39 0702 6566 2760 9316' 34 27 1094 5492 0087 4896 9936 5229 0797 6667 2866 9428: 33 28 1166 5567 0166 . 497« 1.40022 5320 0893 0988 t&l 2972 9541 32 29 1238 5642 0244 5o6o 0109 5410 3079 9653' 3 1 3o I.2l3lO 1-25717 1-30323 I-35I42 1-40195 I -45501 1.51084 1-56969 i.63i8D i.69766|3o 3i i382 5792 0401 5224 0281 5592 1170 1275 7069 3292 9879 20 9992 28 32 1454 5867 0480 5307 0367 5682 7170 3398 33 i526 5943 0558 5389 0454 lul 1370 7271 35o5 I. 70 106 27 34 1598 6018 0637 5472 o54o 1466 7372 3612 0219 26 35 1670 6093 0716 5554 0627 5955 i562 7474 3719 3826 o332 25 36 1742 6169 0695 5637 0714 6046 1658 7575 0446 24 11 1814 6244 0873 5719 0800 6i37 i85o 7676 3934 o56o; 23 1886 63i9 639! 0952 5802 0887 6229 7778 7879 4041 0673 22 0787 21 .39 1959 io3i K 5885 0074 6320 1946 4148 40 I-2203l I -2647 1 I-3III0 1.35968 1-41061 1.46411 1.52043 '•% 1-64256 1.70901 20 41 2104 6546 1190 6o5i 1 148 65o3 2139 4363 ioi5! 19 42 2176 6622 1269 1348 6i34 1235 6595 223d 8184 4471 II29'l8 43 2249 6698 6217 l322 6686 2332 3286 4579 1244! 17 44 2321 6774 1427 63oo 1409 6778 2429 8388 4687 i358l 16 45 2394 6849 6925 1 507 6383 1497 6870 252D 8490 4795 1473 i5 46 2467 1 586 6466 1 584 6962 2622 8593 4903 i588l 14 tl 2539 7001 1666 6549 1672 7053 lU 8695 Soil 1702J i3 2612 7077 1745 6633 1759 7146 8797 5l20 i8i7j 12 49 2685 7153 1825 6716 1847 7238 2913 8900 5228 1932' II 5o 1-22758 1-27230 1-31904 1-36800 1-41934 1-47330 I.530IO 1-59002 1-65337 5445 1.72047I 10 5i 283 1 7306 1984 6883 2022 7422 3107 3205 9105 2163 I 52 .2904 7382 2064 6967 2110 7514 9208 5554 2278 -63 2977 7458 J144 7o5o 2198 2286 7607 3302 9311 5663 2393 7 54 3o5o 7535 2224 7134 7699 3400 9414 5772 262: 6 55 3i23 7611 23o4 7218 2374 ]lll 3497 95,7 5881 5 56 3196 7688 2384 7302 2462 3595 9620 5990 2741 2857 2973 4 57 3270 7764 7841 2464 7386 255o 7977 8070 3693 9723 9826 6099 3 58 3343 2544 7470 2638 3791 6209 2 59 3416 7917 2624 7554 2726 8i63 3888 9930 63i8 3089 1 39* 38' 37° 30° 35- 34° 33° 32° 31° 30° a Natural Co-tangeuts. P. to v; 1-20 1-25 i.3i 1-37 1-44 i.5i 1.59 1-68 1.78 1.88 86 NATURAL TANGENTS. 60° 6r 62° 63° 64° 65° 66° 67° 68° 69° ~\ 1.73205 i.8o4o5 1.88073 1-96261 2.o5o3o 2.i445i 2-24604 2-35535.2-47509 2-6o5o9 6o| 3321 o529 8205 6402 5i82 4614 4780 5776 7716 0736 59 0963, 5§ 3438 o653 8337 6544 5333 4777 4956 5967 3924 8i32 3555 0777 8469 6685 5485 4940 5i32 61 58 1190 57 3671 0901 8602 6827 5637 5io4 5309 6349 8340 1418 56 5 8788 To25 8734 6969 5790 5268 5486 6541 8549 8758 1646 55 6 3905 ii5o 8867 7111 5942 5432 5663 6733 1874 54 I 4022 1274 9000 7253 6094 5596 5840 6925 8967 2io3 53 4140 ,399 9133 7395 6247 7538 6400 5760 6018 7118 l^sl 2332' 52 9 4257 1 524 9266 5925 6196 73ii 256i;5i lO 1-74375 1-81649 1.89400 1-97681 2-06553 2 . 16090 2 • 263742 - 37504 6255 65521 7697 2.49597 2-62791! 5o| II 4492 1774 9533 7823 6706 9807 3o2i! 49 3252 48 12 4610 1899 9667 7966 8110 6860 6420 6730 IX 2-50018 i3 4728 2025 9801 7014 6585 6909 6751 7088 0229 3483 47 14 4846 2i5o 9935 8253 7167 t?,t 0440 37i4i 46 i5 4964 2276 1.90069 0203 8396 7321 6917 7267 o652 3945; 45 i6 5082 2402 8540 7476 7083 7447 8668 0864 4177 44 \l 5200 2528 0337 8684 763o 7249 7626 8863 1076J 4410^ 43 1289 4642 42 53i9 2654 0472 8828 7785 7416 7S06 9o58 19 5437 2780 0607 8972 7939 7582 7987 9253 i5o2 4875,41 20 1.75556 1-82906 I. 90741 I -99116 2 • 08094 2-17749 2-28167 2-39449|2.5i7i5 2-65109 40 | 21 5675 3o33 0876 ■ 9261 825o 7916; 8348i 9645 1929 5342 39 5576 38 22 5794 3i59 I0I2 9406 8405 8084 8528| 9841 2142 23 59T3 3286 1 147 9550 856o 825i 87102-40038 2357 58111 37 24 6o32 34i3 1282 9695 8716 8419 8891 0235 2571 6046, 36 25 6i5i 3540 1418 9S41 8872 8587 9073 0432 2786 6281; 35 26 6271 3667 1554 9986 9028 8755 9254 0629 3ooi 65i6 34 U 6390 3794 1690 2-OOl3l 9184 8923 9437 0827 3217 6752 33 65x0 3922 1826 0277 9341 9092 9619 1025 3432 6989' 32 29 6630 4049 1962 0423 9498 9261 9801 1223 3648 7225,3. 3o 1.76749 1-84177 '■Kt 2-00569 2-09654 2-19430 2-29984|2-4i42i 2 -538652 -67462! 3o| 3i 6869 43o5 0715 0862 9811 95992-30167 1620 4082 7700I 29 79371 28 8175; 27 32 6990 4433 2371 9969 9769 9938 o35i 1819 4299 33 7110 4561 25o8 1008 2.10126 0534 2019 2218 45i6 34 7230 4689 4818 2645 1155 0284 2-20108 0718 4734 8414I 26 35 7351 2782 I302 0442 0278 0902 2418 4952 8653! 25 36 7471 4946 2920 1449 0600 0449 1086 2618 5170 5389 56o8 8802; 24 9i3ii 23 11 7592 5075 3o57 1596 0708 0619 1271 2819 7713 7834 5204 3io5 3332 1743 0916 0790 0961 1456 3019 937I] 22 39 5333 1801 2-02039 1075 1641 3220 5827 9612 21 40 1.77955 1.85462 1-93470 2-11233 2-21132 2.3l826'2-43422 2- 56o46;2- 698531 20 1 41 8077 8198 5591 36o8 2187 1392 l3o4 2012 3623 6266,2.70094' iqI 42 5720 3746 3885 2335 1552 1475 nil 3825 6487 o335 18 43 83i9 585o 2483 1871 2o3o 1647 4027 6707 05771 17 44 8441 5979 4023 263 1 1819 2570 423o 6928 0819! 16 45 8563 6109 4162 2780 1992 2756 4433 7.5o 1062 i5 46 8685 6239 43oi 2929 2190 2164 2943 4636 7371 i3o5 14 % 8807 6369 4440 3078 23 DO 2337 3i3o 4839 3317 5o43 7593 1548 i3 8929 6499 4570 4718 1-94858 3227 25ll 25l0 7815 8o38 1792: 12 2o36| 11 49 905 1 663o 3376 2671 2683 35o5 5246 5o 1.79174 1.86760 2-03526 2-12832 2-22857 2-336932-45451 2.58261 2.72281 10 5i 9296 6891 iV3] 3675 2993 3o3o 388 1 ~ 5655 8484 2526 I 52 9419 7021 3825 3Id4 3204 T.^ 586o 8708 2771 53 9542 7i52 5277 3975 33i6 3378 6o65 8932 3017 7 54 9665 7283 54.7 4125 3477 3553 4447 6270 91 56 32631 6 55 9788 74i5 I 4276 3639 3727 4636 6476 9381 3509 5 3756i 4 56 9911 7546 4426 38oi 3902 4825 6682 9606 5? I -80034 7677 58?8 4577 3963 4077 5oi5 6888 9831 4004! 3 58 01 58 7809 5979 4728 4125 4252 52o5 70952-60057 4i5i 2 59 0281 7941 6120 4879 4288 4423 5395 73oa 0283 4499 _i 29° 2S° 27° 26° 25° 24° 23° 22° 21° 20° 3 Natural C o-tangeuts. P. to] ^;>.oo 2-13 2.27 1-44 i.62 a-82 3-05 331 3-6i 3.95 NATUEAL TANGENTS. 87 s 70° 71° 72° 73° 74° 75° 76° 77° 78° 70° 1 2-74748 2.9042i;3-07768 3-27085 3-48741 3-732054-01078 4-33148 4-70463 5-14455 60 I 4997 0696 8073 7426 9125 3640 1576 3723 1137 5236 5q 2 5246 0971 8379 7767 9509 4075 2074 43oo i8i3 6o58 58 3 5496 1246 8685 8109 9894 45i2 2574 4879 2490 6863! 57 4 5746 i523 8991 8452 3-50279 4930 3076 5459 3170 7671 56 5 5996 1799 9298 8795 0666 5388 3578 6040 3S5i 8480 55 6 6247 6498 67D0 2076 9606 9139 io53 5828 4081 6623 4534 9293 54 1 2354 99 '4 9483 1441 6268 4586 7207 521915. 20107: 53 8 2632 3.10223 9829 1829 6709 5092 lit 5906 0923 52 6395 1744 5i 4-77286;5. 22566 5o 9 7002 2910 o532 3-30174 2219 7102 5599 10 a.77254 2-93i8g;3. 10842 3-3o52i 3-52609 3-77595I4.06107 4-38969 9560 II 7507 3468 ii53 0868 3ooi 8040 6616 7978 3391 49 13 7761 8014 3748 1464 1216 3393 8485 7127 4-40152 8673 4218 48 i3 4028 1775 1 565 3785 8931 &. 0745 9370 5o48 4-T i4 8269 4309 2087 1914 ^^?? 9378 i34o 4.80068 588o, 46 i5 8523 4591 2400 2264 9827 8666 1936 C769 6715 45 i6 8778 4872 27.3 2614 4968 3-80276 9182 2334 1471 7553 44 8393 43 9235 43 \l 9033 5i55 3o27 2965 3J17 5364 0726 9699 3i34 2175 2882 & 5437 3341 5761 "77 4-10216 3735 4338 19 5721 3656 3670 6.59 i63o 0736 359o'5-3co8o 4, 1 20 2-79802 2-96oo4'3' 13972 3-34023 3-56557 3-82083 4-11256 4-44942 5548 4-843oo'5. 30928 40 21 2-80059 6288 4288 4377 %] 2537 1778 5oi3 1778 39 22 o3i6 6573 6858 46o5 4732 2992 23oi 6i55 5727 263i|3S 23 0574 4922 5087 V^o 3449 2825 6764 6444 34871 3- 24 o833 7144 5240 5443 3906 4364 335o 7374 7162 43451 36 25 \Z 7430 5558 58oo 8562 3877 7986 7882 86o5 52o6j 35 26 8004 5877 6.58 8066 9370 4824 44o5 8600 6070 34 27 1610 6197 65i6 5284 4934 9215 9330 6936: 33 28 1870 8292 6517 6875 9775 5745 5465 9832 4-90036 78o5; r,2 8677131 29 2i3o 8580 6838 7234 3-6oi8i 6208 5997 4. 5045 1 0785 30 2-82391 2653 2-98868 3.i7i5, 3.37594 3.6o58S 3-86671 4-i653o 4-51071 4-9I3I6 5-39552 3o 3i 9i58 7481 7935 0996 7i36 7064 1693 2249'5-4o429 2g| 32 2914 '^U 7804 8127 83i7 i4o5 7601 7600 8i37 23i6 2984 i3o9 29 33 3176 8679 1814 8068 2941 3721 2192; 27 34 3439 3-00028 845 1 9042 2224 8536 8675 3568 4460 3077 26 35 3702 o3i9 8775 9406 2636 9004 9215 4196 5201 3966 ,5 4837; 24 36 3965 061 1 9100 977 « 3-40136 3o48 9474 9756 4826 5945 3? 4229 0903 9426 3461 994514-20298 5458 6690 57511 23 38 4494 1489 3.01783 9752 0502 3874 3-90417 0842 6091 7438 6648 22 39 4758 3-20079 0869 4289 0890 1387 6726 8188 7548; 21 40 2-85023 3.20406 3.41236 3-64705 3-91364 4-21933 4-57363 8001 4-98940 5-48451 20 ' 41 5289 2077 0734 1604 5l2i 1839 2481 96951 9356 \t 42 5555 2372 io63 1073 2343 5538 23i6 3o3o 8641 5-00451 '5-50264 43 5822 2667 2963 1392 If^l 2793 358o 9283 1210! 1176 T> 44 6089 1722 2713 3271 4i32 9927 1971 2090 j6 45 6356 3260 2o53 3o84 6796 3751 4685 4-60372 2734 3007 15 46 6624 3556 2384 3456 7217 4232 5239 1868 3499 Itu '4 47 6892 3854 2715 3829 7638 8061 4713 5793 4267 i3 48 7161 4i52 3o48 4202 5196 6332 25i8 5o37 ^111 12 49 7430 4450 338i 4576 8485 56bo 6tjii 3i7ii 5809' 6706 ni 5o 2.87700 3-04749 3-23714 3-44951 5327 3.68909 9335 3-96155,4-27471 4-63825 5-06584 5- 7638 10 73001 8573. 5i 7970 'J049 4049 665 1 8o32 4480 K 8240 5349 4383 5703 9761 7139 8505 9139 5i38 8139! 95i> b sT 85ii 5649 47:9 6080 3.70188 7627 5797 6458 892 r5- 6045 2 7 54 8783 5950 5o5d 6458 0616 8117 9724 97o4i i3q7 6 55 9055 6252 5392 6837 . 1046 8607 4-30291 7121 5.10490 2344 6 56 9327 6554 5729 7216 1476 9099 0860 llf. 1279! 32,5 4 57 9600 6857 6067 7596 2338 9092 i43o 2069! 45 48 c 58 9873 7160 6406 mi 4-000S6 2001 9121 2862 5^05 2 59 2-90147 7464 6745 2771 o582 2573 979' 3658 666 I 19° 18° 17° 1G° 15* 14° 13° 12° 11° 10" ■"1 i i Natural Co-tanareuts. ^ i J 1 "pi to 1^3= . 482 5 36 6 01 6 79 7 73 ' 8 90 10 35 12 20 ! 14.60 J 88 NATURAL TANGENTS. 5 80- 81° 82° 83° 84° 85° 8G° 87° 88° 89° n 5-67128 6.31375 7.ii5378.i4435'9.5i436 1 11.4301 14.3007 19.0811 28-6363*57-2900 60 I 8094 9064 2566 3o42 6398 4106 4685 3607 1879 877i'58.26i2, 59 2 3761 4553 8370 6791 5072 4212 2939 29-1220,59-2659' 58 37ii,6o-3o58 57 3 5.70037 4961 6071,8-20352 0400 5461 4823 4o5i 4 ioi3 6i65 7594 23449-62205 5853 5438 5i56 624561-3829 56 5 1992 nr, 9.25 4345 4935 6248 6059 6273 882362-4992 55 6 2974 7-20661 6355 7680 6645 6685 7403 8546 30-1446 63 -6367 54 I 3960 q8o4 2204 8376:9.70441 7045 7317 41 16 64-8580 53 4949;6-4i026 3754J8-30406 3217 7448 7954 9702 20-0872 6833,66-1055 52 '9 5941 2253 53 10 2446 6009 7853 8D96 959967-4019 5i 10 5.769376-43484 7.2687318-344969-78817 11.8262 14.9244 20-2056 3i.24i668-75oi 5o II Ifs 4720 8442 65d59-8i64i 8673 9898 3253 5284170- 1533 40 82o5|7i-6i5i 48 la 5961 7.30018 8625 4482 9087 i5.o5d7 4465 i3 9944 7206 1600 8.40705 7338 9504 1222 5691 6932 32-1181 73.1390 47 14 5.80953 8456 3190 4786 27059.90211 9923 1893 42i3 74.7292 46 i5 1966 9710 480 3ioi IS -0346 2571 8188 7303176.3900 45 1 i6 29826-50970 6389 7007 6007 0772 3254 9460I33-0452I78-1263 44 1 \l 4001 2234 7999 9128 8931 1201 3943 21.0747 3662 79-9434 43 693581-8470 42 5024 35o3 9616 8.5125910.0187 i632 4638 2049 «9 6o5i 4777 7-41240 34o2 0483 2067 12-25o5 5340 3369 34-027383-8435 4, 20 5-87080 8114 6-560557.42871 8.55555 10-0780 15.6048 21.4704 34-3678:85-9308 40 • 7i5i 88.14^6 39 35-0965,90.4633 38 43 1 3192.9085:37 21 7339 4509 7718 1080 2946 6762 6o56 22 9i5i 8627 61 54 9893 i38i 3390 3838 7483 8211 7426 88i3 23 5-90191 9921 7806 8-62078 1 683 24 I2366-6I219 9465 4275 1988 4288 8945 22.0217 800695:4895:36 25 2283 2523 7.5ii32 6482 2294 4742 9687 1640 36.1776.98.2179 35 26 3335 3831 2806 8701 2602 5i99!i6-o435 3o8i 5627 101.1071 34 27 4390 5i44 4487 8.70931 2913 5660 1952 4541 9360 37.3579 7686 io4-i7il33 28 5448 6463 6176 3172 3224 6124 6020 107.426 32 29 65io 7787 7872 5425 3538 6591 2722 7519 110-892: 3i 114-589^30 3o 5-97576 6-69116 7-59575 7.61287 8.77689 10.3854 12.7062 16.3499 4283 22.9032 38.1885 3i 8646 6.70450 9964 4172 7536 23-0577 6177 118-540! 20 32 9720 '^t^ 3oo5 8-82252 4491 8014 5075 2l3l 371S 5321 39.0568 122-774 2^ 33 6-00797 4732 455 1 4813 8496 gjf 5o59 127.321 27 34 1878 4483 6466 6862 5i36 8981 9655 132-219I 26| 35 2962 5838 8208 9>85 5462 9469 IZ 6945 40.4358 137-3071 251 36 4o5i Ul 9957 8.91520 5780 6118 9962 8D93 9174 143-23T 24 37 5i43 7.71715 3867 13.0458 9i5o 24-0263 41 -4106 149-465 23 38 6240 9936 3480 6227 645o 0958 9990 17.0837 19^7 9158 156-259 22 39 734o6.8i3i2 5254 85o8 6783 1461 3675 42.4335 i63-7oo| 21 1 40 6 •08444:6 -82694 '■']il 9.00983,10.7119 13.1969 17.1693 24-5418 42-9641 43 -308 1, 171-885 20 41 9552 4082 3379 7457 2480 2558 7185 8978 180-932 ;? 42 6-10664 54757-80622 5789 8483 2996 3432 44 -0661 190-984 43 1779 6874 2428 8211 3di5 43i4 25-0798 6386 202-219 n 44 2899 4023 8278 4242 9.10646 4039 5203 2644 45.2261 214-858 16 45 9688 6064 3093 8829 4566 6106 45.7 82941229-182 i5 46 5i5i 6-91104 7805 9734 55D4 9178 5098 5634 7015 6418 46-4489|243.552| 14 47-0853 264-4411 i3 47 6283 2525 8028 9529 ^ii 8348 48 7419 3952i7.9i582 9.20516 9882 6174 26-0307 7395I286.478 12 49 8559 5385 3438 3oi6 11.0237 6719 9802 2296 48.4121 312.521 n 5o 6-19703 6-96823 7.95302 9.25530 11.0594 13.7267 18-0750 26-4316 49. io39';543-774| 10 | 5i 6-20851 8268 9o58 8o58 0904 7821 8378 1708 6367 8157 381.971 § 52 2oo3 9718 9.30599 i3i6 2677 845o 50-5485 429.718 53 3i6o 7-01 1748.00948 3lDD 1681 8940 3655 27.o566 5i.3o32 491.106 7 54 4321 2637 4io5 2848 5724 2048 9007 4645 271552-0807 572.957 6 55 5486 4756 83o7 2417 14-0079 5645 4899 8821 687.349 5 56 6655 5579 6674 9.40904 !I?1 o65d 6656 711753-7086 859.430 4 57 7829 7059 8600 35i5 1235 7678 8711 937254-5613 1145.92 1718.87 3 58 9007 8546 8-10536 .6141 3540 1821 28-i664|55-44i5 2 59 6-30189 7.10038 2481 8781 3919 2411 9755 3994 56-3306 3437.73 ' 9° 8° 7° 6° 5° 4° 3° 2° 1° 0° a Natural C o-tangeuts. P.] tol ^;,i7-8o 23.19 2?8v46 37.33 5.03 7.88 i3.oi .... .... .... 1 '9 UNIVERSITY OF CALIFORNIA LIBRARY Los Angeles This book is DUE on the last date stamped below. RECEIVED L - : 2l 1SS2 OCT 13 iy«2 EMS LIBRARY rely igliA and bvfit srican and City Espe- ricnn f their ?, it is 315 lected ly by rged lv)r Ainericau siuueuts uy ot^^^. ^i. ^tj.Axvx*.'«, j^-^-, -; ^J"*^'* of Philosopliy and Logic in the University of the- City of >ew York. 1 vol. large 12mo, We shall still continue to publish Shfur's Coinjylf'fe Maniifil of EtifflisJi and American Literafifre. Bv Tnos. B. Shaw, M.A.. Wm. Smith, LL.D , author of Smith's Bible and Classical Dictionaries, and Prof. Henp^ T. Tuckerman. With copious notes and illustrations. 1 vol. large 12mo, 540 pp. Sheldon & Co^npany's 2'ext'!SooA:s. OOLTON'S NEW GEOGEAPHIES. Hie tchole subject in Two Jioohs. These hooka are the most simple, the most practical, and beat adapted to the wants of the scJiool-room of any yet published. I. Colton's New Introductory Geography, With entirely new Maps made especially for this book, on the most improved plan ; and elegantly Illustrated. ZZ. Colton's Common School Geography, With Thirty -six new Maps, made especially for this book, and drawn on a uniform system of scales. Elegantly Illustrated. Tliia book is the best adapted to teaching the subject of Geog- raphy of any yet published. It is simple and comprehensive, and embraces just what the child should be taught, and nothing more. It also embraces the general princij)l(^s of Physical Geog- raphy so far as they can be taught to advantage in Common Schools. For those desiring to pursue the study of Physical (geography, wo have prepared Colton'3 Physical Geography. One Vol! 2t6. A very valuable book and fully illustrated. The Maps are compiled with the greatest care by Geo. W. Colton, and repre- sent the most remarkable and interesting features of Physical Geography clearly to the eye. The plan of Cotton's Geography is the bcpt I have ever seen. It meets the exact wants of our Grammar Schools. The Beview is unsurpassed in its tendency to make thorough and reliable scholars. I have learned more Geog- raphy that is practical and available during the t»hort time we have used this work, than in all my life before, including ten years teaching by Mitcheirs plan.— A. B. Hetwood, Prin. Franklin Gram. School, Lowell, Mass. So well satisfied have I been with these Geographies that I adopted them, and have procured their introduction into most of the schools in this county. James W. Thompson, A.M., Prin. of CentreHlle Academy, Maryland. Any of the above fent by mail, post-paid, on receipt of price.