P E R K I N-S' S E R E S. PLANE AND SOLID. X E 0 M E T R T. TO WHICH IS ADDED PLANE AND SPHERICAL T R I G O N OME T R Y AND MENSURATION. ACCOBIPANIED WITH ALL THE NECES-SARY LOGARITHMIC AND TRIGONOMETRIC TABLES. BY GEORGE R. PERKINS, LL.D. NEW YORK: D. APPLETON & CO., 346 & 348 BROADWAY, AND 16 LITTLE BRITAIN, LONDON. 1M DCCC LVI. Entered according to Act of Congress, in the year 1854. BY GEORGE PR. PERKINS, In the Clerk's Office of the District Court for the Northern District of New York. PRE F A C E THIs work is not designed to take the place of the " ELEMENTS OF GEO'mETRY," but rather to give a more extended and com prehensive course, better adapted to the use of Advanced Schools and Colleges. In the First Part of the work, which is confined to the theoretical principles of Geometry, great care has been taken to classify and arrange the various Theorems, by bringing together such as correspond to analogous subjects. We have also arranged the Problems by themselves, and have in no case mixed them up with the Theorems. In the preparation of the Geometrical portion of this work, we have made free use of the admirable work of VINCENT. The edition used is the fifth, revised by its author, with -the assistance of the distinguished BOURDoN. This is a most excellent and valuable French work, but it is not in all respects well adapted to the wants of our Higher Institutions; we have, therefore, preferred to make selections from it, rather than to give a translation of the whole. We have endeavored to arrange and present the different Propositions in the manner which to us appeared best caleulated to impart a thorough knowledge of the most important principles of the science of Geometry. As a general thing, we have adhered to the ordinary method of conducting our demonstrations; still we have not hesitated to:introduce the;algebraic notation, when by so iv PREFACE. doing we could present the subject in a clearer and more satisfactory manner. Throughout the Second Part, which is of a more mixed or practical nature, we have studied to present the whole in a distinct and clear manner, giving in all cases the full and complete work of all examples necessary to illustrate each principle. We would direct the attention in particular to the chapters on Spherical Trigonometry. This subject is generally regarded as intricate and difficult. It is believed the student will find the arrangement and full development of the different cases as here given, nearly as simple and of as easy comprehension as those of Plane Trigonometry. We might particularize other portions of the work, but will content ourselves by remarking that we have spared no pains to make the whole acceptable to the mathematical student of the present age. How well we have succeeded remains to be shown. GEO. R. PERKINS. UTIcA, September, 1854. CONTENTS. PART I. PAGE Introduction.................................................. 3 FIRST BOOK. Definitions.......................................................... 9 Definitions of terms.11...................... 1 Reimarks in reference to signs, symbols, etc............................... 12 Axioms.......................................................... 13 Postulates.................................................. 13 THEOREMS. Of angles.............................................. 14 Of the sides of triangles.................................... 19 Of perpendicular and oblique lines.................... 20 Of parallel lines.......................................... 24 Of triangles.............................................. 8 Of quadrilaterals.......................................... 33 Additional Theorems of triangles...................................... 87 SECOND BOOK, Definitions.......................................................... 40 Of chords, secants, and tangents........................................ 42 Of the measure of angles............................................. 47 Of inscribed and circumscribed polygons................................. 51 Of secant and tangent circles............................... 57 PROBLEMS. Of perpendiculars, angles, and parallels...................... 61 Construction of polygons............................... 66 Of contact............................................... 69 Of common measure...................... 70 THIRD BOOK. Definitions.........3............... 73 Proportional lines........................................ 4 Similar triangles................................................... 77 Yi CONTENTS. PAGE Similar polygons............................................. 80 Proportional lines —properties of the sides of triangles.................... 82 Determination of areas.............................................. 88 Ratios between the areas of similar figures............................. 94 Comparison of squares constructed on certain lines....................... 97 FOURTH BOOK. Definitions.......................................................... 101 Proportional lines connected with the circle............................. 102 Determination of the sides and of the areas of regular polygons............ 107 The determination of the sides and of the areas of regular polygons of a particular kind..................................................... 112 Measure of the circle and of its circumference........................... 116 Ratio of the circumference to its diameter............................... 121 PiOBTLEMiS. Construction of proportional lines........................... 126 Problems of areas................................................... 130 Construction of similar polygons under certain conditions.................. 132 FIFTH BOOK. Definitions.......................................................... 137 Theorems.......................................................... 138 SIXTH BOOK. Definitions...........................1.................. 152 Theorems............................................................. 154 SEVENTH BOOK. Definitions.................................................................. 174 Theorems...................................................... 17 EIGHTH BOOK. Definitions.................................................... 191 Theorems......................................... 192 Regular polyedrons................................................ 208 Application of Algebra to the solution of Geometrical Problems...21..... 210 CONTENTS., Vi PART II. CHAPTER I. First Principles of Plane Trigonometry............................ 1 Definition of an angle................................... 1 Definition of sines, tangents, secants, etc................................. 4 To find the sine and cosine of the sum and difference of two arcs........... 9 Numerical values of sines, tangents, etc.................................. 12 CHAPTER II. Explanation of Table I. of logarithms................................... 17 Arithmetical calculations by logarithms................................ 24 Arithmetical complement............................................. 27 Explanation of Table III. of natural sines and tangents.................... 218 Explanation of Table II. of logarithmic sines and tangents................. 29 CHAPTER III. Solution of right triangles........................................... 83 Additional examples of right triangles.................................. 43 CIHAPTER IV. Solution of oblique triangles.................................... 44 Additional examples of oblique triangles................................ 54 CHAPTER V. SPHERICAL TRIGONOMETRY. General trigonometric formulas....................................... 566 CHAPTER VI. Solution of spherical right triangles................................... 1 7 Of the circular parts.................................................. 71 Examples for practice............................................ 92 CHAPTER VII. Mlensuration of surfaces...................................... 93 CHAPTER VIII. Mensuration of solids......................................... 109 PART FIRST. PLANE AND SOLID GEOMXETRY. INTRODUCTION, 1. ALL bodies occupy, in the indefinite space which embraces the material universe, a determinate, or fjnite ptlace which we properly call a space. This finite space occupied by the body, has limits or bozndaries, which separate it from the rest of space. These boundaries constitute what we call the Swrface of the body. Surface being then the separation of a body from the rest of space, belongs as well to the one as the other; and as an infinite number of bodies may exist, in indefinite space, each having its proper surface, it follows that In space we may conceive of an infizite nzmber of surfaces. When one surface is cut by another surface, the place of their mutual intersection is called a -Line. This line belongs to each of the surfaces. Since the intersection of two surfaces gives a line, and any surface may be intersected by an infinite number of other distinct surfaces, it follows that On any surface whatever, we may conceive an infinite number of lines. The place of meeting, or of intersection of two lines, is called a Point. This point is common to the two lines. Since a point results from the meeting of two lines, and any line may be met by an infinite number of other distinct lines, it follows that Any line may be regarded as having an infniite number of points. 2. Although we acquire the notion of a point by the consideration of lines, the notion of a line by the consideration of surfaces, and that of the surface by the consideration of a body; that is to say, of material things, we must not conclude from this, that points, lines, and surfaces are themselves material objects. In virtue of an inherent faculty of the mind, we may readily conceive of a point without the lines which determine it, 4 GEOMETRY. of a line without the surfaces of which it is the intersection, of a surface independently of the body, or of the space of which it is the limit; in short, we may conceive of space itself, as absolutely immaterial. It is the result of these different abstractions which we call a point, a line, a surface, or a space. It is the same when we speak of the points of a line, of a surface, of a space; the lines of a surface, &c. 3. A space, a surface, and a line, may be considered in two distinct ways. We may consider them in regard to their different forms, which we name in general their Figures. Or, we may consider themn in regard to their relative magnitudes, which is comprehended under the name of Extension. Extension takes the particular name of Volume, of Area, or of -Length, according as it is applied to a space, a surface, or a line. Thus, the length of a line, or its linear extension, is the magnitude of this line, estimated or measured in units of a line. In the same way the area of a surface, or its superficial extension, is the magnitude of this surface, estimated or measured in units' of a surface. In short, the volume or the extension of ispace is the magnitude of this space, estimated or measured in units of space. GEOMETRY is the science which treats of these two principal objects: The properties of different kinds of Jigures, and the measure of extension, considered under the different circumstances, as already noticed. THE POINT. 4. A point has neither figure nor extension; it is this, above all, which distinguishes it from all other objects of Geometry, which are all capable of being described and measured. Nevertheless as it is often necessary to consider one or more isolated points, we represent their position by a dot or distinct mark made on a surface with the pen, pencil, or crayon, and we distinguish them from each other, usually, by letters. Thus we say the point A, the point B, the point C, &c..A.B.C.D The position of a point in reference to any other point is determined by its direction and distance from that point. This distance, which is the length of the shortest line which can INTRODUCTION. 5 unite these points, is obviously mutual to the two points; that is, the distance of the point B from the point A is the same as the distance of the point A from the point B. The direction of the point A from the point B is opposite to the direction of the point B from the point A. THE STRAIGHT LINE. 5. Of all geometrical lines, the simplest is the straight line. Although the idea of a straight line, is the first to which we are conducted by our experience, and the use of our senses, still it is very difficult to define it. The definition usually given is as follows: A straight line is the shortest distance between two points. A more general definition is as follows: A straight line is an indefinite line, such, that any limited portion whatever, is the shortest distance between the points which fix this limit. In any line the direction of one of its points, 13, from another point, A, as has already been noticed, is opposite the direction of A from B; so thatla line has two different directions exactly opposite, either of which may be regarded as the direction of the line. A Straight line is one which has the same direction throughout its whole extent. A curved line is one which changes its direction at every point. Two straight lines are evidently capable of superposition, that is, of being placed the one on the other, so as to coincide. Hence, two straight lines coincide throughout their whole extent when they have two points common. Or, in other words, two points determine the position of a straight line. We also infer that two distinct straight lines can intersect or meet each other in only one point. THE PLANE. 6. The plane surface, or, as usually expressed, the Plane, is the simplest of all surfaces. It may be defined as follows: A plane is an indefinite surface, on which we may conceive that through each of its points a straight line may be made exactly to coincide with it throughout its whole extent. It immediately results from this definition, and the nature of 6 GEOMETRY. a straight line, That any line, having two of its points common with the plane, lies wholly in this plane. Hence, a straight line cannot be partly in a plane and partly out of it. When a straight line has only one point in common with a plane, it is said to meet or pierce the plane, and the plane is said to cut the line, and the segments or portions of the line thus separated will be on different sides of this plane. THE CIRCLE. 7. When a line is not a straight line, or made up of finite portions of straight lines, it is called a curved line. The simplest of all curved lines is the circumference of a circle, which may be thus defined: The circurnference of a circle is a plane curve returning into itself, every point of which is equally distant from a certain point in its plane, which point is called the centre of the circumference. The portion of the plane limited by this circumference is called a circle. Any portion of the circumference of a circle is called an arc. When the arc is equal to one-fourth of the circum- F ference it is called a quadrant. Each of the e straight lines drawn from the centre to the circumference is called a radius. Hence, all E M A radii of the same circle are equal. A line passing through the centre and terminating in both directions by the circumference, is called a diameter. All diameters of the same circle are equal, since each is twice the radius. THE ANGLE. 8. When two straight lines meet, the opening between them is called a plane angle, or simply an Angle. The c magnitude of the angle does not depend upon the lengths of these lines, but only upon the difference of their directions. B If in any circle we draw two radii, the dis- F C tance between their extremities which terminate in the circumference, will embrace an B, A are. When the are between the two radii is E D equal to a quadrant, these radii form with each INTRODUCTION. 7 other an angle, which we call a right angle. And the radii are said to be the one perpendicular to the other. When the arc between the radii is less than a quadrant, the angle is called acute. When the arc is greater than a quadrant, the angle is called obtuse. The magnitude of an angle may be estimated or measured by means of any particular angle, taken as the unit angle. The right angle is generally the angle chosen as the unit angle. THE RULER AND THE COMPASS. 9. The straight line and the circumference of the circle, which are the only lines treated of in Elementary Geometry, are respectively traced or drawn upon a plane, by the aid of the Ruler and of the Compass. These instruments are so simple, and of such general use, as to need no description in this place. With the Ruler we can draw a straight line on a plane from any one point to any other point. With the Compass we can describe on a plane the circumference of a circle having any given point as a centre, and for its radius any given line. METHODS OF DEMONSTRATION. 10. There are two distinct methods employed in Geometrical demonstration: The Direct Method, and the Indirect Method. The most simple process of direct demonstration is the principle of sutperposition, which consists in being able to make two figures exactly coincide, by applying the one upon the other. The demonstration is also direct when we employ, by a direct course of reasoning, axioms, definitions, and principles already established. The indirect method, known under the name of Reducing to an absurdity, consists in first supposing the proposition not to be true; afterwards, by certain deductions, to draw, from truths already recognized as rigorously exact, a result contradictory to some one of these truths, or to the proposition itself. GEOMETRY We will terminate this subject by noticing two kinds of fals6 ieasoingy, very common with beginners, and against which they should be constantly on their guard. The first is called, Reasoning in a circle. The second is called, Begging the question. We are said to reason in a circle when, in the demonstration of a proposition, we employ, either implicitly or explicitly, a second proposition, which cannot, itself, be established without the aid of the first. We are said to beg the question, when, in order to establish a proposition, we employ the proposition itself. GEOMETRY. FIRST BOOK. THE PRINCIPLES. DEFINITIONS. I. GEOMETRY is the science of Position and Extension. II. A Point has merely position, without any extension. III. Extension has three dimensions; Length, Breadth, and Thickness. IV. A Line has only one dimension; length. V. A Surface has two dimensions; length and breadth. VI. A Solid has three dimensions; length, breadth, and thickness. VII. A Straight line is one which has the same direction through its whole extent. In reality a line has two directions, the one exactly opposite the other; either of which may be considered as its dil ecti( rl. VIII. A Brokern line is one which is made up of two or more straight lines. IX. A Curved line is one which changes its direction at every point. X. Parallel lines are those which have the same direction. XI. An Angle is the difference in direction of two straight lines meeting or crossing each other. The Vertex of the angle is the point where its sides meet. XII. When one straight line meets or crosses another, so as to make the adjacent angles equal, each of these angles is called a Right angle, and the lines are said to be perpendicular to each other. 10 GEOMETRY. XIII. An Acute angle is one which is less than a / right angle. XIV. An Obtuse angle is one which is greater than a right angle. XV. When the sum of two angles is equal to a right angle, they are called complementary angles; each being the complement of the other. XVI. When the sum of two angles is equal to two right angles, they are called Supplementary angles; each being the supplement of the other. XVII. A Plane is a surface which is straight in every direction, or one with which a straight line, joining ally two of its points, will coincide. XVIII. When a surface is not a plane surface it is called a curved surface. XIX. A plane figure is a limited portion of a plane. When it is limited by straight lines, the figure is called a rectilineal figure, or a polygon; and the limiting lines, taken together, form the contour or perimeter of the polygon. XX. The simplest kind of polygon is one having only three sides, and is called a triangle. A polygon of four sides is called a quadrilateral; that of five sides is called a pentcagon; that of six sides i8 called a hexagon; one of seven sides is called a heptagon; one of eight sides an octagon; one of nine sides a nonagon, and so on for figures of a greater number of sides. XXI. A triangle having the three sides equal, is called an equilateral triangle; one having two sides equal, is called an isosceles triangle; and one having no two sides equal, is called a scalene triangle. XXII. A triangle having a right angle, is called a right-angled triangle. The side opposite the right angle is called the hypotenuse. XXIII. A triangle having its three angles acute, is called an acute-angled triangle. FIRS'I 1OOK. 11 XXIV. A triangle having an obtuse angle, is called an obtuse-angled triangle. XXV. When the opposite sides of a quadrilateral are parallel, the figure is called a parallelogram. XXVI. When the four angles of a parallelogram are right angles, the figure is called a rectangle. XXVII. When the four sides of a rectangle are equal, the figure is called a square. XXVIII. When the four sides of a parallelogram are equal, and the angles not right, the figure is called a rhombus. XXIX. When only two sides of a quadrilateral are parallel, the figure is called a trapezoid. XXX. A diagonal of a polygon is a line joining the vertices of two angles, not adjacent. DEFINIrTIONS OF TERMS. I. A demonstration is a logical train of reasoning employed to establish an asserted truth. II. A proposition is a statement of a truth to be demonstrated, or of an operation to be performed. III. An axiom is a self-evident proposition, of which the simple mention carries a conviction of its truth. 12 GEOMETRY. IV. A postulate is a proposition, the truth of which is required to be admitted without demonstration, notwithstanding it does not present the same degree of evidence as in the case of the axiom. V. A theorem is a proposition which requires a delnonstration. VI. A corollary is an immediate consequence of one or more propositions. If any new course of reasoning is required to establish it, this reasoning is so simple that it may be supplied without much inconvenience. VII. A problem is a question proposed, which requires a solution. VIII. A lemma is a subsidiary truth, employed for the demonstration of a theorem, or the solution of a problem. IX. A scholiumz is a remark on one or several preceding propositions, which tends to point out their connection, their use, their restriction, or their extension. X. An hypothesis is a supposition, made in the statement of a proposition, or in the course of its demonstration. REMARKS IN REFERENCE TO SIGNS SIYMBOLS ETC. The signs and symbols which we shall employ, have the same signification in Geometry as in Algebra, to which, for their full explanation, we shall refer the student. For the doctrine of ratios and proportions, we will also refer the student to the method explained in the Algebra. There is this difference between geometrical ratios of magnitudes, and ratios of numbers: All numbers are commensurable; that is, their ratio can be accurately expressed: but many magnitudes are incommensurable; that is, their ratio can be expressed only by approximation; which approximation may. however, be carried to any extent we desire. Such is the ratio of the circumference of a circle to its diameter, the diagonal of a square to its sides, etc. Hence many have deemed the arithmetical method not sufficiently general to apply to geometry. This would be a safe inference, were it necessary in all cases to assign the specific ratio between the two terms compared. But this is not the case. Such ratios themselves may be unknown, FIRST BOOK. 13 indeterminate, or ilTational, and still their equality or inequality may be as completely determined by the arithmetical method as by the more lengthy method of the Greek geometers. AXIOMS. I. Things which are equal to the same thing, are equal to each other. II. When equals are added to equals, the wholes are equal. III. When equals are taken from equals, the remainders are equal. IV. When equals are added to unequals, the wholes are unequal. V. When equals are taken from unequals, the remainders are unequal. VI. Things which are double of the same or equal things, are equal. VII. Things which are halves of the same thing, are equal. VIII. Every whole is equal to all its parts taken together, and greater than any of them. IX. Things which coincide, or fill the same space, are identical. X. All right angles are equal to one another. XI. A straight line is the shortest distance between two points. XII. Through the same point only one straight line can be drawn parallel to another. XIII. Only one straight line can be drawn joining two given points. XIV. Straight lines which are parallel to the same line are parallel to each other. POSTULA TE S. I. Let it be granted that a straight line may be drawn from any point to any other point. II. That a terminated straight line may be produced, in either direction, to any extent. III. That a straight line may be bisected or halved. IV. That from a point, either within or without a straight line, a perpendicular may be drawn to the line. 14 GEOMETRY. V. That a line may be drawn, making any given angle with another line. VI. That a line may be drawn from the vertex of an angle, bisecting it. OF ANGLES. THEOREM I. When a straight line meets or crosses another, the adjacent angles are supplements; and the opposite angles are equal. For, drawing FG perpendicular to AB, we see that the angle AFC exceeds the right angle c AFG, by the angle CFG; while the angle BFC is less than the right angle BEFG by the same A 7 B angle CFG. Hence, the sum of AFC, BFC is D equal to two right angles, and they are therefore supplements of each other (D. XVI.).* In the same way by drawing a line perpendicular to CD, it may be shown that the angles CFA and AFD are supplements; consequently, BFC + CFA = CFA + AFD (A. I.): from each taking CFA, we have BFC equal to its opposite angle AFD. In a similar manner we have CFA=DFB. Cor. I. If either of the four angles, formed by the intersection of two straight lines, is a right angle, the remaining three angles will each be right, and the two lines will be mutually perpendicular. When the angles are not right there will be two equal acute angles, and two equal obtuse angles. The acute angle and the obtuse angle will be supplementary. Cor. II. All the angles which can be made E at any point D, by any number of lines on the \\ same side of AB, are together equal to two right A D B angles. * In the references we shall use the following abbreviations: A. for Axiom, B. for Book, C. for Corollary, D. for Definition, P. for Problem, S. for Scholium, T. for Theorem. FIRST BOOK. 15 Cor. III. And as all the angles that can be E C made on the other side of AB are also equal to / two right angles; therefore all the angles that can be made around the point D, by any number of lines, are together equal to four right angles. Cor. IV. If two adjacent angles, ADC and BDC, are supplementary, their exterior sides AD and BD will form one and the same line. D Scholium. Since a straight line has two different directions, exactly opposite each other, it is sometimes considered as making with itself an angle equal to two right angles. THEOREM II. If through the vertex of any angle, lines are drawn perpendicular respectively to its sides, they will formr a new angle, either egqual to the first, or supplementary to it. Let BAC be the given angle, DE perpendic- D ular to AB, and FG perpendicular to AC. F\ Then we shall have the angle DAF equal to A B BAC, since each is the complement of CAD (D. XV.). The angle EAG being opposite DAF o must also equal BAC (T. I). But the angle E DAG, or its opposite angle FAE, is supplementary with DAF, and consequently supplementary with its equal angle BAC. THEOREM III. Two angles having their corresponding sides parallel are either equal or su2pplementary. First. When the sides AB and AC, forming C the angle at A, have respectively the same F directions as the sides DE and DF fbrming / the angle at D. E' Produce CA and DE until they cross each other at G. Now, since AB and DE are parallel, they have the same direction in reference to the line CG (D. X.), consequently the angle BAC is equal to KGC. And since AC and DF are parallel they have the same direc 16 GEOMETRY. tion in reference to the line DG; consequently, the angle KGC is equal to EDF; therefore (A. I.), the angle BAC is equal to EDF. Secondly. When the sides AB and AC are respectively in opposite directions to DE and / DF. E K B As before, we have the angle BAC equal to 7G KGC, and the angle EDF equal to EGII. But F 11 KGC is equal to its opposite angle EGH (T. I.), consequently, the angle BAC is equal to EDF. Thirdly. When one of the sides, as AB, has C F an opposite direction to DE, its corresponding ZA B side. E -— Z K As before, the angle BAC is equal to KGC, and EDF is equal to EGC. But FKGC and E EGC are supplementary (T. I.), consequently BAC and EDF are supplementary. Cor. Two angles having their corresponding sides perpendicular, are either equal or supplementary. For, if we draw through the vertex of the first angle lines respectively parallel to the sides of the second angle, we should thus form a third angle, which will be equal to the second. But this third angle is either equal to the first or supplementary to it (T. II.). Consequently the second angle is either equal to the first or supplementary to it (A. I.). THEOREM IV. When all the sides of a polygonal figure are produced, in the same direction, the sum of all the exterior angles will be equal to four right angles. For, if from any point in the same D plane, straight lines be drawn parallel F respectively to the sides of the figure, the d \ angles contained by the straight lines / f about that point will be equal to the exterior angles of the figure (T. III.), each B to each. Thus the angles a, b, c, etc., are respectively equal to the exterior angles A, B, C, etc.; but the former angles are together equal FIRST BOOK. 17 to four right angles (T. I., C. III.); therefore all the exterior angles of the figure are together equal to four right angles. Scholiunm. This proposition must be restricted to the case in which the polygon is convex. A convex polygon may be defined to be one, such that no side, by being produced in either direction, can divide the polygon. The polygon ABCDFG is not con- G vex, since it may be divided by producing either of the sides CD or FD. This polygon is said to have a re-entering angle at D. A THEOREM V. In any convex polygon, the sum of all the interior angles, taken together, is equal to twice as many right angles as the polygon has sides, wanting four right angles. Let ABCDFG be a convex polygon. F X Conceive the sides to be produced all in the d same direction, forming exterior angles, C which we will denote by the capital letters G A, B, C, etc., while their corresponding in- b terior angles are denoted by the small letters a, b, c, etc. Now any exterior angle, together with its adjacent interior angle, as A + a, is equal to two right angles (T. I.), therefore the sum of all the interior angles, together with all the exterior angles, is equal to twice as many right angles as the polygon has sides; but the sum of all the exterior, angles is equal to four right angles (T. IV.); therefore the sum: of all the interior angles is equal to twice as many right angles as the polygon has sides, wanting four right angles. Cor. I. In any triangle, the sum of the three angles is equal! to two right angles. cor. II. If one angle of a triangle is a right angle, as in the case of a right-angled triangle (D. XXII.), each of the other angles must be acute, and they are complementary, since their sum must equal a right angle. Each of the angles of a triangle may be acute, but only one of the angles can be obtuse. Cor. III. If two triangles have two angles of the one respectively equal to two angles of the other, their third angles will be equal, and the triangles will be mutually equiangular. 2 18 GEOMETRY. Cor. IV. In any quadrilateral, the sum of the four interior angles is equal to four right angles. Cor. V. If two angles of a quadrilateral are right, the other two angles will be supplementary. Cor. VI. If from a point D within a triangle c ABC, we draw two lines to the extremities of one of the sides AB, the angle ADB formed by these B lines will be greater than the angle ACB, opposite this side. For, the sum of the two angles DAB, DBA is obviously less than the sum of the twb angles CAB, CBA. And since the sum of the three angles of every triangle is equal to two right angles, it follows that the remaining angle ADB, of the triangle ADB, must be greater than the remaining angle ACB of the triangle ACB. THEOREM VI. The exterior angle, formed by producing one of the sides of a triangle, is equal to the sum of the two interior and opposite angles of the triangle. In the triangle ABC, if the side AB be pro- c duced to D, the exterior angle CBD will be equal to the sum of the two angles at A and A B D at C. For, the angle CBD, together with its adjacent angle, CBA, is equal to two right angles (T. I.), and the sum of A and C, together with the same angle CBA, is also equal to two right angles (T. V., C. I.); hence, the exterior angle is equal to the sum of the two interior and opposite angles. Cor. The exterior angle is greater than either of the interior and opposite angles. FIRST BOOK. 19 OF THIE SIDES OF TRIANGLES. THEOREM VII. Either side of any triangle is less than the sum of the other two sides, and greater than their difference. First. The straight line AB is the shortest distance between A and;B (A. XI.), and therefore shorter than the broken line AC+ CB. The same reasoning applies to each of the sides. A B Hence, we have AB < AC + CB; AC < AB + CB; BC < AB+ AC. Secondly. Since AC < AB + CB, we have AC - CB < AB (A. V.). That is, AB > AC - CB. In a similar manner we deduce AC > AB- CB; CB > AB- AC. THEOREM VIII. If from any point within a triangle two lines be drawn to the extremities of either side, the sum of these two lines will be less than the sumr of the other two sides. From the point D, suppose the lines DA and c DB to be drawn to the extremities of the side AB; then will DA + DB < AC + BC. For, producing AD until it meets BC at E, we A B have (T. VII.) DB < DE + EB, consequently (A. IV.), AD + DB < AD + DE + EB, that is, AD + DB < AE + EB. Again, we have AE < AC + CE, consequently, AE + EB < AC + CE + EB, that is, AE + EB < AC + CB. Comparing these conditions, we have AD+DB< AC-+CB. 20 GEOMETRY. OF PERPENDICULAR AND OBLIQUE LINES. THEOREM IX. If two straight lines have two points common they will coincide throughout their whole extent. If A and B are points common to two F straight lines, they will coincide between A and B (A. XIII.). A If possible, we will suppose that when produced, they begin to separate at C, the one taking the direction CD, and the other the direction CE. Draw CF perpendicular to AC; then since the lines ACE and ACD are each straight, we have the angle FCD and FCE equal, each being equal to a right angle (T. I., C. I.); that is, the whole is equal to one of its parts, which is impossible (A. VIII.). It is therefore absurd to suppose these lines can separate when produced. Hence, if two straight lines have two points common they will coincide throughout their whole extent. THEOREM X. Through a given point in a straight line only one perpendicular can be drawn to this line. If there could be two perpendiculars, as CD D E and CE, the angles BCD and BCE would be equal, each being a right angle; that is, the whole would be equal to its part, which is im- A c 1 * possible (A. VIII.). Hence, it is absurd to suppose that more than one perpendicular can be drawn to a given line through any one of its points. THEOREM XI. From a given point without a straight line, only one perpendicular can be drawn. Let C be the point, and AB the given line. If possible, suppose we can draw the two perpendiculars CD and CE. Revolve FIRST BOOK. 21 the figure CDE about DE as a hinge, until it returns into its primitive plane on the opposite side of AB, having the position FDE. Since each of the angles FDE and CDE is right, CDF is a straight line (T. I., C. IV.). By the same reasoning we have CEF a straight line. That is, we have two straight A B lines joining the points C and F, which is im- E possible (A. XIII.). Hence, from a point without a straight line only one perpendicular can F be drawn to this line. rHEOREM XII. If from a point without a line, a perpendicular be drawn, and several oblique lines: I. The perpendicular will be shorter than any oblique line. II. Any two oblique lines which terminate at equal distances from the foot of the perpendicular, will be equal. III. Of two oblique lines terminating at unequal distances from the foot of the perpendicular, the one at the greater distance will be the longer. Let A be the given point, BC the given A line, AD perpendicular, and AB, AE, and AC oblique lines. GC First. As in the last Theorem, suppose ADC to revolve about BC into the position F FDC, on the opposite side of BC in its primitive plane. We have FD=AD; FE=AE; and FC=AC. But since each of the angles FDE and ADE is a right angle, ADF is a straight line (T. I., C. IV.), and therefore shorter than the broken line AE + EF; hence, AD, the half of AF, is less than AE, the half of AE + EF. That is, the perpendicular is shorter than any oblique line. Secondly. If we suppose the figure ADB to revolve about AD, the point B will coincide with E, since DB is equal to DE, and the angle ADB is equal to ADE, each being a right angle. Hence, the oblique line AB will coincide with AE. That is, two oblique lines which terminate at equal distances from the foot of the perpendicular are equal. Thirdly. Returning to the figure as first revolved, we have, 22 GEOMETRY. since E is a point within the triangle ACF, AC + CF > AE + EF (T. VIII.); hence, AC, the half of AC + CF, is greater than AE, the half of AE + EF. That is, the oblique line terminating farther from the foot of the perpendicular is the longer. Cor. I. The perpendicular measures the shortest distance from a point to a line. Cor. II. From the same point without a straight line, only two equal oblique lines can be drawn, one on each side of the perpendicular. THEOREM XIII. If through the middle point of a straight line, a perpendicular be drawn: I. Any point in this perpendicular will be equally distant from the extremities of this line. II. Any point without the perpendicular will be uznequally distant from, the extremities of this line. Let AB be the given line, DE a perpen- F dicular through C, its middle point. D G First. Let D be any point in this perpendicular. Drawing DA and DB, we know A B these lines to be equal, since they terminate c at equal distances from C, the foot of the perpendicular DC (T. XII.). That is, D is equidistant from A and B. In the same manner we may show that any other point of this perpendicular, as E or F, is equidistant from the extremities of the given line. Secondly. Suppose G to be a point without the perpendicular. If we draw GA and GB, one of these lines must cut the perpendicular. We will suppose GA to cut it at D. Draw DB, then GB will be less than GD + DB (T. VII.). But, we already have DB = DA, hence GB is less than GD + DA, or less than GA. That is, the point G is unequally distant from the extremlities of the given line. In the same manner we can show that any other point without the perpendicular is unequally distant from the extremities of this line. Scholium. The most distant extremity A of this line will be on the side of the perpendicular opposite the point G. FIRST BOOK. 23 Cor. I. Any point equidistant from the extremities of a straight line is situated in the perpendicular which bisects this line. Cor. II. If a straight line have any two of its points equidistant from the extremities of a second line, it will be perpendicular to the second line and bisect it. THEOREM XIV. If a line be drawn bisecting a given angle, that is, dividing it into two equal angles: I. Any point in this bisecting line will be equidistant from the sides of the angle. II. Any point without this bisecting line will be unequally distantfrom the sides of the angle. First. Let the given angle BAC be bisected A by the line AD, then will any point as D in this line be equidistant from the lines AB E and AC. The distance of D from these lines will be B C measured by the perpendiculars DE and DF (T. XII., C. I.). If we conceive the figure folded about the line AD, so that the portion on the left of AD may be superposed upon the portion on the right, the line AB will then coincide with AC, since the angle BAD = CAD. Consequently the perpendiculars DE and DF will coincide (T. XI.), and are therefore equal. Secondly. Suppose G to be a point without the bisecting line. If we draw the perpendiculars GF and Gi, one of these must cut the bisecting line. Let GF cut it at D. Draw DE perpendicular to AB, and join G and E: then will GH be less than GE (T. XII.). But GE is less than GD + DE (T. VII.); therefore, GH is less than GD + DE, or less than GD + DF, since by the first case of this Theorem DF =DE. Hence, finally, Gi is less than GF. Cor. I. Any point equidistant from the sides of a given angle is situated on the line bisecting this angle. Cor. II. If a straight line have two of its points equidistant from the sides of a given angle, it will bisect this angle. '24 GEOMETRY. OF PARALLEL LINES. THEOREM XV. When two lfnes are perpendicular to the sarne line, they are parallel. If the lines AB and CD are perpendic- G ular to GH they will be parallel. A- E B For, since they are perpendicular to Gi, we have the angles BEG and DFE equal, c D each being a right angle; hence, these l lines have the same direction, and are therefore parallel (D. X.). Cor. A line which is perpendicular to one of two parallels, is also perpendicular to the other. THEOREM XVI. Two _parallel lines are throughout their whole extent equally distant. Suppose AB and CD to be parallel. F L H Through any two points, as E and G, of the line CD, draw EF and GH per- c _.. D E KG pendicular to CD, they will also be perpendicular to AB (T. XV.), and they will measure the distance of E and G respectively from the line AB (T. XII., C. I.). We are now to prove these lines equal. Through K, the middle of EG, draw KL perpendicular to CD, and it will also be perpendicular to AB. Now, suppose the portion of the figure on the left of KL to revolve about KL as a hinge, until it returns into its primitive plane, on the right of KL. The angles at K and L being right, KE will take the direction of KG, and LF the direction of LH, and E will coincide with G, since KE = KG, and since EF and GIl are each perpendicular to CD, they must coincide, which proves them equal. Cor. Two parallels cannot meet, however far they are pro. duced. FIRST BOOK. 25 Scholium. When two parallels are A cut by a third line, the angles formed / receive particular names. c WHEN TAKEN SEPARATELY: 1. The four angles AGF, CHE, BGF, and DIEE, situated within the parallels, are called interior angles. 2. The four angles AGE, CHF, BGE, and DHF, situated without the parallels, are called exterior angles. WHEN COMPARED TWO AND TWO: 1. The two angles AGF and CHE, as well as BGF and DEIE, are called interior angles on the same side; that is, on the same side of the secant or cutting-line. 2. The two angles AGE and CHF, as well as BGE and DHEF, are called exterior angles on the same side. 3. The two angles AGF and DIE, as well as CHE and BGH, being interior angles on different sides of the secant, are called alternate interior angles. 4. The two angles AGE and DIlF, as well as CHF and BGE, are called alternate exterior angles. We have noticed two pairs of each of the four distinct kinds of angles. 5. There are still four pairs of angles, referred to as Corresponding angles, namely: AGE and CHE, AGF and CHF, BGE and DHE, BGHI and DHF. These last angles, which establish the direction of the lines, are sometimes called opposite exterior and interior angles. THEOREM XVII. If two straight lines are cut by a third line, making the sum of the two interior angles on the same side egual to two right angles, the two lines will be parallel. If the two lines AB and CD are cut by the line EF, making BG1H + DHG = 2 right angles. these lines will be parallel. 26 GEOMETRY. For we have BGIH + BGE = 2 right E angles (T. I.), therefore BGH + DHG = BGH + BGE; taking BGH from A- / each we have DHG=BGE, hence the D lines AB and CD have the same di- C rection, and are consequently parallel (D. X.). Cor. If two straight lines are cut by a third line, they will be parallel: I. When the sum of the exterior angles on the same side is equal to two right angles. II. When the alternate interior angles are equal. III. When the alternate exterior angles are equal. IV. When the corresponding angles are equal. first. Suppose EGB + FHD=2 right angles. We have EGB = AGH, and FHD = CHG (T. I.), hence AGH + CHG = 2 right angles, which agrees with the Theorem itself, hence AB and CD are parallel. Secondly. Suppose AGH = GHD. We have AGH = EGB (T. I.), consequently EGB = GiD; that is, the lines AB and CD have the same direction, and are therefore parallel. Thirdly. Suppose EGB = CHF. We have CHF = GHD, consequently EGB = GIHD, and the lines have the same direction, and are therefore parallel. Fourthly. Since the corresponding angles are equal, the lines have the same direction, and are therefore parallel. THEOREM XVm. If two parallels are cut by a third line, the sum of the two interior angles on the same side is equal to two right angles. Since the lines are parallel they / have the same direction, and their G corresponding angles must be equal. A / We therefore have EGB =GHD; to each add BGH, and we shall have EGB + BGH = BGH + GHD. F FIRST BOOK. 21 But EGB + BGH = 2 right angles (T. I.), consequently BGH + GHD = 2 right angles. Cor. It follows immediately from D. X., in connection with T. I., C. I., that if either one of the eight angles formed by the intersection of a straight line with two parallels is a right angle, the other seven will also be right. When these angles are not right, there will be four equal acute angles, and four equal obtuse angles. These angles will be supplementary. Any two of the acute angles, or any two of the obtuse angles taken together, constitute a pair of corresponding angles, and fix the direction of the parallels. Hence it readily follows, thatIf two parallels are cut by a third line, we shall have: I. The sum of the exterior angles on the same side equal two right angles. II. The alternate interior angles equal. III. The alternate exterior angles equal. IV. The corresponding angles equal. THEOREM XIX. If two straight lines are cut by a third line, and the sum of the interior angles on the same side is not equal to two right angles, the two lines will meet, if sufficiently produced. If we suppose the angle AGH + A CIIG > 2 right angles, then, obviously, B will the angle BGIH + DIHG < 2 right D angles, since these angles are respectively supplementary angles. This being supposed, these lines will meet. For if they do not meet they are parallel; but they are not parallel, since the sum of the interior angles on the same side is not equal to two right angles. Hence these lines will meet if produced sufficiently far. Scholiumn. It is evident they will meet in the direction of B and D, on that side of the secant line which has the sum of the interior angles less than two right angles. 28 GEOMETRY. OF TRIANGLES. THEOREM XX. If two triangles have two sides and the included angle of the one, equal to the two sides and the included angle of the other, the triangles will be identical, or equal in all respects. In the two triangles ABC, DFG, if the side CA be equal to the side DG, and the side CB equal to the side GF, and the angle C equal to the angle G, then will the two triangles be identical, or eQual in all respects. C G A B D F For, conceive the triangle ABC to be placed upon the triangle DFG in such a manner that the point C may coincide with the point G, and the side CA with the equal side GD. Then, since the angle G is equal to the angle C, the side CB will take the direction of the side GF. Also CB being equal to GF, the point B will coincide with the point F; consequently the side AB will coincide with DF. Therefore the two triangles are identical, and have all their other corresponding parts equal (A. IX.), namely, the side AB equal to the side DF, the angle A equal to the angle D, and the angle B equal to the angle F. Cor. Two right-angled triangles are equal, when the two sides containing the right angle of the one are respectively equal to the two sides containing the right angle of the other. THEOREM XXI. If two triangles have two sides qf the one respectively equal to two sides of the other, and the included angle of the first greater than the included angle of the second, the third side of the first will be greater than the third side of the second. FIRST BOOK. 29 In the two triangles c ABC and DEF, suppose CA and CB of the first to be respectively equal E to FD and FE of the second, and the included an- D gle ACB > DFE, then will AB > DE. Apply the triangle DEF upon the trian- F gle ABC, making EF coincide with BC, so that the triangle may, in its new position, be represented by GBC. Draw CL bisect- A ing the angle ACG (Post. VI.), and meet- L G D ing AB at L, draw GL. The two triangles ACL, GCL, are equal (T. XX.), and LG =B B LA. Therefore, AB = AL + LB = LG + LB, which is greater than BG (T. VII.). That is, AB > DE. Although the triangle DEF, when ap- c plied to ABC, may have three different positions, as here represented, still our demonstration is alike applicable to A L G B D E either case. Cor. Conversely. If two triangles have two sides of the one respectively equal to two sides of the other, and the third side of the first greater than the third side of the second, the included angle of the first will be greater than the included angle of the second. For, if the included angle of the first is not greater than the included angle of the second, it must be either equal or less. If it were equal, the third sides would be equal (T. XX.). If it were less, the third side of the first would be less than the third side of the second, by the Theorem itself. Both these results are contrary to the hypothesis. Consequently, the included angle of the first is greater than the included angle of the second. THEOREM XXII. If two triangles have two angles and the interjacent side of the one equal to two angles and the interjacent side of the other, the triangles will be identical, or equal in all respects. 30 GEOMETRY. In the two triangles ABC, DFG, if the angle A is equal to the angle D, the angle B equal to the angle F, and the side AB equal to the side DF, then will the triangles be identical, or equal in all respects. C G A B D F For, conceive the triangle ABC to be placed on the triangle DFG, in such a manner that the side AB may coincide with the equal side DF. Then, since the angle D is equal to the angle A, the side AC will take the direction of the side DG; also, since the angle F is equal to the angle B, the side BC will take the direction of the side FG; consequently the point C must coincide with the point G. Therefore the two triangles are identical (A. IX.), havlnb'e two sides AC and BC respectively equal to DG and FG, and the remaining angle C equal to the remaining angle G. THEOREM XXIII. ln an'sosceles triangle, the angles at the base are equal; or, if a triangle have two sides equal, the angles opposite those sides will be equal. If the triangle ABC have the side AC equal to the side BC, then will the angle B be equal to the angle A. For, conceive the angle C to be bisected, or divided into two equal parts, by the line CD, making the angle ACD equal to the angle BCD. Then An B the two triangles ADC and BDC will be identical, or equal in all respects (T. XX.), and consequently the angle B is equal to the angle A. Cor. I. The line which bisects the vertical angle of an isosceles triangle, bisects the base perpendicularly, and divides the triangle itself into two equal parts. Cor. II. An equilateral triangle is equi-angular, that is, has all its angles equal. FIRST BOOK. 31 THEOREM XXIV. If a triangle have two angles equal, the sides opposite those angles will be equal, and the triangle will be isosceles. In the triangle ABC, if the angles CAB, CBA, are equal, the opposite sides BC, AC, will be equal. For, if they are not equal, suppose BC > AC, and take BD =AC. Then in the two triangles BAD A B and ABC we have the side BD = AC by supposition, the side AB common, and the included angle ABD of the first triangle equal to BAC, the included angle of the second triangle. Hence (T. XX.), the triangle BAD is equal to ABC, that is, a part is equal to the whole, which is impossible (A. VIII.). There is, therefore, no inequality of the sides CB and CA, that is, they are equal, and the triangle is isosceles. THEOREM XXV. When two triangles have the three sides of the one respectively equal to the three sides of the other, the triangles will be identical, and equal in all respects. Let the two triangles ABC, ABD, C have their sides respectively equal, namely, AB equal to AB, AC equal to A B AD, and BC equal to BD, then will D these triangles be identical. For, conceive the two triangles to be joined together by their longest equal sides, and draw the line CD; then in the triangle ACD, since AC is equal to AD, we have the angle ACD equal to the angle ADC (T. XXIII.). In like manner, in the triangle BCD, since BC is equal to BD, we have the angle BCD equal to the angle BDC. Hence the angle ACB, which is the sum of ACD and BCD, is equal to the angle ADB, which is the sum of ADC and BDC. Since, then, in the triangle ACB, we have the two sides AC and BC, and their included angle ACB, equal respectively to the two sides AD and BD, and their included angle ADB, of the triangle ADB, it therefore follows that these triangles are identical (T. XX.). 32 GEOMETRY. THEOREM XXVI. Two right-angled triangles are equal, when the hypotenuse and a side of the one, are respectively equal to the hypotenuse and a side of the other. Suppose BC = EF and AB = DE, then C F will the two right-angled triangles, ABC G and DEF, be equal. For, if we apply the triangle ABC upon the triangle DEF, so that AB may i____ coincide with its equal DE, the side AC A B D E will take the direction of DF, since the angle at A is equal to the angle at D, each being a right angle. If, now, BC coincides with EF, there will be a complete coincidence, and the two triangles will be equal. If possible, suppose BC to take the position EG, so that the triangle ABC may be represented by DEG; then, since the hypotenuses of the two triangles are equal, we shall have EG = EF, that is, two oblique lines are drawn from E, terminating at unequal distances from D, the foot of the perpendicular ED, which is impossible (T. XII.). It is, therefore, absurd to suppose BC to take any position different from EF. Hence, the triangles coincide throughout, and are in all respects equal. Cor. Two right-angled triangles will also be equal, when the hypotenuse and an acute angle of the one are respectively equal to the hypotenuse and an acute angle of the other. For, since the sum of the two acute angles of any right-angled triangle is equal to a right angle (T. V., C. II.), it follows that the remaining acute angles of the two triangles will be equal. Hence, the two triangles will have two angles and the interjacent side of the one equal, respectively, to two angles and the interjacent side of the other. Consequently, they will be equal (T. XXII.). THEOREM XXVII. The greater side of every triangle is opposite the greater angle, and, conversely, the greater angle is opposite the greater side. FIRST BOOK. 33 First. Suppose, in the triangle ABC, the angle CBA > CAB, then will the side CA > CB. For, draw BD, making the angle DBA = DAB, and DA will equal DB (T. XXIV.). To each adding DC, we have DA + DC = DB + DC, but A B DB + DC is greater than CB (T. VII.); hence, DA + DC, or its equal CA, is greater than CB. Secondly. Suppose, in the triangle ABC, the side CA>CB, then will the angle F CBA > CAB. For, through the middle point of the side AB draw the perpendicular DE. Now, since A D B CA > CB, the point C must be on the side of DE opposite A (T. XIII., S.); consequently, the perpendicular must cut CA at some point, as F. Draw FB, and we shall have FB =FA (T. XIII.), and angle FBA = FAB (T. XXIII.); consequently, FBA + FBC, which equals CBA. is greater than CAB. OF QUADRILATERALS. THEOREM XXVIII. If the o2pposite angles of a quadrilateral are equal, the figure is a parallelogram. We have the angles A +-B + C +D - 4 right angles (T. V., C. IV.), true for all / quadrilaterals. Now, if we suppose A = C, and B = D, we shall find 2A + 2D A B = 4 right angles, or A + D =2 right angles, hence AB and DC are parallel (T. XVII.). In a similar manner, we find 2A + 2B = 4 right angles, or A + B = 2 right angles, and, as before, AD and BC are parallel. Consequently the figure is a parallelogram (D. XXV.). THEOREM XXIX. If the opposite sides of a quadrilateral are equal, the figure is a parallelogram. 3 34 GEOMETRY. If we suppose AB =DC, and AD = D BC, and draw the diagonal BD, we shall have the three sides of the triangle ABD respectively equal to the three sides of A B the triangle CDB, consequently they are equal, (T. XXV.), and the angle ABD =CDB, which are alternate interior angles in reference to the sides AB and DC, hence AB and DC are parallel (T. XVII., C.). The equality of the two triangles also gives the angle A DB = CBD, which are alternate interior angles in reference to the sides AID and BC, hence these sides are parallel. Consequently the figure is a parallelogram (D. XXV.). THEOREM XXX. If two o2posite sides of a quadrilateral are equal and parallel, the fJgure is a parallelogram. In the quadrilateral ABCD, suppose D C AB equal and parallel to DC, then will // / the figure be a parallelogram. For, drawing BD, the two triangles ABD, A B CDB, have the common side BD, and AB of the first equal to CD of the second; also, the included angles ABD, CDB, equal, being alternate angles with reference to the parallels AB, DC. Hence these triangles are equal (T. XX.), and the angle ADB is equal to CBD, but these are alternate angles with reference to AD and BC, consequently AID and BC are parallel, and the figure is a parallelogram. THEOREM XXXI. The opposite angles of a parallelogram are equal; so also are the opposite sides equal. Since the figure is a parallelogram, we have AB and DC parallel, also AD and BC parallel; hence, if we draw the diagonal BD, we shall have the angle A B ABD = CDB, being alternate interior angles in reference to the parallels AB and DC (T. XVIII., C.). For a similar reason FIRST BOOK. 35 the angle ADB =CBD. Therefore, the two triangles ABD and CDB have a common side BD, and the adjacent angles of the one equal respectively to the adjacent angles of the other, consequently they are equal (T. XXII.). Hence the angle A = C, and ABD + CBD = CDB + ADB, or ABC = CDA. fhat is, the opposite angles of a parallelogram are equal. The equality of the triangles also gives AB = DC, and AD = BC; that is, the opposite sides of a parallelogram are also equal. Cor. I. The diagonal of a parallelogram divides it into two equal parts. Cor. II. Two parallels included between two other parallels are equal. Cor. III. If one angle of a parallelogram is right, the other three will be right also. THEOREM XXXII. The diagonals of a parallelogram mutually bisect each other, that is, they divide each other into halves. In the two triangles EAB and ECD, D C we have the angle EAB of the first equal to ECD of the second, being alternate angles in reference to the parallels AB and A B DC. For a like reason the angle EBA is equal to EDC. We also have the interjacent sides AB and DC equal, consequently these triangles are equal (T. XXII.), and AE = EC, BE = ED. Cor. If the four sides of the parallelogram are equal, as in the case of a rhombus, we have AB = AD, and the two triangles AEB and AED will have the three sides of the one equal to the three sides of the other respectively, consequently they will be equal (T. XXV.), and the angle AEB = AED, that is, in a rhombus the diagonals bisect each other at right angles. Scholium. The point E, where the diagonals intersect, is called the centre of the parallelogram. THEOREM XXXIII. Ithe straight line joining the middle points of the oblique sides of a trapezoid, will be parallel to the other sides, and equal to half their sum. 36 GEOMETRY. Suppose EF to join the middle points C E and F, of the oblique sides AD and F BC of the trapezoid ABCD. Through F draw GH parallel to AD, and meeting A H, DC produced. In the two triangles, FBH and FCG, we have the side FB = FC, the angle FI3H equal to its alternate angle FCG, and the angle BFH equal its opposite angle CFG, hence these triangles have two angles and the interjacent side of the one, respectively equal to the two angles and the interjacent sides of the other, and are therefore equal (T. XXII.). Consequently, FG = FH; that is, FG equals half of HG. But HG is equal to AD, since the figure AHGD is a parallelogram. Therefore, FG is half of AD, but ED is also half of AD, hence FG and ED are equal and parallel, and the figure EFGD is a parallelogram, and EF is parallel to DC, and also to AB, since DC and AB are parallel. Again, since EF = DG = AH; and CG = IIB, by reason of the equality of the two triangles FGC and FHB, we have EF as much greater than DC as it is less than AB. Hence, EF is one half of the sum of DC and AB. Cor. If we suppose the side DC of the trapezoid to be reduced to a point, the trapezoid will then change to a triangle. Hence, the straight line joining the middle points of any two sides of a triangle, will be parallel to the third side, and equal to half of it. THEOREM XXXIV. The four lines joining the middlepoints of the adjacent sides of a quadrilateral form a parallelogram. If we draw the diagonals of the quadri- G lateral, we shall have EF and HG, each D parallel to AC, and equal to half of it H (T. XXXIII., C.); consequently they are equal and parallel. For the same reason EH and FG are each parallel to BD, and equal to half of it. Therefore the figure EFGH is a parallelogram. Cor. The two lines joining the middle points of the opposite sides of a quadrilateral mutually bisect each other, since they are the diagonals of a parallelogram. FIRST BOOK. 37 ADDITIONAL THEOREMS OF TRIANGLES. THEOREM XXXV. In any triangle if a line be drawn from either angle to the middle of the opposite side: I. When this line is equal to the half side, the angle will be a right angle. II. When this line is greater than the half side, the angle will be acute. III. When this line is less than the half side, the angle will be obtuse. In the triangle ABC, suppose the line CD to c be drawn from the angle C to the middle of the side AB. First. When CD = AD = DB, we have, angle A = DCA, angle B = DCB (T. XXIII.), A D B hence A + B =DCA + DCB; that is, the angle ACB is one half the sum of the three angles of the triangle, and is therefore a right angle (T. V., C. I.). Secondly. When CD > AD or DB, we have angle A > DCA, angle B > DCB (T. XXVII.), hence A + B > DCA + DCB; that is, the angle ACB is less than half the sum of the three angles of the triangle, and it is therefore an acute angle. Thirdly. When CD < AD or DB, we have angle A < DCA, angle B < DCB (T. XXVII.); hence, A + B < DCA + DCB; that is, the angle ACB is greater than half the sum of the three angles of the triangle, and it is therefore an obtuse angle. Scholirum. This Theorem affords a very simple method of determining the kind of angle in any given triangle. HEOREM XXXVI. The three lines bisecting the three angles of a triangle, inter. sect each other in the sare point. Every point in the bisecting line AD is equally c distant from AB and AC (T. XIV.). For the same reason every point of the bisecting line BE is equally distant from AB and BC. Hence, the point G, where these two lines intersect, is equally A F B 38 GEOMETRY. distant from AC and BC; consequently, it is in the line CF which bisects the angle ACB (T. XIV., C. I.). That is, this point is common to the three bisecting lines. Cor. If the sides AB and AC are produced, and the exterior angles are bisected F by the lines BF and CF, the point F, where they intersect, will be on the line AF, which bisects the angle at A. A B D Scholium. This common point, equidistant from the three sides of a triangle, is always within the triangle. THEOREM XXXVII. The three perpendiculars bisecting the three sides of a triangle intersect each other ir the same point. Every point in the perpendicu- c c lar DG is equidistant from A and a B (T. XIII.). For the same rea- E F son every point in the perpendicular EG is equidistant from A and A D B A D B C. Hence, the point G, where these perpendiculars intersect, is equidistant from B and C; consequently, it is in the perpendicular which bisects the side BC (T. XIII., C. I.). That is, this point is common to the three perpendiculars. Scholium. This common point, equidistant from the three angles of a triangle, may be either within or without the triangle; or, in the case of a right-angled triangle, it will evidently be at the middle point of the hypotenuse. THEOREM XXXVIII. iThe three lines passing through the three angles of a triangle, and bisecting the opposite sides, will intersect each other in the same point. We will first consider the two lines CD, C BE, which intersect at G. Draw HK F g joining the middle points of GB, GC, and F it will be parallel to BC and equal to half of it (T. XXXIII., C.). For the A D FIRST BOOK. 39 same reason DE, which joins the middle points of AB, AC, will be parallel to BC and equal to half of it; consequently, HIK and DE are equal and parallel. Now in the two triangles GHlK, GED, we have the angle GHK equal to its alternate angle GED, and GKH equal to its alternate angle GDE, also the side EK =DE; hence, these triangles are equal (T. XXII.), and GH = GE, but GH was taken one half of BG, therefore we have EG one third of EB. For the same reason, DG is one third of DC. If we now consider the two lines CD, AF, we can, by the same kind of demonstration already employed, show that the point where they intersect is one third of FA, measured from F on FA; and one third of DC, measured from D on DC, as before. Hence, the three lines intersect at the same point. Scholium. This common point, situated at one third the distance of the middle points of the sides of a triangle from their opposite angles, is always within the triangle. THEOREM XXXIX. LTh/e three lines passing through the three angles of a triangle, perpendicular to the opposite sides, will intersect each other in the same point. Let ABC be the given triangle. Through E C F the vertices draw lines parallel to the opposite sides, thus forming a second triangle DEF. Then will the figures ABCE, ABFC, and D ADBC be parallelograms, and we have AE and AD each equal to BC; hence, A is the middle point of ED. For a similar reason, B is the middle point of DF, and C the middle point of EF. From this we see that lines drawn from A, B, C, perpendicularly to BC, AC, AB, will be the same as the three perpendiculars bisecting the sides of the triangle DEF, which perpendiculars we already know must intersect each other in the same point (T. XXXVII.). Hence, three lines passing through the three angles of a triangle perpendicularly to the opposite sides, will intersect each other in the same point. Scholium. This common point may be either within or without the triangle (T. XXXVII.). SECOND BOOK. THE CIRCLE, AND ITS COMBINATION WITH THE STRAIGHT LINE. DEFINITIONS. I. The circumference of a circle is a curved line, all the points of which are equally distant from a point within called the centre. The circle is the space bounded by the circumference. II. Any straight line drawn from the centre to the circumference is called a radius. Hence, all radii of the same circle are equal. A line passing through the centre and terminating in both directions by the circumference, is called a diakneter. Hence, all diameters of the same circle are equal, each being made up of two radii. III. Any portion of the circumference of a circle is called an arc. One fourth of the entire circumference is called a quadrant. The straight line joining the extremities of an arc is called a chord. The chord is said to subtend the arc. Every chord corresponds always to two arcs, which together make up the entire circumference. It is the smaller are which is referred to as the subtended arc, unless otherwise expressed. SECOND BOOK. 41 The portion of a circle included by an are and its chord is called a segment. The portion included between two radii and the intercepted arc is called a sector. IV. When a straight line cuts the circumference of a circle it is called a secant. When a straight line touches the circumference in only one point it is called a tangent; and the common point of the line and circumference is called the point of contact. Two circumferences are tangent to each other when they have only one point in common. Two circumferences are concentric when they have the same centre. V. A line is inscribed in a circle when its extremities are in the circumference. An angle is inscribed in a circle when its sides are inscribed. A polygon is inscribed in a circle when its sides are inscribed; and under the same circumstances, the circle is said to circumscribe the polygon. A circle is inscribed in a polygon when its circumference touches each side, and the polygon is said to be circumscribed about the circle. By an angle in a segment of a circle, is to be understood an angle whose vertex is in the arc, and whose sides intercept the chord of said arc; and by an angle at the centre, is meant one whose vertex is at the centre. In both cases the angles are said to be subtended by the chords or arcs which their sides include. VI. Any polygonal figure is said to be equilateral when all its sides are equal; and it is equiangular when all its angles are equal. Two polygons are said to be mutually equilateral when their corresponding sides, taken in the same order, sre equal. When this is the case with the corresponding angles, the polygons are said to be mutually equiangular. A regular polygon has all its sides equal, and all its angles equal. If all the sides are not equal, or all the angles are not equal, the polygon is irregular. 42 GEOMETRY. A regular polygon may have any number of sides not less than three. The equilateral triangle is a regular polygon of three sides. The spuare is also a regular polygon of four sides. OF CHORDS, SECANTS, AND TANGENTS. THEOREM I. Every diameter divides the circle and its circumference into two equal parts. Revolve the portion ACB about the diam- eter AB as a hinge, until it returns to its primitive plane, on the opposite side of AB; then will the portion of the circumference A B ACB wholly coincide with ADB. For, if not, there would be points in the circum- D ference unequally distant firom the centre, which is impossible (D. II.). Hence, the diameter divides the circle and its circumference into two equal parts. THEOREM II. A straight line cannot meet the circumference in more than two points. For if it could, drawing radii to these points, we should have more than two equal lines drawn from the same point to a straight line, which is impossible (B. I., T. XII., C. II.).* * When a reference is made from one Book to another Book, the Book referred to will be given as above; but when the Book is not given, the reference is confined to the Book in which the proposition occurs. SECOND BOOK. 43 THEOREM III. The diameter of a circle is greater than any chord. The diameter AB is greater than any D chord, as CD. For, drawing the radii EC and ED, we have EC+ED>CD (B. I., A B T. VII.). But the diameter AB = EC + ED (D. II.); hence, AB > CD. THEOREM IV. 7he radius drawn perpendicular to a chord, bisects the chord, and also bisects its subtended arc. Let the radius CD be drawn perpendicu- F lar to the chord AB. Produce DC to F, and apply the figure DAF to DBF by revolving it about DF. Then the arc DAF will coincide with DBF A B (T. I.). And since the angles DEA and DEB are right, the line EA will coincide with EB. Hence we have AE =EB, arc AD = arc DB. Also we have arc AF = arc FB. Scholium. The straight line CD fulfils four different conditions: Ist, it passes through the centre; 2d, it passes through the middle of the chord; 3d, it passes through the middle of the subtended arc; 4th, it is perpendicular to the chord. Any two of these conditions are sufficient to determine the direction of the line. Hence we also have the following propositions: A radius drawn bisecting a chord is perpendicular to it, and bisects the subtended arc. A radius drawn bisecting an arc, will also bisect its chord perpendicularly. A line drawn bisecting a chord and its subtended arc, will pass through the centre, and be perpendicular to the chord. A perpendicular bisecting a chord, will also bisect its subtended arc, and pass through the centre. A line drawn from the middle of an arc perpendicular to its chord, will bisect the chord, and pass through the centie. 44 GEOMETRY. THEOREM V. A line perpendicular to a radius at its extremity, is tangent to the circumference. If ED is drawn perpendicular to the E A F D radius CA, at its extremity, it will be tangent to the circumference; that is, it will have only the point A in common with the C circumference. For any other point, as F, being joined with the centre, gives an oblique line greater than the radius (B. I., T. XII.). Hence, A is the only point common to the straight line and the circumference; consequently, this line is tangent to the circumference (D. IV.). Cor. I. If a straight line is tangent to the circumference of a circle, it will be perpendicular to the radius drawn to the point of contact. For all other points of the tangent line, except that of contact, are situated without the circumference, and therefore at a greater distance from the centre than the radius. Hence the radius, being the shortest line which can be drawn from the centre to the tangent, is perpendicular to it (B. I., T. XII.). Cor. II. Only one tangent can be drawn through the same point of the circumference. Cor. III. From a point within the circumference no tangent can be drawn. Cor. IV. The perpendiculars drawn at the extremities of a diameter will be parallel tangents, and conversely two parallel tangents will have their points of contact situated at the extremities of the same diameter. THEOREM VI. Parallel secants or tangents intercept equal arcs of the circumference. First. When the parallels are both F secants. Draw the radius EF perpendic- A L B ular to AB, and it will also be perpendic- c D ular to its parallel CD, and we shall have E (T. IV.) are GF = are KF and are HF = are LF. Consequently, are GF - are HF SECOND BOOK 45 -arc KF- arc LF; that is, the intercepted arcs GH and KL are equal. Secondly. When one of the par- F A B allels is a secant and the other a tangent. Draw the radius EF to the c D point of contact, and it will be perpendicular to the tangent (T. V., C. I.), and consequently perpendicular to its parallel CD. Hence the are GF= are KF (T. IV.). Thirdly. When both parallels are tan- A gents. The line FG, joining the points of contact, will be a diameter (T. V., C. IV.), consequently the intercepted arcs will be semi-circumferences, and therefore equal. (or. Conversely, if two secants inter- c D cept equal arcs of the same circumference they will be parallel, provided they do not intersect, within the circumference. If a secant and a tangent intercept equal arcs of the same circumference, they will be parallel. If two tangents intercept equal arcs of the same circumference they will be parallel. THEOREM VII. In the same circle, or in equal circles, if two arcs are equal, they will have equal chords, which will be equally distant from the centre. If the arcs are unequal and each less than the semi-circumference, the greater are will have the greater chord, which will be nearer the centre. Two circles having equal radii, by superposi- At C C tion, may be made exactly to coincide. Hence we may confine our demonstration to the case! F G of chords in the same circle. First. Suppose the are AB = CD. From the centre E draw EF and EG perpendicular D to the chords AB and CD. Also through H, the middle point of the arc AC, draw the diameter HIK. If we 46 GEOMETRY. revolve the portion of the figure on the left of HK over upon the portion on the right, the arcs HA and HAB will coincide with their equal arcs HC and HCD, consequently the chords AB and CD will be equal, and they will have the same perpendicular; that is, EF will equal EG, and the chords will be equally distant from the centre. Secondly. Suppose the are AB less than A H C CD. Applying the portion of the figure on the left of HK upon the portion on the right, F \ as before, we see that the point B will fall at E G ) some point L, between C and D, since the arc B L AB is less than CD. Hence, the middle of D the arc CL, which is in the prolongation of K the perpendicular EM, will be nearer C than the middle of the arc CD which is in the prolongation of the perpendicular EG; consequently the point G is between D and N, the point where the perpendicular EM intersects CD. Therefore CG, the half of CD, is greater than CN, but CN being an oblique line in reference to EM is greater than the perpendicular CM; consequently, CD > CL, or CD > AB. That is, the greater arc has the greater chord. Again, we have EG < EN, and EN < EM, consequently EG < EM, or EG < EF; that is, the chord corresponding to the greater arc is nearer the centre. Cor. Conversely, in the same circle, or in equal circles: 1. If the chords are equal, the arcs will be equal. And these chords will be equally distant from the centre. 2. The greater chord will correspond to the greater arc. And the greater chord will be nearer the centre. We may also add: 3. Chords equally distant from the centre are equal, and subtend equal arcs. 4. Of two chords unequally distant from the centre, the one nearer the centre is the greater, and it subtends the greater arc. Scholium. We must keep in mind that the arcs considered in this Theorem are in each case less than a semi-circum. ference. SECOND BOOK. 47 OF THE MEASURE OF ANGLES. THEOREM VIII. Inb the same circle, or equal circles, equal angles at the centre correspond to equal arcs. Conversely, equal arcs correspond to equal angles. First. Suppose the angle at E E F to equal the angle at F. Apply the angle AEB upon the angle CFD; EA and EB will take, re- A B C D spectively, the directions of FC and FD; and since the radii are equal, the points A and B will coincide respectively with the points C and D. Hence the arc AB = are CD. Secondly. Suppose the arcs AB and CD equal. Then, as before, applying the figure ABE upon CDF, so that the radius AE may coincide with its equal radius CF, then will the arcs AB and CD coincide, since they are equal, and the point B will coincide with the point D; consequently the angle AEB is equal to the angle CFD. C(or. In the same circle, or equal circles, the greater angle at the centre corresponds to the greater arc, and, conversely, the greater arc corresponds to the greater angle. THEOREM IX. In the same circle, or in equal circles, angles at the centre are to each other as their included arcs..First. When the two arcs AB E and CD are commensurablethat is, when they are to each other as two whole numbers, for A / D instance, as 7 to 4. If we divide the arc AB into seven equal portions, and the arc CD into four, and draw radii to these points of division, we shall divide the angle AEB into seven angles and CFD into four, and these angles will be equal, since their arcs are (T. VIII.). Hence the 48 GEOMETRY. angle AEB will be to the angle CFD as 7 to 4; that is, as the arc AB to the arc CD. Secondly. When the arcs E AB and CD are not commensurable. Apply the smaller A B angle CFD upon the larger D AEB, so that it may take the position AEG. If now, these angles are not to each other as their arcs, suppose we have angle AEB: angle AEG: are AB: arc Al greater than arc AG. If we conceive the arc AB to be divided into equal portions, each of which shall be less than arc Gi, there will be at least one point of division between G and H, as at K. Draw the radius EK, and then since the arcs AB and AK are commensurable, we have by the first part of this Theorem, angle AEB: angle AEK:: arc AB: arc AK. From these two proportions, since their antecedents are the same, we deduce angle AEG: angle AEK: arc AH: arc AK, which cannot be, for the antecedent of the first couplet is less than its consequent, while in the second couplet the antecedent is greater than its consequent. Hence the angle AEB cannot be to the angle AEG as the arc AB is to an arc greater than AG. And by a similar process we may show that the angle AEB cannot be to the angle AEG as the arc AB is to an arc less than AG. Consequently we must have in the same circle, or in equal circles, the angles at the centre to each other as their corresponding arcs. Scholiuzm. Since, in the same circle, or in equal circles, the angles at the centre are to each other as their corresponding arcs, we may use the arcs as the measure of their corresponding angles. In the case of a right angle, its measuring arc is a quadrant; in the case of an angle equal to two right angles, the measuring arc is a semi-circumference. We sometimes regard the right angle as the unit angle, so that acute angles would all be less than 1, and obtuse angles would exceed 1 and be less than 2. SECOND BOOK. 49 THEOREM X. An inscribed angle is measured by half the arc included between its sides. First. When the centre of the circle is e within the angle. Draw the diameter CE and the radii DA A and DB. Now, since the triangle DAC is \ isosceles, the exterior angle AIDE is double the angle DCA. The angle ADE is measured E B by the arc AE (T. IX., S.); hence the angle ACE is measured by one half of the arc AE. In the same way we may show that the angle ECB is measured by one half of the arc EB; consequently the angle ACB is measured by one half its included arc AEB. Secondly. When the centre of the circle is C without the angle. As before, draw the diameter CE, and we D shall have the angle ACE, measured by one half of the arc ABE, also the angle BCE measured by half the arc BE; consequently, their difference, the angle ACB, is measured by half the difference of these arcs; that is, by half the included arc AB. Cor. I. Each angle inscribed in a semicircle is a right angle, since it is measured by half the semi-circumference, or by a quadrant. Each angle is a segment greater than a semicircle is acute, since half the included are, which measures it, is less than a quadrant. Each angle in a segment less than a semicircle is obtuse, since half the included arc, which measures it, is greater than a. quadrant. 4 50 GEOMETRY. Cor. II. An angle, ACB, formed by a A C F tangent and chord, has also for its measure one half the arc CB included between its sides. For, drawing the diameter CE, we D have ACE a right angle (T. V., C. I.), and B it has for its measure one half of the semicircumference CBE. The angle BCE has E for its measure one half the arc BE; hence, the difference of these angles; that is, the angle ACB has for its measure one half the arc CB. If we consider the supplementary angle FCB, we have the right angle FCE measured by one half the arc CGE; hence, the sum of the angles FCE and ECB; that is, the angle FCB has for its measure one half the included arc CGEB. THEOREM XI. The angle formed by the intersection of two chords is measured by half the sum of the included arcs. And the angle formed by the intersection of two secants is measured by half the dfference of the igncluded arcs. Drawing the chord AF parallel to CD, we have the arc FD= AC (T. VI.), and the angle BAF=BED (B. I., T. XVIII., C.), but the angle BAF is measured, in the case of the chords, by E A 0 A D B B~ I B B F half the sum of the arcs BD and DF; that is, by half the sum of BD and AC. But in the case of the secants, the angle BAF is measured by half the difference of BD and DF; that is, by half the difference of BD and AC. Consequently, the angle formed by the intersection of two chords IS measured by half SECOND BOOK. 51 the sum of the included arcs; and the angle formed by the intersection of the two secants is measured by half the difference of the included arcs. Cor. The angle AEC formed by the inter- E section of two tangents, is measured by half A the difference of the concave and convex arcs B AFC and AC, comprehended between the points of contact. For, drawing AF parallel to CE, we have the arc FC=AC (T. VI.), and the angle F BAF= BEC, but the angle BAF is measured by half the arc AF (T. X., C. II.); that is, by half the difference of AFC and FC, or which is the same, by half the difference of the concave arc AFC and the convex arc AC. OF INSCRIBED AND CIRCUMSCRIBED POLYGONS. THEOREM XII. All triangles are capable of being inscribed in a circle and of circumscribing a circle. First. The three perpendiculars bisecting c the three sides of the triangle ABC, meet in the same point G, which is equally distant from A, B, and C (B. I., T. XXXVII.). Hence, if with G as a centre, a circumfer- A D B ence be described with a radius equal to the distance from G to either angle of the'triangle, it will circumscribe the triangle, and consequently the triangle will be inscribed in the circle. Secondly. The three lines bisecting the three C angles of the triangle ABC meet in the same point G, which is equally distant from the three sides of the triangle (B. I., T. XXXVI.). A B Hence, if with G as a centre, a circumference be described with a radius equal to the distance from G to either side of the triangle, it will be inscribed in the triangle, and consequently the triangle will circumscribe the circle. 52 GEOMETRY. Scholium I. Since, in the first case, the perpendiculars can meet in only one point, there can be only one circumference circumscribing a triangle. That is, through three points not in the same straight line, only one circumference can be made to pass. Hence, when two circumferences have three points common, they must coincide. If the three points A, B, and C, are all in the same straight line, the perpendiculars drawn bisecting AB, BC and CA will be parallel, and cannot meet, in which case there can be no centre, unless we regard the centre as at an infrnite distance. Scholi~um II. If, in the second case, the sides AB and AC are produced, the point F, where the lines bisecting the exterior angles meet, is also equally distant from the three lines forming the triangle. Therefore, with F as a centre, a new cir- A B D cumference may be described tangent to the three sides. This circle is called the escribed circle. Hence, if lines be drawn bisecting the angles, and the exterior angles of a triangle, they will intersect each other by threes at the centres of the inscribed and escri9bed circles; thus the three lines which bisect the angles will meet at the same point, giving the centre of the inscribed circle. Any one of the lines bisecting an angle of the triangle will intersect, at the same point, two of the lines which bisect the exterior angles, giving the centre of an escribed circle. Scholium III. This Theorem, taken in connection with Theorem XXXV., Book First, shows that, when the triangle is right-angled at C, the centre of the circumscribed circle will be at I), the middle point of the hypotenuse, for in this case the three lines DA, DB, and DC are all equal. / If the triangle is isosceles, the centre of the circumscribing circle as well as that of the inscribed circle will be in the line bisecting the angle formed by the equal sides. If the triangle is equilateral, the centres of the two circles-will coincide, and the circumscribed and inscribed circles will then be concentric. SECOND BOOK 53 THEOREM XIII. In any inscribed quadrilateral, the sum of the opposite angles is equal to two right angles. The angle at B is measured by half the are c CDA, and the opposite angle at D is measured by half the arc ABC; hence the sum of the angles at B and D is measured by half the entire circumference, which is the inceasure of two right angles. Cor. Conversely, a quadrilateral may be inscribed when the sum of its opposite angles is equal to two right angles. If we describe a circumference through the three points A, B, and C, which can always B C be done (T. XII., S. 1.), and it does not pass through D, then the point D must be either within or without this circumference. First D suppose it within. Produce AD until it meet A D the circumference at E, then; by the Theorem itself, we shall have the sum of ABC and CEA = 2 right angles: but by hypothesis ABC + CDA= 2 right angles; consequently CEA = CDA, which is absurd (B. I., T. VI., C.). Hence, the point D cannot be within the circumference. By a similar process, we can show that it cannot be without the circumference. It must therefore be in the circumference. Scholium. The rectangle, which includes the square, is the only parallelogram which can be inscribed in a circle. The diagonals of the rectangle are diameters of the circumscribing circle. THEOREM XIV. Whien a quadrilateral circum- B scribes a circle, the sums of the op- G posite sides are equal. From the centre draw EF and EG/ to the points of contact of the sides AB and BC, and they will be per- K pendicular to those sides (T. V., C. I.). The two right-angled triangles EFB, D 54 GEOMETRY. EGB, will be equal, since they have the hypotenuse and a side of the one respectively equal to the hypotenuse and a side of the other (B. I., T. XXVI.), and BF = BG. In a similar manner we may show that AF=AK; CH = CG; DH = DK. Consequently by adding, we have AB + CD = BC + AD. Cor. When the sums of the opposite sides of a quadrilateral are equal, it is capable of circumscribing a circle. For, if we describe a circumference tangent to AD, AB, and B BC, which can always be done (T. XII., S. I. or II.), it will also be tangent to DC. For, if the c side DC is not tangent to the cir- A cumference, it must be either a secant or lie wholly without the circumference. We will first D suppose it to be a secant. If DE is drawn tangent to this circumference, we shall have AD + BE = AB + DE, but by hypothesis we have AD + BC = AB + DC; hence, by subtraction, we obtain EC = DE - DC; that is, one side of a triangle is equal to the difference of the other two sides, which is impossible (B. I., T. VII.). Hence the side DC cannot be secant. In a similar way we can show that it cannot be wholly without the circumference. It must therefore be tangent. Scholiumn. The rhombus and the square are the only quadrilaterals capable of circumscribing a circle. THEOREM XV. All regular polygons are capable of being inscribed in a circle, and of circumscribing a circle. Let ABCDEF be any regular polygon; B and through three consecutive angles A, B, and C, describe the circumference of a A a, circle (T. XII.). This circumference will / k\ also pass through all the other angles of l G the polygon. For, drawing from the centre G, the lines GA, GB, GC, and GD, we shall form three triangles, GAB, GBC, SECOND BOOK. 55 GCD. In the two triangles GAB and GBC, we have GB common, AB = BC and GA= GC; hence the triangles are equal (B. I., T. XXV.), and being isosceles since GA = GB = GC, we have the angles GAB = GBA = GBC = GCB. That is, each of these angles is one half the angle of the polygon; hence the angle GCB = GCD. Now, comparing the two triangles GBC and GCD, we have the side GC common, BC = CD, and the angle GCB = GCD; these triangles are therefore equal (B. I., T. XX.); and we have GD = GB = GC = GA; hence, the circumference which passes through A, B, and C, will also pass through D. In a similar manner we can show that the circumference which passes through B, C, and D, will also pass through E, and so on, for all the angles of the polygon, which demonstrates the first part of the Theorem. As to the second part, we observe, that the sides AB, BC, CD, etc., are equal chords, in reference to the circumscribed circle, and therefore they are equally distant from the centre G (T. VII., C. I.). If then, with G as a centre, and with the perpendicular GII as a radius, we describe a circumference, it will be tangent to all the sides of the polygon. This circle will be inscribed in the polygon, and the polygon will circumscribe the circle. Scholium. The point F, the common centre of the inscribed and circumscribed circles, is called the centre of the regular polygon. The radius of the circumscribing circle is called the radius of the polygon, and the radius of the inscribed circle is called the apothem. The angles AGB, BGC, CGD, etc., are called angles at the centre. The lines GA, GB, GC, etc., which are the radii of the polygon, bisect the angles of the polygon. The apothems GH, GK, GL, etc., which are the radii of the inscribed circle, bisect the sides of the polygon perpendicularly, and also bisect the angles at the centre, since they would, if produced, bisect the arcs which measure the angles at the centre. 56 GEOMETRY. THEOREM XVI. If through the corners of a regular polygon, already inscribed in a circle, we draw ta/ngents, we shall thus circumscribe a regular polygon of the same number of sides. The tangents thus drawn will form with A' the sides of the inscribed polygon, AB, BC, CD, etc., a series of triangles, AA'B, AA,, " A',, BB'C, CC'ID, etc., all isosceles and equal, W —-- since AB = BC = CD = etc., and the an- E -, — gles A'AB, A'BA, B'BC, B'CB, C'CD,' C'DC, etc., are all equal, having for their measure half of the equal arcs AB, BC, i CD, etc. (T. X., C. II.). Hence, all the angles, A', B', C', etc., of the circumscribed polygon are equal. And since AA'= A'B = BB'=B'C = CC' = etc., we have A'B'= B'C'= C'D'= etc. That is, this circumscribed polygon has all its angles equal, and all its sides equal; it is therefore regular (D. VI.). Scholium. Since the radius OA' of the circumscribed polygon bisects the angle E'A'B' (T. XV., S.), we have, in the two rightangled triangles OAA' and OBA', the acute angle OA'A of the one, equal to the acute angle OA'B of the other; consequently, the remaining acute angles are equal; that is, the angle A'OA = A'OB. In the same way we can show that the angle B'OB = B'OC. But the angle A'OB = B'OB, since the apothem FB bisects the angle A'FB' at the centre (T. XV., S.). Hence, all the angles AOA', A'OB, BOB', B'OC, COC', etc., are equal; and their measuring arcs AI, IB, BK, KC, CL, etc., are all equal. If then we suppose the circumscribed polygon to revolve in its own plane about its centre 0, as a pivot, so that A may reach the point I, B will then coincide with K, C with L, etc. The circumscribed polygon will by this means, without having its relative parts in the least changed, assume a new position, having its sides parallel with the sides of the inscribed polygon. SECOND BOOK. 57 OF SECANT AND TANGENT CIRCLES. THEOREM XVII. If two circumferences of circles have two points in common, the line joining their centres will bisect their common chord perpendicularly. For, the line bisecting a chord at right angles passes through the centre (T. IV., S.); and as this chord is common to both circles, this bisecting line must pass through both centres. THEOREM XVIII. When two circumferences have only one point in common, this point will be situated on the line joining their centres. If A and B are the centres of two C circles having only one point in common, that point will be situated on the A line AB joining their centres. For, if not, suppose it to be situated without D this line, as at C. Draw CE perpendicular to AB, and prolong it until ED = CE; then we evidently have AC -AD, also BC =BD, and since C was a common point of the circumferences, D will also be a common point. These circumferences will then have two points in common, which is contrary to the hypothesis. Hence, the point in common to the two circumferences must be in the line joining their centres. Scholium. Suppose we consider two circles whose centres are at A and B, the one at B being the smaller. Draw CF passing through their centres, and at the extremities of their diameters, CD and EF, draw the perpendiculars GG', HH', KK', LL'. If we suppose the circle, whose centre is at A, fixed, and the other to move towards it, so that the centre B may move in the line AB, we may notice frve distinct positions of these circles. 58 GGEOMETRY. First. When the tangent HK KK' is on the right of HH'. In this case, each circle will c F be wholly without the other, A D E B and they have no common point. G L' H' BK Secondly. When the tangent KK' coincides with HH', G H and becomes a common tangent to the two circumferences. In this case, the point D c B F is the only point common to the two circumferences, and lt the circles are said to touch externally. Thirdly. When the limiting K H tangents of the smaller circle are M on opposite sides of HH'. In this case, there will be a portion c A E of space common to the two circles; hence' their circumferences L' must necessarily cut each other G' K' H' in two points. Fourthly. When the tangent LL' co- G incides with HH', and becomes a common tangent to the two circumferences. In this case, D is the only point common! to the two circumferences. For, drawing c A 1i any radius, AM, of the larger circle, cutting the circumference of the smaller -- circle at N, and we shall have' H' AN R+R', and when they are interior-that is, when the one is wholly within the other —we shall have D < R - R'. THEOREM XX. When two circumferences are tangent to each other, the distance between their centres is equal to the sum of their radii, or equal to their difference. There are two positions of the circumference in which 60 GEOMETRY. they are tangent to each other (T. XVIII., S.). In the first Case, D =R +R', and in the second case, D=R — R'. THEOREM XXI. When two circumferences cut each other, the distance between their centres is less than the sumrn of their radii, and greater than their difference. By the third case under the Scholium of Theorem XVIII. (see figure of the same), we see that the point M, common to the two circumferences, must be necessarily without the line AB which joins their centres; hence the three points A, B, and M[ must form a triangle, consequently we must have D < R + R' and D > R - R'. Scholium. The reciprocals of the foregoing Theorems may readily be demonstrated by the method of reducing to an absurdity. They may be thus stated: When two circumferences are in the same plane: I. If D> R + R', the two circles are exterior the one to the other. 2. If D = R + R', the two circles will touch externally. 3. If D < R + R', and at the same time D > R - R', the two circles will cut each other in two points. 4. If D = R - R', the two circles will touch internally. 5. If D < R - R', the smaller circle will lie wholly within the other. PROB L EMS, WHICH REFER TO THE FIRST AND SECOND BOOKS. OF PERPENDICULARS, ANGLES, AND PARALLELS. PROBLEM I From any given point of an indefinite straight line, to draw a perpendicular to this line. Let A be the given point of the G indefinite line LM. With the com- K\ F pass, take two equal distances, AB and AC. With A and B as centres, and with a radius greater than AB L B A or AC, describe two circumferences intersecting each other at D. Draw l F' DA, and it will be the perpendicular El G' required. For, by construction, the two circumferences described are such that the distance of their centres BC is less than the sum of their radii, since each radius is greater than half this distance, and it is at the same time greater than the difference of their radii, which is zero; hence these circumferences cut each other in some point, as D (T. XXI., S.); and if we draw DA, it must be perpendicular to BC (B. I., T. XIII., C. II.). Scholium. The two circumferences cut each other in a second point D', on the other side of the line LM, which point may be used to verify our work, since the three points D, A, D' must be in the same straight line. In practice, instead of describing the entire circumferences, we trace only the arcs EF, GK, E'F', G'K', in the immediate vicinity of the points of intersection. 62 GEOMETRY. PROBLEM II. From a given point without a straight line, to draw a perpendicular to this line. Let A be the given point without the line LM. With A A as a centre and with a radius sufficiently great, describe an arc cutting LM in two points B C B and C; and with B and C as L / centres, and with a new radius greater than half of BC, describe two new arcs cutting each other in D and D'; draw ADD', and it will be the perpendicular required. For the points A, D, and D' are, by construction, equally distant from B and C, consequently the line ADD' is perpendicular to BC or LM (B. I., T. XIII., C. II.). PROBLEM I1I. To divide a given stracight line of a deterzminate length into two equal parts; that is, to bisect a given line. Let AB be the line which we wish C to bisect. With A and B as centres, and with a radius greater than half of AB, de- A _ B scribe arcs cutting each other in C and D C', draw CC', and it will bisect AB perpendicularly at D. The same demonstration as in preceding problems Cs applies. Cor. The following problems follow immediately from the foregoing: First. To describe on a given straight line as a diameter a semi-circumference, or the entire circumference. For this question is reduced to finding the middle point of this line. Secondly. To describe a circumference through three given SECOND) BOOK. 63 points. For, having joined these points, two and two, the whole is reduced to drawing perpendiculars through the middle points of these lines. Where these perpendiculars meet, will be the centre of the circle. The radius will be found by joining the centre and either of the given points. Thirdly. To find the centre of a circle, or of an arc already described. For, taking at pleasure any three points on the arc and proceeding as in the last case, we shall find the centre. Fourthly.. To divide an arc into two equal parts. For, draw the chord, and then draw a perpendicular through its middle point (T. IV., S.). PROBLEM IV. To draw from a given point a perpendicular to a given line which can be produced only in one direction. First. When the given point A D is in the given line AB, which line we will suppose cannot be pro- C longed to the left of A. Take any point, as C, for a cen- A B tre, and with CA as a radius de- E scribe an arc cutting AB at E. Draw EC, and produce it until CD is equal to CE. Then drawing AD, it will be perpendicular to AB. For, by construction, CA = CE = CD, and the angle DAB is right (B. I., T. XXXV.). Secondly. When the given point D is without the given line AB. Draw any line, as DE, meeting AB and with C its middle point as a centre, and with CE = CD as a radius, describe an arc cutting AB in A. Join A and D, and it will be the perpendicular required. For, by construction we have, as before, CA = CE = CD; consequently, DAE is a right angle PROBLEM V. Through a given point in an indeflnite straight line to draw a second line,forming with thefirst a given angle. Let M be the given angle, and A the given point in the in 64 GEOMETRY. definite line AX. With M and B' A as centres with a radius MN = AB, describe the arc NP comprised between the sides of x the given angle and the indefi- N A B nite arc BB'. Then with B as a centre, and with a radius equal to the chord of the arc NP, describe a new arc cutting the indefinite arc BB' in C. Draw AC, and BAC will be the angle required. For, by construction, the chords of the two arcs BC and NP are equal; hence the two arcs are equal, and consequently the angles are equal (T. VIII.). Scholiurnm I. To divide an angle into two, four, eight, etc., equal parts, it is sufficient to divide the corresponding arc first into two equal parts, and afterwards to divide each half into two equal parts, and so on. Scholiuzn? II. The same problem furnishes the means of determining the supplement of the sum of two given angles, or, in other words: Knowing two angles of a triangle, to determine the third. After forming the angle BAC equal D C to the first of the given angles, draw AD, making the angle CAD equal to the second of the given angles. Then B' A B produce BA to B' and DAB' will be the supplement required. PROBLEM VI. From a point without a straight line to draw a parallel to this line. Let C be the given point, and AB D' the given line. F- G Draw any line from C meeting AB, as CD; and with C and D as centres ABD with CD for a radius, describe the indefinite arc DD', and the definite arc CE terminating at E in the given line AB. Take the arc DF equal to CE, and draw CF, which will be the parallel required. For, by construction, we have angle FCD = angle CDE (T. VIII.), hence FG and AB are parallel (B. I., T. XVII., C.). SECOND BOOK. 6. Cor. Problems V. and VI. afford the means of solving the following: Through a given point, C, without a straight line ABA to draw a second line meeting the flrst under a given angle M. For, at any point, I, of c the line AB, draw IL, L' L making the angle LIB = angle M; then through C draw CD parallel to A D I D' B LI, which is evidently the line required. Since from any point, I, two lines IL and IL' may be drawn, making with AC an angle equal to IM, if M is not a right angle, it follows that the problem admits of two solutions, unless MX is a right angle. PROBLEM VII. On a given straight line, as a chord, to describe an arc of a circle capable of containing a given angle. Let AB be the given line, N and M the given angle. / Draw BL, making the an- x gle ABL= M, and draw BK perpendicular to BL. Through I the middle point A - of AB draw the perpendicu- N, lar IE meeting BK in E. / From the point E with a radius EA = EB, describe a circle. The portion ANB is the are required. For, by construction, BL is a tangent, being perpendicular to the radius EB, and the angle ABL = IM, is measured by half the arc AN'B (T. X., C. II.). But each angle in the segment ANB is also measured by half the' same~ are (T. X.).. 5 66 GEOMETRY. CONSTRUCTION OF POLYGONS. PROBLEM VIII. To construct a triangle when we have given: I. One side and the two acjacent angles. II. Two sides and the included angle. III. The three sides. It is not necessary to give at length the construction of the first and second cases, since they are so readily deduced from the Corollary to Problem VI. We remark, however, that in the first, in order that the construction may be possible, the sum of the two given angles must be less than two right angles. And in' the second case, the construction is always possible, since, after having drawn two lines, forming with each other the required angle, we can always take on these lines portions equal to the given sides. In the third case. Draw AB equal to one of the given sides, and with A as a centre with a radius equal the second of the given sides, describe an arc; also, with B as a centre and with a radius A equal to the third side, describe an arc intersecting the former are in C. Draw CA and CB, and the triangle ABC will evidently satisfy the conditions announced. That this construction may be possible, it is necessary that the first side may be less than the sum of the other two sides, and greater than their difference (B. I., T. VII.). Scholium. If, in the first case, instead of having the two adjacent angles given, we had one of the adjacent angles and the opposite angle, we could, by Prob. V., Scholium II., find the third angle, and then the two adjacent angles would be known. PROBLEM IX. To construct a triangle when two sides are given and an angle opposite one of these sides. First. When the side opposite the given angle is greater SECOND BOOK. 67 than the adjacent side, the given angle being of any magnitude whatever. Draw the two lines AB and AC, forming the angle BAC equal to the given angle; take AD equal to the adjacent side, and with D as a centre, and with a radius equal to the oppo- A E B site side, describe an are cutting AB in E. Draw DE, and the triangle DAE will fulfil the conditions announced. Secondly. When the side opposite the given angle is less than the adja- D cent side, the given angle being acute. As before, make the angle BAC equal to the given angle, take AD A E - i A E' F E B equal to the adjacent side, and with D as a centre, and with a radius equal to the opposite side, describe an arc cutting AB in two points E and E': drawing DE and DE', we have two triangles DAE and DAE' fulfilling the required conditions. If, however, the opposite side is equal to the perpendicular DF, the arc will touch AB, and the two points E and E' will unite with F, and there will be only one triangle, which will be right-angled at F. But if it. is less than this perpendicular, the arc cannot either intersect or touch AB, and the construction will be impossible. It is obvious that the construction would be impossible if the given angle were either right or obtuse, so long as the opposite side is less than the adjacent side. Scholium. It is not necessary to consider the case when the two sides are equal, for the triangle would then be isosceles; and having one angle given, the other angles could readily be found, and then it would be included in Problem VIII. PROBLEM X. A regular inscribed polygon being given, to construct a regular circumscribed polygon of the same number.of sides; and conversely. First construction. Through the vertices A, B, C, etc., of the inscribed polygon, draw tangents to the circumference, and they will form the polygon required 68 GEOMETRY. Second construction. Through the ex-' Bi tremities I, lK, L, etc., of the radii drawn x i\ perpendicular to the sides AB, BC. CD, A,,, etc., of the inscribed polygon, draw tangents which will form the polygon re- EC quired. We have already shown that the two E' polygons thus obtained are equal (T. D, XVI., S.). Reciprocally. To obtain the inscribed polygon, when we already know the circumscribed polygon A'B'C'D'E', we join the consecutive points of contact A, B, C, etc., of the circumscribed polygon. These chords, AB, BC, CD, etc., are equal, since their arcs AIB, BKC, CLD, etc., are equal; and they form with each other equal angles, since the arcs which measure them are equal. Or we could equally as well use the circumscribed polygon A"B"C"D"E", by joining the consecutive points A, B, C, etc., where the straight lines OA", OB", OC", etc., drawn from the centre to the vertices of the circumscribed polygon, meet the circumference. The polygon ABCDE is regular, since the angles at the centre A"OB", B"OC", etc., being equal, their arcs which measure them are equal, and consequently the chords of these arcs are equal. PROBLEM XI. Gavbin g given two regular polygons, the one'tnscribed and the other circumscribed, each of the same number of sides, to inscribe and circumscribe regular polygons of double the number of sides, and of half the number of sides. Let AB and MN be sides of the i n N given polygons. K11 P First. To obtain an inscribed polygon of double the nmnber of sides, it is evidently sufficient to join the points A, B, with the point I, the middle of the are AB, and to repeat the operation for each of the arcs BC, CD, etc. SECOND BOOK. 69 The chords AI, IB, BL, etc., are equal, since they are subtended by equal arcs. Secondly. We obtain the corresponding circumscribed polygon, by drawing through the points A, B, two tangents which terminate at the points in, n, on the tangent MN, and repeating the operation at the points C, D, etc. The line mnn is the side of a circumscribed polygon of double the number of sides; and Am, nB are half sides. In effect, the right-angled triangles OAm, OIm are equal, having the common hypotenuse Om and the side OA= OI; consequently Am = Imn, and the angle AOm = angle IOin. We see also that the angle mOn is half the angle at the centre of each of the given polygons, and it is therefore the angle at the centre of the polygon sought. Thirdly. As to the inscribed and circumscribed polygons of half the number of sides, we may obtain them by drawing through the alternate points A, C, E, etc., chords AC, CE, etc. And through the same points drawing tangents, having no regard to the intermediate points B, D, F, etc. ScAholium. It is evidently necessary in the third case, in order that the problem may be possible, to admit that the given polygons have an even number of sides. OF CONTACT PROBLEM XII. Through a given point without the circumference of a circle to draw a tangent to tiis circumference. Let E be the centre of the given circle, and A the given point. Draw AE, and upon it, as a diameter, describe a circumference which will cut the given circumference in two points M, M', AM, and AM', being drawn, will be the tangents required. For, 70 GEOMETRY. the angles EMA, EMI'A being in a semicircle, are right (T., X. C. I.). Consequently, AM and AM' are respectively perpendicular to the radii EM and EM', and therefore tangent to the circumference (T. V.). Scholium. The case when the given point is in the circumference offers no difficulty, since we then draw a radius to the given point, and through its extremity draw a perpendicular. OF COMMON MEASURE. PROBLEM XIII. To Jind the common measure of two given lines, provided the, have one, and consequently their numerical ratio. Let AB and CD be the given lines. C A From the greater AB cut off parts equal to the less i line CD, as many times as possible; for example, - twice, with the remainder FB. D From the line CD cut off parts equal to FB, as many times as possible; for example, once, with the remainder GD. F From the first remainder FB cut off parts equal to the remainder GD, as many times as possible; for I example, once, with the remainder HEB. From the second remainder GD cut off parts equal to the third remainder HEB, as many times as possible; for example. twice, without a remainder. The last remainder, HE, will be a common measure of the given lines. If we regard HEB as a unit, GD will be 2, and FB = FII + HB = GD + HB = 3; CD = CG + GD = FB + GD = 3 + 2 = 5; AB = AF + FB = 2 CD + FB = 10 + 3 = 13. Therefore the line AB is to the line CD as 13 to 5. If AB is taken for the unit, CD will be -53; but if CD be taken as the unit, AB will be l.3 SECOND BOOK. 71 If AB is 2 of a yard, then CD will be - of I a yard, or - of a yard. Again, if CD is - of a foot, then AB will be 13 of 3 of a foot = 26 of a foot; and so on for other comparisons. Cor. Since arcs of the same circle, or of equal circles, are capable of superposition, this same method may be employed when we wish to find a common measure of two arcs of the same circle, or of equal circles. Seholium. Magnitudes frequently have no common measure; that is, they are incommensurable; in which case, we shall always, by the foregoing method, have a remainder, however far we carry our successive operations. But these successive remainders would finally become so small that they might practically be neglected. Consequently, we can always find two numbers whose ratio shall approximate as close as we please to the ratio of the given incommensurable magnitudes. This being admitted, we say that two incommensurable magnitudes A and B are proportional to two other incommensurable magnitudes A' and B', or that the ratio of A to B is equal to the ratio of A' to B', when we obtain for each the same numerical ratio at each successive step of the approximation. PROBLEM XIV. Show, geometrically, that the side of a square and its diagonal are incommensurable. Let ABCD be the square, and AC its D C diagonal. Cutting off AF from the diagonal equal to AB, a side of the square, we have the el remainder CF, which must be compared with CB. If we join FB, and draw FG perpendicu- A B lar to AC, the triangle BGF will be isosceles. For, the angle ABG = AFG, each being a right angle; and since the triangle ABF is isosceles, the angle ABF=AFB. Therefore, subtracting the angle ABF from ABG, the remainder FBG will equal the angle BFG, found by subtracting the angle AFB from AFG. Consequently the triangle BGF is isosceles, 72'GEOMETRY. and BG =FG; but, since AC is the diagonal of a square, the angle FCG is half a right angle; but CFG is a right angle, and consequently FGC is also half a right angle, and CG is the diagonal of a square whose side is CF. Hence, after CF = FG = BG has been taken once from CB, it remains to take CF from CG; that'is, to compare the side of a square with its diagonal, which is the very question we set out with, and of course we shall find precisely the same difficulty in the next step of the process; so that, continue as far as we please, we shall -never arrive at a term in which there will be no remainder. Therefore there is no common measure of the diagonal and side of a square. THIRD BOOK. THE PROPORTIONS OF STRAIGHT LINES AND THE AREAS OF RECTILINEAL FIGURES. DEFINITIONS. I. Two figures are similar when they resemble each other in all their parts, differing only in their comparative magnitudes, so that every point of the one has a corresponding point in the other. If, in two similar figures, we conceive a line joining any two points of the one, and another line joining the corresponding points of the other, the numerical ratio of these lines, which is always the same, is called the ratio of similitude. When the ratio of similitude is equal to unity, the two similar figures become equal in all respects. II. Two triangles are similar when their corresponding sides are proportional. III. Two polygons are similar when they are capable of being decomposed into the same number of similar triangles, each to each, having the same order of arrangement. IV. In similar figures homologous points, lines, or angles, are those which have like positions, in regard to the two figures. V. The area of a surface, or its superficial extent, is the nu merical ratio of this surface referred to its unit surface. We readily see that figures of very different forms may have the same superficial extent. When we wish to express the property that two surfaces have the same area, without, however, being equal, or capable of superposition, we say they are equivalent. VI. In determining the area of surfaces, it is found convenient to call the bases of a parallelogram the sides which are parallel. The altitude or height is the common perpendicular to these bases, of which one is called the inferior or lower base, and the other the superior or 2pper base. 74 GEOMETRY. In the rectangle, two consecutive sides form the base and the height respectively. In the square, the base and height are equal. In the case of a triangle, either side may be regarded as the base. The height will be the perpendicular, drawn from the opposite angle to the base, or to the base produced. PROPORTIONAL LINES. THEOREM I. If equal portions are taken on a straight line, and through the points of division parallel lines are drawn mneeting a second line, the intercepts of this second line will be equal also. If the portions CD, DE, and EF, of the straight line AB, are equal; A A' and if through the points C, D, E, C / C and F, the parallels CC', DD', EE', D Di and FF', are drawn, meeting the line A'B', then will the intercepts E d 1 C'D', D'E', and EF' be equal. F / -' For, drawing Cc, Dd, and Ee, B parallel to A'B', the triangles CDc, DEd, and EFe will be equal, since the sides CD, DE, and EF are equal, and their corresponding angles are obviously equal (B. I., T. III.). Hence, Cc = Dd = Ee, and consequently C'D'= D'E'= E'F' (B. I., T. XXXI., C. II.). THEOREM II. If a line is drawn parallel to the bases of a trapezoid, intersecting the sides, it will divide them into proportional parts. If EF is parallel to AD and BC, we A D shall have AE: EB:: DF: FC. First. WVhen AE and EB are commensurable; that is, when they are to ~ - THIRD BOOK. 75 each other as two whole numbers, as 7 to 11 for instance, so that AE:EB:: 7: 11, then will DF and FC be to each other in the same ratio. For, if we conceive the straight line AB divided into 7- +11= 18 equal parts, and through the points of division parallels to BC to be drawn, they will divide DC into 18 equal parts (T. I.), of which 7 will belong to DF, and 11 to FC. Hence, DF: FC:: 7: 11. Secondly. When AE and EB are incom- A mensurable. We may conceive AB divided into equal K L parts so small as to bring K, one of tile points of division, as near as we please to E. Then B drawing KL, we have, by the first case, AK: KB:: DL: LC. And as this will continue, however small the divisions are, we shall have (B. II., P. XIII., S.) AE:EB:: DF:FC. THEOREM III. if a line be drawn parallel to either side of a triangle, cutting the other sides, it will divide them into proportional parts. Although this Theorem might have been made a corollary to the last, we have preferred making it one of our principal Theorems, not only on account of its own importance, but also by reason of the important corollaries which it is capable of furnishing. Let DE be drawn parallel to the side BC A G of the triangle ABC, cutting the other sides at D and E; then we shall have AD: DB:: AE: EC. D F Through A draw AG parallel to BC, and / through any point, as G, of this parallel, draw B c K 76 GEOMETRY. GK parallel to AC meeting BC produced. Also produce DE to F. We have (T. II.) AD: DB:: GF: FK; but, by reason of the parallels AC, GK, we have GF=AE, FK = EC; consequently, AD: DB:: AE: EC. Cor. I. Conversely. If a straight line DE divide two sides of a triangle into proportional parts, it will be parallel to the third side. For, if this is not the case, we might draw through D another straight line DI parallel to BC, and we should then have AD: DB:: AI:IC; but by hypothesis, AD:DB:: AE: EC; hence, by equality of ratios, we have AI: IC:: AE:EC, which is absurd, since AIEC. Cor. II. From the proportion AD: DB::AE: EC we have, by composition, AD + DB: AD:: AE + EC: AE, AD + DB:DB::AE + EC:EC, which gives these new propositions: AB:AD:: AC: AE and AB: DB:: AC:EC. Conversely. If a straight line DE is so drawn as to give AB:AD::AC:AE, we shall have, by division, AB-AD:AD:: AC-AE: AE, that is, DB:AD::EC: AE, which is the same as AD: DB:: AE: EC; consequently, the straight line DE is parallel to BC (C. I.). Scholium. Since, in every proportion, the product of the extremes is equal to the product of the means, we derive from the above proportion, AD: DB:: AE: EC, this equation, AD xEC = DB xAE. THIRD BOOK. 77 Now, in order to comprehend the sense which we ought to attribute to the expression, the product of two lines, AD x EC, or DB x AE, it is necessary to suppose that the straight lines AD, EC, DB, and AE, have been referred to the same linear unit; and instead of lines, properly speaking, we have only to consider, in reality, abstract numbers expressing the ratio of these lines to their unit. We ought then, as in Arithmetic, to attach to the wordproduct the idea of the multiplication of two numbers. Hereafter, we may be required to multiply a surface by a line, a surface by a surface, and even a volume by a volume, etc. But in all these cases it must be understood that it is the ratios of these geometrical magnitudes to their respective units, which are thus multiplied. SIMILAR TRIANGLES. THEOREM IV. A straight line drawn parallel to either side of a triangle, intersecting the other two sides, will form a new triangle similar to the first. If DE is parallel to BC, we shall have A AB: AD::AC:AE::BC:DE. D For, since DE is parallel to BC, we have (T. III., C. II.) B F c AB:AD::AC:AE. Drawing EF parallel to AB, we also have AC: AE:: BC: BF or DE; hence, we have, by equality of ratios, AB: AD:: AC: AE:: BC: DE. Consequently, these triangles are similar (D. II.). 78 GEOMETRY. THEOREM V. Two similar triangles, that is to say, which have their sides proportional, have their homologous angles equal, each to each. Let ABC and A'B'C' be the two A similar triangles. Take AB"= A'B', and draw B"C" A' parallel to BC; the two triangles'a A'B'C', AB"C", being each similar to ABC, the one by hypothesis, and the C B C' other by the preceding Theorem, are similar to each other, and as we have, by construction, AB"= A'B', these triangles must be equal, since the ratio of similitude is unity (D. I.). Now, since B"C" is parallel to BC, the angles of the triangle AB"C" are respectively equal to the angles of the triangle ABC; that is to say, we have the angle at A common, B"= B, C"= C. Then, also, we must have A'= A, B'= B C'= C. We see that the equal angles are opposite the homologous sides. tHEOREM VI. Two triangles, which have the angles of the one respectively equal to the angles of the other, are similar, and the sides opposite the equal angles are homologous. In the two triangles ABC, A'B'C', A suppose A=A', B = B', C = C'. As in the preceding Theorem, take AB"=A'B' and draw B"C" parallel B c to BC. The two triangles A'B'C, AB"'C", are equal, having a side | equal, A'B'=AB", and the adjacent B C B' C' angles equal, A'=A, B'= B = B". Now the triangle AB"C" is similar to the triangle ABC (T. IV.); hence its equal A'B'C' will be similar to ABC. Consequently we have this series of equal ratios: AB: A'B':: AC: A'C':: BC: B'C'. THIRD BOOK. 79 The homologous sides, AB and A'B', AC and A'C', BC and B'C', as we see, are opposite the equal angles C'= C, B'= B, A'=A. THEOREM VII. Two triangles are similarc when the sides of the one are respectively parallel or perpendicular to the sides of the other, and the homologous sides are the parallel or the perpendicular sides. From the preceding Theorem we see that the demonstration will be made out if we can show that these triangles have their angles respectively equal. Let AB and A'B', AC and A'C', BC and B'C', represent the respective sides, either parallel or perpendicular, we are to show that C = C', B = B', A=A'. In effect, we know (B. I., T. III., C.) that the angles C and C', B and B', A and A', must be either equal or supplementary. Now, in the first place, it is impossible that three set of angles should be supplementary, since we should then have A + A'+ B + B'+ C + C'=6 right angles, which is absurd (B. I., T. V., C. I.). Neither can two set of these angles be supplementary; as, for example, A and A', B and B', for then we should have A +- A' -+ B + B'= 4 right angles, which is also absurd. It is then necessary that two angles, at least, of the first triangle should be equal respectively to two angles of the second; consequently, the third angle of the first triangle must equal the third angle of the second (B. I., T. V., C. III.). So that we shall have A = A', B = B', C = C'; and consequently the triangles are similar, and we have this series of equal ratios (T. VI.): BC: B'C':: AC: A'C':: AB: A'B'. From the method of demonstration we see that the homologous sides BC and B'C', AC and A'C', AB and A'B', are theparallel or the perpendicular sides. 80 GEOMETRY. THEOREM VIII. Two triangles, which have an angle of the one equal to an angle of the other, and the sides containing these angles proportional, are similar. In the two triangles ABC, A'B'Q', A suppose we have A = A', and AB: A'B': AC: A'C'. BI,, Take on AB and AC, AB" = A'B', AC" = A'C', and draw B"C". The two triangles A'B'C', AB"C", are B C B,' c equal, having an equal angle included between equal sides each to each. Now, by hypothesis, we have AB: A'B':: AC: A'C', consequently, AB: AB":: AC: AC"; and B"C" is parallel to BC (T. III., Cor. II.), and the triangle AB"C" is therefore similar to ABC (T. IV.). Then, also, is A'B'C' similar to ABC. SIMILAR POLYGONS. THEOREM IX. Two similar polygons have their homologous sides proportional, and the homologous angles respectively equal. For, by the similarity of the E triangles ABC and A'B'C', ACD and A'C'D', ADE and F A'D'E', etc., we deduce the c following series of equal ra-' tios: A B A' B' AB: A'B': AC: A'C':: BC: B'C': ~:, etc. AC:A'C':: AD: A'D'::CD:C'D':, etc. AD: A'D': AE: A'E': DE: D'E':, etc. &c. &c. &c. THIRD BOOK. 81 Omitting the ratios which are common in these series, we have AB:A'B'::BO: B'C':: CD: C'D':: DE: D'E'::, etc. Again, from the similarity of these same triangles, it follows that their homologous angles are equal (T. V.), and consequently the respective angles of the two polygons are equal, since they are composed of angles equal each to each. Thus, A=A', B = B', C = C', D = D', etc. THEOREM X. Two polygons are similar when they have their sides, taken in the samne order, proportional, and their angles, also taken in the same order, equal each to each. That is to say, when we have AB: A'B':: BC: B'C':: CD: C'D'::, etc., and A = A', B = B', C = C', D = D', etc. Let the two polygons be D decomposed into triangles by drawing lines from A and A'. / The two triangles ABC, F c A'B'C', are similar, having F' equal angles, B = B', included between proportional sides A B A' B' AB, A'B', BC, B'C' (T. VIII.). Hence, angle ACB = angle A'C'B', and AC: A'C':: BC:'C'. But, by hypothesis, we have BC: B'C':: CD: C'D'. Consequently, AC: A'C':: CD: C'ID'. Now, comparing the two triangles ACD, A'C'D', we see thatthe angles ACD, A'C'D' are equal, being the differences be — tween the equal angles BCD, B'C'D', and ACB, A'C'B', and we have just shown that the sides about these equal angles are proportional, hence these triangles are also similar. The same may be shown for all the couples of triangles, ADE and A'D'E'. AEF and A'E'F'. Hence the polygons are similar (D.. III.). 6 82 GEOMETRY. Scholium. Two parallelograms are similar, when they have an angle of the one equal to an angle of the other, and the sides containing the equal angles proportional. Two rhombuses are similar, when they have an angle of the one equal to an angle of the other. All squares are similar Jigures. All regular polygons of the same number of sides are similar wfigures. PROPORTIONAL LINES.-PROPERTIES OF THE SIDES OF TRIANGLES. THEOREM XI. The intercepted portions of two parallel lines, made by any number of lines proceeding from the same point, are proportional. The comparison of the different couples of similar triangles PAB and PA'B', PBC and PB'C', etc., imme- B C' C L diately gives the following series of A B C D E I. equal ratios: PA: PA'' AB: A'B':: PB: PB', PB: PB':: BC: B'C':: PC: PC', PC PC':: CD: C'D':: PD: PD', &c. &c. &c. Now, all the ratios are equal, however far they may be extended, since the last in each line above is the first of the following line. Then, if we use only the intermediate ratios, we shall have AB: A'B'::BC: B'C':: CD: C'D':: DE: D'E'::, etc., which establishes the proposition. When the point P is within the parallels, the same demonstration will apply. In this case, the corresponding segments of the two lines AL and A'L' will be situated in opposite directions. THIRD BOOK. 83 THEOREM XII. If a line be drawn bisecting either angle of a triangle, it will divide the opposite side into two segments proportional to their adjacent sides. If the angle BAC is bisect- E' ed by AD, we shall have AB:AC: BD: CD. Draw BE and CF perpen- B x dicular to AD. The two triangles ABE, ACF are similar, since they are equiangular; hence AB: AC: BE: CF. But the two triangles BDE, CFD are also similar, and give BE: CF: BD: CD. Consequently, by equality of ratios, we have AB: AC: BD: CD. Cor. The line AD', bisecting the supplementary angle at A, that is to say, the line perpendicular to AD, will also determine, on the prolongation of the side BC, two segments BD', CD', such as to give the proportion, AB: AC: BD': CD'. For, drawing BE' and CF' perpendicular to AD', we have, by reason of similar triangles, AB: AC::BE': CF'; BE': CF':: BD': CD'; consequently, AB: AC: BD': CD'. Scholium. From the two proportions, AB: AC: BD: CD, AB:AC:: BD': CD', just demonstrated, we deduce the following: BD: CD:: BD': CD'; and the straight line BC is said to be divided harmonically at the points D and D'. 84 GEOMETRY. THEOREM XIII. If from the right angle of a right-angled triangle a line is drawn perpendicular to the hypotenuse: I. The two partial triangles thus formed will be similar to the whole triangle, and consequently similar to each other. II. The perpendicular will be a mean proportional between the two segments of the hypotenuse. III. Either side containing the right angle will be a mean proportional between its adjacent segment and the hypotenuse. First. The two triangles ABC, ABD A are similar, since they have the com- mon angle at B, and the angle BAC of the one equal to the angle BDAB of the other, each being a right angle. For the same reason, ACB is similar to ACD, consequently these triangles are similar to each other. Secondly. Comparing the two partial triangles ABD, ACD, we have BD: D:: AD: DC. T7hirdly. Comparing the total triangle ABC with its partial triangle ABD, observing that in these triangles BC and AB are homologous, being the hypotenuses, that AB and BD are also homologous, being opposite the equal angles BCA, BAD, we shall obtain the proportion, BC AB::AB:BD. The comparison of the triangles ABC, ACD will give, in like manner, BC:AC A C: DC. THEOREM XIV. In any right-angled triangle, the square, or second power, of the numerical value of the hypotenuse, is equgal to the sum of the squares of the numerical values of the other two sides. For, taking the product of the means A and extremes of the last two proportions of the preceding Theorem, we have B c THIRD BOOK. 85 AB2 = BC x BD; AC2 - BC x DC; from which, by adding member to member, AB2 + AC2 = BC x (BD + DC) = BC x BC; that is, AB2 + AC2 = 13C2. Cor. When AB = AC, we have BC2 = 2AB2; that is, in any right-angled isosceles triangle, the square of the hypotenuse is double the square of one of the sides. Consequently, the second power of the numerical value of the diagonal of a square is double that of the side. And the diagonal and side must be to each other in the ratio of / 2 to 1; that is, they are incommensurable, as has already been shown (B. II., P. XIV.). Schlium I. From the two equations, AB2 = BC x BD, AC2 = BC x DC, we deduce the proportion, AB2:AC2::BC xBD:BC xDC, or,::BD:DC. In a similar manner, by combining the identical equation 3BC2 = BC2, with the same two equations above, we have BC2: AB2:: BC x BC: BC x BD, or,:: BC: BD; and BC2: AC2::BC x 3C:BC xDC, or,::BC:DC. From this we see that, in any right-angled triangle, the squares or second powers of the sides containing the right angle and of the hypotenuse, are proportional to the segments of this hypotenuse, and the hypotenuse itself. That is to say, we have AB2: AC2: BC2:: B3DI): DC:BC. Scholiurn II. The segments BD, DC, formed by the perpendicular AD on the hypotenuse BC, are called the projections of the two sides on the hypotenuse. In general, the projection of any defi- N nite straight line MIN, on another indefinite straight line AB, is the distance PQ R of the intersection of the perpendiculars drawn through the extremities of the P first line to the second. If through X we draw AIR parallel to PQ, we shall have MR = PQ, and the right-angled triangle MNR will give MN2 = MR + NR2 = PQ2 + (NQ - MP)2. 86 GEOMETRY. That is-The square of a straight line of determinate length is equal to the square of its projection on another line, plus the square of the difference of the perpendiculars which determine this projection. THEOREM XV. In any obtuse-angled triangle, the square of the side opposite the obtuse angle is equal to the sum of the squares of the other two sides, increased by twice the product of either of the sides containing the obtuse angle into the projection of the other side on the prolongation of the first. If the triangle ABC is obtuse-angled at a A, we shall have BC2 = AB2 + AC2 + 2AB x AD. For, since BD = AB + AD, if we square D A B both members of this equation, using the algebraic formula, we shall have BD2 = AB2 + AD2 + 2AB x AD. Adding now, to each member of this, CD2, and observing that BD2 +CD2=BC2, and also that AD2 + CD2= AC2, we shall obtain BC2 = AB2+ AC2 + 2AB x AD. THEOREM XVI. In any triangle, the square of the side opposite an acute angle is equal to the sum of the squares of the other two sides, diminished by twice the product of one of these sides, by the projection of the other on the preceding one, produced if necessary. If the angle A is acute, we C shall have BC2 = AB2 + AC2- 2AB xAD. For, BD = AB - AD, or AD / - AB, according to which di- A D B A B agram we use; but in both cases BD is the difference of AB THIRD BOOK. 87 and AD; consequently, if we take the squares of both members, by the well-known algebraic formula, we shall have BD2 = AB2 + AD2 - 2AB x AD. Adding to each member of this DC2, and observing that BD2 + CD2 = BC2, and also that AD2 + DC2 = AC2, we shall obtain BC2 = AB2 + AC2 - 2AB x AD. This result will hold good even when the perpendicular CD coincides with the side CB, in which case AD = AB, and the above will become BC2 = AC2 - AB2, or, AC2 = AB2 + BC2, which we know to be true, since the triangle ABC is, in this case, right-angled at B (T. XIV.). Scholium. A triangle is right-angled, acute-angled, or obtuseangled, according as the square of the longest side is equal, less, or greater than the sum of the squares of the other two sides. If a triangle has 3, 4, and 5 for the numerical values of its sides, it will be right-angled, since 52=32 + 42. One having 4, 5, and 6 for its sides, is acute-angled, since 62< 42- +52. One having 2, 3, and 5 for its sides, is obtuse-angled, since 52 > 22 + 32 Thus we are furnished with a second very simple method for determining the kind of triangle, when its sides are given (B. I., T. XXXV., S.). THEOREM XVII. In any triangle, the sum of the squares of any two sides is equal to twice the square of half the third side, increased by twice the square of the line drawn from the middle qf this third side to the opposite angle. If CD is drawn bisecting AB, we shall have AC2 + BC2 = 2AD2 - 2CD2. For, the two triangles ADC, CDB, the A D 88 GEOMETRY. one obtuse-angled and the other acute-angled at D, give, by the two preceding Theorems, AC2 =AD2 + CD2 2AD x DE, BC2 =BD2 + CD2 - 2DB + DE; adding these, and observing that AD = DB, we have AC2 + BC2 = 2AD2 + 2CD2. Cor. The sum of the squares of the four sides of any parallelogram is equal to the sum of the squares of its diagonals. This proposition is easily deduced from the above. DETERMINATION OF AREAS. THEOREM XVIII 2Two parallelograms having equal bases and the same altitude, are equivalent. We will suppose the one L F' D I El cF F Ti, figure placed upon the other, so that their inferior bases, which are equal, may coin- A cide. Let ABCD be the first parallelogram, and ABEF or ABE'F' the second. Since these parallelograms have the same altitude, their superior bases will be situated in the same indefinite line LL', parallel to the lower base. If now, we confine our attention to the two parallelograms ABCD, ABEF, we see that the triangles BCE, ADF are equal, having equal angles CBE and DAF (B. I., T. III.) included between equal sides, namely, BE = AF, BC =AD (B. I., T. XXXI.). If, now, from the quadrilateral ABED we subtract the triangle BEC, we shall have left the parallelogram ABCD; but if from the same quadrilateral we subtract the equal triangle AFD, we shall have left the parallelogram ABEF. Hence the two parallelograms are equivalent. We may in the same way prove that the parallelograms ABCD, ABE'F' are equivalent. THIRD BOOK. 8~ Cor. A parallelogram is equivalent to a rectangle having the same base and the same altitude. Two rectangles of the same base and the same altitude are equal, and consequently equivalent. THEOREM XIX. A triangle is equivalent to half of a parallelogram having the same base and the same altitude. For, through B and C, draw BD parallel to AC, and CD parallel to AB; we thus form a parallelogram having the same base and the same altitude as the triangle. A X The triangle is one half this parallelogram (B. I., T. XXXI., Cor I.). Cor. Two triangles of the same base, or of equal bases, and the same altitude, are equivalent. THEOREM XX. Two rectangles of the same altitude are to each other as their bases. Let ABCD, AFGD be the two rectan- G C gles having the common altitude AD; then will they be to each other as their bases AB, AF. A F B We will first suppose the bases AB, AF to be commensurable; as, for example, suppose they are to each other as 7 to 4. If we divide AB into seven equal parts, AF will contain four of these parts; at each point of division draw lines perpendicular to the base, thus forming seven partial rectangles, which will be equal, since they have equal bases and the same altitude (T. XVIII.). Now, as the rectangle ABCD contains seven of these equal rectangles, while AFGD contains only four, it follows, that ABCD: AFGD:: 7:4; but AB:AF:: 7:4; therefore, ABCD: AFGD:: AB: AF. 90 GEOMETRY. We will suppose, in the second place, that 1 G K C the bases AB, AF are incommensurable, still we shall have ABCD: AFGD:: AB: AF. A FH L B For. if this proportion is not true, the first three tenms remaining the same, the fourth term will be either greater or less than AF. Suppose it to be greater, and that we have ABCD: AFGD:: AB: AL. Divide the line AB into equal parts, each of which shall be less than FL. There will be at least one point of division, as at H, between F and L. Through this point H draw the perpendicular HK, then will the bases AB and AH be commensurable and we shall have, by the first part of this proposition, ABCD: AHKD: AB: AH. But, by supposition, ABCD: AFGD:: AB: AL. Since, in these two proportionals, the antecedents are equal, the consequents are proportional, and we have AHKD: AFGD:: AH: AL. But AL is greater than AH; hence, if this proportion is correct, we must have AFGD greater than AHKD; on the contrary, it is less, hence the above proportion is impossible. Therefore, ABCD cannot be to AFGD as AB is to a line greater than AF. By similar reasoning, we can show that the fourth term of the proportion cannot be smaller than AF. Hence, whatever may be the ratio of the bases, the two rectangles ABCD, AFGD, of the same altitude, will be to each other as their bases AB, AF.. THEOREM XXI. Any two rectangles are to each other as the product of their bases multiplied by their altitudes. Let ABCD, AFGH be two rectangles, then will ABCD: AFGH:: AB x AD: AH x AF. THIRD BOOK. 91 Having placed the two rectangles so K D that the angles at A may be vertical and opposite, produce the sides CD, FB GF until they meet in K; the two rectangles ABCD, AFKD, having the G H same altitude AD, are to each other as their bases AB, AF. In like manner, the two rectangles AFKD, AFGH, having the same altitude AF, are to each other as their bases, AD, AF. Hence we have these two proportions: ABOD: AFKD:: AB:AF. AFKD: AFGH:: AD: AH. Multiplying the corresponding terms of these proportions together, and observing to omit AFKD, since it will occur in an antecedent and consequent, we shall have ABCD: AFGH:: ABxAD: AHxAF. Scholiurm. Hence we are at liberty to take for the measure of a rectangle, the product of its base by its altitude, provided we understand by this product the same as the product of two numnbers, which numbers denote the linear units in the base and altitude respectively. This measure, however, is not absolute, but only relative. It supposes that the area of any other rectangle is estimated in a similar manner, by measuring its sides by the same linear unit; we shall thus obtain a second product, and the ratio of these two products is the same as that of the two rectangles, in accordance with this proposition. For example, if the base of the A rectangle A is ten units, and its height three, the rectangle will be represented by the number 10x 3 = 30, a number which, thus isolated, has no signification; but if we have a second rectangle B, whose base is twelve units and height seven, this second rectangle will be represented by the number 12 x 7- 84. From which we conclude that the two rectangles A and B are to each other as 30 to 84. If we take the rectangle A as the unit of measure of surfaces, the rectangle B will have for its measure 3-0; that is, it will be 4 of our superficial units. 92 GEOMETRY. It is more common and more simple to take a square for the unit of surface, and we choose a square whose side is a unit of length. In this case, the measure which we have regarded as relative becomes absolute; for example, the number 30, which measured the rectangle A, represents 30 superficial units, or 30 squares, each side of which is a unit long. In geometry, we frequently confound the product of two lines with that of their rectangle, and this expression is even employed in arithmetic to denote the product of two unequal numbers; but we use the term square to denote the product of a number multiplied by itself. The squares of the numbers 1, 2, 3, etc., are 1i, 4, 9, etc. From which we see that the square formed on a line of double length is quadruple; on a line of triple length it is nine times as great, and so of other squares. THEOREM XXII. The area of any parallelogram is equal the product of its base by its height. For the parallelogram ABCD is equivalent to C D F C the rectangle ABFG, which has the same basei AB, and same altitude BF (T. XVIII., C.). But the rectangle ABFG is measured by AB x BF, A B consequently the parallelogram is measured by AB x BF. Cor. Parallelograms of the same base are to each other as their altitudes, and parallelograms of the same altitude are to each other as their bases; and in general, parallelograms are to each other as the products of their bases multiplied by their altitudes. THEOREM XXIII. The area of a triangle is equal to the product of its base by half its altitude. For, the triangle ABC is half the parallelo- A F gram ABCF (T. XIX.). The parallelogram is measured by the base BC multiplied into the altitude AD; therefore, B D a THIRD BOOK. 93 the triangle is measured by the base BC into half the altitude AD. Cor. I. Triangles of the same altitude are to each other as their bases; and triangles of the same or equal bases are to each other as their altitudes. Cor. II. The area of a trapezoid ABDC is equal to the product of half the sum of the bases, AB, CD, into the altitude IK:; or, it is equal to the product IK into the straight line EF drawn at equal distance between the bases. C K D In effect this trapezoid is composed of two triangles CAB, BCD; and by the Theorem, we have A I B AB xIK CD x IK ABC- —, BCD= 2 2 AB x IK CDxIK AB+CD then ABDC = 2 + x IIK. And as we have proved (B. I., T. XXXIII.)that EF= AB + CD AB + we have ABDC = IK x EF. THEOREM XXIV. Tlhe area of a triangle is eqgual to half the product of its perimeter by the radius of its inscribed circle. Let K be the centre of the inscribed circle. c Draw KD, KE, KF respectively perpendicu- / lar to the sides AB, BC, AC. Also draw KA, KB, KC. A D B We obviously have ABC = AKB + BKC + AKC. But (T. XXIII.) we have AB x KD BC x KE AC x KF AKB=,BKCC,AKC2 2 2 And since KD = KE = KF, we havw ABC = (AB + BC + AC) x KD ABC~ 2 94 GEOMETRY. HiEOREM XXV. The area of a triangle is equal to the product of the three sides divided by twice the diameter of its circumscribed circle. Let K be the centre of the circumscribed C circle. Draw the diameter CKD, and the chord AD, and the perpendicular CH, which K will be the altitude of the triangle. A B The two triangles CAD, CHB are right- H angled, the one at A (B. II., T. X., Cor. I.), and the other at H; moreover, the angles at D and B are equal, being inscribed in the same segment ADBC. Hence these triangles are similar, and their homologous sides being proportional, we have CD: CB: AC: CII. Consequently, C = CB x AC But we also have ABC AB x C 2 in which, substituting the value of CHI just found, we have AB x CB x AC ABC= 2 -. 2CD RATIOS BETWEEN THE AREAS OF SIMILAR FIGURES THEOREM xxVI. The areas of two triangles which have an equal angle, are proportional to the rectangles qf the sides containing the equal angle. A A Referring to either figure, let the two triangles ABC, E ADE have the angle at A common; then we shall D have B ABC:ADE:: AB x AC: AD x AE. THIRD BOOK. 95 For, joining D and C, we have, since triangles of the same altitude are to each other as their bases, ABC:ADC::AB:AD, ADC: ADE:: AC:AE. Multiplying together the corresponding terms of these proportions, and omitting the common term ADC which enters into the antecedent and consequent of the first couplet, we have ABC: ADE::AB x AC:AD x AE. Cor. Whenever the side DE of the second triangle is parallel to BC of the first, the triangle ADC is a mean proportional between the triangles ABC and ADE. In effect we have, in this case (T. III.), AB: AD::AC:AE; so that the two first proportions of the preceding Theorem have equal ratios, and give ABC: AD:: ADC: ADE. THEOREM XXVII. The areas of similar triangles are proportional to the squares of their homologous sides. A These triangles being similar, are A' equiangular (T. V.). Hence, by the preceding Theorem, we have B C B'I C ABC: A'B'C':: AB x AC: A'B' x A'C'; but since the triangles are similar, we also have AC: AC':: AB: A'B'. Multiplying the corresponding terms of these two proportions, omitting the common factor AC which enters into the antecedents, and the common factor A'C' which enters into the consequents, we have ABC: A'B'C':: AB2: A'B'2. 96 GEOMETRY. THEOREM XXVIII. [he perimeters of two similar polygons are proportional to their honzologous sides. And their areas are proportional to the squares of these sides. First. By reason of the similarity of these polygons, we have (T. IX.), AB: A'B':: BC: B'C':: CD: C'D'::, etc. Now the sum of these an- E tecedents, AB + BC + CD +, etc., which makes the perim- E' eter of the first polygon, is to F C the sum of their consequents F' C' A'B' + B'C' + C'D' +, etc., which makes the perimeter A B A/ B' of the second polygon, as any one antecedent is to its corresponding consequent, and therefore as AB is to A'B'. Secondly. Since these polygons are similar, they are each composed of the same number of similar triangles, which, compared together, are each to each, in the same ratio, that of the squares of the homologous sides. It follows, then, that the sum of the triangles which compose the first polygon, is to the sum of the triangles which compose the second polygon, as the squares of the homologous sides of the two polygons. That is, we shall have ABCDEF: A'B'C'D'E'F':: AB2: A'B'2. Cor. The perimeters of similar figures are proportional to their homologous lines, and their areas are proportional to the squares of those lines. For, by the definition of similar figures, the ratio of any two homologous lines, which is called the ratio of similitude, is constantly the same. THIRD BOOK. 97 COMPARISON OF SQUARES CONSTRUCTED ON CERTAIN LINES. THEOREM XXIX. The square constructed on the hypotenuse of a right-angled trriangle, is equivalent to the sum of the squares constructed re-.spectively on the other two sides. This Theorem is not a fundamental one, like Theorem XIV., but its importance and its fecundity have given rise to several demonstrations founded solely upon the equality and equivalence of figures. We shall confine ourselves to the one generally given in works of geometry. The three squares being constructed without the triangle ABC, we draw from the right angle at A the line AL perpen- A dicular to the hypotenuse, and F produce it to meet DE at M, thus dividing the square BCED B L C into the two rectangles BDML, LCEM. We also draw AD and CF. Now, the angles ABD, FBC are equal, being each composed D M E of a right angle and of the common angle ABC; the lines AB and BF are equal, being sides of the same square, so also are BC and BD equal, for a like reason. Hence the two triangles BAD, BFC are equal (B. I., T. XX.). But the triangle ABD is half the rectangle BLMD, since they have the same base BID and the same altitude BL (T. XIX.). For the same reason the triangle BFC is half the square ABFG, for they have the samebase BF and the same altitude AB; consequently, BLMD = ABFG. We can prove in the same manner, by drawing AE and BI,. that LMEC = ACIK; 98'GEOMETRY. and as'{he square BDEC is equivalent to BLMD + LMEC, there will result square BDEC = square ABFG + square ACIK. Cor. I. The square MNPQ, constructed on the Ad A N diagonal BD or AC of a square ABUD, is double the square itself. B D This is obvious from the simple inspection of the figure. In effect, the four squares AKBM, c Pr AKDN, BKCQ, DKCP, are equal, and respectively the doubles of the triangles AKB, AK(D, BKC, DKC, of which the sum is equal to the square ABCD. Cor. II. We have already shown that the squares ABFG, ACIK (see first figure) are respectively equivalent to the rectangles BLMD, LMEC. And since these rectangles and the square BCED have the common altitude BED, they are to each other as their bases BL, LC, and BC (T. XX.). Hence we obtain this relation: AB2: AC2: BC2: BL: LC: BC. Cor. III. Since the two squares ABFG, ACIK are respectively equivalent to the rectangles BLMD, LMEC, we have AB2 = BC x BL, AC2 = BC x LC; from which we deduce these two proportions: BC:AB:: A: BL, BC:AC::AC: LC. These two proportions, together with the relations of the preceding Corollary, correspond in all respects with the third portion of Theorem XIII., and Scholium I. of Theorem XIV. We have thus, by another method, established properties already demonstrated. Cor. IV. In short, if on the three sides of a right-angled triangle ABC, we suppose three similar polygons to be constructed, as these polygons will be proportional to the squares of their homologous sides (T. XXVIII.), and we have this relation between these sides: BC2 = AB2 + AC2; it follows, that, of the three similar polygons, the polygon constructed on the hypotenuse is equivalent to the sum of the polygons constructed on the sides containing the right angle. THIRD BOOK. 99 Scholium. The algebraic formulas, (p + q)2:=2p + q2 + 2pq, (p - q)2 =p2 + q2 - 2pq, which have been made the bases of Theorems XV. and XVI,, and the other well-known formula, (p +) x (p - q) =p2 _ 2) being translated into geometrical language, will give rise to some new Theorems in regard to areas, with which we will close this Book. THEOREM XXX. The square constructed on the sum or on the difference of two lines, is equivalent to the sum, of the squares constructed respectively on these lines, plus or minus twice their rectangle. The simple inspection of the figure is D F nearly sufficient to satisfy one of the truth of this double proposition. G First. Let AE be the greater of the two lines, EB the less. Construct the squares AEIG, ABCD, and produce EI A E B until it meets CD at F. The square constructed on AB, the sum of the two lines, is evidently composed of the squares AEIG, IKCF, constructed on the line AE and on the line IK = EB, increased by the two rectangles BEIK, GIFD, which have for their respective bases GI = AE, EI = AE, and for altitudes GD = EB, IK = EB. Hence, we have square AB = square AE + square EB + twice rectangle AE xEB. Secondly. Let AB be the greater line, BE the less, that which gives AE for the difference of these two lines. Construct the same figure as in the first case, and, in addition, construct the square GDLM equal to IKCF. The square AEIG is equal to the square ABCD plus the square GDLM, minus the two rectangles EBCF, MIFL. Now, these two rectangles have respectively for bases FE =AB, IM = GK = AB, and for altitudes BE = IK = IF. Hence we have square AE = sguare AB + square EB - twice rectangle AB xBE. 100 GEOMETRY. THEOREM XXXI. The rectangle constructed on the sum2n and the difference of two lines, is equivalent to the diference of the squares constructed on these two lines. Let AB be the greater line, BE = BE' D F c the less, so that AE will represent the sum G of these two lines, and AE' their difference. Construct on AE as a base, and AG= AE' as an altitude, the rectangle AENG, also the square ABCD, and draw A E' B E. E'IF perpendicular to AB, forming the square AE'IG. The two rectangles BENK, GIFD are equal, having equal bases and equal altitudes, namely, BK = AG = GI = AE', EB =E'B = IK = IF = DG; hence it follows that the rectangle AENG is equivalent to the figure DFIKBA. But this figure is the difference of the squares constructed on AB and on IK = BE' = BE. Hence, we have rectangle AENG = square AB - square BE. FOURTH BOOK. THE PROPORTIONS OF LINES AND THE AREAS OF FIGURES IN CONNECTION WITH THE CIRCLE. DEFINITIONS. I. Similar arcs, Sectors, or Segments, are those which, in different circles, correspond to equal angles at the centre. A At Thus, if the angles at A and A' are equal, the arcs BC, B'C' will be similar, the sectors BAG, B'A'C' will be similar, so also will the segments BDC, B C' B'D'C'. B C D A II. When two similar sectors are superposed so that their equal angles coincide, their difference is called a circular trapezoid. Thus the space BB'C'C B,' is a circular trapezoid. B C II. The space included between two concentric circumferences is called a circular ring. 102 GEOMETRY. IV. The arc of a circle is said to be rectified, when it is devel. oped; that is, unfolded or drawn out into a straight line. It is only in this rectified form that we can conceive of its length, since it could not otherwise be compared with the linear unit. PROPORTIONAL LINES CONNECTED WITH THE CIRCLE. THEOREM I. When two chords intersect each other within a circle, the segments will be reciprocally or inversely proportional. Let the two chords AA', BB' intersect A at P. Drawing the auxiliary chords AB', A'B, we thus obtain two triangles PAB', PA'B, which are similar, since the angles A, at P are equal (B. I., T. I.); the angles at A and B are equal, so also are the angles at A' and B' (B. II., T. X.), hence these triangles are equiangular, and consequently similar (B. III., T. VI.), and their homologous sides give this proportion, PA: PB::PB':PA'. Scholium I. When one of the chords B is a diameter, and the other is perpendicular to it, we have A A' PA:'PB::PB:PA'. E P The line PB, drawn perpendicular to a diameter, is called an ordinate to this diameter. Hence, we have this condition: In a circle any ordinate to a diameter is a mean proportional between the two segments of this diameter. We can show that this property is one of the consequences of the properties of right-angled triangles; because if we join A, A' with the point B, we shall form a right-angled triangle (B. II., T. X., C. I.), which gives (B. III., T. XIII.) this proportion, PA: PB:: PB PA'. TSA: ~E T ats PA w FOURTH BOOK.. 103 ScAolum II. This same right-angled triangle ABA' gives the proportion, AA' AB:: AB:. AP. That is, Any chord which is drawn through the extremity of a diameter is a mean proportional between its projection (B.. III., T. XIV., S. II.) on the diameter and the diameter itself. THEOREM II. When two secants intersect each other, w5hout a circle, they will be reciprocally proportional to their external segments. Let the two secants intersect each other P at P. Draw, as in the preceding Theorem, A' B' the chords AB', A'B, and the two triangles AB'P, BA'P, will be similar, since the angle at P is common, and. A =B, each being measured by half of the, arc A'B' A (B. II., T. X.), and the homologous sides give the proportion PA: PB: PB':PA'. THEOREM HI. When a secant and tangent are drawn from the same point, the tangent is a mean proportional between the secant and its external segment. This Theorem is in reality only a particu- P lar case of the preceding, when the points B, B' of the secant PB, are united in one AB point. But we will give a direct demonstration, by drawing the chords AB, A'B. We have, / E in effect, two triangles, PAB, PA'B, similar, A since the angle at P is common and the two angles at A and B are equal (B. II., T. X., C. II.). Hence, we have this proportion, PA: PB: PB: PA'. 104 GEOMETRY. Scholium I. As a particular case, let us suppose the secant to pass /A through the centre of the circle, and F. that the tangent is equal to the diameter; that is. PB -- AA' the pro- A portion will then become PA: AA':: AA':PA'. A" B When a straight line, as PA, is thus divided at a point A'. so that the greater segment AA', is a mean proportional between the whole line and the other segment, it is said to be divided into mean and extreme ratio. The above proportion will give AA': PA-AA':: PA': AA'- PA', or, AA': PA'::PA': AA'- PA'. Now if we take PA" equal to PA', and observe that AA'=?PB, and consequently that AA'- PA'- PB - PA"= A"B, we shall have PB: PA":: PA"': A"B. From this we see that the straight line PB is also divided at the point A" into mean and extreme ratio. Scholium II. The three preceding Theorems give immediately these two equations, PAx PA'= PB x PB', PAx PA'= PB2, which may be included in one single proposition, as follows: The product of the distances from the same point, either,within or without a circle, to two points of its circumference, taken in the same straight line, is always the same. In the case of the tangent we must regard the two points of thile circumference as united. THEOREM IV. In a quadrilateral inscribed in a circle, the rectangle of the diagonals is equal to the sum of the rectangles of the opposite sides. FOURTH BOOK. 105 Let ACBD be the inscribed quadrilateral, C and we shall have AB x CD = AC x BD +- AD x BC. A E For, drawing CE, making the angle ACE equal to the angle BCD, and consequently the angle ECB equal to the angle ACD. Cl The two triangles CEA, CBD are similar, since the angle ACE equals the angle DCB, by construction, and CAE = CDB, since each is in the same segment (B. II., T. X.). Their homologous sides give this proportion, AC: CD: AE: BD; consequently, CD x AE = AC x BD. The comparison of the homologous sides of the two triangles BEC, DAC, which are similar, gives AD: BE:: CD: BC, and CD x BE = AD x BC. Adding the corresponding members of these equations we obtain CD x (AE + BE) = AC x BD + AD x BC, or AB x CD = AC x BD + AD x BC. Scholium. This Theorem has many important applications. FIRST. To find the chord AB of the sum of two arcs AC, CB, when their chords are known. For abridgment, let a, b, c denote the three sides BC, AC, AB of the triangle ABC, and R the radius of its circumscribing circle. Draw the diameter COC'= 2R, also the supplementary chords AC', BC'. Now the inscribed quadrilateral ACBC' will give AB x CC'=AC x BC'+ AC'x CB. But the triangles CAC', CBC' being right-angled, since they are each in a semicircle, give AC'= VCC'2 - AC2, BC'= /CC2 - BC2; and since CC'= 2R, AC = b, BC = a, the foregoing expressions will become AC/= V/4R2 _ b2, BC= 1/4R2 - a2 106 GEOMETRY. these values, substituted in the first condition, cause it to become c x 2R = b /4R2 - a2 + a V/4R2 - b2; b aR hence, c = /4R2 a2 + -R V/42 _ b2. (1) SECONDLY. To find the chord AB = c, of double an are, when we know the chord BC = a of the arc itself. C This will evidently be accomplished by A making b = a in the general expression just formed; so that, in this case, we have o c= oV4R2-a2. (2) C, THIRDLY. _To find the chord of half an are, when we know the chord of the are itself. In the triangle CBC', CB2 -= CI x CC' (B. III., T. XIII.). Now, CI = OC - OI = OC -/OB2 - B12 (B. III., T. XIV.); or, CI =R- ViR2 - c2. Hence, CB2 = a2 = 2R x(R- /R2_i4c2)= 2R2R-/4R2_ c2; or, a = 1/2R2 _ R /4R2 _ c2. (3) It would not be difficult to deduce, algebraically, the value of a in terms of c and b, from formula (1); that is, to find the chord of the difference of two arcs, when we know their chords. When found, the expression is c b a =2 g/4R2b2 V4R2 _ _ 2. (4) We would also remark that formula (3) could have been obtained very readily from (2) by the ordinary rules of algebra. FOURTH BOOK. 107 DETERMINATION OF THE SIDES AND OF THE AREAS OF REGULAR POLYGONS. THEOREM V. The area of any regular polygon is ejual to half the product of its perimeter by its apothem,. For, the equal isosceles triangles OAB, 3 OBC, OCD, etc., give A C OAB = AB x i of OG, OBC =BC x x of OG, OCD = CD x of OG, F &c., &C. Consequently, area ABCDEF = (AB + BC + CD + etc.) x -}OG; or, more concisely, A = 2p x r. A denoting the area of the polygon, p its perimeter, and r the apothem, or the radius of the inscribed circle. Scholium. Calling R the radius of the polygon, n the number of sides, and a the length of one of the sides, we have, from the right-angled triangle OGA, OG = /OA2 - AG2; that is, r= ~/R2 _ a2. We also have p=na; na %/4R2 - a2 consequently, A- 4THEOREM VI. The perimeters of two regular _polygons of the same number of sides are proportional to the radii of their inscribed circles, or of their circumscribed circles; and their areas are proportional to the squares of those radii. Since these polygons are regular, and have the same number of sides, they are similar figures. Consequently, the isosceles triangles which have their vertices at the centre, and for bases 108 GEOMETRY. the sides of the polygons, are each to each equiangular and similar. The radii of the inscribed circles and of the circumscribed circles are homologous lines. Therefore the proposition is true (B. III., T. XXVIII., C.). Let R and R', r and r', a and a', p andp', A and A', denote, respectively, the radii, the apothems, the sides, the perimeters, and the areas of these two polygons, and we shall have as follows:.p:2p'::R:R'::r:r'::a:a', A: A':: R: R2::,r2: 1r2: a2: al2. THEOREM VII. In a given circle, knowing the side of a regular inscribed polygon, we can obtain: I. The value of the side of a regular inscribed polygon of half the number of sides. II. The value of the side of a regular inscribed polygon of double the number of sides. III. The radius and the side of a circumscribed polygon similar to the given polygon. ~~b~~m I N Using the same diagram as under!\, Prob. XI., Book Second, let us represent I / the side AB, of the given polygon, by -- a, its radius OA by R. First. Since AC subtends an are double the are subtended by AB = a, it is sufficient to substitute AC for c in formula (2), scholium to T. IV. Which thus becomes AC =R 4R a2. (1) Secondly. We obtain, in the same way, the value of AI, by means of formula (3), by substituting AI for a, and a for c. Making these substitutions we have AI = V2R2- R V4R2 - a2. (2) Thirdly. As to the values of OM and MN, the two similar FOURTH BOOK. 109 triangles OMN, OAB, of which OI and OK are homologous lines, give OM: OA:: OI: OK; hence, OM OA —xOI OK MN:AB:: OI: OK; hence, MN O — but we have OA = 0I = R, AB = a, and OK = V0OA2 AK2= =VR —2 a2= 12 V4R2 - a2 Therefore, 2R2 2ax R OM=,2d (3) ~4R2 -a2 A 4R2 a2 Scholium. Knowing the radius and the side of each of the three new polygons, we are able to deduce the values of their areas by substituting, in the formula (T. V., S.), naO /4R2- aa2 A= for n, a, R, the respective values just obtained. must be used for n, for the first polygon, and 2n for n, for the second. It remains the same for the third. THEOREM yIII. The area of a regular inscribed polygon, and that of a regular circumscribed one of the same number of sides being known, we are always able to obtain the areas of two regular polygons, the one inscribed and the other circumscribed, of twice the number of sides. Let A and B denote respectively the \M N areas of the inscribed and circum-'L scribed polygons, each having a num- \ ber of sides denoted by n, and A' and B', the areas respectively of the inscribed and circumscribed polygons of double the number of sides. 110 GEOMETRY. By inspecting the diagram we see that we have the following relations: A = 2n x OAK, B = 2n x OMI, A' = 2n x OAI, B'= 2n X OAmI; hence, the ratios between the areas A, B, A', B', taken two and two, are the same as those between the triangles OAK, OMI, OAI, and of the quadrilateral OAmI; thus the whole is reduced to the determination of the ratios which exists between these last four figures. This being premised, we have immediately, in comparing the three triangles OMI, OAI, OAK, and observing that AK is parallel to MI, OMI: OAI:: OAI: OAK (B. III., T. XXVI., C.); or, replacing these ratios of triangles by those of the polygons, we have B:A'l::A'.:A; that is, A'= VA x B. (1) In the second place, since in the triangle OMI, the straight line Om divides the angle 0 into two equal parts, we have the proportion Mm: mI:: OMk: OI (B. III., T. XII.). Now, the two triangles OMm, OmI, having the same altitude OI, give OMm: OmI:: Mm: mI (B. III., T. XXIII., C. I.). Also, since the triangles OAI, OAK have the same altitude AK, we have OAI: OAK::OI:OK:: OM: OA::O: OI; hence, we have OMm: OmI:: OAI: OAK; consequently, OMm +OmI: OmI:: OAI + OAK: OAK; or doubling the consequents, and observing that OMm + OmI = OMI, and 2 x OmI = OAmI, we have OMI: OAmI:: OAI + OAK: 2 x OAK. Finally, substituting, in place of the three triangles OMI, FOURTH BOOK. 111 OAI, OAK, and of the quadrilateral OAmI, the polygons of which they are like parts, we shall obtain B:B':::A+A':2A; 2xAxB 2xAxB consequently, B' = A+ A, or B'= - x B (2) A+A' A + -/A x B Scholium. These two formulas show, at a glance, that A' > A and B' < B. Tlhus the areas of the regular inscribed polygons in the same circle will continue to augment as the number of sides, arising from doubling each time, increases. But the areas of the circvumscribed polygons will decrease as the sides increase. However great may be the number of sides of the regular inscribed and circumscribed polygons, it is evident that the area of an inscribed polygon cannot exceed the area of the circle, and the area of a circumscribed polygon, cannot be less than the circle. The same is true in regard to their perimeters as compared with the circumference of the circle. To prove this, it will be necessary only to L consider the portion of the figure corresponding to the arc AB, since the same course of reasoning may be repeated for each of the o corresponding portions BC, CD, etc. Let the arc AB be divided into two, four, eight, etc., equal parts. Draw the chords of these new arcs. From the definition of a straight line, we evidently have chord AB < AI + IB < AL + LI + IK + KB <, etc., < arc AB. Hence, the _perimeters of the regular inscribed polygons, in the same circle, will continue to augment as the number of sides increase, without being able to surpass the circumnference of the circle. Now, let AM and BM be two half sides of the first regular circumscribed polygon corresponding to the arc AB; inn, and Am, Bn a side and half sides of the regular cir- A B cumscribed polygon of double the number of sides. 0 112 GEOMETRY. We shall have miM + Mn > imn; adding Am + nB to each member, we have Am + mM + Mn + nB, or AMi + MB > Am + mn +r nB. Hence, the perimeters of the regular circumscribed polygons will continue to decrease as the number of sides increase, without being able to become less than the circumference. This being premised, we give, in mathematics, the name of LIMIT to a constant and determinate magnitude, towards which a variable magnitude is constantly converging, either by increasing or diminishing. Hence, we say that: The area of a circle is the SUPERIOR limit of the areas of the regular inscribed polygons, and the INFERIOR limit of the areas of the regular circumscribed polygons. The circumference of a circle is the sutperior limit of the perimeters of the first polygons, and the inferior limit of the second. We feel authorized to regard the circle as a regular polygon of an infinite number of sides, infinitely small; these sides are called the elements of the curve. And this new definition of a circle, if it does not at first appear very rigorous, has at least the advantage of giving greater simplicity and precision to our demonstrations. THE DETERMINATION OF THE SIDES AND OF THE AREAS OF REGULAR POLYGONS OF A PARTICULAR KIND. THEOREM IX. The side of a regular inscribed hexagon is equal to the radius. E D Draw OA and OB; the value of the angle O, of the triangle OAB, is 4 = 2 of a right angle, hence the sum of the other F - two angles is 2- = 3 of a right angle, and since OA= OB, it follows that the A angle OAB = angle OBA = - of 4 = 2 of FOURTH BOOK. 113 a right angle; hence the triangle OAB is equilateral, and gives AB=OA= OB=R. Cor. I. ITle side AC of an inscribed equilateral triangle is to the radius as /3: 1. If in formula (2), of Scholium to T. IV., we make a= R, we shall find AC = R V3R= RV3; hence, AC: R:: /3:1. Cor. II. The side of a circumscribed equilateral triangle is double the side of the inscribed one. To show this, it is sufficient to substitute R V 3 for a in formula (3) of T. VII., which thus becomes 2R2 2 3 3 MN = 2R / 3. MN V 4R2 3R2 We may remark, that the altitude EL of the inscribed triangle is equal to 3R. For, we have EL = EO + OL = R + 1R, since the figure OABC is a lozenge, consequently EL = IR. It follows, then, that the altitude of the circumscribed triangle is equal to 3R. C'or. III. If, in formula (2) of T. VII., we substitute R for a, we shall find AI = (2R2 - R v 4 R2 - R2)i = R V2 - 3,. for the side of a regular inscribed dodecagon. Gor. IV. Finally, in formula (3), T. VII., substituting R for a, we find the radius and a side of a regular circumscribed hexagon, as follows: OM =R,/3; MN =-R V/3. Scholium. We are now prepared, by the aid of the general formula (T. V.), na V/4R2 _ a2 A= 4 to calculate the areas of these different polygons. Thus, for example, if we make n = 6, a = R, we shall obtain for the expression of the area of the regular inscribed hexagon, A = -R2 V3. 8 114 GEOMETRY. By making n = 3, a =R 3, we find A = R2 /3, for the area of the inscribed equilateral triangle. In a similar manner we proceed for the other cases. THEOREM X. The side of the inscribed square is to the radius in the ratio of /2 to 1. Draw the two diameters AB, CD per- Q C P pendicular to each other, and draw the chords AC, CB, BD, DA. The figure ADBC is evidently a square; and we A have AC2 = AO2 -+ O(A2 2R2; consequently AC = R V2, and N AC: R:: /V2 1. Scholium. In drawing through the points A, D, B, C, tangents, we form the circumscribed square; and we have MN = AB = 2R. That is-The side of the circumscribed sguare is equal to the diameter of the circle. The areas of these two polygons are expressed respectively by 2R2 and 4R2. THEOREM XI. The side of a regular insc'ribed decagon is equal to the greater segment of the radius, when divided into mean and extreme ratio. Draw the radii OA, OB to the extremities of a side AB. The angle O of the triangle OAB is equal to 4 = - of a right angle; there remains, then, 2 — = of a right angle, for the sum of the other two L angles; and as OA = OB, it follows that the angle OAB = OBA = 4 of a right angle. A B FOURTH BOOK. 115 Thus, each of the angles at the base of the triangle OAB is double the angle at the vertex. This being premised, draw BL bisecting the angle OBA, and we shall have this proportion, AL: LO AB: OB (B. III., T. XII.). But since the angle LBO = LBA = LOB = 2, and consequently ALB = 2LOB =-, the two triangles OLB, ALB are also isosceles, and give OL = LB = AB, and also having OB = OA, the above proportion becomes AL: OL: OL: OA. From which we see that the point L divides the radius OA into mean and extreme ratio (T. III., S. I.), and that the greater segment OL is equal to AB a side of the regular decagon. Scholium I. We will deduce the numerical value of this side algebraically. Denoting it by x, we have (T. III., S. I.), R: x:: x: R -, or, 2 + Rx = R2 and x = ( V5 - 1). That is, the numerical value of the side of a regular inscribed R decagon is - (V5 - 1). Scholiuim II. If for a in formula (2) of Scholium to T. IV. we substitute 2 (V5 - 1), which we have just found for the value of a side of a regular inscribed decagon, we shall find 2 (10 - 2V5) for the side of a regular inscribed pentagon. Now, since [ (10 -2 V5)]=[ (5- 1)] + R2, it follows, that the square of the side of a regular inscribed pentagon is equal to the square of the side of the regular inscribed decaagon increased by the square of the radius. 116 GEOMETRY. THEOREM XII. The side of a regular inscribed pentadecagon is the chord of the difference between the arcs subtended respectively by the sides of a hexagon and a decagon. For, we have 1 — = -5 33 - =2 -; which shows that -The difference between the arcs subtended by the sides of a hexagon and of the decagon is equal to one fifteenth of the entire circumference. If we wish the numerical value of the side of the pentadecagon, we may employ formula (4) of T. IV., in which, for c, we must substitute the value of the side of a hexagon, and for b we must use the side of a decagon. General Scholium. It results from what has been said in the last four Theorems, and of the principles before established, that we have geometrical methods for constructing regular inscribed and circumscribed polygons of 3, 6,12, 24, etc., of 4, 8, 16, 32, etc., of 5, 10, 20, 40, etc., in short, of 15, 30, 60, etc., number of sides. That we also have arithmetical methods for calculating the sides, and afterwards the areas, of all these polygons, if not exactly —since nearly all the numbers which enter into our calculations are incommensurable-yet to as close a degree of approximation as may be desired. MEASURE OF THE CIRCLE AND OF ITS CIRCUMFERENCE. THEOREM XIII. The area of a circle is equal to half the product of the circumrference into its radius. In effect, let us consider a series of regular circumscribed polygons, each having double the number of sides of the preceding one. The superficies of these polygons have for their respective measures half the product of each perimeter into its apothem-that is, into the radius of the circle. Hence, the area of the circle, which is the limit (T. VIII., S.) of these polygons, FOURTH BOOK. 117 has for its measure half the product of the circumference, which is the limit of the perimeters, into its radius. Or, considering the circle as a regular polygon of an infinite number of sides, it immediately follows that its area is equal to half the product of its perimeter, that is, of its circumference, into its radius. Scholium I. Let R, C, and S denote respectively the radius, circumference, and surface of any circle. We shall have s=~C xR=C x MR. We must keep in mind that S is an abstract number, expressing the ratio of the surface of the circle to the unit of surface; C and R are ratios of the circumference and of the radius to the linear unit. Scholium II. We may also say that the area of a circle is equal to that of a triangle having for its base the rectified circumference (D. IV.), and for its altitude the radius of the circle (B. III., T. XXIII.). THEOREM XIV. In two circles, the ctrcumferences are proportional to their radii or to their diameters, and their areas are proportional to the squares of their radii. First. Let us conceive a series of regular polygons, each having double the number of sides of the preceding one, to circumscribe the circumference C; also another series of similar polygons to circumscribe the circumference C'; and let us designate by R, R' the constant apothems of these two series of polygons; by p, p' the perimeters of two similar polygons, taken, the one in the first series, the other in the second. This being supposed, we shall have this proportion, p:p':: R: R', which is applicable to any two similar polygons whatever; it must therefore hold good at the limit of the perimeters, that is, whenp, p' become C, C'; hence, C:C'::"R:R', or,::2R:2R'. 118 GEOMETRY. Secondly. Multiplying this last proportion by the obvious proportion, ~R: R':: R: R', or,:2R:2R', it becomes, C x ~R: C' x,R':: R2 R'2, or 4R2: 4R12. Or, since (T. XIII., S. I.) C x ~R - S, C' x CI' = S', this becomes S:S':: R2,R2, or 4R2' 4R'2. Or, considering the circle as a regular polygon of an infinite number of sides, the above follows immediately from T. XXVIII., B. III. bor. The proportion C: C': 2R: 2R', may be written C: 2R:" C': 2R', and being applicable to any number whatever of circumferences, we have C: 2R:: C' 2R':: C": 2R":: C"':2R"'::, etc., from which we infer that Thie ratio of the circumference to the diameter is a constant znumber. We usually denote by r this constant ratio. Scholium. If R, C, and S denote respectively the radius, the circumference, and the surface of any circle whatever, we shall have this proportion i: C: 2R; consequently, C = 2qrR, from which, by multiplying each term by 12R, we find C x'R =S = nR 2. From this, we see that the circumference of any circle may be found by multiplying its diameter (2R) by r; that its surface may be obtained by multiplying the square of its radius by ~. THEOREM XV. The area of a circular sector is equal to half the product of its are into its radius. In effect, from the definition of a circu- / lar sector, it follows that in the same circle two sectors are proportional to their angles, 0 and consequently to their corresponding arcs. Hence, comparing the given sector AOB A FOURTH BOOK. 119 with the sector AOC, which has its angle AOC right, we have this proportion, sector AOB: sector AOC:: are AB: AC; or, multiplying the two consequents by four, and denoting by C, S, the circumference and surface, or area of the circle, we have sector AOB: S:: are AB: C. Again, multiplying the terms of the second couplet of this last proportion by ~R, we obtain, sector AOB: S:: are AB x -R: C x ~R. But, C x 1R = S (T. XIII., S. I.); consequently, sector AOB = are AB x 1IR = ~ are AB x R. Schowlum. Theareaof a sector is equal to that of a triangle having for its base the rectified arc of the sector, and for its altitude the radius of the circle. THEOREM XVI. The areas of two similar sectors are proportional to the squares of their radii, or of their arcs. The sectors being similar, have their angles 0 equal. We will, therefore, suppose the one A'l//',BI placed upon the other so that their angles A B may coincide. Since (T. XV.) sector AOB = are AB x ~OA, sector A'OB' = arc A'B' x ~OA', we have sector AOB: sector A'OB':: arc AB x IOA: arc A'B' x IOA'; but the arcs AB, A'B' being evidently in the same ratio as the entire circumferences of which they are parts, which circumferences are in the ratio of their radii OA, OA' (T. XIV.), there will result are AB: are A'B':; OA: OA', hence, are AB x 1OA: are A'B' x 1OA':: OA2: OA'2; consequently, sector AOB: sector A'OB':: 0A2: OA'2, also, sector AOB: sector A'OB':: (arc AB)2 (arc A'B')2. 120 GEOMETRY. Scholium I. The difference AA'B'B between the similar sectors OAB, OA'B', which is called a circular trapezoid (D. II.), has for its measure theproduct of half the sum of its bases into the difference of their radii. To prove this, draw the indefinite tangents BL, B'L', and take on BL the portion BK equal to the rectified are BA, and draw KO o meeting B'L' in K'. Then, by reason of the \ \ B similarity of figures, we shall have OB: OB'::BK: B'K'; OB: OB':: are BA: arc B'A'; consequently, L' BK: B'K':: are BA: are B'A'. L But BK = are BA by construction, therefore B'K' = are B'A'. Hence the two sectors OAB, OA'B' are respectively equivalent to the two triangles OBK, OB'K' (T. XV., S.); it therefore follows that the circular trapezoid is equivalent to the right-angled BK + B'K' trapezoid KBB'K', which has for its measure 2 +B/E/ x BB' (B. III., T. XXIII., C. II.). Consequently we have = arc AB + arc A'B' circular trapezoid AA'B'B x BB'. It would be, moreover, easy to prove that the half sum of the bases, or the mean difference between the two bases, is a concentric arc with AB, A'B', at equidistance from the same. Scholium II. The circular ring (D. III.), that is to say, the space between the two concentric circumferences (see last diagram), is only a particular case of the circular trapezoid, consequently its area is expressed by the product of the mean circunmference into its width BB', the difference of the radii. FOURTH BOOK. 121 RATIO OF THE CIRCUMFERENCE TO THE DIAMETER. FIRST METHOD. We know that the side of a regular inscribed hexagon is equal to the radius (T. IX.). If we suppose, then, the radius R = 1, we can find the side of a regular inscribed polygon of twelve sides, by making a =R = 1, in formula (2) of Theorem VII., that which gives,/2- /3 for a side of a regular inscribed dodecagon. Its perimeter is 12./2 -,/3; consequently the ratio of this perimeter to the diameter is 6 /2 -,/3. Were we to substitute anew, in the same formula, 1 for R, and./2- /3 for a, we should find a side of an inscribed regular polygon of twenty-four sides. Thus we might continue to find the sides of the successive polygons, each having double as many sides as the preceding one, to any extent we please. As the number of sides of the polygons increase, their perimeters will approximate more nearly to the circumference of the circle; we are thus enabled, by this method, to obtain the value of t to as close a degree of approximation as we please. SECOND METHOD. We have demonstrated (T. XIV., S.) that the area of a circle which has R for its radius, is equal to yrR2. Now, if we make R = 1 in this expression, it will become a, which shows that The ratio of the circumference to the diameter is equal to the area of a circle whose radius is unity. This being supposed, let us start with an inscribed and circumscribed square, whose areas are respectively 2 and 4 (T. X., S.). And the formulas A'= AxB, B= 2 x A x B A +A' (T. VIII.), if we suppose A = 2, B = 4, will become A'=l=, B'= 1, = 6 4 (/8- - 2), 2+, 8 2 which, converted into decimals, will give 2.8284271, etc., 122 GEOMETRY. 3.3137085, etc., for the areas, respectively, of the inscribed and circumscribed octagons. Using these values in the same formulas, we shall obtain the areas of the inscribed and circumscribed polygons of sixteen sides, which in turn will make known the areas of polygons of thirty-two sides, and so on, till we arrive at an inscribed and circumscribed polygon, differing front each other, and consequently from the circle, so little that either may be considered as equivalent to it. The subjoined table exhibits the area, or numerical expression for the surface of each succeeding polygon, carried to seven places of decimals. Area of the inscribed Area of the circumscribed polygon. polygon. 4 2.0000000 4.0000000 8 2.8284271 3.3137085 16 3.0614674 3.1825979 32 3.1214451 3.1517249 64 3.1365485 3.1441184 128 3.1403311 3.1422236 256 3.1412772 3.1417504 512 3.1415138 3.1416321 1024 3.1415729 3.1416025 2048 3.1415877 3.1415951 4096 3.1415914 3.1415933 8192 3.1415923 3.1415928 It appears, then, that the inscribed and circumscribed polygons of 8192 sides differ so little from each other, that the numerical value of each, as far as six places of decimals, is absolutely the same; and as the circle is between the two, it cannot, strictly speaking, differ from either so much as they do from each other; so that the number 3.141592 expresses the area of a circle whose radius is 1: that is, the value of ~ to six places of decimals is 3.141592. THIRD METHOD. Instead of seeking the approximate value of the circumference, or of the area of a circle, of which the radius is 1, we shall seek the value of the radius corresponding to a given circumference. For this purpose, we will premise the following FOURTH BOOK. 123 LEMM:A. Having given the radius R and the apothem r of a regular polygon, it is required to find the radius R' and the apothemr r' of a second regular polygon having the same perimeter as the Jfrst and double the number of sides. Suppose the given polygon to be circumscribed, and that AB is one of the AB sides, AOB the angle at tile centre, OA =R = the radius, and OP =r = the apothem. This being supposed, the angle at the centre of the second polygon is half the angle AOB. If we produce PO until it meets the circumference at C, and draw the chords CA, CB, the angle ACB, half of AOB is the angle at the centre of the second polygon. Also, if we draw OA' perpendicular to CA, and draw A'B' parallel* to AB, we shall have A'B' = ~AB (B. I., T. XXXIII., Cor.) for the side of the second polygon, and CA' will be its radius, and CP' its apothem. The similar triangles CAP, CA'P', give OFP = -CP ='(CO + OP) = (OA + OP), consequently, r' = ~(R + r). Again, the right-angled triangle OA'C gives CA'2= CO x CP'= OA x CP'; that is, R' = VR/ x r'; and as r' has already been found, the problem is resolved. These formulas are much more simple than those already used in the first and second methods. Now, to make an application of these formulas, we will take for our given polygon a square whose side is 1, and consequently its perimeter is 4. Its radius is -/2, being obviously half its diagonal, and its apothem is 2. If, then, in the formula r'= -(R + r), we make R = /V2 = 0.7071068, etc., r= ~ = 0.5, we shall find r' = 0.6035534, etc. Formula IR' = x r' will now give R' = 0.6532815, etc. We 124 GEOMETRY. thus find the radius and apothem of a regular polygon of eight sides, and whose perimeter is 4. Using these values in the same formulas, they will in turn make known the radius and apothem of a regular polygon of sixteen sides, having the same perimeter, 4. We give the results of these successive operations, to seven places of decimals, in the following table: No. of sides. Apothems. Radii. 4 0.5000000 0.7071068 8 0.6035534 0.6532815 16 0.6284174 0.6407289 32 0.6345731 0.6376435 64 0.6361083 0.6368754 128 0.6364919 0.6366836 256 0.6365878 0.6366357 512 0.6366117 0.6366237 1024 0.6366177 0.6366207 2048 0.6366192 0.6366199 4096 0.6366195 0.6366197 8192 0.6366196 0.6366196 From this table, we see that a circle whose circumference is 4 has for its radius 0.6366196, etc., and consequently for its diameter 1.2732392, etc. Hence, the ratio of this circumference to its diameter, or the value of %r, is 4 — = 3.1415931, etc. 1.2732392, etc., This value of v differs less than a unit in the sixth decimal place from the true value. There are other methods, depending upon a knowledge of the higher branches of mathematics, by which this value of %r has been extended to more than two hundred decimal places. The value to thirty-one decimals is r = 3.1415926535897932384626433832795. The following original geometrical construction is very simple, and gives this ratio sufficiently accurate for all practical purposes: Let ACB be the diameter of the given circle: produce it towards N; take BD and DE each equal to AB; through E draw FOURTH BOOK. 125 EG perpendicular to AE, and take EF and FG each equal to AB; join AG, AF, DG, and DF. Set off on the line EN, from E, the distances EH and HK, each equal to AG; then set off, in the opposite direction, the distance KL equal to AF, and from L set off LM equal to DG; also set off MN equal to DF Then bisect EN at the point P; bisect EP at the point R, and, finally, trisect ER at the point T; then will CT be the circumference of the circle, nearly. G F T R P B D E H L M K N For, by construction, we have, if we call the diameter a unit, CE = 21; EL = 2E -KL = 2,/13 - 10; LM =,5; MN = v2. Therefore, EN = 2 /13 - V10 + V5 + 2; and ET = - (2 /13 - V10 + V5 + V2); and, therefore, CT = 21 + 1 (2V13 - V10 + +5 + V2) = 3.1415922, etc., which is the ratio true to six decimals. For simplicity and accuracy, a better graphic method of finding this ratio can hardly be expected, or even desired. PROBLEMS, WHICH REFER TO THE THIRD AND FOURTH BOOKS. CONSTRUCTION OF PROPORTIONAL LINES. PROBLEM I. To divide a given line into any number of equal parts. To make a definite case, suppose we wish to divide AB into five equal parts. Through A draw any indefinite p line AX, making an angle with AB. Lay off on this line any con- A q R S B venient length five times, as Ap, pg, gr, rs, sb. Join the last extremity b with B, and through the other points of division draw parallel to Bb, lines cutting AB in the points P, Q, R, and S (B. II., P. VI.). The straight line will thus be divided into five equal parts (B. III., T. I.). PROBLEM[ II. 5To divide a given line into parts proportional to given lines. As a definite case, suppose we wish b/ to divide AB into three parts, which r q shall be to each other as the three lines P, Q,R. As in the last Problem, draw any indefinite line AX, making an angle with A P, Q, B AB. Make Ap, pq, qb respectively equal to P, Q, and R. Draw Bb, and parallel to it draw pP', gQ', and the line AB will be divided at P' and Q' as required (B. III., T. III.) FOURTH BOOK. 127 Scholium,. As a particular case, suppose we wish to divide a straight line AB into two parts proportional to the lines M and N. Through the points A X and B draw any two par- c allel lines AX, BY, so that the alternate angles BAX, ABY maybe equal. Take Di on AX and BY portions AC, BC', respectively M N equal to M and N; then Y draw CC' meeting AB in D. The straight line AB will thus be divided at D in the ratio required. For, the two triangles DAC, DBC', are evidently similar, and give AD: DB:: AC:BC':: M: N. If, instead of taking BC', equal to N, in the direction of BY, opposite to AX, we had taken BC", equal to N, in the same direction with AX, then the line CC" would have met AB produced at D', which is called the conjugate point of the point D (B. III., T. XII., S.). PROBLEM III. To0find a fourth proportional to three given lines MI, N, P. Form any angle, as XAY, and E take on AX, AB = M, BC = N, and D on AY, AD=P; then draw BD, and through the point C draw CE A x parallel to BD. The line DE will be the fourth proportional re- M -- N quired. p For, we have (B. III., T. III.), AB:BC::AD:DE, orM:N::P:IDE. Cor. I. If P were equal to N, the above proportion would become M:N::N:IDE. 128 GEOMETRY. That is, DE would in this case be a third proportional to the two lines M and N. Cor. II. We deduce immediately the following problem: A point 0 being given within an angle YAX, to draw through 0 a straight line DOE, such that the segments DO, OE may be to each other in the ratio of M to N. Through the point 0 Y draw OB parallel to AY; N find a fourth proportional to the three lines M, N, AB; make BE equal to X B E A B E this fourth proportional, and draw EOD, which will be the line required. For, since OB is parallel to AD, we have OD:OE::AB:BE:: M:N. When M = N, it is sufficient to take on AX, BE = AB, and then to draw EOD. PROBLEM IV. To find a mean proportional between two given lines M and N. On the indefinite line AX, take D AB = M, BC = N, and on AC, as a diameter, describe the semi-circumference ADC. Through B A X B C draw BD perpendicular to AC, and BD will be the mean pro- M portional sought. For, we have N (T. I., S. I), AB: BD::BD:BC, or M: BD::BD:N. PROBLEM V. To divide a gRven line into mean and extreme ratio. We have already noticed this kind of division (T. III., S. I.). And we have also actually solved this problem algebraically (T. XI., S. I.). FOURTH BOOK. 129 We will now proceed with its geometrical solution. Let AB be the line which we are required to divide. Draw BO perpendicular to AB c and equal to one half of AB; then draw AO, with O as a centre; A D B with OB for a radius, describe the circumference meeting AO in C. Finally, make AD equal to AC, and the line AB will be divided at the point D as required. In effect, from the construction, AB is a tangent to the circle OB; and if AO is produced until it meets the circumference in C', we shall have (T. III.), AC' AB:: AB:AC. Consequently, AC'- AB: AB:: AB - A: AC. Now, AC' = AC + CC' = AC + AB; consequently, AC' - AB=AC=AD, and AB-AC = AB-AD=BD hence the proportion evidently becomes AD: AB: BD:AD, or interchanging means and extremes, AB:AD AD: BD. PROBLEM V1. To draw a tangent common to two circuzmferences. We will suppose the problem solved, and that ]MN, mn arethe two tangents, the one exterior and the other interior, meeting the line of the centres 00' in two points 0, 0'. 9 130 GEOMETRY. It is evident that, if these points were known, it would be sufficient to draw through each of them a tangent to one of the circles (B. II., P. XII.), and it would of necessity be tangent to the other; the problem would thus become resolved. Now, in drawing the radii OM and O'N, Om and O'n, we shall evidently obtain two couple of similar triangles OC and O NC, OmC and O'nC', which give the proportions OC: O'::OM: O'N, OC:O'C'::OM: O'N. But the radii OM, O'N, are given lines. We see then that the points 0, C' are the conjugate points (B. III., T. XII., S.; also P. II., S.), which divide the distance 00' in the ratio of OM to O'N. From this results the following construction: Draw any diameter KOk of the circle 0, and through the point O' draw the radius O'L parallel to OK; join the points K and k with the point L. The points C, C', where the straight lines KL, kL meet the line of the centres, are the points sought; since we have (B. III., T. IV.), OC: O'C:: OM: O'N, OC': O'C':: OM: O'N. Now draw through each of these points a tangent to either of these circles, and it will be tangent to the other. Scholium. This problem is evidently susceptible of four, three, two, or only one solution, or it may not admit of any, according to thefive relative positions of the two circumferences (B. II., T. XVIII., S.). PROBLEMS OF AREAS. PROBLEM VII. To transform a polygon into another having one side less than the first, and jinally into a triangle. Let ABODE, etc., be any polygon whatever, which we here represent by a broken line, in order that the generality of the construction may be the better shown. FOURTH BOOK. 131 Through the point A draw B I the diagonal AC cutting off A the triangle ABC. Draw afterwards, through the point B, the line BI parallel to AC, L meeting DC, produced, in I, then draw AI. The polygon AIDE, etc., is equivalent to the given polygon ABCDE, etc., and it has one side less. For, since BI is parallel to AC, the two triangles AIC, ABC are equivalent (B. III., T. XIX., C.), and if to these two triangles we add the portion of the surface ACDE, etc., we shall have, in the first case, the polygon AIDE, etc., and in the second case, the polygon ABCDE, etc.; consequently these two polygons are equivalent. It is evident, moreover, since the side CI of the triangle AIC is the prolongation of the side DC, that the two sides AB, BC of the first polygon have been replaced by the single side AI; hence the second figure has one side less than the first. Actually operate upon the polygon AIDE, etc., as upon the first; that is to say, draw the diagonal AD cutting off the triangle AID; afterwards draw IL parallel to AD meeting ED produced in L; we shall thus obtain a new polygon ALE, etc., equivalent to the second and having one side less. In continuing, in this way, we shall necessarily reach a polygon of three sides, and the problem will be resolved. PROBLEM VIII. To transfor, a polygon into a square. If the given polygon is a triangle, let b represent its base, and h its height, and x the side of the square sought. We have this condition, 2= b x h, from which we obtain the following proportion: b: x:: x:'h. Thus, the side of the square sought is a mean proportional between the base and half the height of the triangle. 132 GEOMETRY. This line being constructed (P. IV.), we easily obtain the construction of the square. Whatever may be the polygon, we commence by transforming it into a triangle (P. VII.); afterwards we transform, as above, the triangle into a square. When the given polygon is a parallelogram, or a rectangle, or in short, any figure whose area depends immediately upon the product of two lines, all that is necessary in order to obtain the side of an equivalent square is to construct a mean proportional between these two lines. Thus, in the case of a regular polygon, it is sufficient, after having developed on an indefinite straight line the perimeter of the polygon, to seek a mean proportional between the half perimeter and the radius of its inscribed circle. To transform a circle into a square, it is necessary first to rectify the circumference, and afterwards to determine a mean proportional between the radius and the rectiJied semi-circumference. Scholium. The quadrature, of the circle is wholly dependent upon the rectification of a circumference; and thus, up to the present, we have not been able, by the assistance of the Ruler and the Compass, to construct rigorously a square equivalent to a circle, as we can do for rectilinear figures, since all known methods give only approximate values for the ratio of the circurnference to the diameter. CONSTRUCTION OF SIMILAR POLYGONS UNDER CERTAIN CONDITIONS. PROBLEM IX. Upon a given line, to construct a polygon similar to a given polygon. Let ab be the given line, and ABCDEF the given polygon. After decomposing the given polygon into triangles by drawing diagonals from the corner A to the other corners, form at the FOURTH BOOK. 133 points a and b two angles cab, abc, respectively equal to the angles CAB, ABC. We shall thus obtain a triangle abc similar D E A B a b to the triangle ABC. Construct, in the same manner, on ac a triangle acd similar to the triangle ACD; and so on for the other triangles. The polygon abedef thus obtained will be similar to the given polygon ABCDEF. PROBLEM X. Two simtirar polygons being given, it is required to construct a third polygon similar to the two first, and equivalent to their sum or to their difference. The solution of this problem is an immediate consequence of C. IV., T. XXIX., B. III. Let a, a' be two homologous sides of the given polygons A, A'. Upon these sides, regarded as the sides about the right angle, or as the hypotenuse and one of the sides of the right angle, construct a right-angled triangle; afterwards, on the third side a" of the triangle thus obtained, construct (P. IX.) a polygon A" similar to one of the given polygons. The polygon thus constructed will be the polygon required. For, by construction, a,"2 = a2 ~ a, or a112 = a2 - a2; hence, (B. III., T. XXIX., C. IV.), A"= A + A', or A" = A - A'. PROBLEM XI. To find a square which shall be to a given square in the ratio of two given lines. 134 GEOMETRY. On an indefinite straight line A BX take BD = m, one of the given lines, DC = n, the other given line, and on BC = m + n, D as a diameter, describe a semicircumference; having drawn a, DA perpendicular to BC, draw nthe chords AB, AC, and prolong them if necessary beyond B afid C. Upon AB take AB'= a, the side of the given square, which we here suppose to be greater than AB, and draw B'C' parallel to BC. The line AC' will be the side of the square sought. In effect, the two similar triangles ABC, AB'C', give the proportion AB: AC:: AB': AC'. Consequently, AB2: AC2: AB'2: AC'2; but we have (B. III., T. XIV., S. I.), AB2: AC2: BD: DC, or: m: n; hence, AB'2: AC'2:: m: n; or, putting a for AB', and changing the order of the terms of the proportion, we have m: n: a2: AC'2. PROBLEM XII. A polygon P being given, to construct a second polygon X similar to the first, and such that the first may be to the second as m to n. Let a denote the side of the given polygon, and x the homologous side of the sought polygon. We shall have P: X:: a2:', P::: m:n; consequently, m: n:a2:. The question is thus made to depend upon the preceding problem; x being known, it is sufficient then to construct on this line a polygon similar to the given polygon. FOURTH BOOK. 135 PROBLEM X:III. Having given two polygons, P and Q, to construct a third polygon X similar to the first, and such that the second may be to the third in the ratio of m to n. Let a denote a side of the polygon P, and x the homologous side of the polygon X. Then we shall have, by the conditions of the problem, P: X:: a 2) Q:X: m: n. This being supposed, conceive the polygons P and Q transformed into equivalent squares (P. VIII.), having respectively for their sides p and q. In the same manner let y represent the side of a square equivalent to X; the preceding proportions will be changed into the following: p2 y2 a2. 2; hence. p:y::a: and p2: y2::m:n. Now, the last of these proportions will make known y by the construction of Problem XI., and the second will then give z as a fourth proportional to the lines p, y, and a, which will be the side of the polygon sought, homologous to a. PROBLEM XIV. To construct a rectangle equivalent to a given square n2, and such that the sum or the diJference of two adjacent sides may be equal to a given line a. First case. The given line a being the sum of the adjacent sides. Upon a straight line AB = a, as a diam- I eter, describe a circumference; through c A the extremity of the diameter AB, / drawthe perpendicularAC=m. Through A B C draw CL parallel to AB, meeting the circumference in the points E, E'. From YG the points E and E' draw EF, E'F' perpendicular to AB. 136 GEOMETRY.The distances AF and FB, or AF' and F'B, will be the two adjacent sides of the rectangle required to be constructed. In effect, we have (T. I., S. I.), AF x FB = EF2 = AC2= 2 and AF+FB=AB=a. The problem is impossible if AC > OI or OA; that is, when > a. If = la, the points E, E' will unite at I, and the rectangle will become a square. This demonstrates that the greatest rectangle which we can form by dividing a given line into two parts for the adjacent 8sides, is the square constructed on half the given line. Second case. The given line being the difference of two adjacent sides. After having drawn AC = m, as in the first case, we join C and 0, and produce it so as to meet the circumference at the points K and G. The two lines CG, CK will be the adjacent sides of the rectangle sought. We have, by the construction (T. III.), CA2 = CG x CK, or CG x CK = m2, and CG - CK = KG = AB = a. In this case, the problem is always possible for all values of a and m. FIFTH BOOK. OF THE PLANE, AND ITS COMBINATION WITH THE STRAIGHT LINE. DEFINITIONS. I. The intersection of two planes is the line in which they meet to cut each other. It is obvious, from our definition of a plane (B. I., D. XVII.), that this intersection is a straight line. II. A line is perpendicular to a plane, when it is perpendicular to all the lines in that plane which meet it. III. One plane is perpendicular to another, when every line in the one which is perpendicular to their intersection is perpendicular to the other plane. IV. A line is parallel to a plane, when, if both are produced to any distance, they do not meet; and, conversely, the plane is then also parallel to the line. V. Two planes are parallel to each other, when, both being produced to any distance, they do not meet. VI. The inclination of two planes which intersect, is called a diedral angle. The planes forming a diedral angle are A D called the faces of the angle, and their inter- F section is called its edge. H If a rectangle ABCD is revolved about its side AB as an axis, so as to assume in succession the positions ABEF, ABGH, etc., the point C will describe the circumference of a circle. In fact, any point of the plane of this B C rectangle, not situated in the axis AB, will, in a similar manner, describe the circumfer 138 GEOMETRY. ence of a circle having its centre in the axis. During this revolution, the arc CE will increase in the same ratio as the opening or inclination of the planes ABCD, ABEF, increases; that is, as the diedral angle DABE increases. Now, the are CE is the measure of the angle CBE (B. II., T. IX., S.), hence a diedral angle has for its measure the angle contained by two lines drawn from any point of its edge, perpendicular to the same, one in each face. VII. When several planes pass through a common point, the angular space included between these planes is called a polyedral angle. Each plane is called aface; the line in which any two faces intersect is called an edge; and the common point through which the planes pass is called the vertex. Three planes, at least, are necessary for forming a polyedral angle. In the case of three planes, the angle is called a triedral angle. THEOREM I. One part of a straight line cannot be in a plane, and another part out of it. For (B. I., D. XVII.), when a straight line has two points common with a plane, it must coincide with the plane. THEOREM II. Two straight lines which intersect each other, are situated in the same plane, and determine its position. Let AB, AC be two straight lines which intersect each other in A; and conceive a plane passing through one of the lines, as / C AB, and if also AC be in this F plane, then it is clear that the two lines, according to the terms of the proposition, are in the same plane; but if not, let the plane passing through AB be supposed to be turned round AB till it passes through the point C, then the line AC, which has two of its points, A and C, in this plane, coincides with it; and FIFTH BOOK. 139 hence the position of the plane is determined by the single condition of containing the two straight lines AB, AC. Cor. I. A triangle ABC, or any three points not in a straight line, determines the position of a plane. Cor. II. Hence, also, two parallels AB, CD determine the position of a plane; for, drawing the secant EF, the plane of the two straight lines AB, EF is that of the parallels AB, CD. THEOREM m. The intersection of two planes is a straight line. Let DC and EF be two planes cutting each E other, and A, B two points in which the planes D A meet. Draw the straight line AB; this line is the intersection of the two planes. For, since the straight line touches the two planes in the points A and B, it lies wholly in B both these planes, or is common to both of them; that is, the intersection of the two planes F is in a straight line. THEOREM IV. If a straight line is perpendicular to each of two straight lines at their point of intersection, it is perpendicular to the plane of these lines. Let AP be perpendicular to the two lines A PB, PC, at their point of intersection P; then M will it be perpendicular to MN the plane of the lines. Through P draw in the plane MN any line, as PQ; and through any point of this line, as N Q, draw BQC, so that BQ = QC (B. IV., P. III., C. II.); join AB, AQ, AC. The base BC being divided into two equal parts at the point Q, the triangle BPC (B. III., T. XVII.) will give PC2 + PB2 = 2PQ2 + 2QC2. 140 GEOMETRY. The triangle BAC will, in like manner, give AC2 + AB2 = 2AQ2 + 2QC2. Taking the first equation from the second, and observing that the triangles APC, APB, which are both right-angled at P, give AC2 - PC2 = AP2, and AB2 - PB2 = AP2, we shall have 2AP2 = 2AQ2 - 2PQ2, or, AP2 = AQ2- pQ2; that is, AQ2 = AP2 + pQ2. Hence the triangle APQ is right-angled at P, and therefore AP is perpendicular to PQ. Scholiunm. Thus it is evident, not only that a straight line may be perpendicular to all the straight lines which pass through its foot in a plane, but that it always must be so, whenever it is perpendicular to two straight lines drawn in the plane. Cor. At a given point P on a plane, it is impossible to draw more than one perpendicular to that plane. For, if there could be two perpendiculars at the same point P, draw along these two perpendiculars a plane, whose intersection with the plane MN is PQ; then those two perpendiculars would be perpendicular to the line PQ at the same point, and in the same plane, which is impossible. It is also impossible to draw, from a given point out of a plane, two perpendiculars to that plane. For, let AP, AQ be these two perpendiculars, then the triangle APQ would have two right angles APQ, AQP, which is impossible. THEOREM V. Iffrom a point without a plane, a perpendicular and several oblique lines be drawn to this plane: I. The perpendicular will be shorter than any oblique line. II. Any two oblique lines which terminate at equal distances from, thefoot of theperpendicular, will be equal. III. Of two oblique lines, terminating at unequal distances from the foot of the perpendicular, the one at the greater distance uwill be the longer. Let O be the given point, MN the given plane, OP perpendicular, and OA, OA', OA", and OB oblique lines. FIFTH BOOK. 141 First. Since OA is an oblique line, the right-angled triangle OPA gives OP < OA (B. I., T. XXVII.). Secondly. If PA = PA', the two right-angled triangles OPA, OPA' will be equal (B. I., T. XX., C.), and give OA' = -OA. Thirdly. If PB > PA, take on A' PB a distance PA' = PA, and Al draw OA', and we shall have A OB > OA' (B. I., T. XII.); but M we have OA' = OA, consequently OB > OA. Cor. I. The perpendicular drawn from any point to a plane, measures the true distance of the point from the plane. Cor. II. Any point whatever in a line perpendicular to a plane, is equally distant from all the points of the circumference described about the foot of the perpendicular as a centre, with any radius whatever. Scholium I. This perpendicular is called the axis of the circle. All the points of the axis may be regarded as centres of the circle, and the oblique lines as corresponding radii, any one of which may be used for describing the circle. Scholium II. By means of an extended thread, one extremity being fixed at the point 0, and the other extremity attached to a pencil, we may determine any three points, as A, A', A", equally distant from the point 0. If we then describe the circumference of a circle passing through these three points, its centre will be the foot of the perpendicular drawn from 0 to the plane. Scholizum III. When a plane passes perpendicularly through the middle of a straight line: 1. All points of this plane are equally distant from the extremities of this line. 2. All points situated without this plane are unequally distant from these extremities. 142 GEOMETRY. THEOREM VI. If, from a point without a plane, a perpendcicula r be drawn to that plane, and from the foot of the perpendicular a line be drawn perpendicular to a line in the plane, and from the point of intersection a line be drawn to the first point, this last line will be perpendicular to the line in the plane. Let AP be a line perpendicular to the A plane MN, and PD perpendicular to BC; then will AD be perpendicular to BC. Take DB = DC, and join PB, PC, AB, AC. Since PD is perpendicular at D, the middle N point of BC, PB is equal to PC (B. I., T. "E XII.); consequently the oblique lines AB, AC are equal, and the two triangles ADB, ADC have the three sides of the one equal to the three sides of the other; therefore they are equal (B. I., T. XXV.), and the angle ADB =ADC; hence each is a right angle, and AD is perpendicular to BC (B. I., D. XII.). Cor. It is evident, likewise, that BC is perpendicular to the plane APD, since BC is at once perpendicular to the two straight lines AD, PD (T. IV.). Scholium. The two straight lines AE, BC afford an instance of two lines not parallel which do not meet, because they are not situated in the same plane. The shortest distance between these lines is the straight line PD, which is perpendicular both to the line AP and to the line BC. For if we join any other two points, such as A and B, we shall have AB > AD, AD > PD, therefore AB > PD. The two lines AE, CB, though not situated in the same plane, are conceived as forming a right angle with each other, because AE and the line drawn through one of its points parallel to BC would make with each other a right angle. In the same manner, the line AB and the line PD, which represent any two straight lines not situated in the same plane, are supposed to form with each other the same angle which would be formed by AB and a straight line parallel to PD drawn through any point of AB. FIwrn BOOK 143 THEOREM VII. If one of two parallel lines is perpendicular to a plane, the other will also be perpendicular to this plane. Let AP and ED be parallel lines, of E A which AP is perpendicular to the plane / MN; then will ED be also perpendicular to C this plane. B Along the parallels AP, DE extend a N plane; its intersection with the plane MN will be PD. In the plane MN draw BC perpendicular to PD, and join AD. Then BC is perpendicular to the plane APDE (T. VI., Cor.), and therefore the angle BDE is right. But the angle EDP is right also, since AP is perpendicular to PD, and DE parallel to AP (B. I., T. XV., Cor.); therefore the line DE is perpendicular to the two straight lines DP, DB, hence it is perpendicular to their plane MN (T. IV.). Cor. I. Conversely, if the straight lines AP, DE are perpendicular to the same plane MN, they will be parallel. For, if they be not so, draw through the point D a line parallel to AP; this parallel will be perpendicular to the plane MN; therefore, through the same point D more than one perpendicular might be drawn to the same plane, which (T. IV., C.) is impossible. Cor. II. Two lines, A and B, parallel to a third line C, are parallel to each other. For, conceive a plane perpendicular to the line C; the lines A and B, being parallel to C, will be perpendicular to the same plane; therefore, by the preceding Corollary, they will be parallel to each other. THEOREM VIII. If a straight line without a plane isparallel to a line in the plane, it is parallel to the plane itself. Let the straight line AB, without the B A plane MN, be parallel to the line CD of --- this plane; then will AB be parallel to the D c plane MN. For, if the line AB, which lies in the plane ABDC, could 144 GEOMETRY. meet the plane MN, this could only be in some point of the line CD, the intersection of the two planes; but AB cannot meet CD, since they are parallel; hence it will not meet the plane MN, therefore (D. IV.) it is parallel to that plane. THEOREM IX. If two planes are perpendicular to the same line, they are parallel. Let the planes MN and PQ M be each perpendicular to AB; / then will they be parallel. For, if they can meet any- P o where, let O be one of their / common points, and join OA, OB. The line AB, which is B perpendicular to the plane MN, is perpendicular to the straight line AO drawn through its foot in that plane. For the same reason, AB is perpendicular to BO. Therefore OA and OB are two perpendiculars drawn from the same point 0, upon the same straight line, which is impossible; hence the planes MN, PQ cannot meet each other, and consequently they are parallel. THEOREM X. The intersections ofJ two parallel planes with a third plane are parallel. Let the two parallel planes MN and FE M PQ intersect the plane EH; then will EF be parallel to HG. N F For, if the lines EF, GH, lying in the \ 7 same plane, were not parallel, they would meet each other when produced; there- q H fore the planes MN, PQ, in which those lines are situated, would also meet, and the planes would not be parallel. FIFTH BOOK. 145 THEOREM XI. A line wich is perpendicular to one of two parallel planes, is perpendicular to the other also. Let the two planes MN and PQ be A parallel; then if the line AB is perpen- N / dicular to the plane MN, it will also be perpendicular to PQ. Having drawn through B any line BC q in the plane PQ, along the lines AB and BC, extend a plane ABC, intersecting the plane MN in AD; the intersection AD will be parallel to BC (T. X.). But the line AB, being perpendicular to the plane MN, is perpendicular to the straight line AD; therefore, also, it is perpendicular to its parallel BC (B. I., T. XV., Cor.). Hence the line AB, being perpendicular to any line BC drawn through its foot in the plane PQ, is consequently perpendicular to that plane. THEOREM XII. Two parallel lines, included between two parallel planes, are equal. E M Let EG, FI be two parallel lines included between the two parallel planes MN, PQ; then will these lines be N IF equal. Through the parallels EG, FH, draw the plane EGHF to meet the parallel Q H planes in EF and GiH. The intersections EF, GH (T. X.) are parallel to each other, so likewise are EG, FH; therefore the fgure EGHF is a parallelogram, and EG = FH. Cor. Hence it follows, that two parallel planes are everywhere equidistant; for, if EG and FH are perpendicular to the two planes MN, PQ, they will be parallel to each other (T. VII., C.),. and therefore equal. 10 146 GEOMETRY. THEOREM XIII. If two angles, not situated in the same plane, have their sides parallel and lying in the same direction, they will be equal, and the planes in which they are situated will be parallel. Let CAE, DBF be two angles not situated M in the same plane, having AC parallel to BD E A and lying in the same direction, and AE par- N p allel to BF, and also lying in the same direc- m tion; then will these angles be equal, and F B their planes will be parallel. Make AC = BD, AE = BF; and join CE, DF, AB, CD, EF. Since AC is equal and parallel to BD, the figure ABDC is a parallelogram (B. I., T. XXX.), therefore CD is equal and parallel to AB. For a similar reason, EF is equal and parallel to AB; hence, also, CD is equal and parallel to EF. The figure CEFD is therefore a parallelogram, and the side CE is equal and parallel to DF; therefore the triangles CAE, DEBF have their corresponding sides equal, and consequently the angle CAE = DBF. Again, the plane ACE is parallel to the plane BDF. For, suppose the plane parallel to BDF, drawn through the point A, were to meet the lines CD, EF in points different from C and E, for instance in G and H, then (T. XII.) the three lines AB, GD, HF would be equal. But the lines AB, CD, EF are already known to be equal, hence CD = GD, and HF = EF, which is absurd; hence the plane ACE is parallel to BDF. Cor. If two parallel planes MN, PQ are met by two other planes not parallel CABD, EABF, the angles CAE, DBF formed by the intersections of the parallel planes will be equal; for (T. X.) the intersection AC is parallel to BD, and AE to BF, and therefore the angle CAE = DBF. THEOREM XIV. If three straight lines, not situated in the same plane, are equal and parallel, the triangles formed by joining their corresponding opposite extremities will be equal, and their planes will be parallel. FIFTH BOOK. 147 Let ACE, BDF be two triangles formed _ M by joining the opposite extremities of the three equal and parallel lines AB, CD, EF; N then will these triangles be equal, and their D planes will be parallel. / Fl B For, since AB is equal and parallel to CD, Q the figure ABDC is a parallelogram (B. I., T. XXX.); hence the side AC is equal and parallel to BD. For a like reason, the sides AE, BF are equal and parallel, as also CE, DF. Therefore the two triangles ACE, BDF are equal; and, consequently, as in the last proposition, their planes are parallel. THEOREM XV. If two straight lines are cut by three parallel planes, they will be divided proportionally. Suppose the line AB to be cut by the parallel planes MN, PQ, RS, at the N points A, E, B; and the line CD to be P 7 P cut by the same planes at the points C, Q F, D; then. I AE:EB:: CF:FD. B Draw AD meeting the plane PQ in G, and join AC, EG, GF, BD. The intersections EG, BD, of the parallel planes PQ, RS in the plane ABD, are parallel (T. X.); therefore, AE: EB:: AG: GD (B. III., T. III.). In like manner, the intersections AC, GF being parallel, AG: GD:: CF: FD. The ratio AG: GD is the same in both; hence AE: EB:: CF: FD. THEOREM XVI. If a straight line is perpendicular to a plane, then every plane passing through this line will also be perpendicular to taefirst plane. 148 GEOMETRY. Let the line AP be perpendicular to the A plane MN; then any plane, as AB, passing I through this line, will also be perpendicular - to the plane MN. N / D For, let BC be the intersection of the planes AB, MN. In the plane MN draw DE perpendicular to BP; then the line AP, being perpendicular to the plane MN, will be perpendicular to each of the two straight lines BC, DE. But the angle APD, formed by the two perpendiculars PA, PD at their common intersection BP, is the measure of the angle of the two planes (D. VI.); and since, in the present case, the angle is a right angle, the two planes are perpendicular to each other. Scholium. When the three lines, such as AP, 13P, DP, are perpendicular to each other, each of these lines is perpendicular to the plane of the other two, and the planes themselves are perpendicular to each other. THEOREM XVII. If two planes are perpendicular to each other, every line drawn in one of them perpendcicular1 to their conmmon intersection, will be perpendicular to the other plane. Let the planes AB, MN be perpendicular to each other; and in the plane AB, let PA be drawn perpendicular to the common intersection PB; then will it be perpendicular to the plane MN. N For, in the plane MIN draw PD perpendicular to PB; then because the planes are perpendicular, the angle APD is a right angle; therefore the line AP is perpendicular to the two straight lines PB, PD, and consequently perpendicular to their plane AIN. Cor. If the plane AB be perpendicular to the plane MN, and if at a point P of the common intersection a perpendicular be drawn to the plane MIN, that perpendicular will be in the plane AB. For, if not, then in the plane AB we might draw AP perpendicular to PB their common intersection, and this AP at the same time would be perpendicular to the plane MN; therefore, at the same point P there would be two perpendiculars to the plane MN, which is impossible. FTWH BOOK 149 THEOREM XVIII. If two planes be perpendicular to a third plane, their common intersection will be also perpendicular to the third plane. Let AB, AD be perpendicular to MN; then will their common intersection AP be perpendicular to the same plane MN. M For, at the point P draw the perpendicular B to the plane MN; then that perpendicular N C must be in the plane AD; it must also be in the plane AB (T. XVII.), therefore it is their common intersection AP. THEOREM XIX. If a polyedral angle is formed by three plane angles, the -sum of any two of these angles will be greater than the third. The proposition requires demonstration only when the plane angle, which is compared to the sum of the other two, is greater than either of thenm. Therefore, suppose the polyedral angle S to be B A formed by three plane angles ASB, ASC, c BSC, whereof the angle ASB is the greatest: we are to show that ASB < ASC + BSC. In the plane ASB make the angle BSD = BSC; draw the straight line ADB at pleasure; and, having taken SC =SD, join AC, BC. The two sides BS, SD are equal to the two sides BS, SC; the angle BSD = BSC; therefore the triangles BSD, BSC are equal, and BD = BC. But AB < AC + BC; taking BD from the one side, and from the other its equal BC, there remains AD < AC. The two sides AS, SD are equal respectively to the two AS, SC; the third side AD is less than the third side AC; therefore the angle ASD < ASC. Adding BSD = BSC, we shall have ASD + BSD, or ASB < ASC + BSC. THEOREM XX. The sum, of the plane angles which form any polyedral angle, is always less thanfour right angles. 150 GEOMETRY. Conceive the polyedral angle S to be cut by any plane ABCDE; from O a point in that plane, draw to the several angles straight lines AO, OB, 00, OD, OE. The sum of the angles of the triangles ASB, BSC, etc., formed about the vertex S, is equiva- C 1B lent to the sum of the angles of an equal number of triangles AOB, BOC, etc., formed about the point O. But at the point B the angles ABO, OBC, taken together, make the angle ABC (T. XIX.) less than the sum of the angles ABS, SBC. In the same manner, at the point C we have BCO + OCD < BCS + SCD, and so with all the angles of the polygon ABCDE. Whence it follows that the sum of all the angles at the bases of the triangles whose common vertex is in O, is less than the sum of all the angles at the bases of the triangles whose common vertex is in S; hence, to make up the deficiency, the sum of the angles formed about the point O, is greater than the sum of the angles about the point S. But the sum of the angles about the point O is equal to four right angles (B. I., T. I., C. III.); therefore the sum of the plane angles, which form the polyedral angle S, is less than four right angles. Scholium. This demonstration is founded on the supposition that the polyedral angle is convex, or that the plane of no one surface produced can ever meet the polyedral angle. If it were otherwise, the sum of the plane angles would no longer be limited, and might be of any magnitude. THEOREM XXI. If two polyedral angles are composed of three plane angles respectively equal to each other, the planes which contain the equal angles will be equally inclined to each other. T s Let the angle ASC = DTF, the angle ASB= DTE, and the angle BSC = ETF; then will the inclination / of the planes ASC, ASB be equal E /A to that of the planes DTF, DTE. / / Having taken SB at pleasure, draw o 0 BO perpendicular to the plane ASC; from the point O at which FIFTHI BOOK. 151 that perpendicular meets the plane, draw OA, OC perpendicular to SA, SC; join AB, BC; next take TE = SB; draw EP perpendicular to the plane DTF; from the point P draw PD, PF perpendicular to TD, TF; lastly, join DE, EF. The triangle SAB is right-angled at A, and the triangle TDE is right-angled at D (T. VI.), and since the angle ASB = DTE, we have SBA = TED. Likewise SB- TE, therefore the triangle SAB is equal to the triangle DTE; therefore SA = TD, and AB =DE. In like manner it may be shown that SC = TF, and BC =EF. That granted, the quadrilateral SAOC is equal to the quadrilateral TDPF; for, place the angle ASC upon its equal DTF, because SA = TD, and SC = TF, the point A will coincide with D, and the point C with F; and at the same time AO, which is perpendicular to SA, will coincide with PD, which is perpendicular to TD, and in like manner CO with FP; wherefore the point O will coincide with the point P, and AO will be equal to DP. But the triangles AOB, DPE are rightangled at O and P; the hypotenuse AB = DE, and the side AO = DP; hence those triangles are equal-therefore the angle OAB = PDE. The angle OAB is the inclination of the two planes ASB, ASC, and the angle PDE is that of the two planes DTE, DTF; hence these two inclinations are equal to each other. It must, however, be observed, that the angle A of the rightangled triangle AOB is properly the inclination of the two planes ASB, ASC, only when the perpendicular BO falls on the same side of SA as SC falls; for if it fell on the other side, the angle of the two planes would be obtuse, and, added to the angle A of the triangle OAB, it would make two right angles. But, in the same case, the angle of the two planes TDE, TDF would also be obtuse, and, added to the angle D of the triangle PDE, it would make two right angles; and the angle A being thus always equal to the angle at D, it would follow, in the same manner, that the inclination of the two planes ASB, ASC must be equal to that of the two planes DTE, DTF. SIXTH BOOK. BODIES BOUNDED BY PLANES. DEFINITIONS. I. A solid, bounded by polygons, is called a Polyedron. The bounding polygons are calledfaces, the straight lines formed by the intersection of any two faces are called edges, and the points where three or more faces meet are called corners. A straight line joining any two corners not situated in the same face is called a diagonal. II. A polyedron, having two of its faces equal and parallel, and all the other faces parallelograms, is called a Prism. The two equal and parallel faces are called the bases of the prism, usually referred to as the lower base and the upper base; and the parallelograms which make up the other faces, taken together, constitute the lateral surface of the prism. III. A prism is triangular, quadrangular, pentagonal, etc., according as the bases are triangles, quadrilaterals, pentagons, etc. IV. When the edges formed by the intersection of the lateral faces of a prism are perpendicular to its bases, it is a right prism: each edge is then equal to its altitude. In all other cases the prism is said to be oblique, and the edges are greater than the altitude. SIXTH BOOK. 153 V. A prism, whose bases are parallelograms, is called a Parallelopipedon. When all the faces are rectangles it is called a Rectangular Parallelvopipedon. When all the faces are squares it is called a Cube, or regular hexaedron. VI. A Pyramid is a solid formed by several triangular planes meeting in a point, called the vertex, and terminating in the sides of a polygon, forming its base. The triangles meeting at the vertex, taken together, constitute the lateral surface of the pyramid. The perpendicular distance from the vertex to the base is called its altitude. VII. A pyramid is triangular, quadrilateral, pentagonal, etc., according as the base is a triangle, quadrilateral, pentagon, etc. VIII. When the base of a pyramid is a regular polygon, and the triangles forming the lateral surface are isosceles, it is a right pyramid. The perpendicular drawn from the vertex to the base will, in this case, pass through its centre, and it is then called the axis of the pyramid. A line drawn from the vertex of a right pyramid to the middle point of one of the sides of the polygon constituting its base, is called its slant height. IX. A triangular pyramid, having in all four triangular faces, is called a Tetraedron. X. If a pyramid is cut by a plane parallel to its base it will be divided into two portions, one of which will be a second pyramid having the same vertex as the first. The remaining portion, having two parallel bases, is called a truncated pyramid, or frustum of a pyramid. The altitude of a frustum of a pyramid is the perpendicular distance between its bases. The slant height of a frustum of a right pyramid is the portion of its slant height included between its bases. 154 GEOMETRY. XI. Two tetraedrons are similar when they have their edges proportional and arranged in the same order. Consequently the triangular faces of two similar tetraedrons are similar triangles each to each (B. III., D. II.). Any two polyedrons whatever, are similar when they are capable of being decomposed into the same number of similar tetraedrons each to each, having the same order of arrangement. XII. A regular Polyedron is one having all its faces equal and regular polygons; and all its diedral angles equal. THEOREM I. TLhe lateral surface of a riglt prism is equal to the perimeter of its base multiplied by its altitude. For this surface is equal to the sum of the rectangles AFGB, BGHC, CIHID, etc., which F compose it. Now the altitudes AF, BG, CH, etc., of those rectangles, are equal to the alti- H tude of the prism; their bases AB, BC, CD, etc., taken together, make up the perimeter A - E D of the prism's base: hence the sum of these rectangles, or the lateral surface of the prism, B is equal to the perimeter of its base multiplied by its altitude. Cor. If two right prisms have the same altitude, their lateral surfaces will be to each other as the perimeters of their bases. THEOREM II. Two prisms are equal when a polyedral angle in each is contained by three planes, which are respectively equal in both, and similarly situated. Let the base ABCDE be equal to the base abede; the parallelogram ABGF equal to the parallelogram abgf, and the parallelogram BCIG equal to the parallelogram bchg; then will the prism ABCI be equal to the prism abci. For, apply the base ABCDE upon its equal abcde, so that SIXTH BOOK. 155 they may coincide. But the K three plane angles which form the polyedral angle B are respectively equal to the G a three plane angles which form the polyedral angle b; / that is, ABC =abc, ABG = 1 d abg, and GBC=gbc, and A they are also similarly situ- B c ated; therefore the polyedral angles at B and b are equal (B. V., T. XXI.); hence BG will coincide with its equal bg. And it is likewise evident, since the parallelograms ABGF and abgf are equal, that the side GF will coincide with its equal gf: and in the same manner GH will coincide with gh; therefore the upper base FGHIK will coincide with its equal fghik, and the two solids will be identical, since their vertices are the same. Cor. Two right prisms, which have equal bases and equal altitudes, are equal. For, since the side AB is equal to ab, and the altitude BG to bg, the rectangle ABGF will be equal to abgf, and in the same way, the rectangle BGHC will be equal to bghc; and thus the three planes which form the solid angle B will be equal to the three planes which form the solid angle b: hence the two prisms are equal. THEOREM III. In every parallelosppedon the opposite faces are equal and parallel. By the definition of this solid, the bases H E ABCD, EFGH are equal parallelograms, G and their sides are parallel: it remains only to show that the same is true of any two opposite lateral faces, such as AEHID, BFGC. C B Now AD is equal and parallel to BC, because the figure ABCD is a parallelogram; for a like reason, AE is parallel to BF: hence the angle DAE is equal to the angle CBF, and the planes DAE, CBF are parallel; hence also the parallelogram DAEH is equal to the parallelogram CBFG. 156 GEOMETRY. In the same way, it may be shown that the opposite parallelograms ABFE, DCGH are equal and parallel. Cor. Since the parallelopipedon is a solid bounded by six planes, whereof those lying opposite to each other are equal and parallel, it follows that any face and the one opposite to it may be assumed as the bases of the parallelopipedon. Scoliumn. If three straight lines AB, AE, AD, passing through the same point A, and making given angles with each other, are known, a parallelopipedon may be formed on those lines. For this purpose a plane must be extended through the extremity of each line, and parallel to the plane of the other two; that is, through the point B a plane parallel to DAE, through D a plane parallel to BAE, and through E a plane parallel to BAD. The mutual intersections of these planes will form the parallelopipedon required. THEOREM IV. In every prism the sections formed by parallel planes are equal polygols. In the prism ABCI, let the sections NOPQR, K I STVXY be formed by parallel planes; then F will these sections be equal polygons. For the sides ST, NO are parallel, being the skit x intersections of two parallel planes with a third | R-i plane ABGF; moreover the sides ST, NO are N included between the parallels NS, OT, which AF are sides of the prism: hence NO is equal to Fl ST. For like reasons, the sides OP, PQ, QR, A< etc., of the section NOPQR, are respectively B equal to the sides TV, VX, XY, etc., of the section STVXY; and since the equal sides are at the same time parallel, it follows that the angles NOP, OPQ, etc., of the first section are respectively equal to the angles STV, TVX, etc., of the second: hence the two sections NOPQR, STVXY are equal polygons. Cor. Every section in a prism, if made parallel to the base, is also equal to that base. SIXTH BOOK. 157 THEOREM V. If a plane be made to pass through the diagonal and opposite edges of a parallelopvipedorn, so as to divide it into two triangular prisms, those prisms will be equal. Let the parallelopipedon ABCG be divided by the plane BDHF into the two triangular prisms ABDHIIEF, BCDFGH; then will those' prisms be equal. Through the vertices B and F, draw the F planes Bade, Fehg at right angles to the side BF, and meeting AE, DH, CG, the three other c sides of the parallelopipedon, in the points a, -'. c d, c towards one direction, and in e, h, g to- B wards the other: then the sections Bade, Fehg will be equal parallelograms; being equal, because they are formed by planes perpendicular to the same straight line, and consequently parallel; and being parallelograms, because aB, de, two opposite sides of the same section, are formed by the meeting of one plane with two parallel planes ABFE, DCGH. For a like reason, the figure BaeF is a parallelogram; so also are BFgc, cdhg, and adhe, the other lateral faces of the solid BadcFeAg: hence that solid is a prism (D. IV.), and that prism is a right one, because the side BF is perpendicular to its base. This being proved, if the right prism Bh be divided by the plane BFHD into two right-triangular prisms, aBdeFh, B3dcFhg, it will remain to be shown that the oblique-triangular prism ABDEFH will be equal to the right-triangular prism aBdeFlh; and since those two prisms have a part ABDheF in common, it will only be requisite to prove that the remaining parts, namely, the solids BaADd, FeEHh, are equal. Now, by reason of the parallelograms ABFE, aBFe, the sides AE, ae being equal to their parallel BF, are equal to each other; and taking away the common part- Ae, there remains Aa = Ee. In the same manner we could prove Dd = HAh. Let us now place the base Feh on its equal Bad; the point e coinciding with a, and the point h with d, the sides eE, hHI will coincide with their equals aA, dD, because they are perpendicular to the same plane Bad. Hence the two solids in ques 158 GEOMETRY. tion will coincide exactly with each other, and the oblique prism BADFEH is therefore equal to the right one BadFeh. In the same manner might the oblique prism I3DCFHG be proved equal to the right prism BdcFhg. But (T. I.) the two right prisms BadFeh, BdcFhg are equal, since they have the same altitude BF, and since their bases Bad, Bdc are halves of the same parallelogram. Hence the two triangular prisms BADFEH, BDCFHG, being equal to the equal right prisms, are equal to each other. Cor. Every triangular prism ABDHEF is half of the parallelopipedon AG described with the same polyedral angle A, and with the same edges AB, AD, AE. THEOREM VI. T2wo parallelopipedons having a common base, and their uqpper bases in the same plane and between the same parallels, are equivalent to each other. Let the two parallelopipedons AG, AL L M a H II-A E have the common base ABCD; and let K their upper bases EFGH, IKLM be in the same plane, and between the same parallels EK, HL; then will these paral- B A lelopipedons be equivalent. There may be three cases to this proposition, according as EI is greater, less than, or equal to EF; but the demonstration is the same for all. In the first place, then, we shall show that the triangular prism AEIDHM is equal to the triangular prism BFKCGL. Since AE is parallel to BF, and HE to GF, the angle AEI= BFK, HEI = GFK, and HEA = GFB. Of these six angles, the first three form the polyedral angle E, and the last three the polyedral angle F; therefore, the plane angles being respectively equal and similarly arranged, the polyedral angles F and E must be equal. Now if the prism AEM be laid on the prism BFL, the base AEI being placed on the base BFK, will coincide with it, because they are equal; and since the polyedral angle E is equal to the polyedral angle F, the side EH will coincide with its equal FG; and nothing more is required to prove the coinci SIXTH BOOK. 159 denuce of the two prisms throughout their whole extent, for (T. II.) the base AEI and the edge EH determine the prism AEM, as the base BFK and the edge FG determine the prism BFL; hence these prisms are equal. But if the prism AEM is taken away from the solid AL, there will remain the parallelopipedon AIL; and if the prism BFL is taken away from the same solid, there will remain the parallelopipedon AEG; hence these two parallelopipedons AIL, AEG are equivalent. THEOREM VII. Two parallelopipedons having the same base and sam;e altitude, are equivalent to each other. Let ABCD be the common q P G base of the two parallelopipe- N dons AG, AL: since they have M' L,,E the same altitude, their upper \K_ bases EFGH, IIKLM will be in the same plane. Also the sides EF and AB will be equal and parallel, as well as IK and AB; hence EF is equal and parallel; to IK: for a like reason, GF is A B equal and parallel to LK. Let the sides EF, HG be produced, and likewise LK, IM, till, by their intersections, they form the parallelogram NOPQ: this parallelogram will evidently be equal to either of the bases EFGH, IKLMf. Now, if a third parallelopipedon be conceived, having ABCD for its lower base and NOPQ for its upper, this third parallelopipedon will (T. VI.) be equivalent to the parallelopipedon AG; since, with the same lower base, their upper bases lie in the same plane and between the same parallels GQ, FN. For the same reason, this third parallelopipedon will also be equivalent to the parallelopipedon AL; hence the two parallelopipedons AG, AL, which have the same base and the same altitude, are equivalent. 160 GEOMETRY. THEOREM VIII. Any parallelopipedon may be changed into an equal rectangular parallelopipedon having the same altitude and an equivalent base. Let AG be the parallelopip- Q - H G edon proposed. From the o -U points A, B, C, D, draw AI, M\- \L, BK, CL, DiMI perpendicular to K,"', the plane of the base; and we shall thus form the parallelopipedon AL equivalent to AG, and having its lateral faces AK, D.i;'' BL, etc., rectangular. Hence, if the base ABCD be a rectan- A B gle, AL will be the rectangular parallelopipedon equivalent to AG, the parallelopipedon proposed. But if ABCD is not a rectangle, draw AO and BN perpendicular to CD, and OQ and NP perpendicular to the base; then the solid ABNOIKPQ will be a rectangular parallelopipedon; for, by construction, the base ABNO and its opposite IKPQ are rectangles; so also are the lateral faces, the edges AI, OQ, etc., being perpendicular to the plane of the base; hence the solid AP is a rectangular parallelo- M Q LP pipedon. But the two parallelopipedons AP, AL may be conceived as having the same base ABKI, and the same altitude AO; hence i the parallelopipedon AG, which was at first changed into an equivalent parallelopipedon AL, is again changed into an equivalent rect- D --- angular parallelopipedon AP, having the | same altitude AI, and a base ABNO equiva- A 3 lent to the base ABCD. SIXTH BOOK. 161 THEOREM IX. Two rectangaular parallelopipedons having the same base, are to each other as their altitudes. Let the parallelopipedons AG, AL have the H E common base ABCD; then will they be to each r other as their altitudes AE, AI. G First, suppose the altitudes AE, AI to be to m each other as two whole numbers; for example, M as 15 is to 8. Divide AE into 15 equal parts, L. whereof AI will contain 8; and through x, y, z, etc., the points of division, draw planes parallel to the base. These planes will cut the solid c s AG into 15 partial parallelopipedons, all equal to each other, having equal bases and equal altitudes: equal bases, because every section MIKL made parallel to the base ABCD of a prism, is equal to that base (T. IV., C.); and equal altitudes, because these altitudes are the equal divisions Ax, xy, yz, etc. But of those 15 equal parallelopipedons, 8 are contained in AL; hence the solid AG is to the solid AL as 15 is to 8; or, generally, as the altitude AE is to the altitude AI. Again, if the ratio of AE to AI cannot be expressed in whole numbers, it is to be shown that, notwithstanding, we shall have solid AG: solid AL:: AE: AI. For, if this proportion is not correct, suppose we have solid AG: solid AL:: AE: AO, greater than AI. Divide AE into equal parts, such that each shall be less than OI; there will be at least one point of division m between O and I. Let P be the parallelopipedon, whose base is ABCD and altitude Am. Since the altitudes AE, Anm are to each other as two whole numbers, we shall have solid AG: P:: AE: Am. But, by hypothesis, we have solid AG: solid AL:: AE: AO; therefore, solid AL:: AO: Aqa. But AO is greater than Am: hence, if the proportion is correct, the solid AL must be greater than P. On the contrary, how11 162 GEOMETRY. ever, it is less; hence the fourth term of this proportion, solid AG: solid AL::AE: x, cannot possibly be a line greater than AI. By the same mode of reasoning, it might be shown that the fourth term cannot be less than AI; therefore it is equal to AI. Hence rectangular parallelopipedons having the same base, are to each other as their altitudes. THEOREM X. Two rectangular parallelopipedons having the same altitude, uare to each other as their bases. Let the two rectangular parallelopip- H E I edons AG, AK have the same altitude; Q then will they be to each other as their/ bases ABCD, AMNO. A - Produce the plane ONKL till it meets M the plane DCGH in PQ; we shall thus / / N have a third parallelopipedon AQ, which c B may be compared with each of the parallelopipedons AG, AK. The two solids AG, AQ having the same base AEHD, are to each other as their altitudes AB, AO. In like manner, the two solids AQ, AK having the same base AOLE, are to each other as their altitudes AD, AM. Itence we have the two proportions sol. AG: sol. AQ:: AB: AO; sol. AQ: sol. AK:: AD: AM. Multiply together the corresponding terms of these proportions, omitting in the result the common multiplier sol. AQ: we shall have sol. AG:sol. AK::AB xAD:AO x AM. But AB x AD represents the base ABCD, and AO x AM represents the base AMNO; hence two rectangular parallelopipedons of the same altitude are to each other as their bases. SIXTH BOOK. 163 THEOREM XI. Any two rectangular parallelopiyedons are to each other as the products of their bases by their altitudes; that is to say, as the products of their three dimensions. For, having placed the two solids AG, H E I AZ, so that their surfaces have the corn- Q/ L - mon angle BAE, produce the interior Y planes necessary for completing the i D a third parallelopipedon AK, having the - M same altitude with the parallelopipedon l/ N AG. By the last proposition, we shall - B have sol. AG: sol. AK:: ABCD: AMNO. But the two parallelopipedons AK, AZ having the same base AMNO, are to each other as their altitudes AE, AX; hence we have sol. AK: sol. AZ:: AE: AX. Multiply together the corresponding terms of these proportions, omitting in the result the common multiplier sol. AK: we shall have sol. AG: sol. AZ:: ABCD x AE: AMNO x AX. Instead of the bases ABCD and AMNO, put AB x AD and AO x AM; it will give sol. AG: sol. AZ:: AB x AD x AE:AO x AM x AX. Hence any two rectangular parallelopipedons are to each other, etc. Scholium. We are consequently authorized to assume, as the measure of a rectangular parallelopipedon, the product of its base by its altitude; in other words, the product of its three dimensions. In order to comprehend the nature of this measurement, it is necessary to reflect, that by the pro luct of two or more lines is always meant the product of the numbers which represent them; those numbers themselves being determined by their linear unit, which may be assumed at pleasure. Upon this principle, the product of the three dimensions of a parallelopipedon is a 164 GEOMETRY. number, which signifies nothing of itself, and would be dif ferent if a different linear unit had been assumed; but if the three dimensions of another parallelopipedon are valued according to the same linear unit, and multiplied together in the same manner, the two products will be to each other as the solids, and will serve to express their relative magnitude. The measure of a solid, or its magnitude, is called its volume. Hence we say the volume of a rectangular parallelopipedon is equal to the product of its base by its altitude, or to the product of its three dimensions. As the cube has all its three dimensions equal, if the side is 1, the volume will be 1 x 1 x =1; if the side is 2, the volume will be 2 x 2 x2 =8; if the side is 3, the volume will be 3 x 3 x 3 = 27; and so on. Hence, if the sides of a series of cubes are to each other as the numbers 1, 2, 3, etc., the cubes themselves (or their volumes) will be as the numbers 1, 8, 27, etc. Hence it is, that in arithmetic the cube of a number is the name given to the product which results from three factors, each equal to this number. If it were proposed to find a cube double of a given cube, the side of the required cube would have to be to that of the given one, as the cube root of two is to unity. Now, by a geometrical construction, it is easy to find the square root of 2; but the cube root of it cannot be so found, at least not by the simple operations of elementary geometry,. which consist in employing nothing but straight lines, two points of which are known, and circles whose centres and radii are determined. Owing to this difficulty, the problem of the duplication of the cube became celebrated among the ancient geometers, as well as that of the trisection of an angle, which is nearly of the same species. The solutions of which such problems are susceptible, have, however, long since been discovered; and though less simple than the constructions of elementary geometry, they are not, on that account, less rigorous or less satisfactory. THEOREM XII. The volume of a parallelopipedon, and generally of any _prism, is equal to the product of its base by its altitude. SIXTH BOOK. 165 For, in the first place, any parallelopipedon (T. VII.) is equivalent to a rectangular parallelopipedon having the same altitude and an equivalent base. Now the volume of the latter is equal to its base multiplied by its height; hence the volume of the former is, in like manner, equal to the product of its base by its altitude. In the second place, and for a like reason, any triangular prism is half of the parallelopipedon so constructed as to have the same altitude and a double base; but the volume of the latter is equal to its base multiplied by its altitude; hence that of a triangular prism is also equal to the product of its base multiplied into its altitude. In the third place, any prism may be divided into as many triangular prisms of the same altitude, as there are triangles capable of being formed in the polygon which constitutes its base; but the volume of each triangular prism is equal to its base multiplied by its altitude; and since the altitude is the same for all, it follows that the sum of all the partial prisms must be equal to the sum of all the partial triangles which constitute their bases, multiplied by the common altitude. Hence the volume of any polygonal prism is equal to the product of its base by its altitude. Cor. Comparing two prisms which have the same altitude, the products of their bases by their altitudes will be as the bases simply. Hence two prisms of the same altitude are to each other as their bases. For a like reason, two prisms of the same base are to each other as their alttudes. THEOREM XIII. Similar prisms are to one another as the cubes of their honologous sides. Let P and p be two prisms, of which BC, be are homologous'P sides: the prism P is to the prism A p as the cube of BC tothe cube \ a of be.'d From A and a, homologous angles of the two prisms, draw c B c b 166 GEOMETRY. AH, ah perpendicular to the bases BCD, bcd: join BH; take Ba = ba, and in the plane BHA draw ah perpendicular to BH: then ah will be perpendicular to the plane CBD, and equal to ah the altitude of the other prism; for if the solid angles B and b were applied the one to the other, the planes which contain them, and consequently the perpendiculars ah, ah would coincide. Now because of the similar triangles ABH, abh, and the similar figures P, p, we have AH: ah:: AB: ab:: BC: be; and because of these similar bases, the base BCD: base bed:: BC2: bc2. (B. III., T. XXVII.) Taking the product of the corresponding terms of these proportions, we have AH x base BCD: ah x base bed:: BC3: be3. But AH x base BCD expresses the volume of the prism P, and ah x base bed expresses the volume of the other prism p; therefore prism P: prism p:: BC3: bc3. THEOREM XIV. If a pyramid be cut by a plane parallel to its base, the section thus formed will be a polygon similar to the base, and the lateral edges and the altitude will be cut into proportional parts. First, since the planes ABCDEF, abcdef are parallel, it follows that AB and ab, BC T. X.); consequently the angles ABC and abc, / BCD and bed are respectively equal (B. V., f T. XIII.). Moreover the couples of similar / triangles SAB and Sab, SBC and Sbc, etc., give the following equal ratios: SA: Sa:: AB: ab:: SB: Sb, SB Sb:: BC: be:: "SC:, F SC Se:o CD:ed": SD:Sd, &c. &c. &c. A B SIXTH BOOK. 167 Hence, by equality of ratios, we have AB: ab:: BC: b C::D: cd:: etc. Consequently the two polygons ABCDEF, abedef are similar (B. III., T. X.). Now, considering only the ratios between the lateral edges, we have SA:Sa: SB:Sb::SC: Sc:: etc., from which we see that the edges are cut into proportional parts. Finally, if we pass a plane through the edge SB and the altitude SO, its intersection bo with the plane abede will be parallel to BO (B. V., T. X.), and the two similar triangles SBO, Sbo will give SO: So:: SB: Sb:: SA: Sa:: etc., which establishes the theorem. Cor. When two pyramids S-ABCDE, T - MNP have equivalent bases situated in the same plane, and equal altitudes, the sections made by a plane parallel to their bases, are also equivalent. S T C M N F For, since the polygons ABCDE, abode are similar, we have the proportion ABCDE: abode:: AB2: ab2; we also have, by reason of the foregoing relations, ABCDE: abode:: SO2: So2. 168 GEOMETRY. In the same manner the two polygons MNP, mnp give MNP: mnp:: TQ2: Tg2, we have, moreover, by supposition, SO =TQ, So = Tq; hence, ABCDE: abde:: MNP: mnp. Now, by hypothesis, ABCDE- = INP; consequently, abode = mnp. THEOREM XV.'The lateral surface of a right pyramid is equal to the perimeter of its base multiplied by half the slant height. Since the lateral surface is composed of equal isosceles triangles SAB, SBC, SCD, etc., each one of which is measured by its base into one half its altitude, which altitude is the same as e SF the slant height of the pyramid, it follows that the lateral surface is equal to the sum of the bases of all these triangles, or the perimeter\ of the pyramid's base, into one half the slant E\ height. F A B Cor. I. If two right pyramids have the same slant heights, their lateral surfaces will be to each other as the perimeters of their bases. Cor. II. The lateral surface of a frustumn of a regular pyra. mid is measured by half the sum of the perimeters of its two bases multiplied by its slant height. For, since the frustum is formed from a regular pyramid, the two bases are similar and regular polygons (T. XIV.). Consequently the lateral surface of a frustum of a right pyramid is composed of equal trapezoids, the altitude of each being the same as the slant height of the frustum. The area of each trapezoid is measured by half the sum of its parallel bases multiplied into its altitude (B. III., T. XXIII., C. II.). Hence the sum of all these trapezoids, or the lateral surface of the frustum, is measured by half the sum of the perimeters of its bases into its slant height. SIXTH BOOK. 169 THEOREM XVI. Two triangular pyramids having equivalent bases and equal altitudes, are equivalent, or equal in volume. T a d h Let S - ABC, S - abc be those two pyramids; let their equivalent bases ABC, abc be situated in the same plane, and let AT be their common altitude. If they are not equivalent, let S - abc be the smaller; and suppose Aa to be the altitude of a prism, which, having ABC for its base, is equal to their difference. Divide the altitude AT into equal parts Ax, xy, yz, etc., each less than Aa, and let k be one of those parts; through the points of division, pass planes parallel to the plane of the bases: the corresponding sections formed by these planes in the two pyramids will be respectively equivalent (T. XIV., C.), namely, DEF to def, GHI to ghi, etc. This being granted, upon the triangles ABC, DEF, GHI, etc., taken as bases, construct exterior prisms, having for edges the parts AD, DG, GK, etc., of the edge SA. In like manner, on the bases def, ghi, klm, etc., in the second pyramid, construct interior prisms, having for edges the corresponding parts of sa. It is plain that the sum of all the exterior prisms of the pyramid S - ABC will be greater than this pyramid; and, also, that the sum of all the interior, prisms of the pyramid s - abc will be 170 GEOMETRY. less than this. Hence the difference between the sum of all the exterior prisms and the sum of all the interior ones, must be greater than the difference between the two pyramids themselves. Now, beginning with the bases ABC, abc, the second exterior prism DEFG is equivalent to the first interior prism defa, because they have the same altitude k, and their bases DEF, def are equivalent: for like reasons, the third exterior prism GHIK, and the second interior prism ghid, are equivalent; the fourth exterior, and the third interior; and so on, to the last in each series. Hence all the exterior prisms of the pyramid S - ABC, excepting the first prism DABC, have equivalent corresponding ones in the interior prisms of the pyramid s - abe: hence the prism DABC is the difference between the sum of all the exterior prisms of the pyramid S - ABC, and the sum of all the interior prisms of the pyramid s- abc. But the difference between these two sets of prisms has already been proved to be greater than that of the two pyramids, which latter difference we supposed to be equal to the prism aABC: hence the prism DABC must be greater than the prism aABC; but in reality it is less, for they have the same base ABC, and the altitude Ax of the first is less than Aa the altitude of the second. Hence the supposed inequality between the two pyramids cannot exist; hence the two pyramids S - ABC, s- abc, having equal altitudes and equivalent bases, are themselves equivalent. THEOREM XVII. Every triangular pyramid is thle third of the triangular prism. having the same base and altitude. Let F-ABC be a triangular pyramid, B ABCDEF a triangular prism of the same base A C and altitude; the pyramid will be equal to one / third of the prism. I Conceive the pyramid F —ABC to be cut off from the prism by a section made along the plane FAC, and there will remain the solid E D FACDE, which may be considered as a quadrangular pyramid whose vertex is F, and base the parallelogram ACDE. Draw SIXTH BOOK. 171 the diagonal AD, and extend the plane FAD, which will cut the quadrangular pyramid into two triangular ones F - ACD, F - ADE. These two triangular pyramids have for their common altitude the perpendicular drawn from F to the plane ACDE; they have equal bases, the triangles ACD, ADE being halves of the same parallelogram; hence the two pyramids F-ACD, F-ADE are equivalent. But the pyramid F-ADE and the pyramid F - ABC have equal bases, ABC, DEF; they have also the same altitude, namely, the distance of the parallel planes ABC, DEF: hence the two pyramids are equivalent. Now the pyramid F - ADE has already been proved equivalent to F - ACD; hence the three pyramids F - ABC, F - ADE, F - ACD, which compose the prism ABCD, are all equivalent. Hence the pyramid F - ABC is the third part of the prism ABCD, which has the same base and the same altitude. Cor. The volume of a triangular pyramid is equal to a third part of the product of its base by its altitude. THEOREM XVIII. The volume of any pyramid has for its measure the area of its base multiplied into one third of its altitude. Let S - ABCDE be a pyramid, having the altitude SO; then will it be measured by the base ABCDE into one third of the altitude SO. For, extending the planes SEB, SEC through the diagonals EB, EC, the polygonal pyramid S-ABCDE will be divided into several triangular pyramids, all having the same alti- A C tude SO. But (T. XVII., C.) each of these pyramids is measured by multiplying its base B ABE, BCE, or CDE by the third part of its altitude SO; hence the sum of these triangular pyramids, or the polygonal pyramid S- ABCDE will be measured by the sum of the triangles ABE, BCE, CDE, or the polygon ABCDE, multiplied by ISO. Hence every pyramid is measured by a third part of the product of its base by its altitude. Cor. I. Every pyramid is the third part of the prism which has the same base and the same altitude. 172 GEOMETRY. Cor. II. Two pyramids having the same altitude, are to each other as their bases. S&holium. The volume of any polyedral body may be comnputed, by dividing the body into pyramids; and this division may be accomplished in various ways. One of the simplest is to make all the planes of division pass through the vertex of one solid angle; in that case, there will be formed as many partial pyramids as the polyedron has faces, minus those faces which form the polyedral angle whence the planes of division proceed. THEOREM XIX. The volume of a frustum of a pyramid is equivalent to three pyramids having the common altitude of the frustum, and for bases, the lower base of the frustum, the upper base, and a mean proportional between them. We will represent the lower base ABCDE of S the frustum by A, and the upper base abcde. by a; also, we will denote the altitude oO of the le frustum by h. If we denote the altitude SO of the pyramid, whose base is ABCDE by x, we shall have x - h for the altitude So of the pyra- D mid whose base is abode. A c Since the two bases of the frustum are similar polygons (T. XIV.), their areas are to each other as the squares of their homologous sides AB and ab (B. III., T. XXVIII.), which sides are to each other as SO to So; consequently we have A a:: ( — h)2. Extracting the square root of each term, we have 1 1 A2: a2x: x: - h. This proportion immediately gives x = —5- x h, and conA2- a! 1 a2 sequently, x - h = x h. SIXTH BOOK. 173 Hence the volume of the larger pyramid is A x = 3 A 1x I'A. The volume of the smaller pyramid is -a2 3 ax 3 I-. 1 1 A2 _ a2 If we denote the volume of the frustum, which is the differ3 3 ence of these pyramids by V, we shall have V= ----- x h A2 _ a2 which becomes V= (A + a + A+ a) x3 h = A x 3 h +.a x 3h + /A a x a h, which establishes the Theorem. THEOREM XX. Two similar pyramids are to each other as the cubes of their honoloqoous sides. For, two pyramids being similar, the smaller S may be placed within the greater, so that the angle S shall be common to both. In that position the bases ABCDE, abcde will be paral- a lel; because, since the homologous faces are similar, the angle Sab is equal to SAB, and Sbc to SBC; hence the plane ABC is parallel to A o c the plane abc. This granted, let SO be the per- B pendicular drawn from the vertex S to the plane ABC, and o the point where this perpendicular meets the plane abc: from what has already been shown (T. XIV.), we shall have SO: So:: SA: Sa:: AB: ab; and, consequently, I SO: - So:: AB: ab. But the bases ABCDE, abcde being similar figures, we have ABCDE: abcde:: AB2: ab2. Multiply the corresponding terms of these two proportions; there results the proportion, ABCDE x I SO: abode x So:: AB3: ab3. Now ABCDE x SO is the volume of thee pyramid S-ABCDE, and abcde x ~So is that of the pyramid S-abode (T. XVII., C.); hence two similar pyramids are to each other as the cubes of their homologous sides. SEVENTH BOOK. THE THREE ROUND BODIES. DEFINITIONS. I. A cylinder is a solid, which may be pro- P duced or generated by the revolution of a rect- D, angle ABCD, conceived to revolve about the side AB. K~ M In this rotation, the sides AD, BC, continuing L always perpendicular to AB, describe equal circular planes DHP, CGQ, which are called the B bases of the cylinder; the side CD at the same time describing the convex surface. The immovable line AB is called the axis of the cylinder. Every section KLM made in the cylinder, at right angles to the axis, is a circle equal to either of the bases; for, while the rectangle ABCD revolves about AB, the line KI, perpendicular to AB, describes a circular plane, equal to the base, which is a section made perpendicular to the axis at the point I. Every section PQGH passing through the axis is a rectangle, and is double of the generating rectangle ABCD. II. A cone is a solid, which may be pro- s duced or generated by the revolution of a right-angled triangle SAB, conceived to re- H A' volve about the side SA. K In this rotation, the side AB describes a circular plane BDCE, named the base of the B c cone; and the hypotenuse SB describes its D convex surface. The point S is named the vertex of the cone; SA its axis, or altitude. SEVENTH BOOK. 175 Every section HKFI formed at right angles to the axis, is a circle. Every section SDE passing through the axis, is an isosceles triangle double of the generating triangle SAB. III. If, from the cone SCDB, the cone SFKH be cut off by a section parallel to the base, the remaining solid CBHF is called a truncated cone, or the frustum of a cone. We may conceive it to be described by the revolution of a trapezium ABHG, whose angles A and C are right, about the side AG. The immovable line AG is called the axis or altitude of the frustumn; the circles BDC, HFK are its bases, and BH is its slant height. IV. Two cylinders or two cones, are similar, when their axes are to each other as the diameters of their bases. V. If, in the circle ACD which forms the K N base of a cylinder, a polygon ABCDEM be in- F scribed, a right prism, constructed on this base' ABCDEM, and equal in altitude to the cylinder, is said to be inscribed in the cylinder, or the cylinder to be circumscribed about the prism. The edges AF, BG, CH, etc., of the prism, A being perpendicular to the plane of the base, C B are evidently included in the convex surface of the cylinder. Hence the prism and the cylinder touch one another along these edges. VI. In like manner, if ABCD is a polygon Q R circumscribed about the base of a cylinder, a F right prism, constructed on this base ABCD, and equal in altitude to the cylinder, is said to be circumscribed about the cylinder, or the cylinder to be inscribed in the prism.... A Let M, N, etc., be the points of contact in t0o the sides AB, BC, etc.; and through the points c M, N, etc., let MX, NY, etc., be drawn per- B pendicular to the plane of the base: those perpendiculars will evidently lie both in the surface of the cylinder, and in that of the circumscribed prism; hence they will be their lines of contact. VII. A sphere is a solid terminated by a curved surface, all the points of which are equally distant from a point within. called the centre. 176 GEOMETRY. The sphere may be conceived to be D generated by the revolution of a semi- F circle DAE about its diameter DE; Q for the surface described in this move-. ment by the curve DAE will have all B E- A its points equally distant from the centre C. The radius of a sphere is a straight line drawn from the centre to any E point in the surface; the diameter, or axis, is a line passing through this centre, and terminated on both sides by the surface. All the radii of a sphere are equal. All the diameters are equal, and double of the radius. VIII. A great circle of the sphere, is a section which passes through the centre; a small circle, one which does not pass through it. IX. A plane is a tangent to a sphere, when their surfaces have but one point in common. X. A zone is the portion of the surface of the sphere included between two parallel planes, which form its bases. One of these planes may be a tangent to the sphere, in which case the zone has only a single base. XI. A spherical segment is the portion of the solid sphere included between two parallel planes which form its bases. One of those planes may be a tangent to the sphere, in which case the segment has only a single base. XII. The altitude of a zone, or of a segment, is the distance between the two parallel planes, which form the bases of the zone or segment. XIII. While the semicircle DAE, revolving round its diameter DE, describes the sphere, any circular sector, as DCF or FCH, describes a solid, which is named a spherical sector. NOTE. —The cylinder, the cone, and the sphere, are the three round bodies treated of in the elements of geometry. SEVENTH BOOK. 177 THEOREM I. The lateral or convex szurface of a cylinder has for its measure the product of its circumference into its altitude. In the cylinder, suppose a right prism to be K N inscribed, having a regular polygon for its base. F The lateral surface of this prism has for its measure the perimeter of its base multiplied by its altitude (B. VI., T. I.). When the number of sides of the polygon forming the base of the D inscribed prism is indefinitely increased, its limit will become the circle constituting the C B base of the cylinder (B. IV., T. VIII., S.). We may, therefore, regard a cylinder as a right prism having a regular polygon of an infinite number of infinitely small sides for its base; and since the lateral surface of a right prism will always have for its measure the perimeter of its base into its altitude, it follows that the lateral surface of a cylinder has for its measure the circumference of its base into its altitude. THEOREM II. The volume of a cylinder has for its measure the product'of its base into its altitude. As in the last Theorem, if we regard a cylin- K N der as a right prism, having a regular polygon of an infinite number of sides for its base, and recall to mind that the volume of a right prism is the product of its base into its altitude (B. VI., T. XII.), we shall at once see that the volume of D7 E'a cylinder is equal to the product of its base into its altitude. Gor. I. Cylinders of the same altitude are to each other as their bases; and cylinders of the same base are to each other as their altitudes. Cor. II. Similar cylinders are to each other as the cubes of their altitudes, or as the cubes of the diameters of their bases. For the bases are as the squares of their diameters; and the cylinders being similar, the diameters of their bases (D. IV.) are to 12 178 GEOMETRY. each other as the altitudes: hence the bases are as the squares of the altitudes; consequently, the bases multiplied by the altitudes, or the cylinders themselves, are as the cubes of the altitudes. Scholium. Let R be the radius of a cylinder's base, and H the altitude. The area of the base (B. IV., T. XIV., S.) will be R2'; and the volume of the cylinder will be rR2 x I, or ~R2HI, where' = 3.141592, etc. THEOREM III. The convex, or lateral 8surface of a cone, is equal to the prodwct of the circumference of its base into half its slant height. Let the circle whose radius is OA be B the base of the cone, S its vertex, and SA its slant height. Then will the convex A C surface of the cone have for its measure circ. OA x ISA. For, conceive a regular polygon inscribed in the circle OA; and on this E D polygon, as a base, a pyramid having S for its vertex, to be constructed. The lateral surface of this pyramid will have for its measure the perimeter of the polygon, constituting its base, into one half of SF its slant height (B. VI., T. XV.). When the number of sides of the inscribed polygon is indefinitely increased, its perimeter will be limited by the circumference of the circle, its slant height will be limited by the slant height of the cone, and the limit of the lateral surface will be the convex surface of the cone. A cone may thus be regarded as a right pyramid, having a regular polygon of an infinite number of infinitely short sides for its base. And since the lateral surface of a pyramid will always have for its measure the perimeter of its base into half its slant height, however great may be the number of sides in the polygon forming its base, it follows, that the convex surface of a cone has for its measure the circumference of its base into half its slant height. Scholium. Let L be the slant height of a cone, R the radius of its base. The circumference of this base will be 2~R; and the convex surface of the cone will be 2,rR x -L, or'wRL. SEVENTH BOOK 179 THEOREM IV. ihe convex surface of a truncated cone is equal to its side multiplied by half the sum of the circumferences of its two bases. In the plane SAB which passes through the axis SO, draw the line AF perpendicular to SA, and equal to the circumference having AO for its radius; join SF, and draw DH parallel to AF. F M H S From the similar triangles SAOG, SDC, we have AO: DC: SA: SD; and by the similar triangles SAF, SDH, AF: DH:: SA:SD; hence, AF: DH:: AO: DC; or (B. IV., T. XIV.), as circ. AO is to circ. DO. But, by construction, AF = circ. AO; hence DH = circ. DO. Hence the triangle SAF, measured by AF x ISA, is equal to the surface of the cone SAB, which is measured by circ. AO x ~SA. For a like reason, the triangle SDEI is equal to the surface of the cone SDE. Therefore the surface of the truncated cone ADEB is equal to that of the trapezium ADHF; but the latter is measured by AD x (AF D) (B. III., T. XXIII.,. II.). Hence the surface of the truncated cone ADEB is equal to its side AD multiplied by half the sum of the circumferences of its two bases. Scholium. If a line AD, lying wholly on one side of the line OC, and in the same plane, make a revolution around OC, the surface described by AD will have for its measure AD x (circ. AO + circ. D)C the lines AO, DC being perpendiculars, drawn from the extremities of the axis OC. 180 GEOMETRY. For, if AD and OC are produced till they meet in S, the surface described by AD is evidently that of a truncated cone having AO and DC for the radii of its bases, the vertex of the whole cone being S. Hence this surface will be measured as above. This measure will always hold good, even when the point D falls on S, and thus forms a whole cone; and also when the line AD is parallel to the axis, and thus forms a cylinder. In the first case, DC would be nothing; in the second, DC would be equal to AO and to IK. Cor. I. Through I, the middle point of AD, draw IKL parallel to AB, and IM parallel to AF: it may be shown, as above, that IM = circ. IE; but the trapezium ADHF = AD x IM = AD x circ. IK. Hence it may also be asserted, that the surface of a truncated cone is equal to its side multiplied by the circumference of a section at equal distancesfrom the two bases. Cor. II. The point I being the mid- B D dle of AB, and IK a perpendicular A drawn from the point I to the axis, the surface described by AB, by the last Corollary, will have for its measure F M N o - a AB x circ. IK. Draw AX parallel to the axis; the triangles ABX, OIK will have their sides perpendicular each to each, namely, OI to AB, IIK to AX, and OK to BX; hence these triangles are similar, and give the proportion AB: AX or MN:: OI: IK, or as circ. OI to circ. IK; hence AB x circ. IK = MN x circ. OI. Whence it is plain that the surface described by the partial polygon ABCD is measured by (MN + NP + PQ) x circ. OI, or by MQ x circ. OI; hence it is equal to the altitude multiplied by the circumference of the inscribed circle. Cor. III. If the whole polygon has an even number of sides, and if the axis FG passes through two opposite vertices F and G, the whole surface described by the revolution of the half polygon FACG will be equal to its axis FG multiplied by the circumference of the inscribed circle. This axis FG will, at the same time, be the diameter of the circumscribed circle. SEVENTH BOOK. 181 THEOREM V. The volume of a cone has for its measure the product of its base into one third of its altitude. Let the cone have OA for the radius B of its base, and SO for its altitude; then H i will its volume be represented by area A 0 OA x ISO. As in Theorem III., conceive a regular \ / pyramid to be inscribed in the cone. The volume of this pyramid will have for its E D measure the area of the polygon constituting its base multiplied by one third of its altitude (B. VI., T. XVIII.). When the number of sides of the polygon forming the base of the inscribed pyramid is indefinitely increased, the area of the polygon will have for its limit the area of the circle forming the base of the cone, and the limit of the inscribed pyramid will be the cone. HIence, the volume of a cone has for its measure the product of its base into one third of its altitude. Cor. I. A cone is the third of a cylinder having the same base and the same altitude. Whence it follows: 1. That cones of equal altitudes are to each other as their bases; 2. That cones of equal bases are to each other as their altitudes; 3. That similar cones are as the cubes of the diameters of their bases, or as the cubes of their altitudes. Scholium. If R be the radius of a cone's base, and H its altitude, the volume of the cone will be rR2 x -H, or ~-rR2H. Cor. II. If a plane be drawn parallel to the base of a cone, cutting it so as to form a frustum of a cone, it will at the same time cut the inscribed pyramid, forming also a frustum of a regular pyramid inscribed in the frustum of the cone. The limit of the inscribed frustum will be the frustum of the cone. And since the volume of the frustum of a pyramid is equivalent to the volumes of three pyramids having the common altitude of the frustum, and for bases the lower base of the frustum, the upper base of the frustum, and a mean proportional between them (B. VI., T. XIX.), it follows, that the volume of thefrustum 182 GEOMETRY. of a cone is equivalent to three cones having the common altitude of the frustum, and for bases the lower base of the frustum, the upper base, and a mean proportional between them. If the radii of the two bases of a frustum of a cone be represented by R and R', its altitude by h, and its volume by V, we shall have V = - h (R2 +'' + Ifa'). THEOREM VI. Every section of a sphere, made by a plane, is a circle. Let AMB be the section, made by a plane in the sphere whose centre is C. From the Bpoint C draw CO perpendicular to the plane AMB; and draw lines CM, CM to different points of the curve AMB, which terminates the section. The oblique lines CM, CM, CA being equal, being radii of the sphere, they are equally distant from the perpendicular CO (B. V., T. V.); hence all the lines OM, OM, OB are equal; hence the section AMB is a circle, whose centre is O. Cor. I. If the section passes through the centre of the sphere, its radius will be the radius of the sphere; hence all great circles are equal. (or. II. Two great circles always bisect each other; for their common intersection, passing through the centre, is a diameter. Cor. III. Every great circle divides the sphere and its surface into two equal parts; for, if the two hemispheres were separated, and afterwards placed on the common base, with their convexities turned the same way, the two surfaces would exactly coincide, no point of the one being nearer the centre than any point of the other. Cor. IV. The centre of a small circle, and that of the sphere, are in the same straight line perpendicular to the plane of the little circle. Cor. V. Small circles are the less the further they lie from the centre of the sphere; for, the greater CO is, the less is the chord AB, the diameter of the small circle AMPB. Cor. VI. An arc of a great circle may always be made to pass SEVENTH BOOK. 183 through any two given points in the surface of the sphere; for the two given points and the centre of the sphere make three points, which determine the position of a plane. But if the two given points were at the extremities of a diameter, these two points and the centre would then lie in one straight line, and an infinite number of great circles might be made to pass through the two given points. THEOREM VII. A plane perpendicular to the radius at its extremity, is tangent to the sphere. Let PQ be a plane perpendicular to the P A radius OA; at its extremity, A, it will be tangent to the sphere. Q For, taking any other point, as M, in this plane, if we draw OM, it will be lon- 0 ger than the perpendicular OA (B. V., T. V.); consequently the point M is situated without the surface of the sphere. Hence the plane PQ can have only the point A common with this surface; it is therefore tangent to the sphere (D. IX.). THEOREM VIII. When two spherical surfaces cut each other, the line of intersection is a circumference of a circle, of which the plane is perpendicular to the line of their centres, and whose centre is situated on this line. Let O, O' be the centres of two spheres, X a point common to their surfaces, and MP a per- A B pendicular drawn to the line of A B p O' B their centres. If we draw a plane through the three points N 0, 0', M, it will intersect the surface of the two spheres in arcs of great circles. Now, suppose the two semicircles AMB, A'MIB' to make a complete revolution about the common line AB' as an axis, it is evident that the per 184 GEOMETRY. pendicular MP will describe a circle common to the two spheres thus engendered. From which we see that the surface of the spheres cut each other in the circumference of a circle, whose radius is the perpendicular drawn from the point AM to this axis. Scholiumn. Whatever may be the relative positions of two spheres, since all planes passing through the line of their centres give two circumferences whose centres and whose radii are those of the spheres themselves, it follows that the condition of contact and of intersection of their surfaces are, in all respects, identical with those in reference to the two circumferences. So that to enumerate and demonstrate these different conditions, it is sufficient to refer to Book II., Theorems XVIII., XIX., XX., and XXI. We will confine ourselves in this place to noticing the condition relative to contact: WThen two spheres touch each other, the distance between their centres is equal to the sum or the deiference of their radii. They have at their point of contact a common tangent plane. THEOREM IX. The surface of a sphere is equal to its diameter multi1plied by the circumference of a great circle. Let the surface of the sphere be en- E gendered by the revolution of the semi- D circumference ADEFB about its diameter AB; it will have for its measure its diameter multiplied by its entire cir- A L C M B cumference. Suppose a regular semi-polygon to be inscribed in the semicircle, and to revolve simultaneously with the semicircle; then will the surface engendered by the semi-polygon have for its measure the diameter AB multiplied by the circ. of its apothem CK (T. IV., C. III.). Now, when the number of sides of this inscribed semi-polygon is indefinitely increased, the limit of its perimeter will become the semi-circumference, its apothem will become the radius of the semicircle, and the surface engendered by the revolution of this semi-polygon will become the surface of the sphere. Hence, the surface of a sphere has for its measure the product of its diameter into its circumference. SEVENTH BOOK. 185 Cor. I. The surface of the great circle is measured by multiplying its circumference by half the radius, or by a fourth of the diameter; hence, the surface of a sphere is four times that of its great circle. Cor. II. The surface of a zone has for its measure its altitude multiplied by the circumference of a great circle. For, the surface described by any portion of the inscribed polygon, as DE + EF, has for its measure LM into the circ. of its apothem CK (T. IV., C. II.); which, at the limit, gives the surface of the zone equal to its altitude LM multiplied into its circumference CA. Cor. III. Two zones, taken in the same sphere, or in equal spheres, are to each other as their altitudes; and any zone is to the surface of the sphere, as the altitude of that zone is to the diameter of the sphere. THEOREM X. If a triangle and a rectangle, having the same base and the same altitude, turn simultaneously about the common base, the solid described by the revolution of the triangle will be a third of the cylinder described by the revolution of the rectangle. Let ABC be the triangle, and EB E A F the rectangle. To the axis draw the perpendicular AD; the cone described by the triangle ABD is the third part of the cylinder c D B described by the rectangle AFBD (T. V., C. I.); also the cone described by the triangle ADC is the third part of the cylinder described by the rectangle ADCE: hence the sum of the two cones, or the solid described by ABC, is the third part of the two cylinders taken together, or of the cylinder described by the rectangle BCEF. If the perpendicular AD falls without the A F triangle, the solid described by ABC will be the difference of the two cones described by ABD and ACD; but, at the same time, the cylinder described by BCEF will be the difference of the two cylinders described by D C B 186 GEOMETRY. AFBD and AECD. Hence the solid described by the revolution of the triangle will still be a third part of the cylinder described by the revolution of the rectangle having the same base and altitude. Scholium. The circle of which AD is radius has for its measure er x AD2; hence vr x AD2 x BC measures the cylinder described by BCEF, and -r x AD2 x BC measures the solid described by the triangle ABC. THEOREM XI. If a triangle be revolved about a line drawn at pleasure through its vertex, the solid described by the triangle will have for its measure the area of the triangle multiplied by two thirds of the circumference traced by the middle point of the base. Let CAB be the triangle, and CD P the line about which it revolves. A Produce the side AB till it meets B the axis CD in D; from the points A and B, draw AM, BN perpendicular, M to the axis. The solid described by the triangle CAD is measured (T. X., S.) by 3r x AM2 x CD; the solid described by the triangle CBD is measured by ~~r x BN2 x CD: hence the difference of those solids, or the solid described by ABC, will have for its measure ~r (AM2 - BN2) x CD. To this expression another form may be given. From I, the middle point of AB, draw IK perpendicular to CD; and through B draw BO parallel to CD: we shall have AM + BN = 2IK, and AM - BN = AO; hence (AM +- BN) x (AM - BN), or AM2 - BN2 = 2IK x AO. Hence the measure of the solid in question is expressed by ]~r x IK x AO x CD. But if CP is drawn perpendicular to AB, the triangles ABO, DCP will be similar, and give the proportion, AO: CP: AB: CD; hence AO x CD = CP x AB, but CP x AB is double the area of the triangle ABC; hence we have AO x CD = 2ABC; hence the solid described by the triangle ABC is also measured by SEVENTH BOOK. 187 3-Y x ABC x IK, or, which is the same thing, by ABC x 2 circ. IK, circ. IK being equal to 2,r x IK. Hence, the solid described by the revolution of the triangle ABC has for its measure the area of this triangle multiplied by two thirds of the circuntmference traced by I, the middle point of the base. (or. I. If the side AC = CB, the A line CI will be perpendicular to AB, the area ABC will be equal to AB B x ~CI, and the volume F-r xABC xIK will become 1rt x AB x IK x CI. But the triangles ABO, CIK are similar, D c and give the proportion, AB: BO or MN:: CI: IK; hence AB x IK = MN x CI: hence the solid described by the isosceles triangle ABC will have for its measure 2a x MN x CI2. Cor. II. The general solution appears to include the supposition that AB produced will meet the axis; but the results would be equally true though AB were parallel to the axis. Thus the cylinder described by AMNB is B A i equal to r. AMI2. MN; the cone described by ACM is equal to ~rAM2. CMI, and the cone described by BCN to ~r. AM2. CN. Add the first two solids, and take away the third: N N M C we shall have the measure of the solid described by ABC equal to r. AM2. (MN + ICM - CN); and since CN - CM = MN, this expression is reducible to -. AM2. -MN, or ]2rCP2. MN, which agrees with the conclusion above drawn. Cor. III. Let AB, BC, CD be the several successive sides of a regular A polygon, O its centre, and OI the ra- dius of the inscribed circle; if the polygonal sector AOD lying all on one L _ _ o q side of the diameter FG be supposed to perform a revolution about this diameter, the solid so described will have for its measure ~qr. O12. MQ, MQ being that portion of the axis which is included by the extreme perpendiculars AM, DQ. For, since the polygon is regular, all the triangles, AOB, BOC, 188 GEOMETRY. etc., are equal and isosceles. Now, by Corollary I., the solid produced by the isosceles triangle AOB has for its measure 2r. OI2. MN; the solid described by the triangle BOC has for its measure r. OI2. NP, and the solid described by the triangle COD has for its measure 2. OI2. PQ: hence the sum of those solids, or the whole solid described by the polygonal sector AOD will have for its measure 2. O12. (MN + NP + IQ), or Fq. OI2.M Q. Cor. IV. The solid generated by the revolution of the entire polygon will have for its measure Zq-r x OI2 x FG. But the surface generated by the entire polygon has for its measure 27r x OI x FG (T. IV., C. III.), since 2ir x OI is the circumference of the inscribed circle. Hence the solid is measured by 2qr x OI x FG into IOI; that is, the solid generated by the revolution of a semi-polygon inscribed in a semicircle, has for its measure its surface multiplied by one third the apothem of the generating polygon. THEOREM XII. The volume of a sphere is measured by its surface mult'plied by one third its radius. As in Theorem IX., let the sphere E be generated by the revolution of the D f semicircle whose radius is CA, about its diameter AB. The regular inscribed polygon will generate a solid having for A L c A 1 its measure the surface of the solid multiplied by one third the apothem of the polygon (T. XI., C. IV.). When we pass to the limit, the surface of the solid generated by the polygon will be the surface of the sphere generated by the semicircle (T. IX.), and the apothem of the polygon will become the radius. Consequently, the volume of a sphere is measured by its surface multiplied into one third of its radius. Cor. The volume of a spherical sector has for its measure the zone which forms its base multiplied into one third of its radius. For, the solid generated by any sectorial portion of the inscribed polygon, as CDEF, has for its measure -r x CK2= x LMI SEVENTH BOOK. 189 (T. XI., C. III.). Hence at the limit we have, for the measure of the spherical sector generated by the plane sector CDF, 2-r x CA2 x LM' = 2r x CA x LM x ICA. But 2% x CA x LAI is the measure of the surface of the zone forming the base of the spherical sector; hence the volume of a spherical sector is measured by its surface into one third of its radius. Scholium. Let R be the radius of a sphere: its surface will be 4~rR2; its solidity, 4&R2 X iR,. or.. R3. If the diameter is named D, we shall have R = ID, and R3-= -D3: hence the solidity may likewise be expressed by 4r. 1D3, or LrD3. THEOREM XIII. The surface of a sphere is to the whole surface of the circumscribed cylinder (including its bases), as 2 is to 3; and the volumes of these two bodies are to each other in the same ratio. Let MPNQ be a great circle of the sphere, and ABCD the circumscribed c _ square. If the semicircle PMQ and the half square PADQ are at the same time N M made to revolve about the diameter PQ, the semicircle will generate the sphere, while the half square will generate the cyl- B A / inder circumscribed abott that sphere. The altitude AD of that cylinder is equal to the diameter PQ, the base of the cylinder is equal to the great circle, its diameter AB being equal to MN: hence (T. I.), the convex surface of the cylinder is equal to the circumference of the great circle multiplied by its diameter. This measure (T. IX.) is the same as that of the surface of the sphere: hence, the surface of the sphere is equal to the convex surface of the circumnscribed cylinder. But the surface of the sphere is equal to four great circles, hence the convex surface of the cylinder is also equal to four great circles; and adding the two bases, each equal to a great circle, the total surface of the circumscribed cylinder will be equal to six great circles: hence the surface of the sphere is to the total surface of the circumscribed cylinder as 4 190 GEOMETRY. is to 6, or as 2 is to 3, which is the first branch of the proposition. In the next place, since the base of the circumscribed cylinder is equal to a great circle, and its altitude to the diameter, the volume of the cylinder (T. II.) will be equal to a great circle multiplied by its diameter. But (T. XII.) the volume of the sphere is equal to four great circles multiplied by a third of the radius; in other terms, to one great circle multiplied by 4- of the radius, or by I of the diameter. Hence the sphere is to the circumscribed cylinder as 2 to 3, and consequently the volumes of these two bodies are as their surfaces. EIGHTH BOOK. SPHERICAL GEOMETRY. DEFINITIONS. I. A spherical triangle is a portion of the surface of a sphere, bounded by three arcs of great circles. Those arcs, named the sides of the triangle, are always supposed to be each less than a semi-circumference; the angles, which their planes form with each other, are the angles of the triangle. II. A spherical triangle takes the name of right-angled, isosceles, equilateral, in the same cases as a rectilineal triangle. III. A spherical polygon is a portion of the surface of a sphere, terminated by several arcs of great circles. IV. A lune is that portion of the surface of a sphere, which is included between two great semicircles meeting in a common diameter. V. A spherical wedge, or ungula, is that portion of the solid sphere which is included between the same great semicircles, and has the lune for its base. VI. A spherical pyramid is a portion of the solid sphere, included between the planes of a polyedral angle whose vertex is the centre; the base of the pyramid is the spherical polygon intercepted by the same planes. VII. The pole of a circle is a point on the surface of the sphere equally distant from all the points of the circumference. 192 GEOMETRY. THEOREM I. In every spherical triangle, any side is less than the sum of the other two. Let O be the centre of the sphere; and A draw the radii OA, OB, OC. Imagine the planes AOB, AOC, COB; those planes will B form a polyedral angle at the point 0; and the angles AOB, AOC, COB will be measured by AB, AC, BC, the sides of the spherical triangle. But (B. V., T. XIX.) each of the three plane triangles composing a polyedral angle is less than the sum of the other two; hence, any side of the triangle ABC is less than the sum of the other two. THEOREM II. The sum of all the three sides of a spherical triangle is less than the circumference of a great circle. Let ABC be any spherical triangle; C produce the sides AB, AC till they meet again in D. The arcs ABD, ACD will be semi-circumferences- D A since (B. VII., T. VI., C. II.) two great circles always bisect each other. But in the triangle BCD we have (T. I.) the side BCB: A make the angle BAD =B; then we shall have AD = DB (T. XII.); but AD + DC is greater than AC: hence, putting DB in place of B AD, we shall have DB + DC, or D BC > AC. Secondly. If we suppose BC > AC, the angle BAC will be greater than ABC. For, if BAC were equal to ABC, we should have BC = AC; if BAC were less than ABC, we should then, as has just been shown, find BC < AC. Both these conclusions are false; hence the angle BAC is greater than ABC. THEOREM5 XV. The sum of all the angles in any spherical triangle is less than six right angles, and greater than two. For, in the first place, every angle of a spherical A triangle is less than two right angles (see the following Scholium); hence the sum of all the three is less than six right angles. Secondly, the measure of each angle in the spherical triangle (T. VII.) is equal to the semi- C B circumference minus the corresponding side of the polar triangle; hence the sum of all the three is measured by three semni-circumferences minus the sum of all the sides of the polar triangle. Now (T. II.), this latter sum is less than a circumference; therefore, taking it away from three semi-circumferences, the remainder will be greater than one semi-circumference, which is the measure of two right angles. Hence, in the second place, the sum of all the angles in a spherical triangle is greater than two right angles. Cor. I. The sum of all the angles in a spherical triangle is not constant, like that of all the angles in a rectilineal triangle; it 202 GEOMETRY. varies between two right angles and six, without ever arriving at either of these limits. Two given angles, therefore, do not serve to determine the third. Cor. II. A spherical triangle may have two or even three angles right, two or three obtuse. If the triangle ABC have two right angles B and C, the vertex A will (T. IV.) be the pole of the blase BC, and the sides AB, AC will be quadrants. If the angle A is also right, the triangle ABC will have all its angles right, and its sides quadrants. The tri-rectangular triangle is contained eight times in the surface of the sphere, as is evident from the figure in the next proposition, supposing the arc MN to be a quadrant. Sctholium. In all the preceding observations, we have supposed, in conformity with D. I., that our spherical triangles have always each of their sides less than a semi-circumference; from which it follows that any one of their angles is always less than two right angles. For (see the figure of T. II.), if the side AB is less than a semi-circumference, and AC is so likewise, both those arcs will require to be produced before they can meet in D. Now, the two angles ABC, CBD taken together, are equal to two right angles; hence the angle ABC itself is less than two right angles. We may observe, however, that some spherical triangles do exist, in which certain of the sides are greater than a semicircumference, and certain of the angles greater than two right angles. Thus, if the side AC is produced so as to form a whole circumference ACDE, the part which remains after subtracting the triangle ABC from the hemisphere, is a new triangle also designated by ABC, and having AB, BC, AEDC for its sides. Here, it is plain, the side AEDC is greater than the semi-circumference AED; and, at the same time, the angle B opposite to it exceeds two right angles by the quantity CBD. The triangles, whose sides and angles are so large, have been excluded from our definition; but the only reason was, that the solution of them, or the determination of their parts, is always reducible to the solution of such triangles as are comprehended by the definition. Indeed, it is evident enough, that if the sides and angles of the triangle ABC are known, it will be easy to EIGHTH BOOK. 203 discover the angles and sides of the triangle which bears the same name, and is the difference between a hemisphere and the former triangle. THEOREM XVI. The surface of a lune is to the entire surface of the sphere, as the angle of the lune is to four right angles, or as the are which measures the angle of the lune is to the circumference. Suppose, in the first place, the arc MN to A be to the circumference MNPQ as some one rational number is to another, as 5 to 48 for example. The circumference MNPQ being divided into 48 equal parts, MN will contain 5 of them; and if the pole A were joined with the several points of division, B by as many quadrants, we should in the hemisphere AMNPQ have 48 triangles, all equal, because having all their parts equal. IHence the whole sphere must contain 96 of those partial triangles, and the lune AMBNA will contain 10 of them; hence the lune is to the sphere as 10 is to 96, or as 5 to 48; in other words, as the arc MlIN is to the circumference. If the arc AIN is not commensurable with the circumference, we may still show, by the mode of reasoning employed in the case of incommensurable magnitudes, that in this instance, also, the lune is to the sphere as MN is to the circumference. Cor. I. Two lunes are to each other as their respective angles. Cor. II. It was shown (T. XV., C. II.) that the whole surface of the sphere is equal to eight tri-rectangular triangles; hence, if the area of one such triangle is taken for unity, the surface of the sphere will be represented by 8. This granted, the surface of the lune whose angle is A will be expressed by 2 A (the angle A being always estimated from the right angle assumed as unity), since 2 A: A:: 8: 4. Thus we have here two different unities: one for angles, being the right angle; the other for surfaces,. being the tri-rectangular spherical triangle, or the triangle whose angles are all right, and whose sides are quadrants. Scholium. The spherical ungula bounded by the planes AMB, 204 GEOMETRY. ANB, is to the whole sphere, as the angle A is to four right angles; for, the lunes being equal, the spherical ungulas will also be equal; hence two spherical ungulas are to each other as the angles formed by the planes which bound them. THEOREM XVII. Two symmnetrical spherical triangles are equal in surface. Let ABC, DEF be two sym- A D metrical triangles; that is to say, two triangles having their sides AB = DE, Q AC = DF, c, BC = EF, and yet incapable of coinciding B with each other: we are to show that the surface ABC is equal to the surface DEF. Let P be the pole of the little circle passing through the three points A, B, C; from this point, draw (T. IV., S.) the equal arcs PA, PB, PC; at the point F, make the angle DFQ = ACP, the arc FQ = CP; and join DQ, EQ. The sides DF, FQ are equal to the sides AC, CP; the angle DFQ = ACP; hence (T. IX.) the two triangles DFQ, ACP are equal in all their parts; hence the side DQ = AP, and the angle DQF = APC. In the proposed triangles DEF, ABC, the angles DFE, ACB opposite to the equal sides DE, AB being equal (T. VIII.), if the angles DFQ, ACP, which are equal by construction, be taken away from them, there will remain the angle QFE equal to PCB. Also the sides QF, FE are equal to the sides PC, CB; hence the two triangles FQE, CPB are equal in all their parts: hence the side QE = PB, and the angle FQE = CPB. Now, observing that the triangles DFQ, ACP, which have their sides respectively equal, are at the same time isosceles, we shall see them to be capable of mutual adaptation, when applied to each other; for, having placed PA on its equal QF, the side PC will fall on its equal QD, and thus the two triangles will exactly coincide; hence they are equal, and the surface DQF= EIGHTH BOOK. 205 APC. For a like reason, the surface FQE = CPB, and the surface DQE = APB; hence we have DQF + FQE - DQE =APC + CPB - APB, that is, DEF = ABC; therefore, the two symmetrical triangles ABC, DEF are equal in surface. Scholium. The poles P and Q might lie within the triangles ABC, DEF; in which case, it would be requisite to add the three triangles DQF, FQE, DQE together, in order to make up the triangle DEF; and, in like manner, to add the three triangles APC, CPB, APB together, in order to make up the triangle ABC. In all other respects, the demonstration and the result would still be the same. THEOREM[ XVIII. If two great circles intersect each other on the surface of a hemisphere, the sum of the opposite triangles thus formed will be equivalent to the lune whose angle is equal to the angle formed by the circles. Let the circumference AOB, COD in- 0 tersect on the hemisphere OACBD; then will the opposite triangles AOC, BOD be equal to the lune whose angle is BOD. B A For, producing the arcs OB, OD in the other hemisphere, till they meet in N, the are OBN will be a semi-circumference, and AOB one also; and taking OB from each, we shall have BN = AO. For a like reason, we have DN = CO, and BD = AC. Hence the two triangles AOC, BDN have their three sides respectively equal; besides, they are so placed as to be symmetrical; hence (T. XVII.) they are equal in surface, and the sum of the triangles AOC, BOD is equal to the lune OBNDO. whose angle is BOD. 206 GEOMETRY. THEOREM XIX. The surface of any spherical triangle is measured by the excess of the surn of its three angles above two right angles. Let ABC be the proposed triangle: pro- G duce its sides till they meet the great circle DEFGHI, drawn anywhere without the tri- F A angle. By the last proposition, the two triangles ADE, AGH are together equal to the c lune whose angle is A, and which is measured (T. XVI., C. II.) by 2 A; hence we E have ADE + AGH = 2 A; and, for a like reason, BGF + BID = 2 B, and CItI + CFE= 2 C. But the sum of those six triangles exceeds the hemisphere by twice the triangle ABC, and the hemisphere is represented by 4; therefore twice the triangle ABC is equal to 2 A + 2 B + 2 C - 4, and consequently once ABC = A + B + C - 2; hence, every spherical triangle is measured by the sum of all its angles minus two right angles. Cor. I. However many right angles there be contained in this measure, just so many tri-rectangular triangles, or eighths of the sphere, which (T. XVI., C. II.) are the unit of surface, will the proposed triangle contain. If the angles, for example, are each equal to 4 of a right angle, the three angles will amount to 4 right angles, and the proposed triangle will be represented by 4- 2, or 2; therefore it will be equal to two tri-rectangular triangles, or to the fourth part of the whole surface of the sphere. Cor. II. The spherical triangle ABC is equal to the lune whose angle is 2 -1. Likewise the spherical pyramid which has ABC for its base, is equal to the spherical ungula whose angle is A+B+C. Scholium. While the spherical triangle ABC is compared with the tri-rectangular triangle, the spherical pyramid, which has ABC for its base, is compared with the ti-i-rectangular pyramid, and the same ratio is found to subsist between them. The polyedral angle at the vertex of the pyramid is, in like manner, compared with the polyedral angle at the vertex of the tri-rect. EIGHTH BOOK. 207 angular pyramid. These comparisons are founded on the coincidence of the corresponding parts. If the bases of the pyramids coincide, the pyramids themselves will evidently coincide, and likewise the polyedral angles at their vertices. From this, the following consequences are deduced: First. Two triangular spherical pyramids are to each other as their bases; and since a polygonal pyramid may always be divided into a certain number of triangular ones, it follows that any two spherical pyramids are to each other as the polygons which form their bases. Secondly. The polyedral angles at the vertices of those pyramids are also as their bases; hence, for comparing any two polyedral angles, we have merely to place their vertices at the centres of two equal spheres, and the polyedral angles will be to each other as the spherical polygons intercepted between their planes or faces. The vertical angle of the tri-rectangular pyramid is formed by three planes at right angles to each other: this angle, which may be called a right polyedral angle, will serve as a very natural unit of measure for all other polyedral angles; and if so, the same number that exhibits the area of a spherical polygon, will exhibit the measure of the corresponding polyedral angle. If the area of the polygon is A, for example; in other words, if the polygon is 4 of the tri-rectangular polygon, then the corresponding polyedral angle will also be 4 of the right polyedral angle. THEOREM XX. The surface of a sphAerical polygon is measured by the sum of all its angles, MINUS the product of two right angles by the number of sides in the polygon MINus two. From one of the vertices A, let diagonals c AC, AD be drawn to all the other vertices; the polygon ABCDE will be divided into as many triangles, minus two, as it has sides. E B But the surface of each triangle is measured by the sum of all its angles minus two right A angles, and the sam of the angles in all the triangles is evidently the same as that of all the angles in the polygon; hence the sur 208 GEOMETRY. face of the polygon is equal to the sumn of all its angles, diminished by twice as many right angles as it has sides minus two. Scholium. Let s be the sum of all the angles in a spherical polygon, and n the number of its sides; the right angle being taken for unity, the surface of the polygon will be measured by s8- 2 (n — 2), or s- 2 n+4. REGULAR POLYEDRONS. Since. the faces of any regular polyedron are all equal and regular polygons (B. VI., D. XII.), it follows that all its polyedral angles must be equal. There can be only five kinds of regular 2olyedrons. For, we already know that three faces at least are necessary to form a polyedral angle; and moreover that the sum of all the plane angles which form the polyedral angle is less than 4 right angles. It is sufficient then to consider all the cases in which the angles of equal and regular polygons, taken 3 at a time, 4 at a time, 5 at a time, etc., do not equal or exceed 4 right angles. Now, 1. The angle of an equilateral triangle being I- of a right angle, we may construct a polyedral angle by using three,four, or five equilateral triangles for faces. But we cannot use six or a greater number, since - of a right angle x 6 = 4 right angles. 2. The angle of a square being 1 right angle, we may form a polyedral angle by using three squares for the faces. But we could not use four or a greater number, since 1 right angle x 4= 4 right angles. 3. The angle of a regular pentagon being 6, a polyedral angle may be formed with three regular pentagons, which gives 5 x 3- 3. Since the angle of a regular hexagon is 4, and A x 3 =4, it follows, that a polyedral angle cannot be formed with regular hexagons for faces. And of course, regular polygons of a greater number of sides cannot be used in forming a polyedral angle. Thus, the only possible regular polyedrons are those of which each polyedral angle is formed by three, four, orfive equilateral EIGHTH. BOOK. 209 triangles, by three squares, and by three regular pentagons, making in allfive polyedrons. When three equilateral triangles are used in forming each polyedral angle, the whole number of faces of the regular polyedron will befour, and it is called a lietraedron. When four equilateral triangles unite in forming each polyedral angle, the polyedron will have eight faces, and it is called an Octaedron. When five equilateral triangles are used in forming a polyedral angle, the polyedron will have twenty faces, and it is then called an Icosaedron. When three squares are used in forming each polyedral angle of a regular polyedron, there will be six faces, and the polyedron is called a Hexagon, or Cube (B. VI., D. V.). When three regular pentagons are used for each polyedral angle, the number of faces will be twelve, and the solid is called a iDodecaedron. These five regular solids may obviously be inscribed in a sphere, as well as circumscribed about a sphere; or a sphere may be inscribed in each, and circumscribed about each. The inscribed and circumscribed spheres have a common centre, which may be regarded as the centre of the polyedron. If we suppose planes to pass through the centre of a regular polyedron and each of its edges, they will divide the polyedron into as many equal pyramids as the polyedron has faces, since each face of the polyedron will thus become a base of a pyramid, whose altitude will be the radius of the inscribed sphere; and since each pyramid is measured by the area of its base multiplied by one third of its altitude, it follows that the volume of any regular polyedron has for its measure its surface multiplied by one third the radius of its inscribed sphere. Were we to regard the sphere as a regular polyedron of an in — finite number of infinitely small and equal faces, we should; from the above, at once infer that the volume of a sphere has for its measure its surface multiplied by one third its radius. Whether we regard the sphere as a regular polygon or not, itis essentially a regular body. Hence, there are then six regular bodies: the Tetraedron,. Hexaedron, Octaedron, Dodecaedron, Icosaedron, and the Sphere. 14 210 GEOMETRY. APPLICATION OF ALGEBRA TO THE SOLUTION OF GEOMETRICAL PROBLEMS. PROBLEM I. In an equilateral triangle, having given the lengths of the three perpendiculars drawn from a certain point within it to the three sides, to determine its side. Let ABC be the equilateral triangle, and DE, c DF, DG the perpendiculars from the point D upon the sides respectively. Denote these perpendiculars by a, b, c, in order, and the side of the triangle ABC by 2 x. Then, if the perpen- A HGa B dicular CII be drawn, CH = /VAC2 - AH2 = V4 x2- x2=x /3. The area of the triangle ADB = AB.GD = cx. Similarly the triangle BDC = ax, the triangle CDA = bx, and the triangle ACB -= AB.CH =2 3. Also, BDC + CDA + ADB =ABC; that is, in symbols, 2 /3 = (a + b + c) x, and x = —, which is half the side of the triangle sought. Cor. From the resulting equation, we have x V3 =a + b +c: we also had CH = x V3. Hence CII = a + + c; or the whole perpendicular CII is equal to the sum of the three smaller perpendiculars from D upon the sides, whenever the point D is taken within the triangle. Had the point D been taken without the triangle, the perpendicular upon the side which subtends the angle within which the point lies would become negative. Thus, had the point been without the triangle, but between the sides AB, AC produced, then CH = DF + DG - DE. EIGHTH BOOK. 211 PROBLEM IL. A Hay-pole was broken of by the wind, and its top struck the ground twentyfeetfrom the base; and, being repaired, was broken a second time five feet lower, and its top struck the ground ten feet fartherfrom, the base. What was the height of the Iayjpole? B Let AB be the unbroken May-pole, C and H the points in which it was successively broken, and D and F the corresponding points at which the top B struck the ground. Then will CAD and HAF be right-angled triangles. Put. BC = CD = 5, CA = y, AD = a, AF F D A =b, and CH= c. Then AB = a +y, BH =HF = x+c, and HA =y-c; therefore (B. III., T. XIV.), we have 2 + 2=2, (1) (V - C)2 + b2 = A + C)2 (2) Expanding (2) and subtracting (I) from it, we have, after a slight reduction, b2_ a2 $ + y = -2e =50 feet, the required height. PROBLEM Il. A statue eightyfeet high stands on a pedestal fifty feet high, and to a spectator on the horizontal plane, they subtend equal angles; required the distance of the observer from the base, the height of the eye b6eingfive feet. Let AB = a, the height of the pedestal; BC = b, the height of the statue; DE = c, the height of the eye from glound, and DA = EF = x, the distance sought. Then D EC2 = EF2 + CF2 = + (a+ b - c),and EA2 = EF2 + ED2 = X + c2 21.2 GEOMETRY. But, since the angle CEB = BEA, we have (B. III., T. XII.) EC:EA::CB:BA, or, EC2: EA2 CB2: BA2; or, in symbols, 2 + (a + b - c)2 02 c2: b2: a2. From this proportion, we readily deduce V=/(a (a -c)2 ++ ( 02)) the double sign merely indicating that this value of x may be measured either way, from A towards D, or from D towards A. PROBLEM IV. To determine the area of a triangle when the three sides are given. Let ABC be the triangle. From c the angle C draw the perpendicu- / c lar CD, which in the second figure falls without the triangle, and meets the base AB produced. Denote the sides respectively A D B A B D opposite the angles A, B, C, by a, b, c, and the perpendicular CD by p. The right-angled triangles CDA, CDB give b2 = DA2 +p2; a2=DB 2 +p2; consequently, b2 - a2 = DA2- DB2 = (DA + DB) (DA - DB). Hence, in the first figure, b2 a DA-DB= and in the second figure, - b2 _ a2 DA + DB= -. Now, since half the difference of two quantities added to half their sum gives the greater, we have, in both cases, b2 -a2 e b2 +02-aa the greater segment DA -= +- = - 2 2ec 2 2c' Hence, p2 = /-DA2 = D2- c EIGHTH BOOK. 213 Since the difference of two squares is equal to the product of the sum of the two roots into their difference, we have = (b2 + 2be + C2 - a2) (a'2 - b + 2bc -C 2) 2c - [ (b + c)2 _2] X [a2 - (b -e )2] 2c (a + b + c) (- a + b + c) (- b + c) (a + b -c) 2c Multiplying this perpendicular by half the base, we have for the area, V I(a + b + c) (-a + b + c) (a-b + c) (a + b -c), or [(a +b+e )(-a+b+c) a-b c J -ca+b-c)I2 2 2 2 2 ( Hence we may find -the area. of a triangle, when the three sides are known, by this RULE.-Take half the sum of the three sides, and from this half sum subtract each side separately; then take the square root of the continued product of the half sum and the three remainders, and it will be the area. PROBLEM V. Given the three sides a, b, c of a triangle, tolind: 1. The three perpendiculars from the angles upton the opposite sides; 2. The area of the triangle; 3. The radius of the circumscribed circle; 4. The radius of the inscribed circle; 5. The radii of the escribed circles. Let ABC be the triangle; and let a, b, e denote the sides opposite the angles A, B, C respectively, and P1, P2, P3 the perpendiculars drawn from the angles A, B, C; A the area of the triangle; IR the radius of the circumscribing circle; r that of the inscribed circle; and ri, r2, r3 the radii of the three escribed circles, which touch the sides a, b, c externally. An escribed circle has already been defined as a circle which 214 GEOMETRY. touches one of the sides of a triangle exteriorly, and the other two sides produced. (See B. II., T. XHI., S. II.) We have already found (Prob. IV.) the perpendiculars to be PI,= ( + b + c) (- a + b+ c) (a- b + c) (a + b-) (I) 2a (I) -2 IV'(a+b+c)(-a++c)(a-b+c)(a+b -c6) (2) 2b / =(a +b+c) (-a&+o+) (a-b+c) (a+b-c) (3) 2c We have, also, under the same Problem, found the area to be 1 a+b+c -a+b c a-b+c /a+b-c I tAl 2 2 2 2 abc Pla therefore, 4R 2 4-R = 2 Hence, R= b Substituting for P. its value already found, we have =abe /(a+ b + c) (-a + b + c) (a - b + c) (a + b- c) The sum of the areas of the three triangles ADB, BDC, CDA equals the area ABC. But K ADB =~ rea BDC = ra, CDA = rb; A L B hence, ++er=, and (6) a + b + c (a - b + e)( We will now seek the radius r, of the escribed circle, whose centre is at E. EIGHTH BOOK. 215 Area EBA + ECA- EBC is evidently equal to the area ABO. Area EBA is equal to the base AB multiplied by half the perpendicular drawn from E upon AB produced; hence area EBA - c x r= cr. In a similar way, we find area ECA = 2 br1, and area EBC = ~ arn. Therefore we have b+ x r-=A, and (8) r=b+ -a (a- +b (+c)(a- b+) b c) (9) By simply permuting, we obtain { a+b+c)(-a+b+e)(a+b-c) } 1 (10) 4 (a - b + c) r= {t(a+b+c) (-a+b+c) (a-b+c) } (11) (a + b-c) (11) Equation (8) readily gives 1 -a+b+c. r, 2A and, in a similar manner, 1 a-b+c 1 a+b-c. r 2 2,- and - 2_ 216 GEOMETRY. 1 + 1 + 1 - a + b +c therefore, + - 2 r, r2 3 2 A But we have already from equation (7), ra -+ob+c'r 2' therefore, we have + - + - (12) ri r2 r3 Since any side of a triangle, multiplied by the perpendicular which meets it from the opposite angle, gives double the area of the triangle, we have 2 A = a P1, or 2A 2 =. (13) In a similar manner, we have 2 A (14) 2 A 2p3=c. ((15) P3 Taking the product of (13), (14), and (15), we have 8A3 = abc. (16) PIP2P3 Again, we have 2 RP, = be, (17) 2 RP2 = ac, (18) 2 RP3 = ab. (19) Taking the product of (17), (18), and (19), we have' 8 R3P1P2P3 = a2b2c2. (20) Extracting the cube root of the product of (16) and (20), we find 4RA = abc. (21) Dividing (20) by the square of (16), we find,2-13= 1, or R P1P2P3 =2 A2. (22) 8A6 By taking the continued product of (7), (9), (10), and (11), we have rr3 =(a + b + c) (-a + b + c) (a-b+c) (a b -c) _ 2. (23) By multiplying (5) and (7) together, or (7) and (21), we have abc 2 Rr = ab (24) a — b T-c EIGHTH BOOK. 217 By a similar multiplication, we find abc 2 R r- = 1 21Rr2= abc+ (25) a —b+ abc 2I r3 = a + b - J By combining the values of (9), (10), and (11), by two and two, we find rlr2+ rlr3+ r2r3= (a +b + )2 (26) We may also deduce 1- 1- + - +- (27) PI P2P3 rl r2 r3 r A2 IRr= (28) - Rr = p1P2 + P2P3 + P3P1 (28) 2 r2r3 r2 + r3 -P2= 2 rr3 (29) r, + r3 1 1 1+1 $- = 2 rr2 P1 P2 P3 r3 1 1 1 1 P1 P3 P2 r2 r 4- r2+r3= 4R +r. (31) Any of the foregoing expressions, when properly translated into common language, leads to a theorem. We will translate some of the most interesting ones. Equation (12) gives the following THEOREM. T7h/e reciprocal of the radius of the inscribed circle is equal to the suum of the reciprocals of the radii of the three escribed circles. 218 GEOMETRY. Equation (21) yields the following THEOREM. Four times the radius of the circumscribed circle, into the area of the triangle, is equal to the continued product of the three sides. Equation (22) gives this THEOREM. The radius of the circumscribed circle, into the continued product of the three perpendiculars, is equal to twice the square of the area. Equation (23) gives this THEOREM. The radius of the inscribed circle, into the continued product of the radii of the three escribed circles, is equal to the square of the area. Equation (27) gives this THEOREM. ThAe sum of the reciprocals of the three perpendiculars is equal to the sum of the reciprocals of the three radii of the escribed circles. By combining the equations already formed, new ones would arise, which might still afford interest. Thus, by a comparison of equations (22) and (23), we find R P1P2P3 = 2 rrlr2r3; (32) which gives this THEOREM. The radius of the circumscribed circle, into the continued product of the three perpendiculars, is equal to the diameter of the inscribed circle, into the continued product of the three radii of the escribed circles. PROBLEM VI. Given the three sides of a triangle, to jind the three lines drawn from the angles to the middle points of the opposite sides. Let ABC be the triangle. Denote the sides opposite the angles A, B, C, by a, b, c respectively; also denote the lines drawn G from the angles A, B, C to the middle points of the opposite sides, by m,, m2, m3 respectively. Then (B. III., T. XVII.), we shall have B A D EIGHTH BOOK. 219 2 AF2 + 2 BF2 =AB2+ AC02, or 2m + - a2 = C2 + b2; which gives 4 mn2= - a2 +2 b2 + 2. (1) In a similar manner, we find 4 m =- b2 + 2 c + 2 a2, (2) 4 m3 = - c2+2 2+2b2. (3) Equations (1), (2), and (3), readily give m-= Va2+2b2+2c2 m2 -- L b2 + 2 2 + 2 a2, 4) m3 = -2 + a2+2b2, J We will now deduce a few remarkable relations, which, when properly translated, will give some beautiful theorems. Taking the sum of (1), (2), and (3), we obtain 4 (m + n2 + m = 3 (a2 + b2 + ). (5) If we take the product of (1) and (2), we shall have 16 mm =- 2a4+5 a2b2+2a2o2 2 b4 + 2 b22 + 4 o4. (6) By simply permuting, we find 16 2m = - 2 b4 + 5 b22 + 2 b2 _ 2 C4 + 2 2a2 + 4 a4 (7) 16 m2m2 = - 2 c4 + 5 a2 + 2 c2b2 - 2 a4 + 2 2b2 + 4 b4. (8) Taking the sum of (6), (7), and (8), we obtain 16 (,m,2i + =, m' + mm) =9 (a'2b2 + b22 + c2a) (9) If, from the square of (5), we subtract twice (9), we shall have 16 (ml + m* + m4)= 9 (a4 + b4 + c4). (10) The point where these lines trisect each other, is the centre of gravity of the triangle. If we denote the distances AK, BK, CK, by d1, d2, d3 respectively, we shall have m, = 3 d1, m2 = 3 d2, m3 = 2 d3 (B. I., T. XXXVIII.). These values substituted in (5), (9), and (10), cause them to become 3 (d2 + d 22+ db = a2 + b2 + c2, (11) 9 (d(2 1 + d32 )/ + /3 d) = - a2 b2 + b2 c2 + C2 a2 (2) 9 (d, + d2 + d,3) = a4 + b4 + el. (13) 220 GEOMETRY. Equation (5) is equivalent to the following THEOREM. Four times the sum of the squares of the lines drawn from the angles of a triangle to the middle points of the opposite sides, is equal to three times the sum of the squares of the sides. Equation (9) gives this THEOREM. Sixteen times the sum of the products, taken two at a time, of the squares of the lines drawn from the angles of a triangle to the middle points of the opposite sides, is equal to nine times the sum of the products, taken two at a time, of the squares of the sides. Equation (10) gives this THEOREM. Sixteen times the sum of the fourth powers of the lines drawn from the angles of a triangle to the middle points of the opposite sides, is equal to nine times the sum of the fourth powers of the sides. Equations (11), (12), and (13) would lead to beautiful theorems in reference to the distances of the centre of gravity of three equal bodies, and the mutual distances of the bodies themselves. For it is evident that, instead of considering the triangle as material, we may suppose equal weights placed at its vertices; and, under this point of view, equation (11) will give the following THEOREM. Three times the sum of the squares of the distances of three equal bodies from their common centre of gravity, is equal to the sum of the squares of their mutual distances. PROBLEM VII. Given the three sides of a triangle, to find the lines bisecting the angles, and terminating in the opposite sides. Let the sides opposite the angles A, B, C be denoted by a, b, c respectively; also C denote the lines which bisect the angles A, F K B, C respectively by l,, 12, 13. Pass a circumference through the three A G B points A, B, C, and produce 11 to meet it at K; join BK. EIGHTH BOOK. 221 The two triangles ABIK, ADC are similar, and give AB:AD::AK: AC; consequently we have AB x AC =AD x AK =AD x (AD + DK) = AD2+ AD xDK; which finally becomes AB xAC =AD2+BD x DC, since BD x DOC = AD x DK (B. IV., T. III., S. II.). In symbols this becomes be =2+ BD xDO. (1) Again, we have (B. III., T. XII.) AB: AC::BD: DO. Consequently, AB + AC:B:: BD+DC: BD, AB + AC: AC::BD+DO:DC; or, symbols, c+b: c:: a: BD, c + b: b:: a:DC. Hence BD -a DO=c These values of BD, DC, cause (1) to becoine cb= + a'b (c + b)2 This readily gives l2 = cb-(, or (c+ b)2 2 =b(a+b+c) (-a+b+c) (2) (C+ b)2 In a similar manner, we find =2 ac (a + b + c) (a- (3)+ c) (2a + c)2' = ba(a + b +c) (a+ b-c) (4) 13 (6b+ a)2 Equations (2), (3), (4), give Vcb(a+- b -+c) (-a+6bc) c+b 2= — V'ac a+ c (5) 3 (a+ b + c)(a + b -) ~~~~~=~~~t..t.. t,. 222 GEOMETRY. Taking the continued product of these values given by (5), we find Zabe (+b+c) /(a+b+c) (-a+b+c) (a-b+c) (a+b-c) (a + b) (b+c) (c+a) The expression V(a+b+c)(-a+b+c)(a-b+ c)(a+b -- c), is equal to four times the area of the triangle (see P. IV.), which we will denote by 4A; so that we shall have 4 abe (a + b + c) A (6) (a + b) (b + c) (c + a) PROBLEM vIU. To determine a right-angled triangle, having given the hypotenuse and difference of two lines drawn from the two acute angles to the centre of the inscribed circle. Let AC=h; AO=x+d; c CO = x - d; d being half the difference of the lines AO and CO. Produce AO, and draw CD perpendicular to AO thus produced. Then, since the an- AI gle COD is equal to the sum of CAo and ACO, and since CAO is half the angle CAB, and ACO the half of ACB, it follows that COD is half a right angle; consequently OCD is also half a right angle: therefore OD is to OC as the side of a square is to its diagonal; that is, as 1 to./2. Hence OD=CD= /2 and AD=x+ - d +=,2',-V2 Again, AC2 = AD2 + CD2; or which is in symbols A2 c + d + $-d)2 + %($ )2 (1) 2-=(h (2 -/-2)d 2- This readily gives = ( 2 - 2) (2) 2+ %/2 Having found x, we of course know AO and CO. Let AO be denoted by a, and CO by b; also let the radius of the in-.scribed circle be denoted by r. EIGHTH BOOK. 223 Then AF = VAO2 - OF2 = /Va2 - r/, CE = V/CO2 -OE2 = l/b2 _- A. But AF = AG, and CE = CG; therefore, /Va2 _ r2 + Vb'2 _- v = h. (3) This readily gives r = { 6' ( +b2 b2 )2 }a 2 or /(a +b + h)(- a+ b + h)(a - b +h) (a + b-) (4) 2h (4) This value of r might have been found as follows: In the triangle AOC, all the sides are known; and it is required to find the perpendicular OG, which is r, the radius of the circle. Under P. IV. we have found the perpendiculars when the three sides are known. The perpendicular OG found in this way will be precisely the same as the above value of r. Now, having found r, we will denote AB by x, and CB by y; then AF =x - r, and CE = y - r. Hence, AF + CE = =x +y - 2r. (5) Again, my = double the area of the triangle ABC; also, (x + y + Ah)r = double the area. Therefore, Xy = (x + y + h)r. (6) Equations (5) and (6) readily make known x and y, or the sides of the triangle; these values are h + 2r+ Vh2 _- 4hr- 4r yA + 2r - -2- 4hr- 4r J PROBLEM IX. In a right-angled triangle, having given the perimeter, and the perpendicular drawn from the right angle upon the hypotenuse; to find the sides of the triangle. Let S = sum of sides, or perimeter; P = perpendicular CD; AC = x, and BC = y. Then will + 2y2= AB. xy=double the A, B 224 GEOMETRY. area of triangle ABC; also, p/x2 + y2 = double the area of triangle ABC; therefore, xy = P V2 + y2. (1) We also have x + y + VX2 + y2 = S. (2) Equation (2) becomes, by transposing and squaring, (X + y)2 = (S - /X2 + y2)2 or x2 + 2xy + y2 = S2 _ 2S V2 + y2 + y2, or 2xy = S2 - 2s VX2 + y2. (3) Taking the double of (1), we have 2xy = 2P /I2 + y2. (4) Equating right-hand members of (3) and (4), we have 2P V/2 + y2 = S2 - 2S VX2 + y2, or ~- +y2 = S s) (5) 2(P + S)( Using this value in (1), we find PS2 xY = 2 (P+ S)' (6) Squaring (5), we have X2 + y2= 4(S2 ('7) Adding twice (6) to (7), and also subtracting twice (6) from (7), we have = S2(S2 + 4PS + 4P2) 2 + 2xy + y2 (8) 4 (P + S)2 -s (s2 _ 4PS - 4P() -2 - 2xy + y2 (9) - (P + S)' Extracting the square root of (8) and (9), we have s+ (S + 2P) (10 + 2 (P+S)) S /S2 - 4PS - 4P2 2(P + S) (1) Hence, S (S + 2P) + S VS2 - 4PS - 4P2 (12) 4(P+S) (12) S (S - 2P) - S Vs2 - 4p - 4 (13) 4(P +S) EIGHTH BOOK. 225 PROBLEM X. To determine a right-angled triangle; having given the hypotenuse, and side of the inscribed square. Let h= hypotenuse; s = side of inscribed square; AB =, and BC = y. Then AE: AB:: EF: BC; that is, - s: X::s: y; or, xy = (y)s. (1) /_ A E B Again, z2 + y2 = h2. (2) Add twice (1) to (2), and we obtain 2 + 2xy + y2 = 2s(x + y)+ h2; or (X + y)2 - 2s(x + y) = h2, (3) which is a quadratic in terms of x + y: hence we find x+ y =s+ V s2h+ h2. (4) Substituting this value of x + y in (1), we shall obtain xy = s2 +s Vs2 + h2. (5) From the square of (4) subtracting four times (5), we obtain X2 2 2xy + y2 = h2 - 2s2 - 2s s2 +h2 (6) Extracting the square root of (6), we have x- y = Vh2- 2s2 2s Vs2 + h2. (7) Taking half the sum of (4) and (7), and also half their differ ence, we obtain 15 s + 4s2 + 2 + J (h2 _ 2s- 2s 4s~ + A2) (8) a +,/82 + h2 _ (h2-22 - 2s,sZ + h2)~ 226 GEOMETRY. PROBLEM XI. To determirne a right-angled triangle; having given the hypotenuse, and the radius of the inscribed circle. Let h = hypotenuse; r = radius of the inscribed c circle; AB = x, and AC = y. Then BF=BE = x -r, and CF = CD = y - r; F therefore BC = BF + CF = x + y - 2r = h, G D or, a + y = 2r + h. (1) B E A Again, xy = double the area; (x + y + h) r = (x + y)r + hr = double the area: therefore xy = (x + y) r + hr. (2) In (2), for x + y substitute its value given by (1), and it will become xy = 2r (r + h). (3) Equations (1) and (3) readily give 2r + hA+ V/2 - 4hr - 4r2 X= 2 (4) 2r +hA - h-h2 - 4hr - 4r; 2 (5) PROBLEM XII. To determine a right-angled triangle; having given the lengths of the lines drawn from the acute angles to the middle points of the opposite sides. C Let AE=a, and CD=b; also AB=x, and BC = y. Then will AB2 + BE2 = a2, or X2 + y'2 = a2 (1) In a similar way we find y~ + x,=2. (2) A D B EIGHTH BOOK. 227 Equations (1) and (2), when cleared of fractions, become. 4x2 + y2 = 4a2, (3) X2 + 4y2 = 4b2. (4) These equations give 2= 1 (5) y 1= f/~ __ (6) PROBLEM XIII. To determine a triangle; having given the base, the perpendicular, and the difference of the two other sides. Let p = the perpendicular; d = half the difference of sides; x + d = AC; x- d =BC. b=halfthebase; b+y=AD; b —y=DB. Then, from the right-angled triangles ADC, A B D BDC, we have (b + y)2 +p2, = (X + d2 (1) (b - y)2 +p2 = ( _- d)2. (2) Expanding (1) and (2), and then taking half their sum and one fourth of their difference, we obtain b2 + y +p2= + d2 (3) by = dx. (4) These equations readily give b ( -- d2-). (5) b2 - d2, (5) y=d 2d~ )* (6) Hence, AC=((b - d2 p2) d (7) B =b(b d P )2_d (d8) b2 -- d 2)~ 228 GEOMETRY. PROBLEM XIV. To determine the radii of three equal circles, described in a given circle, to touch each other, and also to touch the circumference of the given circle. Let D be the centre of the given E circle, whose radius we will denote by R; also, let A, B, C be the centres of the three equal circles, whose common radius we will denote by r. Then joining A, B, and C, we have the triangle ABC \ A K equilateral, each of whose sides is G 2r. Drawing CK perpendicular to AB, the right-angled triangle AKC gives CK2 = AC2 - AK2, or, in symbols, CK' = 4r2 - r2 = 3r2; consequently, CK = r /3. (1) Now (B. I., T. XXXVIII.), CD = CK =r V3. (2) But ED = EC + CD; that is, R = r + r V/3: (3) from which we readily find 3R R r = (4) 3 +2 /3 1 +2,/(4) PROBLEM XV. Given the radius, r, of a circle, to find the sides of the inscribed and circumscribed pentagons and decagons. 1. The inscribed pentagon. Let N A M_ ADBCE be the pentagon inscribed in the circle, and let O be the centre of \ the circumscribing circle. Join AB, V AC, and draw AF perpendicular to r L BC; then, by known properties, BC is bisected in G, and the line AF passes Q R through the centre O of the circle. K EIGHTH BOOK. 229 Since the angle ADX is measured by half the are AEC (B. II., T. X.), and the angle AXD is measured by half the sum of the arcs AD and BC (B. II., T. XI.), it follows that these angles are equal, and the triangle DAX is isosceles. Again, the angles DAX and BAC are equal, being measured by halves of the equal arcs DB, BC; hence the triangles ABC and DAX are similar. But the triangle DAX is obviously equal to BCX; therefore, AB: BC:: BC: BX; but BX = AB - AX = AB - BC; therefore, AB: BC:: BC: AB - BC. (1) Put BC=2x, or BG==x; BA=y, and BF=z; and the above proportion will become y: 2x:: 2x: y — 2x; which gives y = (1 + 5) x. (2) Now, we have OG = Vr2 - x2, AG = Vy2 - x2; and hence AG = AO + OG gives Vy2- 2 =r + r2 -X2, or, squaring, y2 - 2r2 = 2r /r2 - X2. Again, squaring, we have y4 - 4ry2 + 4rX2 = 0. (3) Substituting the value of y as given by (2), we find the side of an inscribed pentagon (see B. IV., T. XI., S. II.), 2x = -r V10 - 2 V5. (4) We have y = x( + /5) = r V10 + 2 V5, (5) OG = / r2 = -r2 ( 5)r2 = r (1 + 5.) (6) 2. The inscribed decagon. Join BF; then, since AF bisects the line BC at right angles, it bisects the are BFC in F; and hence BF is the side of the inscribed decagon. But ABF being a right angle, since it is in a semicircle, we have BF2 = FA2 - AB2; or, in symbols, z2 =4r2 _ y2 =4r2 _ - (10 + 2 /5)r2 =2 (3- /5)r2; or, extracting the square root, we have, for the side of the inscribed decagon (see B. IV., T. XI., S. I.), 2Ir (,/5 - 1).(7 230 GEOMETRY. 3. lThe circumscribing pentagon. The inscribed and circumscribed pentagons being regular, are similar figures, and their sides are as the perpendiculars from the centre upon the sides. That is, if PK be a side of the circumscribing pentagon, we shall have OG: OB:: BC: PK; or, in symbols, OBxBC r. r,/1 —2/5 P = OBxG = B - 2r,/5-22,/5. (8) OG - r (1 + V5) 4. Tlke circumscribing decagon. Let QR be one of the sides; and draw OH perpendicular to BF, which it bisects in H. Also by similar triangles ABF, OHF, we have OH ='AB = ~y = =r V10 + 2,/5. Also, as in the last case, OH: OF:: BF: QR; which gives OF x BF r r(/S 1) = 2r- 2/5 R OH -r V10+2/5 5 PROBLEM XVI. Given the lengths of three lines drawn from a point to the three angles of an equilateral triangle, to find its side. Let ABC be the triangle, and D the point. Put AD = a, c BD = b, CD =c; G D AB = 2x, EF = y, FD = z. Then will AF = x + y, A E F B A E B F and BF = x- y or y - x. CE = /AC2- AE2 = /4x2 - x2 =x V 3; CG = xV3 - z, or z - x /3. Hence we have the following relations true, whether the point D is within the equilateral triangle, or without it. (x + y)2 + z2 = a2, (1) (X - y)2 + 2 = b2, (2) (z /3 - Z)2 + y2 = 2. (3) EIGHTH BOOK. 231 Subtracting (2) from (1), we find a2 - b2 4xy = a - b2, ory = -; (4) consequently, y2 162 (5) Equation (1) gives immediately z = V/a2 (x + y)2; in which, substituting for y its value given by (4), we have z =Ia2 ( +ab2)2 (6) This value of z, and the value of y2 given by (5), being substituted in (3), will cause it to become {43- $ V a_ a2-;b2)2 = 2 ('2)2 This equation contains only the unknown x, which value may therefore be found: the reduction leads to 16x4 - 4 (a2 + b2 + c) x2 = - a4 -b4 - c4 + ab2 b2c + caa. (8) This equation, when solved by the rule for quadratics, gives 2x (a2 +~~ + +c2 + /36(a2b + b2ca + c2a) - 3 (a4 + b4 + c4) } (9) We must use the + sign when the point is within the triangle, as in the first figure; and the - sign when the point is without, as in the second figure. If we suppose a triangle to be formed with the three lines a, b, c, and denote its area by A, expression (9) will become 2= (a2 + b2 + c2 ~ 4A 3). (10)'2 PROBLEM XVII. Suppose AB to be a tree, standing on the horizontal _plane AC; it is required to find at what point it must break, so that, by falling, the top may strike the ground at C. If we join BC, and bisect it with the perpendicular DF, we shall have F for the point sought. 232 GEOMETRY. Let the height of the tree be denoted by h, and B the distance AC by b; then BC will equal /b2 + h2; and, consequently, D CD = Be = /b2 + 2. F Again, the two triangles BDF and BAC are similar, having the angle B common, and the angle BDF equal to the angle BAC, each being a A right angle; therefore, BA: BC: BD: BF, A: Vb2 + A-2:: b2 + h2:BF. b2 + 42 Consequently, BF= b24 -' b2 ~- h2 h2 — b2 and AF=AB - BF = h 2 + 2 2 2h 2h PROBLEM XVIII. Suppose AC and BD to represent two trees standing on the horizontal plane AB; it is required to find a point in this oplane, situated on the line AB, equally distant fronm the tops C and D. Join CD, and bisect it with the perpen- F dicular FG; then G is the point sought. This point may be found arithmetically as follows: Draw DK parallel to AB, and FL perpendicular to AB. Let the height of the A B tree AC be denoted by h1, the height of the tree BD by h2, and the distance AB by d; then will KC = AC- BD = - h2, FL= (hl + h2). Since the sides of the two triangles DKC, FLG are respectively perpendicular to each other, they are similar, and we have DK: KC:: FL: LG; or, in symbols, d': A1-:: ~ (Al + A2): LG. Consequently, LG A2 2d EIGHTH BOOK. 233 ad G = AL-LG = h2 _ 2 d' _ -2 + h2 and AG=AL-LG=Kd - h'h d2- 1+ 2h 2 2d 2d BG = BL + LG=2d+hh2 d I -h PROBLEM XIX. Suppose three trees to stand vertically utpo? a horizontal plane at the points A, B, and C. It is required to jid a point in this.plane equally distant from their tops. Pass a plane through the two trees at A and B, and suppose this plane to revolve about the line AB until it coincides with the horizontal plane, the trees being then represented by AD and BE. Join DE and bisect it with the perpendicular HL; G D H E then will the point L be equally distant from the points D and E. Also, pass a plane through the two trees at B13 and C, and suppose this plane to be revolved about the line BC until it coincides with the horizontal plane, the trees then taking the position of BF and CG. Join FG and bisect it with the per 234 GEOMETRY. pendicular 1KM; then will the point M be equally distant from the points F and G. Draw LN and VIN respectively perpendicular to AB and BC, meeting at the point N. Then will N be the point sought. For, suppose the planes ADEB and BFGC to be brought back to the first position, so as to be perpendicular to the horizontal plane ABC. Then join ND, NE, NF, and NG; also, join LD, LE, MIF, and MG. Since BE is equal to BF, each representing the height of the tree standing at B, it follows that NE is equal to NF. NL will be perpendicular to the plane ADEB, consequently the triangles DLN, ELN are right-angled, and DIN and LN are respectively equal to EL and LzN; hence their hypotenuses DN and EN are equal, which proves N to be equally distant from the tops of the trees standing at A and B. In a similar manner it may be shown that the right-angled triangles FMN, GMN are equal, and consequently the side FNi is equal to GN. But FN has already been shown to be equal to EN, consequently DN: EN, FN, and GN are each of the same length; that is, N is a point in the horizontal plane equally distant from the tops of the three trees standing at the points A, B, and C. As an illustration, let us suppose AD= 114 feet, BE-= 110 feet = BF, CG = 98 feet. Also, suppose AB = 112, BC = 104, AC = 120. Then we find BL = 60, B3M=-40 (see P. XVIII.). The perpendicular CP we find (P. IV.) to be 96, and BP =40. If we draw MR perpendicular to AB, and NS perpendicular to AIR, we shall have the triangles CPB, MRB, and NSM rightangled and similar. Hence, by the similar triangles CPB, MRB, we have BC: BM:: BP: BR, or 104: 40::40:BR= 15T. And LR = BL - BR = 60 -Il-5y = 44l. Again, by the similar triangles CPB, NSM, we have CP: CB::NS(= LR): NM, or 96: 104:: 448s ~: NM = 483. EIGHTH BOOK. 235 Since the triangle NMB is right-angled, we have BN2= NM2 + BM2= 23361- + 1600 = 3936-1. Again, since the triangle NBE, or its equal NBF, is right-angled at B, we have NE2 = NF2 = BN2 + BE2 = 3936- + 12100 = 16036}. And NE = NF = NG = ND = /16036- = 126.63, nearly, for the number of feet from the point N to the top of each tree. END OF PART L PART SECOND. PLANE AND SPHERICAL TRIGONOMETRY, TRIGONOMETRY. CHAPTER I. FIRST PRrNCIPLES OF PLANE TRIGONOMETRY. ~ 1. TRIGONOMETRY is that science which investigates the relations of the sides and angles of triangles. When confined to plane triangles, it is called Plane Trigonornetry. Definition of an Angle. ~2. When two straight lines, AB, AC, c meet each other, the opening between these lines is called an angle. This angle or opening, which is read BAC or CAB, or simply the angle at A, does not depend upon the A B lengths of these lines, but only upon the difference of their directions. ~ 3. If any straight line, as Cs AB, revolve about A as a cen- C4 tre, in such a manner that the part ACO shall take the succes- ci sive positions AC,, AC3, A, eB &c., then will the arc described by the point Co increase at the Cs same rate as the angle increases; c6 so that the arc COC1 is to the arc C COC2, as the angle CoAC1 is to the angle CoACa2; and the same for other arcs and their corresponding angles. Hence, the arcs thus described may be taken as measures of their corresponding angles. ~ 4. When the line ACo has reached the position AC3, so that the arc CoC3 is a quadrant, or one-fourth of the entire cir 2 TRIGONOMETRY. [CHAP. I. cumference, the angle C0AC3 measured by this quadrant, is called a right angle. The right angle is usually taken as the unit angle; and all angles less than a right angle are called acute angles, and are each measured by an are less than onefourth of an entire circumference. When the line ACO has reached the position AC0, the point Co has passed over an arc equal to one-half of the entire circumference, or over an are which is the measure of two right angles. Hence, when two lines meet from opposite directions, they are said to form an angle equal to two right angles. An angle such as COAC4, which is greater than a right angle and less than two right angles, is called an obtuse angle. ~ 5. All the foregoing relations hold good, whatever be the length of the radius ACO. For simplicity, we shall hereafter use such arcs, for the measure of their corresponding angles, as are described with a unit radius. The Greek letter ~r is usually employed to denote the semicircumference of a circle whose radius is a unit, or, which is the same thing, it represents the ratio of any circumference to its diameter. HIence is the length of an arc called a quadrant, which is the measure of a right angle. The numerical value of r is as follows::r=3-1415926535897932384, &c.=semi-circumference, when the radius=l. (See Geom. B. IV., Lemma.) ~6. If the whole circumference be divided into 360 equal parts, one of those parts is called a degree. Degrees are also divided into 60 equal parts, called minutes; minutes are similarly divided into 60 equal parts, called seconds, and so on for other subdivisions. Thus 43~ 14' 12" denotes an angle whose measuring are consists of 43 degrees, 14 minutes, and 12 seconds. In the same way 90~ denotes an angle whose measuring are consists of I of 360 degrees, and whose length is therefore = 1'5707963267948966192, &c. From the above value of r, which is the are of 1800, it is easy to find the length of an arc corresponding to any angle expressed in degrees, minutes, and seconds. Thus, we find, ~ 7.1 TRIGONOMETRY. 3 y- of jr=10~ =001745329251994329577, &c. -6 of -o of nr=1' =0'00029088820866572160, &c. 6 of 0 of r-~ of n =1"=000000O484813681109536, &c. Modern French writers, instead of using the sexagesimal division, use the centesinzal; and it is to be regretted that the latter is not universally used, on account of the great ease with which all calculations are made in that division. If E denote the number of English degrees in a given angle, and F denote the number of French grades in the same angle, we shall obviously have E F E F 9= 100, or - = 1. Consequently, E=? -ofF=F-f -1 ofF; andF = -- ofE=E + -ofE. ~ 7. When the sum of two angles is equal to a right angle, those angles are mutually complements of each other. In such case, the sum of their measuring arcs is 2, and the sum of the degrees which these arcs contain is 90~. When the sum of two angles is equal to two right angles, these angles are mutually supplement of each other: the sum of the arcs which measure these angles is equal to %r, and the sum of the degrees which Cs they contain is 1800. Thus the c4 angles CoACI, C1AC3, are mutually complements of each other. ci Also if we denote any arc by a, C5 B its complement will be - -a and its supplement will be r-a. c/ If an angle is expressed in de- c7 grees, as AO, where A~ denotes the number of degrees in the arc which measures the given angle, then will 90~ — A denote the number of degrees in the arc which measures its complement. For brevity, we say that a and — a are complementary arcs, and that A~ and 90~-A~ are complementary degrees. For a like reason, a and'K-a are supplementary arcs, and A~ and 180~-A~ are supplementary degrees. 16 4 TRIGONOMETRY. [CHAP. 1. Definition of Sines, Tangents, Secants, &c. ~ 8. In a right triangle the sine of either of the acute angles is the quotient obtained by dividing the side opposite the given C angle by the hypotenuse. Thus, ini the right triangle ABC, if we represent the base, perpendicular, and hypotenuse, by b, p, / and h respectively, we shall have A b sine A = sine C b The tangent of either of the acute angles is the quotient obtained by dividing the side opposite the angle by the adjacent side. Thus, tangent A =; tangent C -. The secant of either of the acute angles is the quotient obtained by dividing the hypotenuse by the side adjacent the given angle. Thus, k h secant A= secant C= -. 6' jp The cosine, cotangent, and cosecant of any angle are respectively the sine, tangent, and secant of its complement. The two acute angles of a right triangle being complements of each other, it follows that we shall have these relations: sin. A cos. C=. (1) b cos. A=sin. C= A-. (2) tang. A =cotan. C = -. (3) b ***(A) cotan.A= tang. C = b-. (4) sec. A = cosec. C = b. (5) cosec. A = sec. C =. (6) ~~g.] TRrGONOMETRY. 5 From the above relations, we immediately find sin. A _p cos. A = -=tang. A. (1) cos. A b =os. A =cot. A. (2) sin. A p 1 h oses. A. (3)..( B). cos. A =s 1 /, - = cosec. A. (4) sin. A sin.2 A + cos.2 A = A +2 1(5) The exponent 2, equation (5) of (B), which is used to denote the square of the sine and cosine of the angle A, is placed immediately after sin. and cos., so that sin.2 A is the same as (sin. A)2, cos.a A is the same as (cos. A)2. From the above relations, we see that the sine and cosecant are reciprocals of each other; cosine and secant are also reciprocals; so also are the tangent and cotangent. ~ 9. If from one extremity of the arc which measures an angle, a perpendicular be drawn to the diameter which passes through the other extremity, such perpendicular will be the sine of that angle. Thus, CID, is the sine of the angle CoAC1. For we have (~ 8) defined the sine of this angle to be the quotient obtained by dividing C1Dl by AC1, which quotient becomes CD,, since the divisor is the C3 C2 radius of the circle, and therefore, (~ 5) equal to unity. In the same c4/ way it may be shown that C2D2 is / the sine of the angle CoAC2, that c/ D _ D C4D4 is the sine of the angle CoAC4. D / As the arc is the measure of angles, / we say that C1Di, C2D2, C4D4, are respectively the sines of the arcs c OCI,, CoC,, c00. c Passing from the angle to the arc, we also say, when the arc,exceeds the semi-circumference, that C(,D, C8Ds, are respectively the sines of the arcs COCA,5 COC0C7C8. 6 TRIGONOMETRY. [CHAP. I. In all these cases we suppose the arc to be estimated fiom the point C0, at the extreme right of the circle, upwards and around towards the left. When angles are estimated fiom C, downwards around towards the left, they are considered as lnerative arcs. Thus, CoAC8 is considered as a negative angle, measured by the negative are CoC8. By reference to the liagram, it will be readily seen that so long as a positive are does not exceed Ir or 180~, the sines are all situated above the horizontal diameter, and are considered as positive; for arcs greater than 180~ and less than 360~, their sines are situated below the horizontal diameter, and are considered as negative. Di ~ 10. If from one extremity of an are a tangent line be drawn, and limited by its intersection with the / radius drawn through the other extremity of the are, then will this portion of the tangent line thus in- C4 eluded between the radius and the radius produced, be the trigonomet- C. / rical tangent of the are. And the radius thus produced, or simply the - produced part, will be the secant of the arc. Thus, CoD1 is the tangent of the arc COC, and CoD3 is the tangent of the are C0(3, in the same way COD5, CoD7 are respectively the tangents of the arcs C0C2C5, COC2C4C7. For, by the former definition (~ 8), the tangent of the angle CoAD1, which is measured by the arc CoC1, is the quotient obtained by dividing CoD, by ACo, which quotient becomes CoD1, since ACo, the radius, is a unit. Also the secant of this angle is the quotient obtained by dividing AD, by ACo, which quotient becomes AD1. When the arc terminates in the first and third quadrants, the tangents are counted on the line D7D1 fiom Co upwards, and are considered as positive. But in the second and fourth quadrants, the tangents are counted from Co downwards, and are considered as negative. The secants of the arcs 0C1o C, 03, CoCc5, 0C0204C0, are respectively AD, AD3, AD5, AD7. By inspecting the diagram it will be seen that the secants represented by the dotted lines ~11.] TRIGONOMETRY. 7 consist of the produced part of the radius, while those represented by the full lines consist of the radius together awith the produced part; the latter are considered as positive and the foriner as negative. HIence in the first and fourth quadrants the secants are positive, and in the second and third quadrants they are neoative. ~ 1 1. These different lines may be clearly exhibited in the four quadrants as follows: First quadrant. Second quadrant. E F G ~K~~ B D //L i iA D K B L E In the above diagrams the dotted lines are considered as negative and the full lines as positive. The letters in the four diagrams are so arranged that in each case BC counted in the positive direction (~ 9) is the arc, CD the sine, CH the cosine, BE the tangent, GF the cotangent, AE the secant, and AF the cosecant. 8 TRIGONOMETRY. [CHAP. I The algebraic sines of these lines will be as follows: In In In In 1st quad. 2d quad. 3d quad. 4th quaY Sine and cosecant,. + + Cosine and secant,. + - - + Tangent and cotangent, + + ~ 12. By carefully inspecting the diagrams of ~ 11, we see that denoting any arc when considered as positive by a, it will,. when counted in an opposite direction, be denoted by -a, and the following relations will hold good for negative arcs: sin. (-a)=-sin. a; cos. (-a)=cos. a. tang. (-a)=-tang. a; cotan. (-a)=-cotan. a. see. (-a)=sec. a; cosec. (-a)=-cosec. a. ~ 13. If the are KC' is equal to the are BC, then will the arc BC' be the supplement (~ 7) of the arc BC. De- K B noting the arc BC by a, we have ob- \ viously the following relations: sin. a=sin. (180~-a); cos. a= —cos. (1800 -a). tang. a= -tang. (180~- a); cotan. a= — cotan. (1800~-a). sec. a=-sec. (180~-a); cosec. a=cosec. (180~-a). ~ 14. The following particular values should be carefully noted by the student: 0~ 900 1800 2700 360~ same as 0~ Sine.. 0 + 1 0 — 1 0 Cosine. +1 0 -1 0 +1 Tangent. 0 o X0 0 Cotangent 0o 0 0 Secant. +1 mo — 1 -1 Cosecant o + 1 o - 1 i 15.1 TRIGONOMETRY. 9 In general, if n denote any integral number, including also the value of n=0, we shall have, sin. 0~ =sin. 180~ = sin. 2 n x900 = 0 sin. 90~ =sin. (4n+1)x90~ =1 sin. 2700 ~- sin. (4 n + 3) x 900 = -1 cos. 00 - cos. 4 n x 900 = 1 cos. 90~ =cos. 270~ = cos. (2 n + 1) x 900 = 0 cos. 1800 = cos. (4 n + 2) x 900 = -1 tang. 0~ = tang. 1800 = tang. 2 n x 90o = 0 tang. 900 = tang. 270 = tang. (2 n + 1) x 900 = oo cotan. 00 = cotan. 180~ = cotan. 2 n x 900 = oo cotan. 900 = cotan. 2700 = cotan. (2 n + 1) x 900 = 0 sec. 00 = sec. 4 n x 900 = 1 sec. 90~ = sec. 270~ = sec. (2 n + 1) x 900 = 0o sec. 1800 = sec. (4 n + 2) x 90o = -1 cosec. 00 = cosec. 180~ = cosec. 2 n x 900 = oo cosec. 90~ = cosec. (4 n + 1) x 90O = 1 cosec. 270~ = cosec. (4 n + 3) x 900 = - 1 ~ 15. To find the sine and cosine of the stum and difference of two arcs. Let the angle BAC=a; that is, let D a denote the length of the arc which measures this angle; also let the angle CAD be denoted by b; then will BAD be denoted by a +b. Draw EF perpendicular tc AC, and \ C from the points E and F draw EG B and FH perpendicular to AB; also A G H draw FK perpendicular to EG. Since the sides of the triangle EFK are respectively perpendicular to the sides of the triangle AFH, these triangles are similar. (Geom. B. III., T. VII.) Hence, the angle FEK is equal to FAH, equal to a. EG HF + KE HF KE sin. (a + b)A AE AE + AE HF AF KE EF AF — x A —-E Al? AIE EF A` E 10 TRIGONOMETRY. [CHAP. L HF. AF KE By our definitions (~ 8), AlF = sin. a; = cos. b ~~ —i.-=o;EF cos. a; A = Sin. b. Hence, we have sin. (a + b)= sin. a cos. b + cos. a sin. b. Again, we have AG AH - KF AHI KF AEcos.(+) AE AE AE AH AF KF EF A- x AE -E x ] —- = cos. a cos. b- sin. a sin. b. For the sine and cosine of a-b, let the angle BAC = a, and the angle CAD = b; then will BAD = a —b. Then, as in the former diagram, we have the triangle EFK similar F7 to the triangle AFH, and the angle KFE equal to the angle FAH, equal D to a. A - AHJ G EG HF-KF HF AF AFE FE AE-A AE AAE FE A-E sin. a cos. b-cos. a sin. b. AG AH + KE AI AF KE FE cos. (a-b) A A X + A AE AE AF E - -x A-Ecos. a cos. b + sin. a sin. b. Collecting these results, we have sin. (a + b) = sin. a cos. b + cos. a sin. b. (1.) sin. (a- b) = sin. a cos. b - cos. a sin. b.. (2.) cos. (a + b) = cos. a cos. b - sin. a sin. b. (3.) cos. (a - b) = cos. a cos. b + sin. a sin. b.. (4.) If we make b = a, equations (1) and (3) will give sin. 2 a= 2 sin. a cos. a,... ) cos. 2 a = cos.2 a-sin. a. (6.) From (5) of (B), ~ 8, we have 1=sin.2 a + cos.2 a.... (7.) p16.] T TcO:'(TM'fTRY~. 11 The sum and difference of (6) and (7) give 1 + cos. 2 a =2 cos.a,... (8.) 1-cos. 2 a = 2 sin..... (9.) Hence, sin. a =.! /2-2 cos. 2a,. (10.) cos. a = 2 + 2 os. 2. (11.) Equations (5) and (6) make known the sine and cosine of a double arc in terms of the sine and cosine of the are itself. Equations (10) and (11) give, conversely, the sine and cosine of half an arc in terms of the cosine of the arc itself. Dividing (1) by (3), we have sin. (a + b) _ sin. a cos. b + cos. a sin. b cos. (a + b) cos. a cos. b - sin. a sin. b' Dividing numerator and denominator of the right-hand member each by cos. a cos. b, we have sin. a + sin. b sin. (a + b)_ cos. a cos. b cos.(a+b) sin. a sin. b' COS. a cos. b which becomes (~ 8), b) tang. a +- tang. b (1.) tang. (a - tang. a tang. b' ~ 16. Equations (1), (2), (3), and (4) (~ 15), give as follows: sin. (a + b) +- sin. (a-b) = 2 sin. a cos. 4. (1.) sin. (a + b) - sin. (a —b) = 2 cos. a sin. b. (2.) cos. (- cos. ( + ) = 2 cos. a cos. b. (3.) cos. (a - b) - cos. (a + b) 2 sin. a sin. b. (4.) If in these results we substitute p for a + b, and q for a-b, which gives a = - (p + q) and = (p-q), they will become sin. 2p + sin. q =2 sin. 2 (p + q) cos. I (p-q). (5.) sin.p - sin. q= 2 cos. I (p +- ) sin. I (p-q). (6.) cos. q + cos. p =2 cos. - (p + q) cos. ~ (p-q). (T.) cos. q - cos. p = 2 sin. ~ (p + q) sin. ~ (p-q). (8.) 12 TRIGONOMETRY. [CHAP. 1. Dividing (5) by (6), we have sin. p+sin. q sin. 2 (p - q) cos.~ (p - ) sin. p-sin. cos. (p + q) sin. (p-q)' hence, sin. p +-sin. _ sin. ~ (p + ) _sin. I (-q) _ tang. - (p+q) sin. p-sin. cos. (p + q) cos. (p-q) tang. (p-q)' (9.) Numerical Values of Sines, Tangents, &ec. ~ 17. Having deduced a few of the most simple relations of the trigonometrical functions of angles, we will now proceed to determine their numerical values. Sines of very small angles are obviously very nearly equal to the arcs which measure these angles. We have found (~ 6) that the length of an arc of 1' is equal to 0'0002908882, &c. If we regard this as the sine of 1', from which, as will be hereafter shown, it does not differ even in its eleventh decimal figure, we may find the cosine of 1' as follows (~ 8): cos. 1'=V1 - sin. 1'=0'99999995'7, &c. Having thus found the sine and cosine of 1', we may continue our work for larger angles by the aid of equations (1) and (3), (~16), which give sin. (a + b)= 2 sin. a cos. b-sin. (a-b). cos. (a + b) -- 2 cos. a cos. b-cos. (a —b). Putting b=l' and a-=l', 2', 3', 4', &c., successively, we shall obtain sin. 2' =2 sin. 1' cos. 1'-sin. 0'. sin. 3' =2 sin. 2' cos. 1' —sin. 1'. sin. 4' =2 sin. 3' cos. 1' —sin. 2'. sin. 5' = 2 sin. 4' cos..' —sin. 3'. &c. &c. &c. cos. 2' = 2 cos. 1' cos. 1' —cos. 0'. cos. 3' = 2 cos. 2' cos. 1'- cos. 1'. cos. 4' = 2 cos. 3' cos. 1' - cos. 2'. cos. 5' = 2 cos. 4' cos. l' —cos. 3'. &c. &c. &c. ~ 18.] TRIGONOMETRY. 13 Using the above value of sine and cosine of 1', executing the work above indicated, and preserving only. nine decimal places, we find sin. 1'=0'000290888. cos. 1'=0'999999958. sin. 2'=0'000581776. cos. 2'=0 999999831. sin. 3'=0'000872665. cos. 3'=0'999999619. sin. 4'=O0001163553. cos. 4'=O0999999323. sin. 5'=0'001454441. cos. 5'=0'999998942. &c. &c. &c. &c. The work might be continued in this way to any desired extent. ~ 18. We will now seek formulas F which shall give the sine and cosine c of any are without our being obliged first to compute those of a smaller arc. G Denote the arc BC, which measures the angle BAC, by x, the radius being unity. Then assulne, H CD =sin. =a+a x +a2 x +a3 + a4 4+ax5 +, &c. (1.) DA =cos. x -- bo+- bl x +b2 + ba 3+b4x4+- b5+,&c. (2.) When x = 0, the sine of x becomes zero, which requires a0 to = 0. Since the sine of a negative arc (~ 12) is the same as qninus the sine of the same arc taken positively, it follows that the expression (1) for the sine of x must be of such a form as to change the algebraic signs of all its terms when -x is written for +x, consequently it cannot contain any even powers of x. Therefore a2 = 0; a4 = 0, &c. Again, when x = 0, the cosine of x becomes 1, so that bo = 1. And, since the cosine of a negative arc is precisely the same as the cosine of the same arc taken positively, it follows that the expression for the cosine of x must be of such a form as not to change the algebraic signs of any of its terms when -x is written for + x, consequently it cannot contain any odd powers of x. Therefore b, = 0; b3 = 0; b5 = 0, &c. 14 TRIGONOMETRY. [CHAP. i. Equations (1) and (2) must therefore assume the following forms: sin. x = a + a3 x3 + a5 x5 + a7 x7 +, &c... (3.) cos. x -= 1 + b2 +2 + b 4 +- b..6 6 &c. (4.) If we suppose the arc x to be exceedingly small, we may obtain a very close approximate value for the sine of x, in (3); by retaining only the first power of x, we thus find sin. x = a, x, very nearly. But in very small arcs the sine and the arc are very nearly equal; hence, when x is very small, we must have sin. x = x, very nearly. Consequently a, = 1. Using this value of a,, equations (3) and (4) become sin. x =x -+a3x 3 +a5 x' + a7 x7 +-, &c... (5.) cos. =1 + b2 x2 + b4 X4 + +b6 6 +, &c.. (6.) It now remains to find the other coefficients, a3, a3, a7, &c., b2, b4, b6, &c. For this purpose we proceed as follows: In (5) write 2 x for x, and we obtain sin. 2x=2 x +- 8 3 - 32 a5 x5 + 128 a7 x7 +&c. (7.) Dividing (7) by 2, we have sin. 2 x = x + 4 a 3 x +16 as s5 + 64 a7 x7 &c. (8.) Taking the product of (5) and (6), we find sin. x cos. x = x + a3 3 + a + at + b2 -+b a3 X5 + b2 a, 7+, &c. (9.) +b4 + b4 a+ - b6 The left-hand members of (8) and (9) being equal [~ 15, eq. (5)], their right-hand members must also be equal; hence equating the coefficients of like powers of x, we obtain 4 a3 = a3 + b2 16a5=a5+b2 a3+b4 ~ * (10.) 64 a7 = a7 + b2 a, + b4 a3 + b6 &c. &c. ~ 18.1 TRIGONOMETRY. 15 Equations (5) and (6), when squared, become respectively sin.2 x = X2 + 2 C3 X4 + a} } + 45> c.,. (11.) cos.x-=1 +2b2x+ b} ~ +2 b6 }x6+,&c. (12.) + 2 b4 2 b2 b4, If from the sum of (11) and (12) we subtract 1, we shall obtain sin.2x a+cos.2' x1=+ _ +2 } a 2 +a3 4- 2 b2;- X4I-2 &C. +2b 4+25 x6+, &c. (13.) +2 b4 +2 b6 Since the left-hand member of (13) is equal to zero [~ 8, eq. 5 of B], the right-hand member must also equal zero; hence the coefficients of the different powers of x must equal zero, which leads to the following conditions: 1 + 2 b62 =0 2 a3 + b22 + 2 b4 = 0 (14.) a3 + 2 a5 + 2 b6 + 2 b4 = O0 &c. &c. Equations (10) and (14) give immediately these results: b -= I i2- 2 Ix2 a3- 1- 1 2x 3 1 b= S lx2x3x4 a5- 120 x2X3x4x5 1 -- =a - 1 Ix 2 x 3 xx 5x 6 I- 1 a7 = — 04 ix 23 x 4x 5 x 6x 7' &c. &c. &c. Hence, equations (5) and (6) become sin.x=x -- -+ I &C. lx2x3 1x2x3x4x5 lx2x3x4x5x6x7+&c (15) xa2 x4 z cos. =1- + - 4-, &c. (16.) 1x2 1x2x3x4 1x2x3x4x5x6 16 TRIGONOMETRY. [CHAP. I If we divide the right-hand member of (15) by the right-hand member of (16), we shall obtain an infinite series, giving the value of tang. x, as follows: X 2 x 17w7 tang. x -x +- + +~ +, &c. tag + 15 + 315 Other series might be found by combining (15) and (16) for the cotan. x, sec. x, and cosec. x. The are of 1' has already been found [~ 6] to be equal to 0-0002908882086, &c. Putting this value for x, in the above series, we find 1 x 2 0'0000000423079, = O0000000000041. 1 x 2x3 If we confine ourselves to thirteen decimals, we shall obtain sin. 1'= x- =x 0-0002908882045; 1x2x3 cos. 1'= 1 - x 09999999576921. ix2 From this we see that the sine of 1' does not differ from its corresponding arc until after we pass the 11th decimal place, as was asserted under ~ 17. For ordinary purposes, seven decimal places is as accurate as is desired. Having already ( 17) computed the sine and cosine of each minute as far as 5', we will now, by the aid of equations (15) and (16), continue the computation, limiting ourselves to seven decimals, and shall obtain as follows: arc of 6' - 0'0017453, sin. - 0'0017453, cos. = 0'9999985. " 7' = 00020362, " = 0'0020362, "c = 0'9999979. " 8' = 0'0023271, " = 0'0023271, " = 0'9999973. " 9' = 0'0026180, " = 0'0026180, " = 0'9999966. " 10' = 0-0029089, " - 0'0029089, " = 0'9999958. " 11' = 0'0031998, " = 0'0031998, " = 0'9999949. " 12' - 0'0034907, " = 00034907, " = 0'9999939. " 13' = 0'0037815, " = 0'0037815, " = 0'9999928. " 14' = 0'0040724, " = 0-0040724, " = 0'9999917. " 15' = 0 0043633, " = 0'0043633, " = 0'9999905. ~19.] TRIGONOMETRY. 17 In the same way, for degrees, we find arc of 10~ 0'0174533, sin. = 0'0174524, cos. = 0'9998477. " 2~ = 0'0349066, " - 0'0348995, " = 0'9993908. " 3~ = 0'0523599, " = 0'0523360, " = 0'9986295. " 40 = 0'0698132, " =- 0'0697565, " = 0'9975641. " 5~ = 0'0872665, " = 0'0871557, " = 0'9961947. " 60 - 0'1047198, " = 0'1045285, " - 0'9945219. " 7~0 01221730, " = 0'1218693, " = 0'9925462. " 8~ = 0'1396263, " - 0'1391731, " _ 0'9902681. " 9 = 01570796, " - 01564345, " = 0'9876883. " 100 0'1745329, " = 0'1736482, " = 0'9848078. Having in this way computed the sines and cosines, we may [see equations (B), ~ 8] obtain the tangents by dividing the sines by their corresponding cosines; the cotangents will be obtained by dividing the cosines by the sines. The reciprocal of the cosines will give the secants, and the reciprocal of the sines will give the cosecants. The values of the sines, cosines, tangents, and cotangents are given in Table III. CHAPTER II. EXPLANATION OF TABLES. TABLE I. LOGARITHMS. ~ 19. TaIS table gives the logarithms of all numbers from 1 to 10000. For the method of calculating these logarithms, the student is referred to my " Treatise on Algebra," chap. IX. The common logarithm, or Brigqean logarithm, as sometimes called, in honor of the inventor B7riggs, has 10 for its base. Hence, for common logarithms, we have 18 TRIGONOMETRY. [CHAP. II 100=1, that is, 0 is the logarithm of 1 101=10 " " 1 " cc cc " 10 102=00 cc cc " 2 " "' 100 103=1000 c 3 " " " 1000 104=10000, c' 4 " " cc' 10000 105=100000, " " 5 " " " a 100000 &c. &c. &c. If we use negative exponents, we shall have 10-1=0.1,that is, -1 is the logarithm of 01 10-2=0.01 c, c " -2 c" " " " 001 10-3=0.001, " " -3 " " "c 0"001 10-4=00001, " -4 " " " " 00001 10-5=000001, " " -5 c" " "C 0.00001 &c. &c. &c. From the above, we see that the logarithm of any numbesr between 1 and 10 is a proper fraction between 0 and 1. The logarithm of any number between 10 and 100 is 1 plus a prop r fraction. The logarithm of any number between 100 and 100( is 2 plus a fraction. We also see that the logarithm of any number between 1 and 0-1 is some number between 0 and -1; it may therefore be represented by -1 plus a fraction. For a similar reason, the logarithm of any number between 0!1 and 0-01 may be represented by -2 plus a fraction. The logarithm of ally number between 0 01 and 0 001 is -3 plus a fraction. Hence, we see that logarithms of all numbers greater than 10 or less than 1 consist of an integral part, positive or negative, plus a proper fraction. The integral part is called the characteristic, and may readily be assigned by the following RULE. The characteristic of the logarithm of any number greater than 1 is one less than the number of places of fgures which express the integral part of the given number. The characteristic of the logarithm of a decimal fiaction is a negative nuzmber, and is equal to the nuimber of places by which the first significeantfgure is removedfronm the units' place. ~ 21.] TRIGONOMETRY. 19 Thus, we have as follows: Number. Characteristic. Number. Characteristic. 4.. 0 0'4.. -1 48.. 1 0-46.. -1 138.. 2 0375.. -1 2304. 3 005. -2 47.. 0 0'0578.. -2 37'04.. 1 0006.. -3 120'005.. 2 000675.. -3 2178'88.. 3 0'00035.-4 The characteristic being so readily obtained by means of the above rule, has been omitted in the tables, except in the small table on page 1, where are found the logarithms of all numbers from 1 to 100, with their characteristics. But for all the other numbers of our table only the decimal part of the logarithms are given. The decimal part of all logarithms are regarded as positive. The characteristic may, however, be considered as negative. When this is the case, it is usual to place the negative sign immediately over the characteristic. Thus the logarithm of 0'75 is 1'875061, or 0'875061-1. ~ 20. To find, in the table, the logarithm of any number from 1 to 100. Seek for the given number in one of the columns headed N, of the first page of the tables of logarithms, and against it on the right in the next column, will be found the logarithm. We thus find log. 7 = 0'845098, log. 39 = 1591065. ~ 21. To find the logarithm of any number consisting of three places of figures. Seek the given number in one of the left-hand columns headed N, and against it in the column headed 0 will be found the decimal part of the logarithm, observing that if only four figures 17 20 TRIGONOMETRY. [CHAP. II. are given in column 0, they are the last four: the first two are to be taken from the number directly above which contains six ulaces of figures. Thus, we find log. 617 = 2'790285, log. 635 = 2'802774. ~ 22. To;find the logarithm of a number consisting of four 7places of figures. Seek the first three figures in column headed N, then passing horizontally across the page until you reach the column headed with the fourth figure of the given number, and you will have the last four decimal figures of the logarithm. The first two figures are to be taken from the column headed 0; observing, if the four figures found, stand opposite to a row of six figures in the column 0, to use the first two. But if the four figures found are opposite a line of only four figures, you are then to ascend the column till you come to a line of six figures, and then to use the two left-hand figures. The six figures thus obtained will be the decimal part of the logarithm. The characteristic is to be found by the Rule under ~ 19. Thus, we find log. 3657 = 3'563125, log. 7641 = 3-883150. In some portions of the table, in the columns headed 1, 2, 3, 4, &c., the character * is to be found in place of a figure, which indicates that the two figures of the first column, which are to be prefixed, have changed, and are to be taken from the horizontal line directly below. This change of the two left-hand figures of the logarithm is indicated by an asterisk placed in the column 0. The place occupied by this character is to be supplied with a zero. The two figures from the column 0 must also be taken from the line below the asterisk, if the character. has been passed over while crossing the page horizontally. Thus, we find log. 2138 = 3'330008, log. 2298 = 3'361350, log. 3237 = 3'510143. ~ 2:3.] TRIGONOMETRY. 21 ~ 23. To find the logarithm of a number consisting of ftve or more plaees of figures. For example, let us seek the logarithm of 734582. We readily find as follows: log. 734500 = 5'865992 log. 734600 = 5-866051 Diff. of numb. =100 Diff. of logs.=59 The difference between the first number 734500 and the given number is 734582-734500 = 82. Hence, if we suppose the difference of logarithms to be to each other as the difference of their corresponding numbers, which supposition is very nearly correct, we shall have 100: 82: 59: 48'38, for the difference between the logarithm of 734500 and the logarithm of 734582. Hence, we have log. 734500 = 5'865992 + 48'38 log. 734582 = 5'86604-o Had there been only five figures in the given number, that is, only one beyond the fourth place, the first term of the above Iproportion would have been 10 instead of 100; had there been three additional figures beyond the fourth place, the first term would have been 1000. The difference between the logarithms of consecutive numbers of four places of figures is given in the table in a column headed D. Thus by turning to the table, we find 59, our third term of the above proportion, immediately opposite the loga rithm of 7345. From what has been done, we see that we may find the logarithm of a number consisting of more than four places of figures by the following method: Consider all the figures after the fourth as zeros. Then find the decimal part of the logarithm of the number given by the first four figures, observing to give a characteristic for the whole tnumnbe of figures by rule under ~ 19. Take from the column 22 TRIGONOMETRY. [CIHAP. I] D the number which is found directly opposite tihe logarithm already taken out, and multiply it by the figures which were regarded as zeros, pointing off in the product as though these figures were all decimals; add the result thus obtained to the logarithm already found, and it will give the logarithm of the given numiber. In this way we find log. 365365 = 5-562727. log. 635536 = 5'803140. log. 704307 = 5847762. NoTE. —Since the foregoing process of finding the logarithm of a number of more than four places of figures is founded on the supposition that tlhe differences of logarithms of different numbers are to each other as the differences of the numbers, which supposition is not strictly true, it follows that this method can be used only to a limited extent. It ought never to be employed for a number consisting of more than six places of figures. ~ 24. To find the logarithm of a V7ulgar F raction. Since a vulgar fraction is the quotient of the numerator dl vided by the denominator, we may obtain its logarithm by subtracting the logarithm of the denominator from the logarithm of the numerator. Thus, the logarithm of 37 is log. 37-log. 53 = 1-568202- 1'724276 1-843926. In a similar manner we find log. o5 = 1'960233. log. 3 = - 1'908450(. ~ 25. To,fnd the logarithmn of a De,~mal n F'action. Since we have, by the property of logarithms, log. 4'5 =log. -4 -=log. 45-log. 10 =log. 45 —1; log. 36'5 =log. 3 50 =log. 365-log. 100=log. 365-2; log. 3-754 log. 34 5 =log. 3754-log. 1000 =log. 3754-3; it follows that the decimal part of the logarithm of a decimal fraction is the same as though the number was wholly integral, the only difference between the logarithm of a decimal numnber and of the number considered as integral is in the characteristic. Hence, take out the logarithm as though the number were integral, and fix a characteristic according to, rule under ~ 19. 26.] TRIGONOMETRY. 23 We thus find, log. 476 = 2'677607 log. 47'6 = 1'677607 log. 4'76 = 0'677607 log. 0476 = 1'677607 log. 0'9476 = 2'677607 log. 0-00476 = 3'677607 ~ 26. To find the natural number corresponding to any logarithm. Seek in the table, in the column 0, for the first two figures of the decimal part of the logarithm; the other four figures are to be sought for in the same column, or in any one of the columns 1, 2, 3, &c. If the decimal part of the logarithm is exactly found, then will the first three figures of the corresponding number be found in the column N, and the fourth figure will be found at the top of the page. This number must be made to correspond with the given characteristic of the given logarithm by annexing ciphers, or by pointing off decimals. Thus the logarithm 5-311754 gives 205000, " 4'341237 " 21940, 3759970 " 5754, 2'759970 575'4, 1'759970 57'54. When the decimal part of the logarithm cannot be accurately found in the table, take out the four figures corresponding to the next less logarithm. Then for the additional figures, subtract this less logarithm from the given logarithm, and divide the remainder, with naughts annexed, by the corresponding number taken from column D. For example, let us seek the number whose logarithm is 1'234567. We find the next less number to the decimal 0'234567 to be 0'234517, which corresponds to 1716. We also find the number in column D to be 253. Hence 0 of 34 /T/ 0-234567 a 23/ 0-23451T 3 50; and 50 + 253 = 0'198, nearly. So that the number answering to the logarithm 1'234567 is 17'16198, nearly. 24 TRIGONOMETRY. [CHAP. II. ARITHMETICAL CALCULATIONS BY LOGARITHMS. ~ 27. Xzultiplication by Logarithms. Since the logarithm of the product of two or more factors is equal to the sum of their logarithms, we deduce, for lmultiplication by logarithms, this RULE. Add the logarithms of the factors, and the sum w~ill be the logarithm of the product. 1. What is the product of 3'65 by 56'3? log. 3'65 = 0'562293 log. 56'3 = 1'750508 log. of product = 2-312801, which gives 205'495 for the product. 2. What is the product of 7'8 by 353? Ans. 275'34 3. What is the product of 2413 by 0'57? Ans. 1'2141. NoTE.-When any of the characteristics of the logarithms are negative, we must observe the algebraie rule for their addition. 4. What is the continued product of 53'7, 0'12, and 0'004. log. 5'37 =1'729974 log. 0'12 = 1'079181 log. 0'004 = 3'602060 product=0'025776, whose log. = 2'411215 5. What is the square of 3'7; that is, what is the product of 3'7 by 3'7 a Ans. 13'69. 6. What is the cube of 3'8; that is, what is the continued product of 3'8, 3'8, and 3'8? Ans. 54'872. ~ 28. For Divisior by Logarithms, we obviously have this RULE. Subtract the logarithm of the divisor fron the logarithm of the dividend. ~ 29.] TRIGONOMETRY. 25 EXAMPLES. 1. What is the quotient of 365 by 73? log. 365 = 2'562293 log. 7'3 = 0'863323 quotient is 5, its log. = 1'698970 2. What is the quotient of 2'456 by 1-47? Ans. 1'67075, nearly. 3. What is the quotient of 7'4 by 12'58? Ans. 0'588235. NoTE.-When either or both of the characteristics of the logarithms are negative, we must observe the algebraic rule for the subtraction of the one from the other. 4. What is the quotient of 0'378 divided by 0'45? log. 0'378 = 1'577492 log. 0'45 = 1i653213 quotient = 0'84, whose log. = 1-924279 5. What is the quotient of 0'10071 by 0'00373 log. 0'10071 = 1'003072 log. 0'00373 = 3'571709 quotient = 27, whose log. = 1'431363, nearly. ~ 29. Involution by Iogarithlms. Since the exponent denoting any power of any number expresses how many times this number is used as a factor to produce the given power, it follows that the logarithm of any power is equal to the logarithm of the number repeated as many times as there are units in the exponent. Hence we have this RULE. fultip7ly the logarithm of the number by the exponent denoting the power. EXAMPLES. 1. What is the 5th power of 1'234? log. 1'234 = 0'091315 5 multiply. power = 2'86137, whose log. = 0'456575 26 TRIGONOMETRY. [CHAP. Il. ". What is the 7th power of 0'5? log. 0'5 - 1698970 7 multiply. power = 0'0078125, whose log. - 3892790 NoTE.-It must be kept in mind that the decimal part of every logarithm is positive, so that, as in the last example, whatever is to carry to the product of the characteristic by the exponent is positive. 3. What is the 30th power of 107? Ans. 7'6123, nearly. 4. What is the 11th power of 0'11? Ains. 00000000000285313. ~ 30. Evolution by Logarithms. Since the exponent denoting a root indicates that the number is to be separated into as many equal factors as there are units in the exponents, we obviously have this RULE. Divide the logarithm of the number by the number denoting the root. EXAMPLES. 1. What is the 11th root of 11? log. 11 = 1'041393, which divided by 11 gives 0'094672 for the log. of the root. Hence the root = 1'24357, nearly. NOTE.-When the characteristic is negative, and not divisible by the exponent, we must put with it a sufficiently large negative integer to make it divisible, and then connect with the decimal portion of the logarithm an equally large positive number. This will be best illustrated by the following example. 2. What is the 5th root of 0'00567? log. 0'00567 = 3'753583 = 5 + 2753583, which divided by 5 gives 1'550716 for the logarithm of the root. Hence the root = 0'3554, nearly. 3. What is the 3d root of 0'365? Ans. 0'714657, nearly. 4. What is the 7th root of 7? Ans. 1'32047, nearly. 5. What is the 5th root of 0'5? Ans. 0'87055, nearly. ~ 32.] TRIGONOMETRY. 27 ~ 3 1. Proportion by Logarithrms. Since the fourth term of a proportion is found by dividing the product of the second and third terms by the first, we obviously have this RULE. From the sum of the logarithms of the second and third terms, subtract the logarithm of the first term. What is the fourth term of a proportion of which the first, second, and third terms are respectively 0'0146, 4'5, and 1'07? log. 4'5 = 0'653213 log. 1'07 = 0'029384 0'682597 subtract log. 0'0146 = 2'164353 log. of fourth term = 2'518244 Consequently the fourth term = 329'794, nearly. ARITHMETICAL COMPLEMENT. ~ 32. The subtraction of a logarithm fiom the sum of two or more logarithms is usually made to depend upon addition, by making use of its Arithmetical Complement, which is the differ-' ence between 10 and the given logarithm, and is readily taken from the table, by beginning at the left hand and subtracting each figure from 9, except the last significant figure on the right, which must be taken from 10. Now it is obvious, that if instead of subtracting a given number from another, we first subtract the number from 10, and then add the result, we shall, after rejecting 10, obtain the true difference. HIence to work a proportion by logarithms, we may use this second RULE. Take the arithmetical complement of the logarithm of the first term, and the logarithms of the second and and third terms, andfrom the sum of the three reject 10 from the characteristic. 28 TRIGONOMETRY. [CHAP. Ii. EXAMPLES. 1. What is the fourth term of tile proportion 6'5: 30'5:: 1'25? Arith. comp. log. 6'5 = 9'187087 log. 30'5 = 1'484300 log. 1'25 = 0-096910 sum, after rejecting 10, = 0'768297, which is the log. of 5'86539, the fourth term. 2. What is the fourth term in the proportion 3456: 105:: 6543? ar. co. log. 3456 = 6'461426 log. 10'5 = 1'021189 log. 6543 = 3'815777 fourth term = 19-8788, log. = 1-298392 3. In the proportion 1'23: 2'34: 345, what is the fourth term? Ans. 6'56342, nearly. TABLE III. NATTMAL SINES AND TANGENTS. ~ 33. It is necessary to explain the arrangement of this table before explaining that of Table II., since the values of the latter have been obtained from those of the former. Table II., being logarithmic values, is with propriety placed immediately after the table of logarithms. Table III. contains the natural sines and tangents to every minute of the first quadrant, and consequently those of the cosines and cotangents. The first 9 pages are devoted to sines and cosines, the remainder of this table gives the tangents and cotangents. When the number of degrees does not exceed 45, they are to be found at the top of the page, and the additional minutes are given in the first or left-hand column of the page. But if the degrees exceed 45, they are to be found at the bottom of the page, and the additional minutes are then given in the last, or right-hand column of the page. ~ 34.] TRIGONOMETRY. 29 It will be seen by examining this table that the number of degrees at the top of any page added to the number at the bottom of the same, gives 89, to which adding the 60 intermediate minutes makes just 90 degrees. Thus, on page 70 we have 3~1 at the top and 550 at the bottom, the sum of which is 89~. Since the cosine and cotangent (~ 8) of an angle is the same as the sine and tangent respectively of its complement, it follows that by this arrangement of the table those columns which are marked sine and tangent at the top of the page will be marked cosine and cotangent respectively, at the bottom of the page. As the sine and cosine of any angle cannot exceed the radius, which is a unit, it follows that their values are always expressed in this table by a decimal. The tangent and cotangent may in some cases exceed a unit, and will then be expressed by a decimal joined to an integral part. Thus, turning to this table, we find sin. 170 13'1=0'29599; cos. 17~ 13'=0'95519. tang. 17~ 13'=0'30987; cotan. 17~ 13'=3'22715. sin. 55'43'=0'82626; cos. 550 43'=0'56329. tang. 55~ 43'-=146686; cotan. 550 43'=0'68173. TABLE II. LOGARITHMIC SINES AND TANGENTS. ~ 34. This table is constructed by taking the logarithms of the values given in Table III., and increasing each logarithm by 10. The arrangement is the same as given in the Table of Natural Sines, Tangents, &c.; that is, the degrees are to be found at the top of the page and minutes in the first column, when the angle is less than 45~. When the angle is greater than 450, the degrees are to be found at the bottom of the page and the minutes in the last column. As in the last table, the column marked sine at top is cosine at the bottom; the column marked tangent at the top is cotangent at the bottom, and conversely. Obtuse angles are also provided for in this table: when found at the top of the page they are at the right-hand side, and the 30 TRIGONOMETRY. [CHAP. I1 minutes are then to be found in the right-hand column. But when found at the bottom of the page they are at the left-hand side, and the minutes are then to be found in the left-hand column. As most cases of Trigonometry are solved by the assistance of proportions, which are most readily wrought by the help of logarithms, it has been found convenient to prepare beforehand the logarithms of the sines and tangents of all angles expressed by degrees and minutes within the limit of the first quadrant. Since the sines, as well as tangents, of many arcs are less than a unit, the characteristic of their logarithms will be negative. By adding 10 to each logarithm, the characteristics will all be positive, and we shall not be as liable to commit errors in using them, as we should if some had positive characteristics, and others negative characteristics. 3 5. To find, in the table, the logavtthmic sine, cosine, tangent, or cotangent of any given angle. If the angle is less than 450, we seek the degrees at the top of the page and the minutes in the first column on the left; then passing across the page horizontally until we come to the column having the appropriate heading of sine, cosine, tangent, or cotangent, as the case may require, we shall find the logarithm sought. Thus,on page 53, we find sin. 350 53'=9-767999; cos. 35~ 53'= 9'908599. tang. 350 53'=9'859400; cotan. 35~ 53'=10-140600. If the angle is greater than 45~, we seek for the degrees at the bottom of the page, and the minutes on the right-hand side of the page; then, as before, passing horizontally across the page till we reach the column having at the bottom the appropriate designation of sine, cosine, tangent, or cotangent, as the case may require, and the number thus given will be the one sought. If the angle is obtuse, we must make use of its supplement (~ 13). ~ 36. When the angle is expressed in degrees, minutes, and seconds, we proceed as follows: Let it be required, for example, to find the logarithmic ~ 36.] TRIGONOMETRY. 31 sine of 13~ 14' 17". By what has already been said, we readily find sin. 13~ 14 = 9'359678 sin. 13~ 15' - 9'360215 537 = diff. for 1. Hence, 537 -. 60 = 8'95 = diff. for 1", and 895 = diff. for 100". If we suppose the logarithmic sines to increase in the same ratio as the increase of their corresponding arcs, which supposition is very nearly correct, we shall be able to find the difference corresponding to 17" by multiplying 8-95 by 17, which gives 152'15. sin. 13~ 14' = 9'359678 152 add. 9'359830 = sine of 130 14' 17". The column D, immediately at the right of the column of sines, gives the differences of the logarithmic sines for 100". Thus, in the tables opposite the sine of 13~ 14' in the column D, we find 895, which we have already shown to be the increase for 100". In the same way, the column D, immediately after the column of cosines, gives the difference of the logarithmic cosines for 100". The column D, between the tangent and cotangent columns, answers for both, giving also the difference for 100". Hence, to find the logarithmic sine of an are given in degrees, minutes, and seconds, we must first seek the value corresponding to the degrees and minutes; then multiply the corresponding tabular number in column D by the number of seconds. cutting off two places to the right for decimals; add the product to the value already found, and the result will be the logarithmnic sine sought. Thus, find the sine of 37~ 31' 13". Sin. 37~ 31' = 9784612. Tabular number D = 274, which multiplied by 13, the number of seconds, gives 274 x 13=3562; pointing off two decimals, we have 35'62, or 36 nearly, which added gives sin. 37~ 31' = 9'784612 36 sin. 37~ 31' 13"- 9784648 32 TRIGONOMETRY. [CHAP. II The tangent of an are where there are seconds is found by the same method as has been employed for finding the sine. In finding the values of the cosine and cotangent, we must bear in mind that the cosines and cotangents decrease as the arcs increase, consequently the corrections found by aid of columns D must be subtracted instead of being added. Find the cosine of 23~ 6' 47". cos. 23~ 6' = 9'963704 subtract 42'30=-A-4- of 90 (D.) cos. 23~ 6' 47" = 9'963662 Find tangent and cotangent of 30~ 30' 30". tang. 30~ 30' = 9'770148 add 144=,0~- of 481 (D.) tang. 30~ 30' 30"= 9'770292 cotan. 30~ 30' =10'229852 subtract 144= 3,0 of 481 (D.) cotan. 30~ 30' 30" —=10229708 ~ 37. To find the are expressed in degrees, minuutes, and seconds, when its logarithmic sine, cosine, tangent, or cotangent is given. Seek for the given value in the column having the proper designation of sine, cosine, tangent, or cotangent, as the case may require, keeping in mind (~ 34) that the columns designated by sine and cosine at the top of the page are marked cosine and sine respectively at the bottom of the page. Also, those marked tangent and cotangent at the top are respectively cotangent and tangent at the bottom. If the exact value is found in the table, the arc will be accurately given in degrees and minutes. The degrees are to be taken from the top of the page, and the minutes from the lef;hand column, when the proper prfece2 of the column is found at the top. But when the precept is found at the bottom, the degrees are to be taken from the bottom of the page and the minutes from the right-hand column. If the number cannot be exactly found in the table, then find, as above, the degrees and minutes corresponding to the nearest ~ 38.] TRIGONOMETRY. 33 less logarithm. Divide the difference of these logarithms, with two ciphers annexed, by the tabular difference taken from column D, the quotient will give seconds, which seconds are to be added to the degrees and minutes already found, in the case of a sine or tangent, but to be subtracted in the case of a cosine or cotangent. EXAMPLES. 1. Find the arc whose logarithmic sine is 9'365365. 9'365365 = given logarithmic sine. sin. 13~ 24' = 9'365016 349 = difference, and 34900 *- 883 = 39, the number of seconds. Hence, the arc 130 24' 39" corresponds with the logarithmic sine of 9-365365. 2. Find the arc whose cosine is 94142857. 94142857 = given logarithmic cosine. cos. 82~ 11' = 94142655 20200 + 1500 = 13". subtract 13" 82~ 0' 47" = arc required. 3. Find the arc whose tangent is 10'528017. 10'528017 = given logarithmic tangent. tang. 730 29' =10'527931 8600. 772 = 11". add 11" 730 29' 11"= arc required. 4. Find the arc whose cotangent is 9'555555. 9-555555 = given logarithmic cotangent. cotan. 700 14' =9'555536 1900 + 661 = 3", nearly. subtract 3" 70~ 13' 57"= arc required. ~ 38. The secants and cosecants are omitted in this table, 34 TRIGONOMETRY. [CHAP. II. since they are so readily found by the aid of the cosines and sines. By equationf B, ~ 8, we have 1 1 sec. A=; cosec. A= cos. A' sin. A' Clearing of fractions, we have sec.A x cos. A=1; cosec. A x sin. A = 1. Taking the logarithms, and observing to add 10 to each logarithm, we have log. sec. A + log. cos. A=20; log. cosec. A + log. sin. A=20. Hence, log. sec. A = 20 —log. cos. A, log. cosec. A = 20 -log. sin. A. From this we see that Th2e logarithmic secant is found by subtracting the logarithmic cosine from 20; and the logarithnmic cosecant is found by subtracting the logarithmic sine from 20. EXAMPLES. 1. Find the logarithmic secant and cosecant of 41~ 41'. 20' 20sin. 41~ 41'= 9'822830; cos. 41~ 41'= 9'873223. cosec. 410 41'=10'177170; sec. 41~ 41'=10'126777. 2. What is the secant and cosecant of 15~ 15' 15"? 20- 20' sin. 15~ 15' 15"= 9'420123; cos. 15~ 15' 15"= 9-984423. cosec. 15~ 15' 15"=10'579877; sec. 15~ 15' 15" —10'015577. ~ 39. Since we have (equation B, ~ 8) tang. A = cotan. A' it follows that tang. A x cotan. A = 1, and log. tang. A + log. cotan. A = 20. Hence, the logarithmic cotangent may be found by subtracting the logarithmic tangentfrom 20; and, conversely, the logarithmic tangent may be found by subtracting the logarithmic cotangent from 20. ~41.] TRIGONOMETRY. 35 EXAMPLES. 1. The logarithmic tangent of an arc is 9'545454; what is the logarithmic cotangent of the same arc? Ans. 10'454546. 2. The logarithmic cotangent is 9'333444; what is that of the tangent of the same angle? Ans. 10'666556. ~ 40. In any plane triangle there are three sides and three angles, making in all six parts to be considered. Any three of these six parts, provided one at least is a side, being given or known, the other three parts can be found. Thus, in Book I.. Geometry, we have seen that when two triangles had three parts of the one respectively equal to the three corresponding parts of the other, provided the three parts compared were not all angles, the two triangles were identical, or equal in all respects. The case in which three angles of a triangle are respectively equal to three angles of a second triangle, does not lead of necessity to an equality of these two triangles, but simply to their similarity. Hence, when in a triangle three parts, provided one at least is a side, are given, the remaining parts can be found. ~ 41. In the case of right triangles, one angle, that is, the right angle, is always given; consequently two parts in addition to the right angle must be given, one of which must be a side, in order that we may find the remaining parts. The geometrical or graphic method of solving a triangle when a sufficient number of parts are known, has already been exhibited in the Problems at the end of Book Second of Geometry. The method of calculating the numerical values of these parts —which we now proceed to explain-belongs to Trigonometry. 18 36 TRIGONOMETRY. [CHAP. III. CHAPTER III. SOLUTION OF RIGHT TRIANGLES. CASE I. ~ 42. Given, the hypotenuse and o one of the angles, to find the other parts. Suppose the given angle to be A. Then will the angle C be found by subtracting A from 900. A b By our definitions (~ 8), we have sin. A=~p; cos. A which, cleared of fractions, give p = h sin. A... (1.) b = h cos. A. (2.) By using logarithms, we have log. p = log. h + log. sin. A.. (3.) log. b = log.h + log. cos. A.. (4.) As an example, suppose the hypotenuse to be 125, and the angle at A to be 37~ 30'. Required the other parts. h = 125; A = 37~0 30'. We find C = 900-370 30' = 520 30'. Br NATURAL NUMBERS. (TABLB III.) sin. 37~ 30' = 0'60876; cos. 37~ 30' = 0'79335 125 125 304380 396675 121752 158670 60876 79335 76'09500 = p. 99416875 = b. ~ 48.] TRIGONOMETRY. 37 BY LOGARITHMS. log. 125 = 2'096910... 2'096910 log. sin. 37~ 30' = 9'784447; log. cos. 37~ 30' = 9'899467 log. p = 1'881357 log. b = 1'996377 Hence, P = 76'095, and b = 99'169. NOTE.-In adding the log. sin. and log. cos. respectively to the log. 1'25, we rejected 10 from the sum to correct for the 10 which had been added to all logarithms of the sines, cosines, &c. (~ 34). CASE II. ~ 43. Given, a side, that is, ei- C ther the perpendicular or the base, and the angle opposite that side, to l find the other parts. / Suppose the given side to be p, A B and the given angle A. Then will b the angle C be the complement of A. By definitions (~ 8), we have b sin. A=_; cot. A= - h p which readily give sin. A b =p cot. A.. (2.) By using logarithms, we have log. h = log. p-log. sin. A.. (3.) log. b = log. p + log. cotan. A.. (4.) For example, suppose the perpendicular to be 75, and the angle at A to be 39~ 15'. Required the other parts. P = 75; A = 39~ 15' We find C = 900-39 15' =- 50~ 45'. 38 TRIGONOMETRY. CHAP. III. BY NATURAL NUMBERS. (TABLE III.) sin. 39~ 15' = 0'63271; cotan. 39~ 15' = 1'22394 0'63271 ) 75.00000 ( 118.53 = A. 63271 117290 1'22394 63271 75 540190 611970 506168 856758 340220 91'79550 = k 316355 238650 189813 48837 BY LOGARITHMS. log. 75 = 1'875061... 1875061 log. sin. 39~ 15' = 9'801201; log. cotan. 39' 15' = 10'087760 log. A = 2'073860 log. b = 1'962821 Hence, A = 118'54, nearly, and b- = 91'795, nearly. NOTE.-In this example, before subtracting the log. of the sine of 390 15' from the log. of 65, it was necessary to increase this logarithm by 10, since the logarithm of the sine had been increased by the same quantity. In the second operation, after adding the log. of the cotangent to the log. of 75, we rejected. 10 frown the sum, for the reason assigned in the last note. CASE III. ~ 44. Given, a side, and the angle adjacent that side, to find the other parts. Suppose the given side to be b,/ and the given angle A. Then will the angle C be the complement A of A. ~ 44.] TRIGONOMETRY. 39 By definitions (~ 8), we have cos. A = -; tang.A = which give b cos. A.) p = b tang. A.... (2.) Using logarithms, we have log. h = log. b-log. cos. A... (3.) log. p = log. b +log. tang. A... (4.) For example, suppose the base to be 90-5, and the angle at A to be 50~ 30'. Required the other parts. 6 = 90 5; A = 50~ 30'. We find C = 90~- 50~ 30' = 390 30'. BY NATURAL NUMBERS. (TABLE III.) cos. 50 30' = 0-63608; tang. 50~ 30' = 1'21310. 0'63608) 90'50000 ( 142-27 = 63608 268920 1'2131 254432 90'5 144880 60655 127216 109179 176640 109'78555 =P. 127216 494240 445256 48984 BY LOGARITHMS. log. 90'5 = 1'956649.... 956649 log. cos. 500 30' = 9'803511; log. tang. 50~ 30' = 10'083896 log. A = 2153138 log. p= 2a040545 Hence, A = 142'27, and p = 109'78. 40 TRIGONOMETRY. [CHAP. HL CASE IV. ~ 45. Gi'ven, the hypotenuse and c one of the sides, to find the other parts. Suppose the given side to be b, then (~ 8) we have sin. C = cos. A- (1. or, using logarithms, log. sin. 0 = log. cos. A =log. b - log. h.. (2.) Again, from the fundamental property of the right triangle, we have ba2 +p2 = h2, hence _ = hS - b' = (h + b) (A- b), p = ~(h + b) (A-b)... (3.) Using logarithms, we have log. p = - [log. (A + b) + log. (A - b)].. (4.) For example, suppose the hypotenuse to be 112, and the base to be 97. Required the other parts. = 112; b = 97. 112) 97'00000 ( 086607 = sin. C= cos. A. 896 740 672 680 672 800 784 16 Hence (by Table mI.), C = 60~ and A = 30~. Again, h + b = 209, and A - b = 15. ~ 46.] TRIGONOMETRY. 41 Extracting Square Root. 209 5 3135 (55'99 =p. 15 105 25 1109 1045 11189 635 209 525 3135 11000 9981 101900 100701 199 BY LOGARITHMS. log. 97 = 1986772 log. 112 = 2049218 9'937554 =log. sin. C =log. cos. A. HIence, C = 60~ 0' 19"; A = 29~ 59' 41'. Again, log. 209 = 2'320146 log. 15 = 1'176091 2) 3496237 log.p = 1'748118 Hence, p = 55'991. CASE V. C ~ 46. Given, the two side, to find the other parts. A B By ~ 8, we have tang. A = cotan.. C (1- or, using logarithms, log. tang. A = log. cotan. C = log. p-log. b. (2). 42 TRIGONOMETRY. [CHAP. III. The hypotenuse is immediately deduced from hid = b2 +p2, which gives = v/b +p.... (3.) Since this last formula is not well adapted to the use of logarithms, we will remark that after having found the angles, by means of equations (1) or (2), we may then determine the hypotenuse by Case II., equation (1), which gives log. I7 = log. -log. sin. A... (4.) As an example, suppose the base to be 83-5 and the perpendicular to be 62-25. Required the other parts. b = 83-5; p = 62-25. BY NATURAL NUMBERS. (TABLE III.) 83-5 ) 62-25 ( 074550 = tang. A = cotan. C. 5845 3800 Hence, A = 36~ 42' 3340 C = 530 18' 4600 4175 4250 4175 750 62-25 =p 83-5 = b 62-25 83-5 31125 4175 12450 2505 12450 6680 37350 6972-25 = bl p2 = 3875-0625 b2 = 6972-25 10847-3125 b2 +_2.'/10847-3125 = 104-15 = k. ~ 4'.] TRIGONOMETRY. 43 BY LOGARITHMS. log. 62-25 = 1'794139 log. 83'5 = 1'921686 log. tang. A = log. cotan. C = 9-872453 Hence, A = 36~ 42' 18"; C = 530 17' 42". Again, log. 62'25 = 1'794139 log. sin. 36~ 42' 18" = 9'776480 log.A = 2'01'7659 Hence, h 104'15. ~ 47. Additional Examples of Right Triangles. 1. Given, the hypotenuse, equal 365, and one of the acute angles, equal 33~ 12', to solve the triangle. The other angle =56~ 48'. Ans. Thesides 199'86. 305-42. 2. Given, one of the sides equal to 33'33, and the angle opposite this side = 83~ 33', to solve the triangle. The other angle = 6~ 27. Ans. The other side = 3'768. The hypotenuse = 33'542. 3. Given, one of the sides equal to 105'5, and the acute angle adjacent this side equal to 46~ 3', to solve the triangle. Ans i The other angle = 430 5'. Ans. ~ The other side = 109'44. The hypotenuse = 152'01. 4. Given, the hypotenuse equal 111'1, and one of the sides equal 3'75, to solve the triangle. The other side = 104-58. Ars. The angles = 190 43' 36". 70~ 16' 24". 5. Given, the two sides equal 29'37 and 37'29, to solve the Ams. TThe angles - 380 13' 28". 510~ 46' 32" The above examples have all been wrought by the use of logarithms, which is, as a general thing, more simple than by 44 TRIGONOMETRY. [CHAP. IV. the use of natural numbers only. It will be well, however, for the pupil to apply both methods, as was done in the examples which followed immediately after each case; by so doing, he will be the better prepared to comprehend the true nature of the trigonometrical values as well as their logarithmic values. ~ 48. Hereafter, when we use the word sine, cosine, tangent, or cotangent, in connection with logarithmic calculations, we wish to have it understood as meaning the logarithms of those values, increased by 10, as given in Table II. In the usual analytical investigations of trigonometrical formulas, we make use of their real or natural values, as given in Table III. CHAPTER IV. SOLUTION OF OBLIQUE TRIANGLES. ~49. ALL the different cases of oblique triangles may be solved by the aid of the following theorems: THEOREM I. The sides of any plane triangle are to each other as the sines of their opposite angles. And conversely, the sines of the angles of any plane triangle are to each other as their opposite sides. In the triangle ABC denote c the sides opposite the angles b A, B3 C, respectively by the letters a, b, c. Draw CD per- / pendicular to AB, and we A C ID shall have (~ 8) CD CD sin. A- —, sin. B =; a dividing the first equation by the second, we have sin. A a sin. B b' (1 which gives a: b:: sin. A: sin. B.. (2.) In the same way it may be shown that a: c:: sin. A: sin. C... (3.) ~ 49.] TRIGONOMETRY. 45 THEOREM II. In any plane triangle, the sum of any two sides is to their difference as the tangent of half the sum of the opposite angles is to the tangent of half their difference. By Theorem I., we have b: c:: sin. B: sin.; C; B hence, by composition and division, we have b+c: b- c: sin. B + sin. C: sin. B - sin. CO, which gives b+c _ sin. B +sin. b-ce sin. B- sin. (1 By equation (9), ~ 16, we have sin. B + sin. _ tang. I (B + C) sin. B - sin. C tang. a (B — C) Hence, we have b + c tang. I (B- +C) b - c tang. - (B- C)' (2.) or, b+c: b - c:: tang. (B+C): tang.~(B-C). (3.) THEOREM III If from an angle of a plane c triangle a perpendicular bo drawn so as to meet the oposite\ side or base, the whole base will be to the sum of the other two sides as the difference of those sides is to the difference of the A/ B segments of the base. Drawing CD perpendicular to AB, we have ba = AD2 + CDa2. e (1.) a2 = BD2 + CD'... (2.) 46 TRIGONOMETRY. [CHAP. IV Subtracting (2) from (1), we have 2 -_ a2= AD2-BD2, or, ( + a) (b - a) = (AD + BD) (AD- BD),. (3.) which, converted into a proportion, will give, observing that AD + BD = c, e: b+a:: b-a: AD-BD.. (4.) If the triangle has an obtuse angle, then the perpendicular should be drawn from this obtuse angle, otherwise it will not meet the opposite side. The theorem might be so modified as to embrace this case; but as it is used only in the solution of a triangle when the three sides are given, such modification is unnecessary. ~ 50. The solution of all oblique triangles may be included in four cases, as follows: CASE I. Given, a side and two angles, toficd the other parts. Since the suim of the three angles of any plane triangle is equal to 1800, c the third angle may be found by subtracting the sum of the two given angles from 1800. Having all the angles, the two remaining sides may be found by Theorem I. Thus, if we suppose the side c to be given, we A B shall have sin. C: sin. A:: c: a, sin. C: sin. B:: c: b, which give c sin. A sin.C-' * (1.) c sin. B sin. (.) Using logarithms, we have log. a = log. c + log. sin. A-log. sin. C.. (3.) log. b = log. c + log. sin. B -log. sin. C.. (4.) ~ 50.] TRIGONOMETRY. 47 By using arithmetical complements (~ 32), we shall have log. a= ar. co. log. sin. C + log. sin. A + log. c. (5.) log. b = ar. co. log. sin. C + log. sin. B + log. c. (6.) As an example, suppose the angle A = 30~ 20', the angle B = 500 10', and consequently the angle C = 990 30'. If the side c is 93'37, what will be the lengths of the other sides? Using equations (5) and (6), we have ar. co. sin. 99~ 30' (80~ 30') = 0'005997.. 0'005997 sin. 30~ 20' = 9703317; sin. 50~ 10'=9'885311 log. 93'37 = 1'970207.. 1'970207 log. a=1-679521 log. b=1'861515 Hence, a = 47-81; b = 72'697. NoTE.-Since the angle C -= 990 30' is an obtuse angle, we subtract it from 180, and obtain its supplement 800 30', the sine of which is 9-994003. Now subtracting this from 10, we have 0'005997 for the arithmetical complement of the logarithmic sine of 990 30'. The arithmetical complement may be readily taken from the table, by beginning at the left hand, and subtracting each figure from 9, except the last significant figure on the right, which must be subtracted from 10. (~ 32.) CASE II. Given, two sides and an angle opposite one of them, to find the other parts. c c b b A A A c D B c B D Suppose the sides a and b to be given, and the angle A opposite the side a. By Theorem I., we have a: b:: sin. A: sin. B, orh sin. A sin. B = sin.. (1.) Using logarithms, we have log. sin. B = ar. co. log. a + log. b + log. sin. A. (2.) Having thus found the second angle, the third angle at C may be obtained by subtracting the sum of these two from 180~. 48 TRIGONOMETRY. [CHAP. IV Then by Theorem I., we have sin. A: sin. C' a: c, a sin. C or, sin.A(3) which by logarithms, becomes log. c = ar. co. log. sin. A + log. sin. C + log. a. (4.) ScioLIuM. In equation (1), the numerator, b sin. A, is the perpendicular CD; now if a is less than this perpendicular, 6 sin. A' the value of the fraction, which gives the sine of B, will exceed a unit, which is impossible. But while a is greater than this perpendicular, and does not at the same time exceed 6, there will be two solutions, as represented in the two diagrams. When there are two solutions, the angle at B in the second triangle will be the supplement of the angle at B in the first. As an example, suppose the side a = 75'5 and the side b = 98'5; and suppose the angle at A to be 370 37', required the other parts. By equation (2), we have ar. co. log. 75'5 = 8'122053 log. 98'5 = 1'993436 log. sin. 370 37 = 9'785597 log, sin. B = 9'901086 Hence, B = 520 46' 48", consequently the angle at C= 890 36' 12". Now, by equation (4), we have ar. co. log. sin. 370 37' = 0'214403 log, sin. 890 36' 12" = 9'999989 log. 75'5 = 1'877947 log. c = 2'092339 and c= 123'69. The above solution corresponds with the first figure; if we take 1270 13' 12" for the angle at B, which is the supplement ~ 50.] TRIGONOMETRY. 49 of 520 46' 48", we shall then find 150 9' 48" for the angle at C. Then, by equation (4), we have ar. co. log. sin. 37~ 37' = 0'214403 log. sin. 15~ 9' 48" = 9'417590 log. 75-5 = 1'877947 log. c = 1'509940 and c = 32'355. This second solution corresponds with the second figure. CASE III. Given, two sides and the included angle, to Jind the other parts. Suppose the sides b and c, with the included angle A, to be given. By subtracting the angle A from b V 1800, which is the sum of the three angles, we have B +C =1800-A. and 4 - (B + C) = 90 ~- A = complement of i A. Then by Theorem II., we have b + c: b-c: cotan. ~ tang. I}(B —C), which gives tang. I (B - C) cotan. (1) Using logarithms, we have log. tang. 1 (B- C) = ar. co. log. (b + c) + log. (b-c) + log. cotan. A A.... (2.) Having thus found half the difference of the angles B and C, we add it to half their sum, and thus obtain the greater angle, which must be opposite the greater side. The smaller angle is found by subtracting half their difference from their half sum. Again, having found all the angles, we have, by Theorem I., sin. B: sin. A:: b: a, or, b sin. A a sin. B 50 TRIGONOMETRY. [CHAP. IV Using logarithms, we have log. a = ar. co. log. sin. B + log. sin. A + log. b. (4.) As an example, suppose the side b = 50'24, and the side c = 43'25, and suppose the angle A to be 40~ 15', required the other parts. A = 40~ 15'; b = 50'24; c = 43'25, consequently, b+c = 9349; b-c- =699; A = 207'30". Equation (2) gives ar. co. log. 93-49 = 8-029235 log. 6'99= 0'844477 log. cotan. 200 7' 30" = 10'435993 log. tang. - (B-C) = 9'309705 And (B - C) = 11 31' 55", (B + C) = 690 52' 30/ B = 81~0 24' 25" = sum. C = 580 20' 35" = difference. By equation (4), we have ar. co. log. sin. 58~ 20' 35" = 0'069965 log. sin. 40~ 15' = 9'810316 log. 43"25 = 1'635986 log. a = 1-516267 And a = 32'829. CASE IV. Given, the three sides, to o find the angles. b If we conceive CD to be drawn perpendicular to the side c, we shall have, by The- A D B orem III., c: b+a:: b-a: AD-BD, AD-BD = (b +a) (b-a). Using logarithms, we have log. (AD-BD)=ar. co. log. c+log. (b +a) + log. (b-a). (2.) ~ 50.] TRIGONOMETRY. 51 This gives half the difference of the segments AD and BD; if we add this half difference to the half sum, that is, add it to half of c, we shall obtain AD, the greater segment. The half difference subtracted from the half of c will give BID, the shorter segment. Then solving the two right triangles ADC and BDC, by Case IV. of Right Triangles, we have AD BD cos. A=D; cos. B =-. a Using logarithms, log. cos. A = log. AD —log. b,.. (3.) log. cos. B = log. BD —log. a... (4.) Having found the angles A and B, subtract their sum from 180~ for the third angle C. For example, suppose we have a= 50'25; b= 605, and o = 68'4. Required the angles. b + a = 110'75; b-a = 10'25. By equation (2), we have ar. co. log. 68'4 = 8'164944 log. 110'75 = 2'044344 log. 10-25 = 1-010724 log. (AD —BD) = 1'220012 AD-BD = 16.596 c + (AD-BD) = 42.498 = AD. 2 c- (AD-BD) = 25.902 = BD. By equations (3) and (4), we have log. 42'498 = 1'628369; log. 25'902 = 1.413333 log. 60'5 = 1'781755; log. 50'25 = 1701136 log. cos. A = 9s846614; log. cos. B = 9'712197 Hence, A = 450 22' 35"; B = 58~ 58' 18". Consequently, C = 750 39' 7". Perhaps a better method for this case would be as fol lows: 52 TRIGONOMETRY. [CHAP. IV If to equation (1), which is AD-BD = b — awe add AD + BD = c, we shall have 2 AD + a hence, AD b2 + c2- a cos. A=- 2 (5.) Adding 1 to each member of (5), we have b2 + C2 -A a2 b2 + 2 bc + ac- a 2 1+cos. A=1+ -2 c be 2 be or, 1 + cos. A = (b + c + a) (b + c- a) 2. (6.) Denoting half the sum of the three sides by s, we have b+c+a=2s; b+c —a=2s-2a=2(s-a). Hence equation (6) will become 1 + cos. A = 2 s (s- a) (7) be By equation (8), ~ 15, we have 1 + cos. A = 2 (cos. ~ A)2. Therefore equation (7) will become (cos. A)= 2 s (s - a) 2 (cos. A be or, (cos. 1 A)2 = (s - a) ~2 " be consequently, COS. 1 A= V/S (8- (a) (8.) be( In a similar manner, we find cos. ~ B= - a (9.) Cos. C - ) (0.) 2 (ab ~ 50.1 TRIGONOMETRY. 53 By using logarithms, equations (8), (9), and (10) become log. cos. A=: [ar. co. log. b+ar. co. log. c-+log. s-log. (s-a)] log. cos.B= [ar. co. log. a+ar. co. log. c+log. s-log. (s-b)] (A) log. cos. I C=4 [ar. co. log. a+ar. co. log. b+log. s+log. (s-c)] J Applying equations (A) to the case already given, we have a=50'25; b=6Q'5; c=68'4; consequently, ( + b + c) = s 89'575, and s —a-= 39'325; s-b = 29'075; s- - 21'175. ar. co. log. b 8'218245; ar. co. log. a = 8-298864 ar. co. log. c = 8-164944 ar. co. log. c = 8-164944 log. s = 1'952187 log. s= 1'952187 log. (s-a))= 1-594669 log. (s-b) = 1-463520 2) 19'930045 2) 19-879515 log. cos. A = 9=965022 log. cos. -1 B = 9'939757 ar. co. log. a = 8'298864 ar. co. log. b = 8-218245 leg. s= 1'952187 log. (s-c) = 1-325823 2 ) 19.795119 log. cos. ~ C= 9'897559 Hence 2 A = 220 41' 17"; 4 B = 290 29' 9"'; C-370 49' 34", and A = 450 22' 34:"; B = 580 58' 18"; C- 750 39' 8". When the sides of a triangle are expressed in small numbers, equation (5) may be advantageously employed: this equation, when the letters are properly permuted so as to apply to each angle, gives b2 +- C _ aea Nat. cos. A -= b -a B aa a- ca _ b Nat. cos. 2a = a.. (B.) Nat. cos. C= a+b 2 2 ah' 54 TRIGONOMETRY. [CHAP. IV As an example, suppose the sides to be 5, 7, and 9. Required the angles. a-=5; b=7; c=9. 49 +- 81-25 105 Nat. cos. A = 2 x 7 x 9 126' 25 + 81-49 57 2x5x9 90' 25 +49-81 -7 Nat. cos. C = 25 -7 7 2x x7 70' log. 105 = 2-021189 log. 57 = 1-755875 log. 126 = 24100371 log. 90 = 1-954243 cos. A = 9-920818 cos. B = 9-801632 log. -7 = 0-845098n log. 70 = 1-845098 cos. C = 9'000000n Hence, A = 33~ 33' 27"; B = 500 42' 13"; C = 95~ 44' 21". NOTE.-In taking the logarithm of -7, we took it as though it had been +7, and at the right annexed the letter n to indicate that the number is negative. So in the value of cos. C we write the n, thus indicating that the natural cosine of this angle is negative. We then seek in the tables for the angle as though its cosine was positive, and find 840 15' 39", which taken from 1800 gives the obtuse angle 95~ 44' 21" for the true value of C. ~ 51. Additional Examples of Oblique Triangles. 1. Given, two angles of a triangle equal 47~ 13' and 63~ 36', and the side opposite the first angle equal 27-5, to solve the tri angle = 69~ 11'. A ( The other sides = 35-0246. 35'024. 2. Given, two sides of a triangle equal 47'13 and 63'36, and the angle opposite the first side equal 27~ 50', to solve the triangle. The other side = 92 - 19..First As. The other angles = 138 1' 3". The other side = 19341. Second Ans. The other angles = 1410 7' 13". ~ 51.] TRIGONOMETRY. 55 3. Given, two sides of a triangle equal 49'5 and 101'5, and the angle opposite the first side equal 30~ 25', to solve the triangle. Ans. The question is impossible. 4. Given, two sides of a triangle equal 630 and 800, and the angle opposite the first side equal 100~, to solve the triangle. Ans. The question is impossible. 5. Given, two sides of a triangle, 77'5 and 90'25, and the included angle equal 830 38', to solve the triangle. The other side = 112'25. Ans. The other angles = 43~ 19' 38".,t~~ ~ 53~ 2' 22". 6. Given, the three sides of a triangle, 40, 50, and 60, to find the angles. 410 24' 35". Anrs. The angles =. 550 46' 16". 82~ 49/ 91" 7. Given, two sides of a triangle, 100 and 90, and the angle opposite the first side, 1110 15', to solve the triangle. The other side = 21-823. Ans. The other angles = 110 44' 7" 57~ 0' 53". 8. In an obtuse-angled triangle one of the acute angles is 290 15', the other acute angle is 2~ 45'; the side opposite the first angle is 6'3. What are the other parts? ( The obtuse angle = 148~. Ans. The other sides = { 0619. 6'832. 9. Given, two sides of a triangle, 332'21, 237'61, and their contained angle equal to 720 29' 48", to find the other parts. As. | The other angles 40 6659 35". Ans. 660 301 36~ The other side = 345'46. 10. The three sides of a triangle are 10, 15, and 20. What are the angles? a 280 5' 18". Anes. 46~ 34' 3" 104~ 28' 39". 11. The three sides of a triangle are 123'48, 135-61, 140'91. What are the angles? 530 0' 10". Ans. 610 17' 50". 650 42' 0 56 TRIGONOMETRY. [CHAP. V 12. Two sides of a triangle are 59'34 and 12'135 and their ineluded angle is 1500 38'. What are the other parts? 240 30' 12". Ans. Angles-= 4:0 51' 48". Side =70165. CHAPTER V. SPHERICAL TRIGONOMETRY. ~ 52. SPHERICAL TRIGONOMETRY treats of the methods of computing the unknown parts of a spherical triangle, when certain parts are given. But before proceeding to the investigation of the relation of the different parts of a spherical triangle, we will give some GENERAL TRIGONOMETRIC FORMULAS. ~53. Under ~ 15 and ~ 16 we have already given several general and useful formulas which were needed in Plane Trigonometry. We now propose to continue these formulas to a greater extent, giving such additional ones as are needed in Spherical Trigonometry, as well as others which will be found convenient in the reduction and simplification of trigonometric expressions in general. Dividing (10) by (11), ~ 15, we have sin. a a/1- cos. 2a - tan.a=... (1.) cos. a V/1 +cos. 2a Multiplying both the numerator and denominator of the fraction constituting the right-hand member of (1) by the numerator, we have 1 —cos. 2a 1-cos. 2a tan. a = - =- s.. (2.) /1 - cos. 2a sin. 2a If we multiply both the numerator and denominator of the right-hand member of (1) by the denominator, we shall obtain /1- cos.2 2a sin. 2a tan. a =. (3.) 1 + cos. 2a 1 + cos. 2a' ( ~ 53.] TRIGONOMETRY 57 Taking the reciprocals of (2) and (3), we have sin. 2a 1 +- cos. 2a4 cot. a- = (4.) 1- cos. 2a sin. 2a( We have already, ~ 15, deduced tan. a + tan. b tan. (a-+6) = 1- tan. a tan. (5.) By a similar process, we find tan. a - tan. b tan. (-b) 1 + tan. a tan. b' Reciprocating, we have 1- tan. a tan. b cot. (a - b) - tan. a + tan. b(7 1 - tan. a -tan. b cot.(a - 6)= tan. a - tan.b. (8.) If we take b = a in (5) and (7), we shall have 2 tan. a tan. 2a = (9 1- tan.2 a cot. 2a = 2 -tana. (10.) 2 tan. a If we take a = (p + q), equation (5) of ~ 15 will become sin. (p + 2) = 2 sin.~ (p + q) cos. (p + q) (11.) Using this in connection with (5), (6), (7), and (8) of ~16, we readily obtain, by division and reduction, as follows: sin.p + sin. q tan. ( + )(12.) 2~V (12.) sin.p - sin. q tan. ( p-q)' cos.p + cos. q _ cot. I (p + q) cos. q - cos.p tan. I (p - )' sin.p + sin. q - cos. - (p - q). sin. (p + q) cos. ( (p ) + () sin.p - sin. g sin. I (p - q). 2Z YI. (15.) sin. (p + q) sin. 2 ( + )' sin.p + sin. = tan. (p + q); cos.p + cos. q 58 TRIGONOMETRY. [CHAP. V sin.p + sin. q = cot. (p - q);. (1.) cos. g - COS.do sin. p - sin. g = tan. 1 (p - -);. (18) cos.2 + cos. q sin. - sin. = cot. (p + ). (19.) cos. q - cos.p The following are readily obtained, and are of frequent use: sin. (p - q) = cot. q - cot.p;. (20.) sin.p sin. q cos. (y =k g) _cot. cot. g = 1;* sin.p sin. q sin. (p g) = tan. P - tan. q; (22.) COS.(.p ). (25.) cos.p cos. cos. (P ) = 1 q: tan. tan.; (23.) Cos.P cos. g sin. (p - q) = cot. v tan. q; sm.(P i f)-1 zt ota.ptan~q;.. (24.) sin.p cos. q cos. (p = q) = cot.p qF tan. q; (25.) sin.p cos. g ~ 54. We now proceed to investigate the relation of the several parts of a spherical triangle. We shall confine our attention to such triangles only as have been treated of in Book VIII. of Geometry, namely, those whose sides and angles are each less than 180~. ~ 55. To express the cosine qf an angle of a spherioal triangle in terms of the sines and cosines of the sides. Let ABC be a spherical triangle E described upon the surface of a sphere whose centre is at O. We will denote the angles of this triangle by the large letters A, B, C, and the sides opposite them by 0 the corresponding small letters a, b, c. A Draw AD tangent to the arc AC, and AE tangent to AB. ~ 55.] TRIGONOMETRY. 59 Then will the spherical angle A be equal to the angle DAE. (Geom., B. VIII., T. V.) Draw the radius OC and produce it to meet AD in D; also, draw the radius OB and produce it to meet AE in E. The radius OA being drawn, will be perpendicular to AD, AE. Now, by Plane Trigonometry, equation (5), Case IV., ~ 50, the triangle DAE gives DE2 = AD2 + AE2 - 2AD x AE x cos. DAE. (1.) The triangle DOE, in like manner, gives DE2 = OD2 + OE2 - 20D x OE x cos. DOE. (2.) Equating the right-hand members of (1) and (2), and dividing each term by the square of the radius of the sphere, we have AD2 AE2 AD AE O A + -2 - 2 x OA x O x cos. DAE GA2 ~A2 GA OD2 OE2 OD OE (3.) = OA —— + OA- 2 x OsX-Ax cos I0E. - OA2+OK GA GA J Now, since the triangles OAD, OAE are right-angled at A (Geom., B. II., T. V., C. I.), we have [(A), ~ 8] AD AE AD = tan. AOD = tan. b; =A tan. AOE -tan. c; GA OA OD OE O OAsec. AOD = sec. b; OA sec. AOE = sec. c. Hence (3) gives, by observing that the angle DAE = A, and that the angle DOE = a, tan.2 b + tan.2 c - 2 tan. b tan. c cos. A (4.) = sec.2 b + sec.2 c - 2 sec. b sec. c cos. a. Transposing, and observing that sec.2 b - tan.2 b and sec.2 ctan. c are each equal to 1, we have 2 + 2 tan. b.tan. c cos. A = 2 sec. b sec. c cos. a. (5.) Dividing by 2, and substituting for the tangents and secants their values in terms of sines and cosines [(B), ~ 8], we have sin. b sin. c 1 1 + - x ~ x cos. A= x x cos. a, cos. b cos. c cos. b cos. c which, cleared of fractions, becomes cos. b cos. c + sin. b sin. c cos. A = cos. a.. (6.) 60 TRIGONOMETRY. [CHAP V. Hence, Cos. a - Cos. b cos. c cos. A = sin. b sin. c By permuting, c B = cos.bcos. c cos. cos. a. sin. c sin. a cos. c - cos. a cos. b COs. C sin. a sin. b We shall not again have occasion to refer to a diagram for the purpose of obtaining any new relation of the respective parts of a spherical triangle, but shall be able to draw from equation (6) all other relations which may be required. Hence, (6) may be regarded as one of the fundamental relations of Spherical Trigonometry. ~ 56. To express the cosine of a side of a spherical triangle, in terms of the sines and cosines of the angles. Let A, B, C, a, b, c, be the angles and sides of a spherical triangle, and A', B', C', a', b', c' the corresponding quantities of its polar triangle. Then, by (A), ~ 55, we have Cos. a' - Cos. b' cos. c' cos.A'=-.. Al =.. (7.) sin. b' sin. c' Now we have these relations (Geom., B. VIII., T. VII.): A'=180~ —a; B'=180~ —b; C'=180~-c; a' = 180~-A; b' = 180~ - B; c' = 180~ - C; hence, (7) becomes cS. (1800 - a) = cos. (180~ - A) - cos. (180~ - B) cos. (180~ - C) sin. (180 - B) sin. (180~ - C) which gives cos. A cos. A + B cos. C ] cos.a= sin.B sin. C and by permuting, cos. B + cos. C cos. A sin. C sin. A cos. C + cos. A cos. B COS. e sin. A sin. B By the application of this property of the Polar Triangle, any formula of a spherical triangle may be changed into a similar ~ 57.] TRIGONOMETRY. 61 formula, where the sides will take the place of angles, and the angles will take the place of sides. This principle of change may be thus stated: F]or the sides write the supplements of the opposite angles, and for the angles write the supplements of the opposite sides. ~ 57. To express the sine of an angle of a spherical triangle, in terms of the sines of the sides of the triangle. By (A) ~ 55, we have cos. a - cos. b cos c cos. A= sin. b sin. c Therefore -1+ cos. A = cos. a - cos. b cos. c + sin. b sin.c cos. a- cos. (b+c) sin. b sin. c sin. b sin. c 2 sin. I (a + b + c)sin. 1 ( —a + b + c) sin. b sin. c. [See (8),'16. If s=-,(a+b+c) we shall have s-a=~ (-a+b+c); (s-b)=!-(a-b+c); s-e =(a+b-c). We therefore have 2 sin. s sin. (s- a) sin. b sin. c (1.) By a similar process we find cos. b cos. c+sin. b sin. c-cos. a cos. (b —c) — cos. a 1 —cos. A= = sin. b sin. c sin. b sin. c'2 -b). (-+c) sin. (a+b-c) _2 sin. (s sin.in. (s-c) sin. b sin. c sin. b sin. c Taking the product of (1) and (2) we have ~1- _cos.2 A= Sin.2 A= 4 sin. s sin. (s-a) sin. (s-b) sin. (s —c) sin. b sin. c and sin.A sin. sin. sin.ssin. (s-a)sin.(s- b)sin. (s-c) } By permuting, sin.B =sin. sin. s sin. (s-a)sin.(s- b)sin.(-c) } ( sin. C = sin. sin. 2 s sin. (s-a) sin.(s-b) sin. (s-c) 2 sia-i.- 62 TRIGONOMETRY. [CHAP V. By (1) we have 1+cos. A=2 cos.2 A=2 sin. ssin. (s-a). 2 sin. b sin. c consequently, cos. -~ A = { sin. s sin. (s —a) ~. sin. b sin. c And permuting, COS. B = in. sin. s sin(s-b) ( sin. csln.a a sin. ssin. (s-c) i I cos. I C = 2 2COS* C { sin. a sin. b By (2) we have 1-cos. A = 2 sin.2. A _ 2 sin. (s —b sin. (s —c) sin. b sin. c Consequently, (in., {sin. (s-b)sin.(s-c); ~~2 -msin. b sin. c And permuting, _ _ (C".) sin. B = sin. (s-c)sin.(s-a) } sin. c sin. a sin. sn a sn C sin. (s-a) sin. (s-b) 1 sin. a sin. b J Finally, dividing the expressions (C") by those of (C') we have tan. A= {simn (s —) sin. (s-c) } sin. s sin. (s -a) tan.~B = sin.(s-e) sin. (s-a) (C"') sin. s sin. (s-b) J tan. ~ C = { sin. (s-a) sin. (s-b) I 2sin. ssin. (s-c) J ~ 58. To express the sine of a side of a spherical triangle, in terms of the sines and cosines of the angles. By (B) ~ 56, we have cos A+ cos. B cos. C cos. sn.Bsn. sin. B sin. C ~ 58.1 TRIGONOMETRY. /13 consequently, cos..A+cos. Eos. C+sin. B sin. C cos. A+cos.(B-C) 1+ cos. a= sin. B sin. C sin. B sin. C 2cos. (A+B- C) cos. (A-B + C) sin. B sin. C If S= ~(A+B+C) we shall have S-A= (- A+B+C); S-B= (A-B+C); S-C= (A+B-C). We therefore have 1+cos~a= 2 cos. (S- -B) cos. (S- C) sin. Bsin.C (1.) By a similar process we find cos. B cos. C - sin. B sin. C + cos. A 1 -cos. a= sin. Bsin. C cos. (B+C)+cos. A 2 cos. -(A+B+C) cos. I (-A+B+C) sin. B sin. C sin. B sin. C 2 cos. S cos. (S-A) sin. B sin.C (2.) Taking the product of (1) and (2), we have 2 * 2 4 cos. S cos. (S- A) cos. (S- B) cos. (S-C) 1-cos. a=si. a=- sin.2 B sin.2 C And sin. sinsin. B in C -os. S cos.(S-A) cos. (S-B) cos.(S-C) } 2; Perniutin'g, sin.b=si-. Ci A g-cos.Scos.(S —A)cos.(S-B)cos.(S —C) }2 (D) sin.. - -cos. S cos. (S-A) cos. (S-B) cos. (S-C) 2 i sin.CAsin.B J By (1) we have 2cos.2 cos. (S - B) cos. (S - C) sin. B sin. C Consequently, cos.1a{ cos. (S -B) cos. (S- C) }; 1 cOS. 2 sin. C sin. A cosb= { cos. (S - C) cos. (S - A)}L (D') cos. ~ b = si2 0 co c { cos. (S-A) cos. (S-B) } 64 TRIGONOMETRY. [CHAP. V By (2) we have - cos. 2 sin. a 2 cos. S cos. (S-A) 1 -cos. a -= 2 s.in2 ~a -- -- sin. B sin. C Consequently, i t.1.={ {-cos. S cos. (S-A) } (D. {_c o sin. B sin. C; sin. {- 2; (D".) 2ssin. sin. -{ cos. S cos.(S-C) } j 2 n. ~c ~sin. A sin. B' Finally, dividing the expressions (D") by those of (D'), we have tan.a= {i- cos. S cos. ((S - I) 2;1 t lan. cos. (Sj- C) cos. (S — h (D"'.C) - cos. S cos. (S - B~ tan.c= j c Gr oup s marked (D), (D'), (D"), and (D"B) might have been Groups marked (D), (D'), (D"), and (D"') might have been immediately deduced from those marked respectively (C), (C'), (C"), and (C"'), by the application of the principle of the Polar Triangle, as indicated under ~ 56. Since the negative sign precedes cos. S in the expressions under the radicals of (D), (D"), and (D"') it might, at first view, be supposed that these values were under an impossible form. It is however easily shown that this is not the case. For (Geom. B. VIII., T. XV.) we know that the sum of all the angles of a spherical triangle is greater than two right angles, and less than six right angles. Consequently, A +CB C > 1800 and < 540~. And - (A+B+B) orS> 90~ and<270~. lience, the cosine of S is always negative, and -cos. S is therefore always positive. Again, if a', b', c' be the three sides of the polar triangle, since the sum of any two sides of a spherical triangle is greater ~ 60.] TRIGONOMETRY. 65 than the third side (Geom. B. VIII., T. I), we have b'+ c'> a'; that is, 1800-B + 180~- C> 180~-A; consequently, -A+ B + C < 180~, and 2(-A+B + C)<90~; hence, cos. (S-A) is always positive, and in like manner cos. (S-B), cos. (S-C), are always positive; hence the expressions (D), ( (D"), (D"') are in every case possible. ~ 59. The sines of the angles of a spherical triangle are to each otlher as the sines of their opposite sides. Expressions (C) immediately give by division, sin. A sin. c sin. a sin. a.' sin.-B sin. bsin.c sin. b' sin. A sin. a sin. b sin. aE.) sin. C sin. b sin. c sin. c sin. B sin. a sin. b sin. b sin. C sin. c sin. a sin. c ~ 60. 0To express the tangent of the sum and difference of two angles of a spherical triangle, in terms of the sides opposite to these angles, and the third angle of the triangle. By (A) ~ 55, we have from the third expression, cos. c = cos. a cos. b + sin. a sin. b cos. C.. (1.) Substituting this value of cos. c in the first expression of (A), we find cos. a - cos. a cos.2 b - cos. b sin. a sin. b cos. C cos. A - sin. b sin. c cos. a (1- cos.2 b) - cos. b sin. a sin. b cos. C sin. b sin. c cos. a sin. b - cos. b sin. a cos. C (2 sin. c If we substitute the value of cos. c, given by (1), in the second expression of (A), we shall obtain cos. b sin. a - cos. a sin. b cos. C ( cos. B3 =.. (3.) sin. c 66 TRIGONOMETRY. [CHAP. V. Adding (2) and (3), we have sin. a cos. b - cos. a sin. b- (sin. a cos. b+ cos. a sin. b) cos. C cos.A+cos. B=sin. c sin. (a + b) - sin. (a + b) cos. C sin. (a - b) (1 - cos. C) (4.) sin. c sin. c Again, we have, by the first equation of (E), sin. A sin. a sin. B sin. b' consequently, sin. A i sin. B sin. a i: sin. b sin. B sin. b Hence, sin.B sin. C sin. A + sin. B=(sin. a sin. ) in. b=(sin. a sin. b) sin.' (5.) Dividing (5) by (4), using first the positive sign, we have sin. A + sin. B sin. a + sin. b sin. C cos. A + cos. B- sin. (a + b) 1- cos. C; hence, by equations (16), (14), and (4) of ~ 53, this becomes tan. a(A + B) =cos 2i ) x cot. I C. (6.) In a similar manner, by dividing (5), with the negative sign, by (4), we find sin. A -sin. B sin. a - sin. b sin. C cos. A + cos. B sin. (a + b) 1- cos.C' which, by equations (18), (15), and (4) of ~ 53, becomes tan. ~ (A- B) =sin. ~ (7a-b) tan. -1 (A - o. _ 2 x cot.-~ C. ~ 61.] TRIGONOMETRY. 67 Equations (6) and (7) and their permuted values give as follows: tan. I (A + B) = c ot. ~ a cos. I-( + b) x cot. ~ C; tan. ~ (A-B) - sin. (a -) tan ( - B)= + cot. C. sin.z (a +b) tan. (B + C) cos. (b - c)cotA. cos. I (b + c) ot. sin. 1 (b - c) tan. ~ (B - C) = 2 x cot. IA. 2i tasin. -~(b + c) cos. ~ (c a) i tan. (C - A) = sin. 1 ( a) sin.I (c) - a) If we convert equations (1) and (2) of (F) into proportions, we have cos. I (a + b): cos. } (a - b)":: cot. I C tan. ~ (A + B), sin. I (a + b): sin. (a - b):: cot. I C tan. I (A - B); which proportions are known as NXapier's first and second Analogies. They are used in solving a triangle when two sides and the included angle are given. ~ 61. To express the tangent of the sum and difference of two sides of a spherical triangle, in~ terms of the angles opposite to them, and the third side of the triangle. Let A, B, C, a, b, c be the angles and sides of a spherical triangle, A', B', C', a', b', c', the corresponding parts of the polar triangle; then by (F), we have cos. ~ (a' - / cot. tan. 1 (A' + B')-= x cot. C' 2 cos. 1 (a'~ + 2 Therefore, tan. [(180~-a)+(180 -b)] cos. [(180~- A)(180~-B)] t.(180-c). cos. j [(180~ -A)+(180~ -B)] This becomes tan. (a+b) cos. (A - B) os. (a ( + B) x tan... (.) 20 68 TRIGONOMETRY. [CHAP. V. Similarly we have tan. (A' -B') = in (a' - b) x cot. - C'. t -- ]' = sin. 1 (a' + b') Therefore, tan. [(180~-a) —(180~-b)] sin. [(180~-A) -(180-c- B)] which becomes sin. I (A - B) tan. I (a - b)= i (A + B) x tan c. c. (2.) Equations (1) and (2), together with their permuted values, give as follows: -cos. (A -- V ~ctanos. I (A + B) x tan. 1 c; ltan. (a - b) sin. I (A- B) tan. (b+c) = 2(( + B x tan. Ic. 2 ta n. I (b + c) = 2a. (B + C). tan. (b + c) = co) x tan. 2 a ) cos. ~ (B - C) l tan. (c a) = (c + A) x tan. I a tan. I (c,- a)= sin.-( - tA) i l The first and second conditions of (G) being converted into proportions, gives Napier's third and fourth Analogies, as follows: cos. I (A +B): cos.~(A - B):: tan. c: tan. (a +b), sin. (A + B): sin. (A - B):: tan. c: tan. (a - b). These analogies are employed in the solution of a triangle when two angles and the interjacent side are given. ~ 62. To express the cotangent of an angle of a spherical triangle, in terms of the side opposite, one of the other sides, and the angle included between these two sides. Equation (2) of ~ 60, when cleared of fractions, becomes cos. A sin. c = cos. a sin. b - cos. b sin. a cos. C.. (1.) ~ 64.] TRIGONOMETRY. 69 By (E) we have sin. A sin. c = sin. C sin. a... (2.) Dividing (1) by (2), we find cot. A = cot. a sin. b cosec. C -cos. b cot. C.. (3.) By interchanging B and C, b and c, (3) will give cot. A = cot. a sin. c cosec. B - cos. c cot. B.. (4.) Proceeding in like manner for the other angles, we shall obtain the following group: cot. A = cot. a sin. b cosec. C - cos. b cot. C 1 = cot. a sin. c cosec. B - cos. c cot. B; cot. B = cot. b sin. c cosec. A - cos. c cot. A = cot. b sin. a cosec. C - cos. a cot. C; cot. C = cot. c sin. a cosec. B - cos. a cot. B = cot. c sin. b cosec. A - cos. b cot. A. These six equations are of such a nature that they do not yield any new relations by the application of the _principle of the Polar Triangle (~ 56). ~ 63. By aid of the eight groups of formulas designated by (A), (B), (C), (D), (E), (F), (G), (H), we shall be enabled to solve all the cases of spherical triangles, whether right-angled or oblique-angled. We shall, in the next chapter, proceed to apply these formulas. ~64. Before, however, passing to the solution of spherical triangles, we will deduce some other general formulas of s8perics, which will frequently be found useful in Astronomy. Equation (6) of ~ 55, becomes, by permuting, cos. cos. a + sin. c sin. a cos. B = cos. b... (1.) Multiplying (6) of ~ 55, by cos. c, it becomes cos. c cos. a = cos. b cos.2 c + sin. b sin. c cos. c cos. A. (2.)'Subtracting (2) from (1) and substituting sin.2 c for 1 - cos.2e, we have sin. c sin. a cos. B = cos. b sin.2 c - sin. b sin. c cos. c cos. A; which, divided by sin. c, becomes sin. a cos. B = cos. b sin. c - sin. b cos. c cos. A. (3.) 70 TRIGONOMETRY. [CHAP. VI Interchanging B, C, and b, c, (3) becomes sin. a cos. C = cos. c sin. b - sin. c cos. b cos. A. (4.) Equations (3) and (4) with their permutations, give sin. a cos. B = cos. b sin. c - sin. b cos. c cos. A; sin. a cos. C = cos. c sin. b - sin. c cos. b cos. A. sin. b cos. C = cos. c sin. a - sin. c cos. a cos. B; sin. b cos. A = cos. a sin. c - sin. a cos. c cos. B. sin. c cos. A = cos. a sin. b - sin. a cos. b cos. C; sin. c cos. B cos. b sin. a - sin. b cos. a cos. C. J If we apply the principle of the Polar Triangle to equations (I) they become sin. A cos. b = cos. B sin. C + sin. B cos. O cos. a; sin. A cos. c = cos. C sin. B + sin. C cos. B cos. a. sin. B cos. c = cos. C sin. A + sin. C cos. A cos. b; sin. B cos. a = cos. A sin. C + sin. A cos. C cos. b. s sin. C cos. a = cos. A sin. B + sin. A cos. B cos. c; sin.C cos.b =cos. B sin. A + sin.B cos. A cos.c. CHAPTER VI. SOL'UTION OF SPHERICAL RIGHT TRIAiNGLES. 65. We shall confine ourselves to such triangles as have only one right angle; those that have two or three right angles will be considered hereafter. (See ~ 74.) A spherical triangle consists of 6 parts, 3 sides and 3 angles; and any three of these being known the others may be found. In the present case, one of the angles being a right angle, it follows that any other two parts being known the other three may be found. The number of combinations of 5 things taken 3 and 3 at a 5x4x3 time, is =10; therefore, ten different cases present themselves in the solution of right-angled spherical triangles. 66.] TRIGONOMETRY. 71 ~66. The solution of these ten cases may all be comprised in two rules first given by Napier, and known under the name of NXaper's Rulesf br Circular Parts. OF THE CIRCULAR PA.RTS. The right angle is not taken into consideration. The two sides, the complements of the two angles, and the complement of the hypotenuse constitute the five circular parts. Let ABC be a spherical triangle, rightangled at C. If we arrange the five circulating parts upon the circumference of its circumscribed circle, we shall readily { discover which are adjacent, and which a are opposite, when either of the five parts A is chosen as the middle part..O.-. NA-PIER'S RULES. I. Tke sine of the middle part is equal to the product of the tangents of the adjacent parts. II. The sine of the middle part is equaGo to tAe product of the cosines of the opposite parts. If now we take in turn each of the five parts as the middle part, and apply these Rules, we shall obtain ten formulas, as follows. 90 - -B First. Let a be the middle part, then B\ will b and (90~-B) be the adjacent parts, and (90~- c) and (90~- A) the opposite / parts. A RULE L sin. a = tan. b tan. (90 — B) = tan. b cot. B. (1.) RULE II. sin. a = cos. (900~-c) cos. (900~-A) = sin. c sin. A. (2.) Secondly. Let b be the middle part, then will a and (90~- A) be the adjacent parts, and (90~- c) and (90~ — B) the opposite parts. RULE I. sin. b = tan. a tan. (90' - A) = tan. a cot. A. (3.) RULE II. sin. b = cos. (900 - c) cos. (90 — B) = sin. e sin. B. (4.) 72 TRIGONOMETRY. [CHAP. VI Thirdly. Let (90~- c) be the middle part,. then will (90~- A) and (90~- B) be the adjacent parts, and a and b the opposite parts. RULE I. sin. (900 — c) = tan. (90~ —A) tan. (90~- B), or cos. c = cot. A cot. B.... (5.) RULE II. sin. (900- c) = cos. a cos. b, or cos. c=cos. a cos. b. (6.) Fourthly. Let (900-A) be the middle part, then will b and (90 — c) be the adjacent parts, and a and (900- B) the opposite parts. RULE I. sin. (900 - A) = tan. b tan. (90- c), or cQs. A = tan. b cot... (7) RULE II. sin. (900~- A) = cos. a cos. (90~- B), or cos A = cos. a sin. B... (8. Fifthly. Let (90~-B) be the middle part, then will a and (90~- c) be the adjacent parts, and b and (90~ — A) the opposite parts. RULE I. sin. (90~ -B) = tan. a tan. (90 — c), or cos B = tan. a cot. c... (.) RULE II. sin. (90~-B) = cos. b cos. (90~ —A), or cos.B = cos. b sin. A... (10.) Collecting these ten results, we have sin. a = tan. b cot. B,.. (.) = sin. c sin. A... (2.) sin. b = tan. a cot. A,.. (3.) = sin. c sin. B... (4.) cos. c = cot. A cot. B.. (5.) = cos. a cos. b. (6.) cos. A= tan. b cot. c,.. (7.) = cos. a sin. B... (8.) cos. B= tan. a cot. c,.. (9.) = cos. b sin. A... (10.) Since the acute angles A and B, as well as their opposite sides a and b, admit of being interchanged, it follows that there are in reality only six distinct formulas in the group (hK). Thus (1) and (2) are immediately changed into (3) and (4) respectively. Also (7) and (8) give, by this change, (9) and (10) respectively. ~ 67.] TRIGONOMETRY. 73 Hence, (1), (2), (5), (6,) (7), and (8) give all that is required for the solution of all spherical right triangles. That these six formulas, as drawn from lYapier's Rules, are correct, may be shown as follows: Equation (4) of (H) ~ 62, gives, when C = 90~, cot. B = cot. b sin. a, which immediately reduces to sin. a = tan. b cot. B, which is (1) of (K). Equation (2) of (E) ~ 59, gives, when C = 90~, sin. aoi sin. A =.', or sin. a = sin. c sin. A, which is (2) of (K). Equation (3) of (B) ~ 56, gives, when C = 90~, cos. A cos. B COs' c =sin. A sin. B' or cos. c=cot. A cot. B, which is (5) of(K). Equation (3) of (A) ~ 55, gives, when C=90~, cos. c - cos. a cos. b 0= sin, a sin. b' or cos. c=cos. a cos. b, which is (6) of (K). Equation (6) of (H) ~ 62, gives, when C=90~, 0 = cot. c sin. b cosec. A-cos. b cot. A, or cos.A=tan. b cot. c, which is (7) of (K). Equation (1) of (B) ~ 56, gives, when C=90~, cos. A cos. a -sin. -B or cos. A = cos. a sin. B, which is (8) of (K). Having proved the formulas (K), which were derived from Napier's two Rules, we are at liberty to use them in solving the six cases of spherical right triangles, or we may in all cases make the direct application of those Rules. ~67. That we may, in the foregoing formulas, distinguish the trigonometric functions of parts less than 90~ from those greater than 90~, we must pay particular attention to the rule for the algebraic signs. Since the sine of an angle and the sine of its supplement are identical, it follows that there will be two solutions when the required part is determined by means of its sine, unless the ambiguity can be removed by the application of the following propositions: 74: TRIGONOMETRY. [CHAP. VL PROPOSITION I. In a spheric pright triangle, an angle and its opposite side are always both less than 90~, or else both greater than 90~. To establish this, take equation (8) of (K), cos. A = cos. a sin. B, which immediately gives cos. A sin.B... (1.) COS. a Now, since B is less than 1800, its sine must be positive, hence the fraction which is equal to sin. B must also be positive; therefore the numerator and denominator must be either both positive or else both negative-that is, A and a must be either both less than 90~, or else both greater than 90~. PROPOSITION 11. When the two sides of a spherical triangle including the right angle are both less or both greater than 90~, the hypotenuse will be less than 90~. But when one side is less than 90~ and the other greater than 900, the hypotenuse will be greater than 90~. This may be established by equation (6) of (K), which is cos. c = cos. a cos. b. If a and b are both less than 90~, their cosines will be positive; if they are both greater than 90~, their cosines will be negative, and in both cases the product of their cosines, which gives cos. c, will be positive; consequently e will be less than 90~. If a and b are in different quadrants-that is, if one is less than 90~ and the other greater than 900, the product of their cosines will be negative; consequently c will be greater than 90~. CASE I. ~ 68. Given the hypotenuse and one angle. For example, suppose c and A given. B To find a. Equation (2) of (K) immediately gives a sin. a = sin. c sinA. A C There will be two values of a correspond ~ 68.] TRIGONOMETRY. 75 ing to the same sine. We must, ~ 67, use the one which is in the same quadrant as A (Proposition I.). To find b. Equation (7) of (K) readily gives tan. b = tan. c cos. A. Tofind B. Equation (5) of (K) readily gives cot. B cos. c tan. A. The quadrants in which b and B are situated are determined by the algebraic sign of tan. b and cot. B, ~ 67. EXAMPLES. 1. Given c = 1040 30', A = 75~ 15'. To solve the triangle, log. sin. c - 9985942 log. sin. A = 9'985447 log. sin. a = 9'971389 Therefore a = 69~ 25' 48", or 1100 34' 12". By Prop. I., under ~ 67, it follows that a must be in the same quadrant with A; consequently we must take 690 25' 48" as the value of a. The supplement of this is not applicable to this triangle. log tan. c =10587342n log.cos.c = 9'398600n log. cos. A = 9'405862 log. tan.A = 10'579585 log. tan. b = 9'993204n log. cot. B= 9'978185n b = 135~ 26' 54". B = 133~ 33' 42". In the above work, since c is an arc greater than a quadrant, its tangent and cosine are both negative, hence at the right of their logarithms we have placed the letter n. (See Note under Case IV., ~ 50.) Now, since the addition of logarithms corresponds to multiplication, and the subtraction of one logarithm from another corresponds to division, it follows that when both logarithms, in the case of addition or of subtraction, are derived from negative values —that is, when both logarithms have the letter n —the result must be positive, and its logarithm will not be marked with n. But when one of the logarithms is distinguished with n and the other is not, the resulting logarithm is also to be marked with n, and the corresponding numerical value will be negative. Our results for the logarithms of tan. b and cot. B above are both marked with n, consequently their numerical values are both negative. We seek in the Table in the usual manner for the arcs whose logarithmic tangent and cotangent are respectively 9'993204 and 9-978185, and find them to be 44~ 33' 6" and 46~ 26' 18"; we then take the supplements of those arcs for the values sought. Or, we may at 76 TRIGONOMETRY. [CHAP. VI. once take out of the Tables the obtuse angles, since our Tables have been so arranged as to give the supplementary angles in all cases. By this arrangement, the logarithmic work with obtuse angles is performed with the same ease as in the case of acute angles. 2. Given c= 65~ 5', A = 480 12', to solve the triangle. a = 420 32' 20". Ans.,b = 550 7' 32". B = 64~ 46' 13". CASE II. ~69. Given the hypotenuse and a side: for example, c and a. To find b. Equation (6) of (K) gives cos. b = CS.o a/_ cos. a To find A. Equation (2) of (K) gives c sin. a sin. A= sin. c We must take A in the same quadrant as a, ~ 67, Prop. I. To find B. Equation (9) of (K) is cos. B = tan. a cot. c. EXAMPLES. 1. Given c= 94~ 5', a = 1000 45', to solve the triangle. cos. c = 85852525 n sin. a = 9-992311 tan. a = 10'721576 n cos. a= 9-270735n sin.c = 9-998896 cot. c = 8-853628n cos. b = 9'581790 sin. A = 9-993415 cos. B= 9-575204 b= 67~ 33' 26", A = 990 57' 8", B = 670 54' 47". In this example, since c and a are each greater than 90~, their cosines, as used in finding the side b, are negative; this we have indicated by the letter n at the right of their logarithms. (See Note under Case IV., ~50.) Also their tangents as well as cotangents are negative, and the letter n indicates this in the work for finding B. Since the sines of all angles not exceeding 180~ are positive, it follows that the logarithms of the sines of a and c, as used in finding A, do not require to be distinguished by the letter n. But the value of A being given in a sine leads to two arcs 80~ 21 52", 99~ 57' 8", supplements of each other. Prop. I., ~ 67, requires us to use the obtuse angle for the value of A. ~ 70.] TRIGONOMETRY. 77 2. Given c = 66~ 32', a = 37~ 48', to solve the triangle. 1 -9~44'13" Ans. A = 41~ 55' 34". B = 700 18' 42". CASE III. ~ 70. Given one side and its opposite angle: for example, a and A. Equations (3), (2), and (8) of (K) readily c give sin. b = tan. a cot. A, sin. a A A' sin. = sin. A' sin. B = cos. A c cos. a\ A A In this case there are always two solutions. And we also remark, see Proposition I. under ~ 67, that a and A are both B less than 90~, as in firstfigure, or else both greater than 90~, as in second figure. If AB and AC be produced to meet at A', we shall have ABA' and ACA' each a semi-circumference, and the angle A =A'. The two triangles ABC, A'BC both have the same given parts a and A = A', but the required parts, b', c', B', of the second triangle are respectively the supplements of b, c, and B of the first triangle. EXAMPLES. 1. Given a = 110~ 4, A = 98~ 30', to solve the triangle. tan. a = 10'437364n sin. a = 9'972802 cos. A = 9'169702 n cot. A = 9'174499 n sin. A= 9'995203 cos. a = 9'535438 n sin. b = 9'611863 sin. c = 9'977599 sin. B = 9'634264 b7 J 240 9'1 1"> C W71~ 45' 19", B 250 31' 3" 1550 50' 59". 1080 14' 41". 154~ 28' 57". 78 TRIGONOMETRY. [CHAP. VI By paying attention to the Propositions under ~ 67, we shall have for the two solutions as follows: = b 240 9' 1" (b = 155 50' 59") Ans. c =-108~ 14'41" or c = 71~ 45' 19" B = 25~ 31' 3" (B = 1540 28' 57" 2. Given a = 50~ 12', A = 75~ 30', to solve the triangle. b = 18 5' 0") b = 1610 55' 0" Ans. c = 52 31' 10" or c = 127~ 28' 50" B = 23~' 35" B = 156~ 58' 25" CASE IV. ~ 71. Given one side and its adjacent angle: for example, a and B. Equations (1), (9), and (8) of (K) readily give tan. b = sin. a tan. B, a cot. c = cot. a cos. B, A cos. A = cos. a sin. B. EXAMPLES. 1. Given a = 101~ 30', B = 42~ 15', to solve the triangle. sin. a = 9'991193 cot. a - 9'308463 n cos. a = 9'299655 n tan. B = 9'958247 cos. B = 9'869360 sin. B = 9'827606 tan. b = 9'949440 cot. c = 9'177823 n cos. A = 9'127261 n b = 410 40' 21", c=98033' 52", A = 970 42' 13". Since the logarithms of cot. c and cos. A have the letter n, indicating that the cot. c and cos. A are negative (see remarks under Case II.), it follows that we must take from the Tables the arcs which are greater than 90~. We thus find b = 41~ 40' 21". Ans. c = 98~ 33' 52". A = 97~ 42' 13". 2. Given a = 48~ 30', B = 400 20', to solve the triangle. b = 320 27' 10". Ans. c =56~ 0'13". A = 64~ 36' 15". ~ P73.] TRIGONOMETRY. 79 CASE V. ~ 72. Given the two sides, a and b. Equations (6), (3), and (1) of (K) readily B give Cos. c = Cos. a cos., a cot. A = cot. a sin. b, A cot. B = sin. a cot. b. EXAMPLES. 1. Given a = 75 15', b = 1200 15', to solve the triangle. cos. a = 9'405862 cot. a = 9'420415 sin. a - 9'985447 cos. b = 9'702236 n sin. b =9'936431 cot. b =9'765805n cos. c = 9'108098 n cot. A = 9'356846 cot. B - 9.751252 n c= 970 22' 9", A = 770 11' 14", B = 119~ 25' 17". 2. Given a = 120~, b = 30~, to solve the triangle. c =115~ 39' 32". Ans. A =106~0 6' 8". B - 33~ 42' 36". CASE VI. ~ 7 3. Given the two angles, A and B. Equations (8), (10), and (5) of (K) readily give cos. A cos. a - cosn. B' cos.B a o. sin. A' cos. c = cot.Acot.cot.. EXAMPLES. 1. Given A = 62~ 15', B =- 56 30', to solve the trangle. cos. A = 9'668027 cos. B = 9'741889 cot. A - 9'721089 sin. B = 9'921107 sin. A = 9'946937 cot. B = 9'820783 cos. a = 9.746920 cos. b = 9.794952 cos.e =9-'541872 a=56~ 3' 25", b = 510 24' 56", c = 690 37' 14". 80 TRIGONOMETRY. [CHAP. VI 2. Given A = 99. 30', B = 1500 15', to solve the triangle. (a = 109~ 25' 39". Ans.. b = 1510 40' 30". c = 72~ 58' 30". ~ 7 4. When a spherical triangle has two or even three of its angles right, it is then isosceles, and may be divided into two equal right triangles, each having only one right angle (except the unique case of three right angles in the given triangle), and the solution may be obtained by the foregoing cases. When all the three angles are right, the sides are all equal, and each a quadrant. ~ 75. The foregoing relations of spheric right triangles may be associated with the corresponding relations of plane right triangles as follows: In plane right triangles. In spheric right triangles. B 1B A b C b b sin. a sin. b Sin A= a; sin. B. Sin. A= -; sin. B c c sin. c sin. c b a tan.b tana Cos. A = -; cos. B Cos.A tacos. B c tan. c tan. e' a b tan.a tan.b Tan. A =-; tan. B-. Tan. A -= tan. B = b' a sin.b' sin. a Sin. A = cos. B; sin: B = cos. A Sin. A= co sin. B o cos. b' cos. a Cos. c = cos. a cos. b, or,2 = a12 + b'. ~ log. cos. c = log. cos. a + log. cos. b o 1 - cot. A cot. B. Cos. c = cot. A cot. B. ~VW.] TRIGONOMETRY. 81 CHAPTER VII. SOLUTION OF SPHERICAL OBLIQUE TRIANGLES. CASE I. ~ 76. When two sides and the included angle are given. For example, suppose a, b, and C to be given. Using formula (F), and wri a ting B —A for A-B, and A B b - a for a - b, since b is greater than a, and consequently B > A, the greater angle being opposite the greater side, we have tan. 1 (B + A)= 2( cot. C; cos. -(b + a) 2 sin. i (b + a) cot. tan.I (B - A)= cot. a C.2 u sin. (b - a) 2 ct ~. Knowing the tangents of (B + A) and I (B - A), we readily find from the Tables 2 (B + A) and I (B - A), and consequently, by addition and subtraction, we have B and A. The first equation of (G) gives cos. I (B + A) tan. 1 c- (B A) tan. (b + a). t2 cos. (B+A - A)~ And, in like manner, if any two sides and the included angle be given, the remaining parts may be found. EXAMPLES. 1. Given a=-70~ 35'; b = 120~ 15'; C = 540 40, to find the remaining parts. (b + a) = 95~ 25'; (b - a) 24~ 50'; 2 C = 27~ 20'. cos.(b - a)= 9957863 sin. 1 (b - a)= 9-623229 ar. co. cos. co. os. (b + a)= 1025038n ar. co. sin. (b + a)= 0 001944 cot. ~ C =10'286614.... 10286614 tan. I (B+A)=11 269515n tan. 1 (B-A)= 9'911787 ~(B+A)=930 4' 39". (B-A)=390 13' 14". 82 TRIGONOMETRY. [CHAP. VII. Consequently, A = 53~ 51' 25"; B = 132~ 17' 53", observing that the greater angle is opposite the greater side (Geom., B. VIII., T. XIV.). cos. 4 (B+A)= 8-729865n ar. co. cos. (B-A)= 0 110856 tan. I (b + a)=11'023094n tan. Ic = 9'863815 ~c= 36~ 9' 37", and c = 72~ 19' 14". Collecting results, we have. A= 53~ 51' 25". Ans. B=132~0 17' 53". c = 72 19' 14". In this case there can be no ambiguity —only one solution can be obtained. 2. Given a = 100~ 50'; b = 990 30'; C = 68~ 40', to find the remaining parts. A=970 51' 59'". Ans. B=950 52' 59". ( c =67~ 27' 10". CASE II. ~ 77. When two angles and 0 the interjacent side are given. b a For example, suppose A, B, A c to be given. By formula (G) we have tan. 2 (b +a) ctan. 2 C, 2' COS. I (B + A) 2 sin. 1 (B - A) tan. - (b-a) - s ~ (B + A) tan. I c. sin. 4 (B + A) t e Having found a and b by the above, we use the first equation of formula (F), which readily gives cot. 2D C * 2c (b + a) tan. I (B + A). And, in like manner, if any two angles and the interjacent side be given, the remaining parts may be found. ~ 17.] TRIGONOMETRY. 83 EXAMPLES. 1. Given c = 110~; A = 84~; B = 88~, to determine the remaining parts. (B + A) = 86~; (B -A) = 2; ~ c = 55o. cos. ~(B-A)= 9'999735 sin. ~(B-A)= 8'542819 ar. co. cos. (B+A)= 1'156415 ar. co. sin. (B+A)= 0'001059. tan. c =10'154773... =10'154773 tan. (b + a-)=11-310923 tan. ~(b - a)= 8'698651 (b + a) = 8'7 12' 7". (b - a) = 20 51' 37". Consequently, a = 84~ 20' 30"; b = 90~ 3' 44". Observing that the greater side is opposite the greater angle (Geom., B. VIII., T. XIV.) cos.~(b + a)= 8'688563 ar. co. cos. I (b - a)= 0'000541 tan. I (B+A)=11'155356 cot. ~ C = 9'844460 ( C = 55~ 2' 51", and C = 1100 5' 42". Collecting, we have a= 840 20' 30". Ans. b = 900 3' 44". C=110~ 5' 42". This case can never admit of more than one solution. 2. Given c=700; A=1400; B=50~, to find the other parts. a= 1260 24' 41". Ans. b= 73~ 33' 19". C= 48~ 38' 18". 21 84 TRIGONOMETRY. [CHAP. VII. CASE III. ~ 78. When two sides and angle opposite one of them are given. For example, suppose a, 6, and A to ba a be given. We first find B by (E), A D' sin. A sin. a sin. B csin. b' which gives b a sin. b sin. A A. (1) sin. a a I Having found B, we proceed to find B C and c by means of the following, drawn from (F) and (G): cos. (a + b) tan. (A + B) COS. 2(-b) ( tan. = S(A + B) tan. (a, (3+ b) tan... (3.) cos. I (A - B) Equation (1) will give two values for B supplementary to each other. If these two values of B, when used in (2) and (3), do not cause either to become negative, there will be two solutions. Since the arcs C and c are each less than 180, ~ C and 2c must each be less than 90~, and their tangents and cotangents must be positive. Hence their values given by the righthand members of (2) and (3) must also be positive. If we draw the perpendicular CD, we see that it will be the shortest arc that can be drawn from C, in case the angle A (see first figure) is acute; but this perpendicular will be the longest arc that can be drawn from C, when the angle A is obtuse (see second figure). From this we see that a may frequently take two positions, as indicated in the diagrams, on opposites sides of this perpendicular, thus giving two triangles ABC, AB'C, both fulfilling the required conditions. ~ 78.] TRIGONOMETRY 85 EXAMPLES. 1. Given a =40~; b = 118~; A = 300, to solve the triangle. sin. b = 9'945935 ar. co. sin. a = 0'191.933 sin. A = 9'698970 sin. B = 9'836838 B -43~ 22' 42", or 136~ 37' 18". B = 43~ 22' 42",fJrst value of B. A = 30~. (B + A) =360 41' 21"; (B - A)= 6~41' 21". 1(b + a) = 790; (b - a)= 39~. cos. I (b + a)=9'280599 cos. 2(B+A)= 9'904114 ar. co. cos. ~(b - a)=0'109497 ar. co. cos. (B-A)= 0'002966 tan. (B+A)=9'872204 tan.. (b + a)=10'711348 cot. 2C =9'262300 tan. ec =10'618428 IC== 790 37' 59", e= 76~ 27' 48", and C =159~ 15' 58"; and c = 152~ 55' 36". B =136~ 37' 18", second vatue of B. A= 30~. (B + A)= 83~ 18' 39"; (B - A) = 53 18' 39". cos. (b + a)= 9'280599 cos. (B+A)= 9'066264 ar. co. cos. f (b - a)= 0'1094;97 ar. co. cos. ~(B-A)= 0'223681 tan. 2 (B+A)=10'930770 tan. ~ (b + a)=10'71134-8 cot. I C =10'320866 tan. Ce =10'001293 ~ C=25~ 31' 58", 2C=450 5' 7", and C = 51~ 3' 56"; and c - 9f0 10' 14". Collecting results, we have these two solutions: B= 430 22' 42") (B =136~ 37' 18". Ans. C a=159~ 15' 58" or C = 510 3' 56". = 152 55' 36" J; = 900 I'O 14". 86 TRIGONOMETRY. [CHAP. VIi 2. Given a = 100~; b = 64~ 30'; A = 95~ 30', to find the other parts. There is only one solution, as follows: - B = 650 49' 25". Ans. C= 98~ 32' 24". c =101 55' 58". CASE IV. ~ 79. Whien two angles and the side opposite one of the m are given. For example, suppose A, B, and a to be given. ~, a We first find b by (E), A -B''A' sin. A sin. a sin. B -sin. b' which gives sin. a sin. B sin. (1A) A Having found b, we proceed to find C and c, by using precisely the same formulas as in Case III., which are o +. I (a - b) tan. (A +B) ] ) cot. C = ~ (S cos. (a - b) cos. 4 (A + B) tan. 4 (a + b) ( tan. I G=. = I - 2 (3.) tan24e= cos.'. (A - B) As in the last case, equation (1) will give two values for b supplementary to each other. If these two values of b, when used in (2) and (3), do not cause either to become negative, there will be two solutions. These two triangles, which thus satisfy the conditions of the problem, will be indicated in the above diagrams by ABC, ArB'C. ' 7-9.3 TRIGONOMETRY. 87 EXAMPLES. 1. Given A = 140~; B = 600; a = 150~, to solve the triangle. sin. aC = 9'698970 ar. co. sin. A = 0'191933 sin. B = 9'937531 sin. b = 9'828434 b = 42~ 20' 58" or 137~ 39' 2". = 150~ b = 42~ 20' 58", first value of b. ~ (a + b) = 960 10' 29"; 2 (a - b) = 53~ 49' 31". (A + B) = 100; (A - B) = 40o. cos. I (a + b)= 9'031654n cos. -(A+B)= 9'239670n ar.co.cos.~(a - b)- 0'228964 ar.co.eos.~(A-B)= 04115746 tanA(A+B)=10'753681n tan. (a + b)=10'965820m cot. 1 C=10'014299 tan. 4 c=10'321236 C =44" 3' 25", lc= 640 29' 11", and C = 880 6' 50". and c - 128~ 58' 22". a = 150~ b = 137~ 39' 2", second value of b. (a + b)= 143~ 49' 31"; 4 (a - b)= 6~ 10' 29". cos. (a + b)= 9 906993n cos. ~ (A+B)=9'239670n ar.co.cos. (a - b)= 0'002525 ar. co. cos. I (A-B)=0'115746 tan. (A+B)=10'753681n tan.- (a + b)=9'864043n cot. 2 C=10'663199 tan. 1 c=9'219459 1 C = 12~ 15' 10", Ic= 9~ 24'41", and C = 24~ 30' 20". and e = 18~ 49' 22". Collecting results, we have these two solutions: { b = 420 20' 58" ) 6 = 1370 39' 2". Ans. c = 128 58' 22" or c = 18~ 49' 22". C= 88 6' 50" (C = 240 30' 20". 88 TRIGONOMETRY. [CHAP. VIL 2. Given A= 80~ 20'; B = 115~ 30'; a = 84~ 20', to find the other parts. There is only one solution, as follows: b = 1140 20' 26". Ans. c = 820 39' 34". C= 790 17' 8". CASE V. ~ 80. When the three sides are given. Either of the groups (C'), a (C'), (C"') will give the angles. If'we use (C"') we have A tan. A= sin. (s - b) sin. (s - c) 2 sin. ssin. (s - a) tan. -2 B= 4 Bin. (s - 0_ Sil. (s-a) - a sin. s sin. (s- b) 2 sin. (s-a) sin. (s-b)} 2 sin. s Smn.( - c) where s = (a + b + c). Since the values of the half angles are given by these forma las, there will be no ambiguity in this case. EXAMPLES. 1. Given a = 1000; b= 800; c = 750, to fi-Jd the arIgles. a=100~ b = 80~ c = 750 -(a+br+c)=s=127' 30'; a -a-'7 3 c, - b - 473, s - c 56 30'. ~ 81.] TRIGONOMETRY. 89 sin. (s- b)= 9'867631 sin. (s- c)= 9'899467 sin. (s- c)= 9-899467 sin. (s - a)= 9-664406 ar. co. sin. (s - a) = 0 335594 ar. co. sin. (s - b) = 0'132369 ar. co. sin. s = 0100533 ar. co. sin. s = 0'100533 2)20-203225 2) 19-796775 tan. I A = 10'101612 tan.l B = 9'898387 ~ A = 380 21' 27", ~ B = 38~ 21' 27", and A = 760 42' 54". and B = 76~ 42' 54". sin. (s - a)= 9'664406 sin. (s - b) = 9'867631 ar. co. sin. (s - c) = 0'100533 ar. co. sin. s = 0'100533 2)19'733103 tan.a -= 9866551 - C = 36~ 19' 57" and C = 72~ 39' 54". 2. Given a=490 8'; b=57~ 16'; c=96~ 12', to find the angles. ( A = 310 32' 42". Ans. B -= 350 35' 15". C = 136~ 32' 48". CASE VI. ~ 81. When the three angles are given. 0 As in last case, we have three b groups (D'), (D"), (D"'), either A of which will give the sides. If we use (D"'), we have tan..:b=j -cos.Scos.(S -B) cos. (S - ) cos. (S - C) ( 2 cos. (s - C) cos.(S- A); tan. ~c 1 - Cos. S Cos. (S - E) 2; * * ( ) 2rcos. (S - A) cos. (S - B) where S = (A + B + C). 90 TRIGONOMETRY. [CHAP. VII. As in last case there can be no ambiguity in the values of a, b, c as here found. EXAMPLES. 1. Given A = 110~; B = 100~; C = 90~, to find the sides. A =1100 B =100~ C= 90~ i(A+B+C)=S=150~; S-A=40~; S-B=500; S-C=60~. -cos. S = 9'937531...... = 9'937531 cos. (S - A) = 9'884254 cos. (S - B) = 9'80S067 ar. co. cos. (S - B) = 0'191933 ar. co. cos. (S - C) = 0'301030 ar. co. cos. (S - C) = 0'301030 ar. co. cos. (S - A) = 0'115746 2)20-314748 2) 20'162374 tan. I a = 10'157374 tan. b = 10'081187 a= 550 9 40", ~ b= 50~ 19' 28", and a = 1100 19' 20". and b = 1000 38' 56". - cos. S = 9'937531 cos. (S- C)= 9'698970 ar. co. cos. (S - A) = 0'115746 ar. co. cos. (S - B)= 0'191933 2) 19.944180 tan. c= 9'972090 ac =430 9' 37" and c = 86~ 19' 14". Since the angle C is right, the solution might have been obtained by Case VI. of Right Triangles. (See Ex. 6, ~ 82.) 2. Given A = 85~; B = 60~; C = 50~, to find the sides. a =510 59' 16". Ans. b = 430 13' 48". c =37~ 17' 26". ~ 82. The methods which we have given for the solution of the six cases of spherical oblique triangles, as well as for the six ~ 82.1 TRIGONOMETRY. 91 cases of spherical right triangles, are all direct and always applicable; but frequently other methods may be used for particular cases, which would lead to simpler methods of solution. We can, however, hardly expect any thing more simple or more easy to be retained by the memory, in the case of right triangles, than the Rules of Nacpier; but in oblique triangles we can frequently obtain simpler solutions than by the methods already given. We may obviously divide any oblique triangle into two right triangles, and thus make all the cases of oblique triangles depend, for their solution, upon the principles of lvrapier's Rules. As an example of this meth- C od, let us suppose in the triangle ABC, that we have the arcs b a AB, AC, and the angle A A given, which corresponds with B Case I. If we draw the per- c D pendicular arc CD, we shall have, by applying Napier's Rules to the right triangles ADC, BDC, as follows: cos. b = cos. AD cos. CD..(1.) cos. a= cos. BD cos. CD.... (2.) Dividing (1) by (2), we have cos. AD c- os., or cos. AD cos. BD:: cos. b: cos. a. (3.) That is, if from an angle of a spheric oblique triangle an are be drawn perpendicular to the opposite side, dividing it into two segments, we shall have the cosines of these segments to each other as the cosines of their adjacent sides. Returning to the triangle ADC, we have tan. AD = tan. b cos. A..(4.) Thus we know AD, which subtracted from AB = c, gives BD. The segments are then known, and condition (3) gives at once Cos. a. It is obvious that, by the aid of the great variety of formulas which we have given, the method of solving many of the cases may be varied to almost any extent we please. We will leave 92 TRIGONOMETRY. [CHAP. VII. this kind of exercise wholly to the student, remarking that in our general solutions we have given those methods which, under all circumstances, seemed to be the safest and best. ~ 83. EXAMPLES FOR PRACTICE. 1. Given, in a right spherical triangle, A = 91~ 11'; B = 111~ 11', to find the remaining parts. a-= 910 16' 8". Ans. b —111~ 48' 43".. c = 89~ 32' 28". 2. Given, in a right spheric triangle, a=35~ 44'; A=37~ 28', to find the other parts. b = 690 50' 24" b = 1100 9' 36". Ans. C = 730 45' 15" or. = 106~ 14' 45". -B = 770 54' 0" B = 102~ 6' 0". 3. Given, in a spheric oblique triangle, a= 138~; A= 95~; C = 1040, to find the other parts. b = 160 34' 19") b 7 0' 21". Ans. c = 139~ 19' 40" or c =40 40' 20". B = 250 7' 38") (B =10~ 27' 42". 4. Given a = 810 17'; b =114~ 3'; c= 59~ 12', to find the angles.' A= 620 39' 43". Ans. 1 B = 1240 50' 50". C = 50~ 31' 43". 45. Given, in a right spheric triangle, a=1180 54'; B =12~ 19', to find the other parts. [ b = 100 49' 17". Ans. 4c = 118~ 20' 20". A= 95~ 55' 2". 6. Given, in a right spheric triangle, A = 1100; B = 1000, to find the other part. (This is the same as first example under Case VI. of Oblique Spherical Trigonometry.) a= 1100 19' 20". Ans. b = 100~ 38' 56". c = 86~ 19' 14". ~ 85.1 MENSURATION. 93 CHAPTER VIII. MENSURVATION OF SURFACES. ~ 84. THE area of a surface is found by comparing it with some known and fixed surface, which is taken as the unit surface, which unit surface is usually in the form of a square; as, a square inch, a square foot, a square yard, &c. The super3fcial unit receives its name from that of the linear unit, which measures its side. Thus, a square whose side is one inch is called a square inch; one whose side is one foot is called a square foot; one whose side is one yard is called a square yard. There are some superficial units which nave no corresponding linear unit, such as the acre, in land measure, and its fourth part, called a ~rood. There is no such linear measure as an acre or a rood. It would be absurd to speak of an acre long or a rood long. The usual units of square measure are given by the following TABLE OF SQUARE IMEASURE. sq. inches. sq.feet. 144 - 1 sq. yd. 1296 = 9 = 1 sq. rd. 39204 = 272 -= 30 1 sq. cA's. 627264 -= 4356 = 484 = 16 - 1 Acres. 6272640 = 43560 = 4840 = 160 = 10 = 1 S.M. 4014489600 = 27878400 = 3097600 = 102400 - 6400 - 640 = 1 ~ 85. In measuring land, Gunter's Chain is most commonly employed. Its length is 4 rods, or 66 feet, and is divided into 10( links, so that we have as follows: tI MENSURATION. [CHAP. VIII. 7-T92 inches, or 7'92 inches =.... 1 link. 100 links=792 inches=66 feet=22 yards=4 rods= 1 chain. 80 chains..1 mile. 10000 sq. links =- 16 sq. rods =.. 1 sq. chain. 100000 sq. links = 10 sq. chains = 160 sq. rods. 1 acre. The unit of measure for land is the acre. It cannot be given in an exact square, since it is made to consist of 160 square rods, and 160 is not a square number, which proves that the side of a square containing 160 sq. rods cannot be accurately expressed in rods; and if not in rods, then not in any multiple or submultiple of rods. One-fourth of an acre is called a rood, and the square rod in land measure is frequently called a perch. If links be multiplied by links, the product will be square links, which may be reduced to acres by dividing by 100000, or by simply pointing off five decimal places. If chains be multiplied by chains, the product will be square chains, which may be reduced to acres by dividing by 10, or pointing off one decimal figure. If chains are multiplied by links, the product will be converted into acres by pointing off three decimals. We will also add, that to convert square feet into acres, we must divide by 43560. To convert square yards into acres, we must divide by 4840; and to convert square rods into acres, we must divide by 160. PROBLEM 1. To find the area of a square or rectangularpiece of land. The obvious rule in this case will be as follows: Take the product of two adjacent sides for the area. [Geom., B. III., T. XXI., Scholium.] EXAMPLES. 1. How many acres in a piece of ground in the form of a square, each of whose sides is 13 chains, 25 links, or 13'25 chains? 13'25 x 13'25 = 1T755625 sq. chains. ~ 85.] MENSURATION. 95 Or, since 13-25 chains = 1325 links, we have 1325 x 1325 = 1755625 square links, and 17'55625 = the number of acres. Converting this decimal of an acre into roods and perches, we have 0'55625 4 roods = 2'22500 40 perches = 9'00000 Ans. 17A. 2R. 9P. 2. How many acres in a rectangular piece of land whose adjacent sides are 30'25 chains and 25'21 chains? 2521 x 3025 = 7626025 sq. links = 76A. 1R. 1 6-4%P. 3. How many rods in a rectangular piece of ground 282 feet wide and 325 feet long a Ans. 336'639 sq. rods. PROBLEM II.'o find the area of a parallelogramn. FIRST RULE. IFhen the altitude is given, D ultitply it by the base, for the A area. [Geom., B. III., T. XXII.] If we multiply AD by the sine of the angle at A, we shall ob- A E B tain DE. (Case I., Right Triangles.) That is, DE = AD x sin. A, and area ABCD = DE x AB = AD x AB x sin. A. (1.) Using logarithms, log. area = log. AD + log. AB + log. sin. A —10. (2.) We subtract 10 to correct for the 10 added to log. sin. of A. Hence we have this SECOND RULE. fult'hply the product of two adjacent 8ides by the natural sine of the included angle. 96 MENSURATION. [CHAP. VIIL OR BY LOGARITHMS. To the sum of the logarithms of two adjacent sides add the logarithmic sine of the included angle, and subtract 10, and the result will be the logarithm of the area. EXAMPLES. 1. How many acres in a field in the form of a parallelogram, having a base of 13 chains 14 links, and an altitude of 10 chains 37 links? 1314 x1037=1362618 sq. links=13A. 2R. 20P. 2. How many acres in a field in the form of a parallelogram, the adjacent sides being 25-17 chains and 30-25 chains, and having 700 30' for the included angle? BY LOGARITHMS. log. 2517 = 1'400883 log. 30'25 = 1'480725 sin. 70~ 30' = 9'974347 log. area = 2'855955 Hence, area = 71772 sq. chains, which is Ans. 71A. 3R. 3}P. 3. A field, in the form of a parallelogram, has 10'21 chains, 12'12 chains for the adjacent sides, and 30~ 45' for the included angle. How many acres does it contain Ans. 63-27 sq. chains. PROBLEM IIL ToJi0nd the area of a trioanle. FIRST RULE. When the altitude is given, multiply its half by the base. [Geom., B. III. T. XXIII.] Or multiply the altitude by D I half the base, or take half the product of thie altitude and base. / Since the triangle ABD is ob- / viously just one-half of the par- A E B ~ 85.] MENSURATION. 97 allelogram ABCD, whose measure is given by equation (1) of Prob. II., we obviously have DE x AB AD x AB x sin. A area ABD = (1.) Or, using logarithms, log. area = log. AD + log. AB + log. sin. A-10'301030. (2.) The log. of 2, the denominator of (1), being, 0'301030, is to be subtracted in addition to the 10. Hence, we have this SECOND RULE. XMuqltiply theproduct of any two adjacent sides by the s'ne of the included angle, and one-half this lastproduct will be the area. BY LOGARITHMS. To the sum of the logarithms of any two adjacent sides add the logarithmic sine of the included angle, and subtract 10'301030; and the remainder will be the logarithm of the area. We have also given [Geom., B. VIII., P. IV.] a demonstration to this THIRD RULE. Take half the sum of the three sides, andfrom this half sum subtract each side separately; then take the sguare root of the continued product of the half sum and the three remainders, and it will be the area. BY LOGARITHMS. Find the half sum of the three sides, and the three remainders as before; take half the sum of their logarithms, and it will be the logarithm of the area. EXAMPLES. 1. How many square feet in a triangle whose base is 110'5 feet and altitude 97-25 feet? Ans. I of 110'5 x 97'25 = 5373'0625 sq. feet. 2. If an angle of a triangle is 330 33', included by sides whose 98 MENSURATION. [CHAP. VIII. lengths are 45'25 feet and 37'5 feet respectively, what will be its area in square rods? log. 45'25 = 1'655619 log. 3'715 = 1'574031 sin. 330 33' = 9'742462 12'972112 10'301030 2'671082=log. of sq. feet. subtract 2'434968=log. of 2721, s. ft. in a sq. rd. 0'236114=log. of sq. rods. Ans. 1'7223 sq. rods. 3. How many acres in a triangular field whose sides are 70, 110, and 120 chains, respectively? a of (70 + 110 + 120) = 150, log. = 2'176091 150- 0 = 80, log. = 1'903090 150 — 110 = 40, log. = 1'602060 150 —120 = 30, log. = 1'477121 2) 74158362 log. area = 3'579181 Ans. 3794'7 sq. chains = 379'47 acres. 4. How many sg. feet in a triangle, each of whose sides is denoted by a? a. Ans. V2 a x aa x I ax -1-a = 0.3 5. Referring to the last example, how many sq. feet in an equilateral triangle whose side is 10 feet? Ans. 43'301 sg. feet. PROBLEM IV Tof nd the area of a trapezoid. D E C RULE. fuilttply hatf the sum of the parallel sides by the perpendicular [ distance between them. [Geom., A F B. III., T. XXIII., Cor. II.] ~ 85.] MENSURATION. 99 EXAMPLES. 1. What is the area of a trapezoid, whose parallel sides are 20 and 30 chains, and which are perpendicularly distant 12'5 chains. l of (20 + 30) = 25; 25 x 12'5 = 312'5 sq. chains, or 311 acres. 2. If the parallel sides of a trapezoid, whose lengths are 25 and 35 feet, are at the distance of 12 feet from each other, what will be its area? Ans. 360 sq. feet. 3. What is the area of a trapezoid, whose parallel sides are 3 and 4'90 chains, and the perpendicular distance between them 166 chains? Ans. 6-557 sq. chains. 4. How many square feet in a trapezoidal table whose parallel sides are 5'5 feet and 4'25 feet, the width being 3'5 feet a Ans. 17'0625 sq. feet. SduOLIUJM.-In land surveying, we frequently have occasion to measure long irregular portions of a field, which are bounded on one side by a straight line. At equal intervals along the b straight line AF, as at B, C, D, and r E, measure the perpendiculars Bb, Cc, Dd, &c.; also measure the extreme perpendiculars Aa and Ff. F E D C B A When the equal intervals AB, BC, CD, &c., are short, the figures ABba, BCcb, CDdc, &c., may, without much error, be regarded as trapezoids; hence, calling s one of the equal distances, AB, BC, CD, &c., we have Aa ~ Bb area ABba = 2- x s, Bb + Cc area BCcb = 2 x s, area oDde = Cc + Dd x 8s Cc + Dd area CDdc = 2 X, Dd + Eo area DEed - x s, area EFfe = +Ff S. The total area is (} Aa + Pb + Co + Dd + Ee + ~ Ff) x 8. 100 MENSURATION. [CHAP. VIIL Hence this RULE. To half the suem of the extreme breadths add all the i~ntermediate breadths; multiply this sum by one of the equal distances, and the product will be the area. As an example, suppose seven breadths of an irregular field, taken at equal intervals, to be 32 links, 47 links, 63 links, 13 links, 17 links, 5 links, and 2 links; if one of the equal intervals is 10 links, what will be the area? 32 + 2+ 47 + 63~+ 13 - 1T + 5 = 162, and 162 x 10 = 1620 for the number of square links in the area. PROBLEM V. To find the area of any irregular polygonal figure. RULE. Draw a sufficient number of diagonals to divide the whole figure into triangles. Then compute each triangle separately, and take their sum. As an illustration, suppose ABCD D EF to be the polygon. Draw the diagonals EA, EB, EC, and suppose by c E / actual measurement we find AB= 12'5; BO = 11'25; D = 7; DE = \ 6'75;EF=575; FA =105; EA= / 10'75; EB = 10; EC = 975. Now computing each triangle by A B the Third Rule under Problem II., we shall find Area EFA = 29'3795 " EAB = 51'6642 EBC = 45'6431 " ECD = 23'6247 area ABCDEF = 150'3115 ~s 8.] MENSURATION. 101 PROBLEM VI. To fUnd the area of a quadrilateral in terms of its diagonals ancd their included angle. Let ABCD be the quadrilat- B F eral figure, of which the diagonals AC, BD, and their angle K of intersection are given. - c Through A, B, C, D, draw lines parallel to the diagonals, a D G which will together form the parallelogram EFGH, whose angles are equal to the angles at K; and whose area is obviously double that of the quadrilateral ABCD. But, Prob. II., EFGH = HE x HG x sin. H = BD x AC x sin. K.; and we have ABCD = BD x AC x sin. K. Hence we have this RULE. Jfultiply the product of the two diagonals by the-sine of their included angle, and one-half this last product will be the area. BY LOGARITHMS. To the sum of the logarithms of the two diagonals add the ogarithmic sine of their included angle., and subtract 10301030; and the remainder will give the logarithm qf the area. NOTE.-This rule will agree with the second rule of Prob. III., if the word diagonals is changed into sides. EXAMPLES. 1. The diagonals of a quadrilateral are 37'5 and 46'25, and they make with each other an angle of 470 47'. What is the area? log. 37'5 = 1'574031 log. 46-25 = 1665112 sin. 47~ 47' = 9'869589 13-108732 10'301030 log. area = 2'807702 &nd area = 642'247. 102 MENSURATIOiN. [CHAP. VIII. 2. The diagonals of a quadrilateral are 90 and 100, and make an angle of 75~ with each other. What is the area?. Ans. 4346'68. PROBLEM VII. lTo fnd the area of any regular polygon. FIRST RULE. Take one-hayf the product of the perimeter and perpendi cular drawn from its centre to one of its sides. [Geom., B.IV., T. V.] When the perpendicular is not given, it may be found as follows: the angle ACB will be found by dividing 360~ by the number of sides in the polygon; and the angle BCD, which is one-half C of ACB, can be found by dividing 180~ by the number of sides. Hence if we denote the number of sides by n, we shall have -80 for the angle BCD. A D B Now BD, which is one-half of a side of the polygon, being multiplied by cotan. BCD, will give CD. If, then, s denote a side andp the perpendicular, we shall have 1800 p _ of s x cotan. 8... (1.) And, the perimeter being n x s, we shall obtain 180~ area = l- ofp x g x 8 Of 8f x n x cotan.. (2.) 2 VL~r /\ In If the side of the polygon is taken as a unit, then s = 1, and 1800 area = ~ of n x cotan. (3.) If, in formula (3), we take successively n = 3, 4, 5, 6, 7, &c.; we shall obtain the areas of regular polygons, whose sides are unity, as given in the following . 85.] MENSURATION. 103 TABLE. Names. Sides. Areas Logarithms. Triangle........ 3.... 0'4330127...... 16365007 Square......... 4...... 1-00000 00.....0'0000000 Pentagon. 5...... 1-7204774......0'2356490 Hexagon.. 6.... 2-5980762.....0'4146519 Heptagon....... 7...... 3'6339124...... 0-5603744 Octagon........ 8...... 48284271...... 06838057 Nonagon........ 9. 6'1818242......0'7911166 Decagoa....... 10...... 7-6942088...... 0-8861640 Undecagon......11...... 9'3656399.....0'9715375 Dodecagon...... 12.... 111961524... 1... 0490687 Since the areas of -similar figures are to each other as the squares of their homologous sides, we may find the area of any regular polygon whose number of sides is not greater than 12 by this SECOND RULE. ifultiply the square of the length of a side of the polygon by the area of a similar polygon whose side is unity, as found in the preceding table. BY LOGARITHMS. To twice the logarithm of a side of the given polygon add the logarithm of the corresponding tabular number. EXAMPLES. 1. What is the area of a regular dodecagon whose side is 7'75. log. 7'T5 = 0'889302 log. 7T75 = 0'889302 tabular log. = 1'049069 log. area = 2'827673 Area = 672'469. 2. What is the area of an octagonal room, each side of which is 10'5 feet? Ans. 532'334 sq. feet. 3. What is the number of square inches in a pentagon, each of whose sides is 10 inches? Ans. 172'048 sq. inches. 4. What is the area of an equilateral triangle, having 100 for a side? Ans. 4330'127. 104 MENSURATION. [CHAP. VIIl PROBLEM VIII. To jind the circumference of a circle zwhen its diameter is known, and conversely. If the diameter of a circle is a unit, its circumference will be 3-141592653589793, &c. (~ 5). Denote this ratio by s,r and put R = radius, D = diameter, C = circumference. Now since circumferences are to each other as their diameters, and consequently as their radii, we shall have C =D x D =Dx 3414159265,. (1.) O = R x 2~r = R x 6'28318531, (2. ) D= x - a x 0'31830989,.. (3.) R=C x-=C x 015915494.. (4.) Using logarithms, we have log. C = 0-4971499 + log. D,... (5.) log. C = 0'7981799 + —log. R,... (6.) log. D = 1'5028501 + log. C,... (7.) log. R = 1i2018201 + log. C... (8.) EXAMPLES. 1. What is the circumference of a circle whose diameter is 12_5 feet? An8. 3'1416 x 12'5 = 39'27 feet, nearly. 2. If the radius of the earth is 3956 miles, what is its circumference? Ans. 24856'281 miles, nearly. 3. If the circumference of a circle is 100 feet, what is its diameter? Ans. 0'31830989 x 100 =- 31'83 feet, nearly. 4. If the circumference of a wheel is 15'5 feet, what is its radius? Ans. 2'4669 feet. 5. What is the circumference of a circle whose diameter is 100 feet? A.ns. 3144159 feet. ~ 85.1 MENSURATION. 105 PROBLEM IX. Tofind the area of a circle when its diameter is known, and conversely. Also when the circumference is known. Denoting the area by A and continuing the notation as under Prob. VIII., we have [Geom., B3. IV., T. XIII.] A=- of D x C. Substituting the value of C as already given, we have A = D2 x 2 = D2 x 0'78539816, (1.) A = R2 x r = R2 x 3-14159265,.. (2.) 1 A = C2 x C2 x 007957747,.. (3.) D = /A x 24- = VA x 1412837917, (4.) R = /A x V/ =,/A x 0.56418958, (5.) C = VA x 2v/' = V/A x 3.54490770.. (6.) Using logarithms, we have log. A = 1'8950899 + 2 x log. D,. (7.) log. A = 0'4971499 + 2 x log. R, (8.) log. A = 2'9007901 + 2 x log. C,. (9.) log. D = 0'0524551 + I x log. A,. (10.) log. R = 17514251 + x log. A,. (11.) log. C = 0.5496049 + ~ x log. A.. (12.) EXAMPLES. 1. What is the area of a circle, whose diameter is 125 feet? By equation (7), we have 1-895090 2 x log. 125 = 4-193820 log. A = 4'088910 A = 12271'8. If more decimals are required, they may be obtained by using equation (1), which gives 0O78539816 x 125 x 125=12271'84625. 106 MENSURATION. [CHAP. VIII. 2. What is the number of square inches in a circular standtop, whose radius is 15 inches? Ans. 3'1416 x 15 x 15 -= 70686 sq. inches. 3. What is the area of a circular fish-pond, whose circumference is 25 rods? Ans. 0'07957747 x 25 x 25 = 49'734 sq. rods. 4. What is the number of rods in the diameter, the radius, and the circumference of a circle, which shall contain just 1 acre? Using equations (10), (11), and (12), we have 0o052455 1.751425 0o549605 I of log. 160 = 1-102060.1..102060.1. 1102060 P154515 0'853485 1P651665 D = 14'273. R = 7'1365. C = 44i84. NoTz.-This last example would have required far more labor had we used equations (4), (5), and (6), which are independent of logarithms. 5. How many chains must the diameter of a circular farm be which shall contain just 100 acres? Ans. 35'68 chains. PROBLEM X. To determineformnulas for the area andfor the are of a sector when the angle at the cente is known, and conversely. Denote the number of degrees which measures the angle at the centre by n, the length of the corresponding arc by c, and the area of corresponding sector by a. Then, since arcs of the same circle are to each other as their corresponding angles at the centre; also, since sectors are to each other in the same ratio, we shall have 360: n C c, 360: n: A' a. And = 360 x c,... (1.) a = 360 x A. (2.) EXAMPLES. 1. What is the length of an are of 15~ of the earth, its entire circumference being 24856-28 miles? Ans. -15 of 24856'28 = 1035'68 miles, nearly. ~ 85.] MENSURATION. 107 2. What is the area of a sector of 10 degrees of a circle, whose diameter is 10 feet? If, in equation (2), we substitute the value of A as given by equation (1) of Prob. IX., it will become'7t r,' ] 2,'I' a= 6-0 x D2 - 360 X4 x 100 = 2'18 sq. feet, nearly. 3. What is the length of an arc of 1~ of the moon, its diameter being 2160 miles? In equation (1), substitute the value of C as given by (1) of Prob. VIII., and we have = 360x x = 36 x x 2160 = 18'849 miles, nearly. 4. An arc on the earth of 500 miles corresponds to how great an angle at the centre? Equation (1) gives immediately [see equation (1), Prob. IX.] c 500 n = — x 360 248 x 360 = -7~242=7~ 14' 31". 5. A sector whose area is 10 square feet corresponds to how great an angle at the centre, provided the diameter of the circle is 10 feet? Equation (1) gives immediately [see equation (1), Prob. IX.] n=3,a x 60 3600 -=~ x- 3660 = 450~8366 = AIY-a 50' 12'.6 450 50' 12' PROBLEM XI. F To find the area of a segment of a H B circle. If n denote the number of degrees in the angle BAC, we shall have [Prob. A X.] area of sector ABFC = 360 x 2 x since A = R2 x r, by equation (5), Prob. IX., area of triangle ABC = I R2 x sin. n. [Prob. III., Second Rule.] 108 MENSURATION. [CHAP. VIII. Hence the area of the segment BHCF, being the difference between the sector and triangle, is (3 Sx ~r — sin. n x.. (1.) And the area of the segment BI{CG, which is greater than a semicircle, is the sum of the sector CBG and triangle ABC. Hence, to determine the area of a segment, we have this RULE. Find the area of a sector which has the same are as the segment; also, the area of the triangle formed by the chord of the segment and the radii of the sector. Then take the sum of these areas when the segment exceeds the semicircle, and their difference when it is less. EXAMPLES. 1. In a circle whose radius is 10 feet, find the area of a segment whose arc corresponds with an angle of 10~ at the centre. Area of sector = - x x R = -o x 314159 x 102 = 8'727. "triangle = sin.10~ x 102 of 0'17365 x 102 = 8683. The difference = area of segment = 0'044 of a sq. foot. 2. In a circle whose radius is 15 rods, it is required to find the area of a segment whose arc corresponds to a central angle of 200~. Area of sector = 392'699 " of triangle = 38'477 of segment 4314176 sq. rods. PROBLEM XII. To jtnd the area of an ellipse. A line drawn through the centre of D an ellipse is called its diameter. The longest diameter is called the trans- A verse diameter; the shortest is called the conjugate diameter. Thus AB is C the transverse diameter, and CD is the conjugate diameter. These diameters mutually bisect each other at right angles. ~ 86.] MENSURATION. 109 The area of an ellipse may be found by this RULE. iMu7ltiply the product of the transverse and conjugate diameters by 0'7854: =- q. A demonstration of this rule cannot be given by the elementary principles of Geometry. EXAMPLES. 1. How many square feet in the surface of an elliptical pond, whose transverse diameter is 100 feet, and conjugate diameter 60 feet? Ans. 100 x 60 x 017854 = 4712'4 square feet. 2. How many square inches is an elliptical table, whose transverse diameter is 5 feet 3 inches, and conjugate diameter 3 feet 6 inches? And how many square feet? Ans. 2078-1684 square inches. 14'4317 square feet. CHAPTER IX. MENSURATION OF SOLIDS, Which includes the measure of their surfaces as well as that of their volumes. ~ 86. TIE common unit for measuring solids is a cube; as, a cubic inch, a cubic foot, a cubic yard, &c. The values of the usual units of solid measure are given in the following TABLE. 1728 cubic inches = 1 cubic foot. 27 cubic feet = 1 cubic yard. 4492- cubic feet = 1 cubic rod. 128 cubic feet = I cord of wood. 231 cubic inches = 1 gallon (liquid measure). 2684 cubic inches = 1 gallon (dry measure). 21502 cubic inches = 1 bushel. 110 MENSURATION. [CHAP. IX. ~ 87. By an act of Parliament of Great Britain, which took effect on the 1st of January, 1826, the Imperial gallon of 277'274 cubic inches was adopted as the only gallon. This value was found to be the measure of 10 pounds, avoirdupois, of distilled water. Estimating 8 gallons to the bushel, we have for the Imperial bushel 2218'192 cubic inches. New York as well as the other States of the Union still continue to use the old English gallons as given in the foregoing table. PROBLEM I. To find the surface of a rightprism, or of a right cylinder. RULE. lMfultiply the perimeter of the base by the altitude,for the convex surface. To which add the areas of the two bases if the entire su/fa~ce is irequired. [See Geom., B. VI., T. I. Also B. VII., T. I.] EXAMPLES. 1. What is the entire surface of an octagonal prism, whose height is 25 feet, and each side of its base being 1 foot 3 inches? 1'25 x 8 x 25 = convex surface 250 sq. feet. area of bases 7'544 257-544 sq. feet. 2. What is the convex surface of a cylinder, whose length is 40 feet, and its diameter 2 feet? Ans. 251'328 sq. feet. PROBLEM IL /Tofind the volume of a rightprisrm, or of a right cylinder. RULE. lflaltiply the area of the base by the altitude. [See Geom., B. VI., T. XII., and B. VII., T. II.] EXAMPLES. 1. What is the volume of a triangular prism, whose height is 20 feet, and each side of whose base is 2 feet? Ans. 34'641 cubic feet. 86.] MENSURATION. 11 2. How many Imperial gallons in a cylindrical oil-can, the diameter being 20 inches, and depth 40 inches? Denoting the diameter, in inches, by d, and height by A, and putting g for the number of inches in a gallon, we have Number of gallons = x d2 x h... (1.) For Imperial gallons, g = 277'274; for United States, or old English gallons, g= 231. Hence Imperial gallons = 0'0028326 x d2 x A, (2.) United States gallons = 0'0034 x d2 x h. (3.) USING LOGARITHMS. log. [Imp. galls.] = 3'4521808 + 2 x log. d + log. Ah, (4.) log. [U. S. galls.] = 3'5314779 + 2 x log. d + log. h. (5.) Now, by equation (2), we find Ans. 0-0028326 x 20a2 x 40 = 45'82 Imp. gallons. 3. How many cubic feet in a cylindric log 14 feet long, and 14 inches in diameter? Ans. 14'966 cubic feet. 4. How many cubic inches in a cylindric,measure of 1Sinches diameter, and 8 inches deep? Ans. 2150'425 cubic inches. NorE. —The dimensions of the measure as given in the last example are those of the old English bushel, usually called the Winchester bushel; so named from the place where it was deposited for safe keeping. 5. How many United States gallons of oil in a can of 18 inches diameter, the oil being only 13 inches deep? Ans. 14'32 U. S. gallons. PROBLEM III. Tofid the s8urface of a regular pyramid, or of a conc. RULE. ifultiply the perimeter of the base by hcalf the slant height. Add the surface of base when the entire surface is required. [See Geom., B. VI., T. XV. Also 1B. VII., T. III.] 112 MENSURATION. [CHAP. iX. EXAMPLES. 1; What is the convex surface of a cone whose slant height is 20 feet, and the diameter of the base 6 inches? Ans. 15-708 sg. feet. 2. What is the convex surface of an octagonal pyramid, its base being 10 inches on each side, and the slant height being 35 feet Ans. 1162 sq. feet. PROBLEM IV. ToJfind the volume of a reguzlarpyramid, or of a cone. RULE. 2fultiply the area of the base by one-third of the altitude. [See Geom., B. VI., T. XVIII., and B. VII., T. V.] EXAMPLES. 1. How many cubic feet in a conical stick of timber 40 feet long, and 20 inches diameter at the base? Ans. 29'0888 cubic feet. 2. How many cubic yards in a conical stack of hay, which is 18 feet in diameter and 27 feet high? Ans. 84-823 cubic yards. 3. How many cubic feet in a regular octagonal pyramid of marble, whose height is 30 feet, and a side of the base 2 feet? _Ans. 193'136 cubic feet. PROBLEM V. To fin the surface of a/frstum of a regular pyramid, or of afrustum of a cone. RULE. i.pltiply half the sum of the perimeters of the two bases by the slant height. When the entire surface is required, add the areas of the two bases. [See Geom., B. VI., T. XV., Cor. II. Also B. VII., T. IV.] ~ 87.1 MENSURATION. 113 EXAMPLES. 1. What is the entire surface of a frustum of a cone, the slant height of which is 15 feet, and the diameter of one base 2 feet 6 inches, and that of the other 1 foot 3 inches? Ans. 944934 sq. feet. 2. What is the convex surface of a frustum of a regular hexagonal pyramid, the slant height being 12 feet, each side of the larger base being 4 feet, and each side of the smaller base 3 feet? Ans. 252 sq. feet. PROBLEM VI. To find the volume of afrustum of a regular pyramid, or of a cone. If we denote the volume of the frustum by V, we shall have (Geom. B. VI., T. XIX.), (See also B. VII., T. V., 0. II.) V=(A + a a+VAx a)x h=A x I h + ax Ih~+ VAx ax1 h. Hence the following RULE. Add together the areas of the two bases, and a mean p2roportional between them, and multiply this sum by one third of the altitude. In the case of the cone, if D and d denote the diameters of the bases, we shall have A=D2x; a=d2xI,andA22a=Ddx-. 4 4' 4 And V= (D2 + d2 + Dd) x h x 0'26179939.. (1.) If D, d, and h are estimated in inches, we shall have the number of United States gallons, which is the same as the number of old English gallons, in this volume, by dividing by 231. If we divide by 277'274, we shall obtain the number of Imperial gallons. Thus, U. S. gallons = (D2 + d2 + Dd) x h x 0'0011333, (2.) Imp. gallons = (D2 + d2 + Dd) x h x 0'0009442. (3.) Using logarithms, we have log. [U. S. galls.] = 3 0543566 + log. (D2' + d2 + Dd) + log. h, (4.) log. [Imp. galls.] =49750595 + log. (D2 + d2 + Dd) + log. h. (5.) ~~' b "'- "''' I~~~~~~~~~~~~~ 1.14 MENSURATION. [CHAP. IX. EXAMPLES. 1. A block of marble is in the form of a frustum of a square pyramid: the larger base is 4 feet each way, and the smaller base is 3 feet on a side; the height is 6 feet. How many cubic feet does it contain? 11 A=42=16; a= 32= 9; and A2 a2 = 12. 1 1 V = (A + a + A aa) x h = (16 + 9 + 12)x2 = 74 cubic ft. 2. A cistern in form of a frustum of a cone is 9 feet deep, having for diameters 8 feet and 10 feet. How many U. S. gallons of 231 cubic inches will it contain? Using equation (4), we have D2 + d2 + Dd= 35136; h = 108. 3'054357 log. 35136 = 4'545752 log. 108 =2033424 log. [U. S. galls.] = 3.633533 Ans. 4300'6 U. S. galls. 3. How many Imperial gallons in a cistern in the form of a frustum of a cone, the bottom diameter being 12 feet and top diameter being 10 feet, the depth also 10 feet? Using equation (5), we have D2 + d2 + Dd = 52416; h = 120. 4'975060 log. 52416 = 4'719464 log. 120 = 2'07918&1 log. [Imp. galls.] = 3'773705 Ans. 5938'9 Imp. galls. 4. How many cubic feet in a block of marble in the form of the frustum of a regular hexagonal pyramid, each side of the larger base being 4 feet, and each side of the smaller base being 3 feet, and the altitude being 9 feet? Ans. 288'386 cubic feet. ~.87.] MENSURATION. 115 PROBLEM VII. T) frnd the volumee of a wedge. F Definition.-A wedge is a solid bounded by five planes; a rectangle ABCD, called the base of the wedge; two trapezoids ABEF, DCEF, called the sides c ~-I-l Dc of the wedge, and which intersect in the line EF, parallel to the base, B M A which is called the edge of the wedge; and two triangles ADF, BCE, which are called the ends of the wedge. If through E we pass the plane EMN parallel to the end FAD, the portion EMBCN thus cut off is a pyramid, and the remaining wedge having AMIND for its base is evidently a triangular prism, and has for its measure onehalf of a parallelopipedon, having the same base and same altitude. Hence if L = AB, the length of base, I = EF, the length ot the edge b = AD, the breadth of base, and A = EK, the altitude. then we shall have Volume of the wedge whose base is AMIND = of AM x AD x EK = Ibh... (1.) Volume of pyramid EMBCN = ~ of MIB x BC x EK = (L - ) b = Lh -. Ib.. (2.) Taking the sum, we have for the volume of the entire wedge 3Lb + - 1bA = 6 (2 L + ) h (3.) If the length of the wedge is less than the length of the edge, the volume sought will be the difference between the prism and pyramid. And in this case the volume of the pyramid will be (I - L) bh = I MbA- LbA, which subtracted from I ibhA, the; volume of the prism, we have 1 lbh + - Lbh = (2 L + ) bh,. (4.) which is precisely the same as expression (3). 23 116 MENSURATION. [CHAP. IX. Hence the following RULE. To twice the length of the base add the edge, multiply the su2me by the breadth of the base, and that product by - of the height of the wedge. EXAMPLES. 1. What is the solidity of a wedge, whose base is 10 inches long, 4 inches wide, whose height is 8 inches, and whose edge is 9 inches a Ans. 154- cubic inches. 2. The length and breadth of the base of a wedge is 70 inches and 30 inches, the edge is 110 inches, and the altitude 42 inches. What is the solidity? Ans. 30'3819 cubic feet. PROBLEM VIII. To find the volume of a rectangular prismoid. Definition.-A rectangular pris- H moid takes its name from its sem- / blance to a prism. It may also be / said to resemble a frustum of a E F quadrangular pyramid. The upper and lower bases are rectangles, having their corresponding sides parallel, and the convex surface consists of four trapezoids. The altitude of the prismoid is the perpendicular distance between the bases. Put L = AB, the length of lower base. B = AD, the breadth of " " I = EF, the length of upper base. b = EH, the breadth of " " h = the altitude of the prismoid. If we pass a plane through EF and DC, the prismoid will be Jivided into two wedges, whose bases are the bases of the pris ~S7.] MENSURATION. 117 moid, and whose common altitude is the same as the altitude of the prismoid. The solidity of these wedges will be (Problem VII.) 6 (2 L +1) Bh, - (2 1 + L) bh. Adding, we have for solidity of prismoid, 6- (2 BL + 2 bl - B1 + B ).... (1.) If we putp-2 (L + ); q = (B + b), we shall have solidity of prismioid = A h (BL + bl + 4pq)... (2.) In this last expression pg is the area of a parallel section equally distant from the two bases. Hence the following RULE. To the sum of the two bases add four times the area of a parallel section equally distant from the bases, and multiply the asum by I of the height. EXAMPLES. 1. How many cubic feet in a block of marble in the form of a rectangular prismoid, the lower base being 4 feet by 3, the npper base 3 feet by 2, and the height being 5 feet? Ans. 44- cubic feet. 2. How many cubic feet in a stick 36 feet long, in the form of a rectangular prismoid, the one end being 20 inches by 16 inches, and the other end being 14 inches by 12 inches? Ans. 60 cubic feet. PROBLEM IX. To find the inclination of two adjacent faces of a, reyular )pd(!.edro~n, and the radii of the inscribed and circumscribed spheres; also, the area of its surface, and its volume. Let AB13 be the edge common to two adjacent faces, C and E the centres of those faces. Draw CO and EO perpendicular 118 MENSURATION. [CHAP. IX to these faces, meeting at O; also draw A CD, ED perpendicular to the common edge AB. The angle CDE will be the inclination of these faces. Let I = inclination of the faces = angle | CDE; a = one of the edges = AB; r= radius of the inscribed sphere c = OC= OE; R = radius of the circumscribed sphere = OA= OB; 0 m = number of faces forming a polyedral angle; n = number of sides in each face; f = whole number of faces in polyedral; S = whole surface of polyedron; V- the volume. If, now, a sphere be described about O as a centre with a radius equal to unity, its intersection with the planes OAC, OAD, ODC will form a spherical triangle acd, right-angled at d. By Napiier's Rules, this triangle gives cos. acat cos. d ( sin. acd' cos. ac = cot. cad cot. accd... (2.) Since ed measures the angle COD, and COD = 90~- I, we have cos. cd = sin. I. We also have angle cad =-; angle acd =-; hence (1) becomes CoS. — sin. I=.... 3.) sin. - rhe triangle ACD gives CD = a1 cot. -, but triangle COD gives qr = CD tan. - I; hence we have r = a tan. 1 I cot.-. (4.) r=2 2Iv~ !~ 87.] MENSURATION. 119 Triangle AO gives R cos. AOC = RGcos. ac = r; using the value of cos. ac given by (2), we find R = r tan.- tan. - = a a tan. ~ I tan.-.. (5.) 7)n n 2n 2 The area of one of the faces is n x AD x CD = it x 2 a x a cot. -=a2 cot. - consequently we have S = c2 x -cot-.... (6) 4 n Since the volume is equal to S x r, we find V = a3 x n- tan. I cot.2... (7.) These equations are general and apply to the five Platonic bodies. The values of n and n are as follows: Tetraedron, mn = 3, n = 3. Hexaedron, m = 3, n = 4. Octaedron, n = 4, n = 3. Dodecaedron, mn = 3, n = 5. Icosaedron, Mn = 5, n = 3. Using these values in (3), we find as follows: Cos. 60~ Tetraedron, sin. ~-I = 60 1 - 3. osin. 60 Hexaedron, sin. 1 I = COS6= 0 sin. 450 2. cos. 45~ Octaedron, sin. 1 I = 60~ -- 6. 2 sin. 60_ 3 /. Dodeeaedron, sin. ~ I —in 36~ =5 cosin. 360 Icosaedron, sin. I = sin. 60~ =- (5 + V5). sin. I; tand. 21 1- sin21 I and then equations (4), (5), (6), and (7) will give, after considerable reduction, as follows: 120 MENSURATION. [CHAP. IX ( r =ax- V/6; RR=ax-V6; Tetraedron, S=a2 x /3; V = ad3x-V,/2. (r=axl; R=axl -/2; Hexaedron, r S=a2x6; V=a3xl. { r=axl V6; R=axl-/2; Octaedron, S=a2x2 /3; V=a3x /2. Dodecaedron, r=a 250 + 11OV5; R=a x-4 (/15 + v/3); D S=a2x3/25 + 10/5; V=-a3x(15 + 7 V5). - r - -axi~(3 3 + /15); R = ax/10 + 2/5; Icosaedron, r X(33 5); XVo 25; S=a2x5 /3; V=a3x (3 + 5). We give, in the next table, to seven decimal places, the values, and their logarithms, of the surfaces and volumes of the five polyedrons, having unity for the edges of each: TABLE. Name& No. of Faces. Surfaces. Volumes. Tetraedron.. 4 7320508 0o1178513 log. [0'2385607] log. [1-0713344] Hexaedron. 6 60000000 1'0000000 log. [0-7781513] log. [0'0000000] Octaedron.. 8 3'4641016 0'4714045 log. [0.5395907] log. [1.6733937] Dodecaedron. 12 20'6457288 7'6631189 log. [1'3i48302] log. [0.8844056] Icosaedron.. 20 8'6602540 2'1816950 log. [0.9375307] log. [0.3387940] The numbers included within the brackets are the logarithms of those numbers immediately above them. Since surfaces of similar solids are to each other as the squares of their like dimensions, we may find the surface of any regular polyedron by this ~ 87.] MENSURATION. 121 RULE. 3fultiply the square of one of the. edges by the tabuLlar surface of the similar unit solid, as given in the foregoing table. Or using logarithms. To twice the logarithm of the edge add the tabular logarithm, of the similar solid, as given in the table, and the sum, will be the logarithm of the surface. Again, since the volumes of similar solids are to each other as the cubes of their like dimensions, we may find the volume of any regular polyedron by this RULE. Mfultiply the cube of one of the edges by the tabular volurme of the similar unit solid, as given in the table. Or using logarithms. 2To three times the logarithm of the edge add the tabular logarithm of the similar solid, as girven in the table, and the sum will be the logarithim of the volume. EXAMPLES. 1. What is the surface and the volume of a regular icosaedron, each edge of which is 10 inches? Ans. Surface = 866'025 sq. inches. Volume = 2181-695 cubic inches. 2. What is the surface and the volume of a regular dodecaedron, each edge of which is 3 feet? Ans. Surface = 185'811 sq. feet. Volume = 206'902 cubic feet. 3. What is the surface and the volume of a regular tetraedron, each edge being 4 inches? Ans. Surface = 27'T12 sq. inches. Volume =- 7542 cubic inches. 122 MENSURATION. [CHAP. IX PROBLEM X. So find the surface and voluBme of any parallelopipedon. Let DE be a parallelopipe- A don, having the edges DB =a, DC =, bE DA =c / and their inclinations BDC= a, D 0 CDA=i3, B P ADB = y, given. The area of any face, as the B face BC, = ab sin. cx; hence, for the whole surface, we have S = (ab sin. a + be sin./ +- ca sin. y). (1.) To find the volume, draw AP perpendicular to BC, and pass a plane through AID and AP, intersecting BC in DP. Conceive the surface of a sphere whose radius is unity to be described about D as a centre, and the arcs formed by its intersection aw ith the planes passing through the centre D will be as follows:'C' = a, C'A' =/, A'B'= y. The are A'P' measures the angle ADP. The altitude AP = AD sin. ADP = c sin. A'P'; this, multiplied by the base BC = ab sin. a, gives, for the volume, V = abe sin. ia sin. A'P'... (2.) The right spherical triangle A'B'P' gives sin. A'P' sin. y sin. A'B'P'; hence (2) becomes V = abe sin. a sin. y sin. A'B'P'... (3.) In the spherical triangle A'B'C' we have, by formula (C), ~ 57, if s =(a +IC + y), sin. B'P' sin. sin. 8 sin. (s-a) sin. (8s-) sin. (s-); sin. a sin. y hence (3) becomes V = 2 abe a/sin. s sin. (s —o) sin. (s —3) sin. (s-y). (4.) ~ 87.] MENSURATION. 123 Since the volume of the tetraedron formed by passing a plane thriough the three points ABC is one sixth of the parallelopipedon, it follows, that if a, b, c denote the three edges of a polyedral angle of a tetraedron, and a,, y are their inclinations, the volume of the tetraedron will be V-3 abe v/sin.s sin. (s —a) sin. (s -) sin. (s-). (5.) PROBLEM XI. To Jind the surface of a sphere. RULE. _ibdtiply the square of the diameter by -r = 3'1415926. [See Geom., B. VII., T. XII., S.] EXAMPLES. 1. Required the number of square inches in the surface of a ball 6 inches in diameter. Ans. 113'0976 sq. inches. 2. How many square miles on the surface of the earth, it being T912 miles in diameter. Ans. 196662896. PROBLEM XII.'To fid the volume of a sphere. RULE. zfultiply the cube of the diameter by'- ~ = 0'5235988. [See Geom., B. VII., T. XII., S.] EXAMPLES. 1. What is the volume of a sphere 6 inches in diameter? Ans. 113'0976 cubic inches. 2. How many cubic miles in the earth, its diameter being 7912 miles? Ans. 259332805350 cubic miles. PROBLEM XIII. To find the area of a portion of the surface of a sphere included between three arcs of great circles. That is, to find the area of a spherical triangle. 124 MENSURATION. [CHAP. IX RULE. From the sum of the three angles of the triangle subtract 180~, divide the remainder by 90~ and multiply the quotient by one-eighth of the entire surface of the sphere of which the triangle forms apart. [Geom. B. VIII., T. XIX., S.] EXAMPLES. 1. How many square inches in a spherical triangle, described on a sphere 6 inches in diameter, and whose angles are 90~, 1000, 1100? Ans. 47'124 sq. inches. 2. What part of the entire surface of a sphere is the spherical triangle, each of whose angles is 120~? Ans. -. ~ 88. The following multipliers with their logarithms will be very convenient for reference. TABLE OF MULTIPLIERS. Log. of multipliers. 1. Radius of a circle X 628318531=Circumference............... 0...07981799 2. Square of the radius of a circle X314159265=Area............... 04971499 3. Diameter of a circle X 3 14159265=Circumference.................04971499 4. Square of the diameter of a circle X0'78539816=Area............ 18950899 5. Circumference of a circle X 015915494=Radius.................. 1'2018201 6. Circumference of a circle X0 31830989=Diameter......... 1.......15028501 7. Square root of area of a circle X 056418958=Radius..............1.'7.514251 8. Square root of area of a circle X 112837917=Diameter........... 00524551 9. Radius of circle X 1 73205081=Side of inscribed equilateral triangle...02385606 10. Side of inscribed equilateral tiriangle X 0 577 35027=Radius of circle...1 7614394 11. Radius of a circle X141421356=Side of inscribed square..........0.1505150 12. Side of inscribed square X 070710678=Radius................... 1'8494850 13. Square of radius of a sphere X 1256637061=Surface............. 1'0992099 14. Cube of radius of a sphere X4'18879020=Volume................. 06220886 15. Square of diameter of a sphere X 314159265=Surface............. 04971499 16. Cube of diameter of a sphere X 052359878=Volume..............17189986 17. Square of circumference of a sphere XO031830989=Surface........ 15028501 18. Cube of circumference of a sphere X0'01688686=Volume......... 2'2275490 19. Square root of surface of a sphere X 0'28209479=Radius.......... 14503951 20. S quare root of surface of a sphere XO 56418958=Diameter -........ 1'7514251 21. Square root of surface of a sphere X 1'77245385=Circumference...0 2485748 22. Cube root of volume of a sphereX0'62035049=Radius............ 17926371 23. Cube root of volume of a sphere X 1l24070098=Diamneter.........0'0936671 24. Cube root of volume of a sphere X 389777707=Circumnference.....0'5908170 25. Radius of a sphere X 1'15470054=Side of inscribed cube........... 00624694 26. Side of inscribed cube X 0 86602540=Radius...............1......9375306 TA BLE I.,, OF LOGARITHMS OF NUMBERS FROM 1 TO 10000. N. Log. N. Log. N. Log. N. Log. I oooooo000000 26 I.414973 5I 1.707570 76 I.88o8I4 2 o.3oo3o 27 I 43 1364 52 1.716003 77 1.88649I 3 0.477I2I 28 1.447158 53 1-724276 78 1.892095 4 0.602060 29 1.462398 54 I 1.732394 79 I'897627 5 0.698970 30 1.47712I 55 I.740363 80 1.903090 6 0-778I5I 3I I.49I362 56 1.748188 81 I.908485 7 0O845098 32 I 505150 57 I.755875 82 1.913814 8 0.903090 33 I.518514 58 1.763428 83 1.919078 9 0.954243 34 1.531479 59 I'770852 84 1-924279 10 I.000000 35 I.544068 60 1.778151 85 1 929419 11 I.041393 36 I.556303 6I 1.785330 So i 93/4498 I2 I 079181 37 1.568202 62 1.792392 87 1.939519 I3 I1Ii3943 38 I.579784 63 I 79934I 88 1.944483 I4 II46128 39. I 591065 64 I 806i80 89 1.949390 I5 1.176091 40 I 602060 65 1.8129I3 90 I.954243 I6 1.204120 41 1.6I2784 66 1.819544 91 1.959041 17 1.230449 42 1.623249 67 1'826075 92 I.963788 i8 1*255273 43 I 633468 68 1.832509 93 I' 968483 19 1.278754 44 I 643453 69 1.838849 94 I.973128 20 I 3oio3o 45 I 653213 70 1.845098 95 I.977724 21 1.32221I 46 1.662758 7I I.851258 96 1.982271 22 I-342423 47 I-672098 72 I.857333 97 1.986772 23 1.361728 48 1.681241 73 I.863323 98 1.991226 24 I-3802II 49 I-690196 74 1-869232 99 1-995635 25 I.397940 50 I*698970 75 I.8756I 10oo 2.000000 N. B. In the following table, in the last nine columns of each page, where the first or leading figures change from 9's to O's, the character + is introduced instead of the O's, to catch the eye, and to indicate that from thence the annexed first two figures of the Logarithm in the second column stand in the next lower line, directly under the asterisk. 1 2 LOGARITHMS OF NUMBERS. TABLE I. oN. -0 1 2 3o 4 5 6 7 8 9 D. 100 000000 0434 o868 [301 1734 2166 2598 3029 346.3891 432 101 4321 4751 51i8i 5609 6o38 6466 6894 7321 7748 8174 428 o102 86oo00 9026 94)j 9876 *300oo 0724 1147 1570 1993 2415 424 I03 01 2837 3259 3650o 4i00, 4521 4940 536o 5779 6197 66i6 419 104 *7033 7451 7868 8284 8700 9116 9532 9947 36oI 0775 416 105 02 1189 i603 2016 2428 2841 3252 3664 4075 4486 4896 412 io6 5306 5715 6125 6533 6942 7350 7757 8164 8571 8978 408 107 9384 9789 +195 o6oo 1004 1408 1812 2216 2619 3021 404 io8 03 3424 3826 4227 4628 5029 543o 583o 6230 6629 7028 400 109 * 7426 7825 8223 8620 9017 9414 981 +207 0602 0998 396 I10 0 i41393 1787 2182 2576 296 3362 3755 4148 4540 4932 393 III 5323 5714 1605 6495 6885 7275 7664 8o53 8442 883o 38 1 02 9218 96o6 9993 *38o 0766 JiS3 I538 1924 2309 2694 386 113 05 3078 J3463 J3846 4230 463 4996 5378 5760o 6142 6524 382 114 * 6905 7286 7666 8046 8426 8805 9185 9563 9942 j320 379 iiS5 060698 1075o 1452 1I829 2206 2582 2958 3333 3709 4083 376 ii6 4458 4832 5206 5580o 5953 6326 699 7071 7443 785 372 117 -8i86 8557 8928 9298 9668 *o38 0407 0776 1,45 1514 369 ii8 07 1882 2250 2617 2985 3352 3718 4o85 4451 4816 5182 366 119 5547 5912 6276 6640 7004 7368 7731 8094 8457 8819 363 120 *918.-954,13 9904.266 0626 0987 1347 1707 2067 2426 360o 121 08.2785 3i44 3o33 386i 4219 4576 4934 5291 5647 6004 357 122 6360 6716 7071 7426 7781 836 8490 8845 9198 9552 355 123 9905,.v58 o6ii o963 i3i5 I667 2018 2370 2721 3071 351 124 09 3422 3'172 4122 4471 4820 5169 5518 5866 6215 6562 349 125 *6910 72571 7604 7951 8298 8644 8990 9335 9681 +026 346 126 0ioo371 0715 1059 i4o3 1747 2091 2434 2777 319 3462 343 127 3804 4146 4487 4828 5169 55io 585i 6191i 653I 6871 34o 128 *72IO 7549 7888 8227 8565 8903 9241 9579 9916.253 338/ 129 o110590 0926 1263 1599 1934 2270 265 2940 3275 36 335 i3o 3043 4277 46i1 4944 5278 56iI 5943 6276 66o8 6940 333 131 * 7271 7603 7934 8265 8595 8926 9256 9586 9915.245 330o 132 120574 0903 123 I 56o 1888 2216 2544 2871 3198 3525 328 133 3852 4178 4504 4830 5,56 5481 58o6 6131 6456 6781 325 134 *7105 7429 7753 8076 8399 8722 9045 9368 9690 +012 323 135 i3 0334 o655 0977 1298 1619 1939 2260 2580 2900 3219 321 1i36 3539 3858 4177 4496 4814 533 5451 5769 6o86 64o3 3i8 137 6721 7037 7354 7671 7987 83o3 86i8 8934 9249 9564 315 i38 *9879 +i94 ojo8 0822 1ii36 1450 1763 20761 2389 2702 314 139 14301o5 3327 3639 3951 4263 4574 4885 5196 5507 58i8 3ii 140 6128 6438 6748 7058 7367 7676 7985 8294 86o3 891 I 309 141 * 9219 9527 9835.142 0449 0756 io63 370 1I676 1982 307 142 15228 2594 2900 3205 35,o 38i5 4120 44241 4728 5032 305 143 5336 5640 5943 6246 6549 6852 7154 7457 7759 8o6i 303 144 x 8362 8664 18965 9266 9567 9868.i68 0469 0769 Io68 3oi I45 i6,368 1 667 1 967 2266 2564 2863 3i6i 3460 3758 4055 299 46,4353 4650 4947 5244 554i 5838 634 643o 6726 7022 297 I47 7317 763 79o 8203 8497 8792 9086 9380 9674 9968 295 I4 17 0262 o555 o848 1 434 1726 2019 231 263 295 293 49 3i86 3478 3769 4060 4351 464, 4932 5222 5512 5802 291 I50 6091 638i 6670 5959 7248 7536 7825 8ii3 84oi 8689 289 i5i * 8977 9264 9552 9839 +126 04 069 0985 1272 JI558 287 152 I8 1844 2129 24I5 2700 2985 3270 355 3839 4123 4407 285 153 4691 4975 5259 5542 5825 6io8 6391 6674 6956 7239 283,54 * 7521 7803 8084 8366 8647 8928 9209 94909771,o51 281 155 19 0332 o61i 2 0892 1171 1451 1730 2010 2289 2567 2846 279 i56 3125 3403 D368 3959 4237 45i4 4792 5069 5346 5623 278 157 5900 6176 6453 6729 7oo005 7281 7556 7832 8io 7 8382 276 i58 *8657 8932 9206 9481 9755 +029 o3o3 0577 o85o 1124 274 iS9 20 1397 1670 1943 2216 2488 2761 3033 33oS 3577 3848 272 N, 0 1/2 3 4 I 5/6<7 8, 9 1D. TABLE I. LOGARITHMS OF NUMBERS. 3 N. 0 1 2 3 4 5 6 7 S 9 D. 160 20 4120 4391 4663 4934 5204 5475 5746 6016 6286 6556 271 i6i 6826 7090 7365 7634 7904 8173 844I 8710o 8979 9247 269 I62 *9515 9783 +o51 0319 o586 o853 I21 i388 I654 1921 267 i63 2I 2188 2454 2720 2986 3252 35i8 3783 4049 4314 4579 266 I64 4844 5o09 5373 5638 5902 6i66 6430 6694 6957 7221 264 I65 7484 7747 80Io 8273 8536 8798 9060 9323 1 9585 9846 262 i66 22 OIo8 o370 o631 o892 1153 I414 I675 1936 12196 2456 26I I67 27I6 2976 3236 3496 3755 40I5 4274 4533 4792 5051 259 i68 5309 568 5826 6084 6342 6600 6858 7I15 7372 7630 258 169 * 7887 8I44 8400 8657 8913 9170 9426 9682 9938 0193 256 170 23 0449 0704 0960 1215 1470 1724 I979 2234 2488 2742 254 171 2996 3250 3504 3757 4011 4264 41. 1 2 5023 5276 253 172 5528 578i 6033 6285 6537 6789 o04I 7292 7544 7795 252 I73 *8046 8297 8548 8799 9049 9299 9550 9800 *050 o300 250 174 24 0549 0799 1048 1297 1546 1795 2044 2293 254I 2790 249 175 3038 3286 3534 3782 4030 4277 4525 4772 5019 5266 248 176 55I3 5759 60o6 6252 6499 6745 699I 7237 7482 7728 246 177 *7973 8219 8464 8709 8954 9198 9443 9687 9932 +176 245 I78 25 0420 o664 ogo8 1151 1395 I638 I88I 2125 2368 26IO 243 179 2853 3096 3338 3580 3822 4o64 4306 4548 4790 503i 242 I80o 5273 5514 5755 5996 6237 6477 67I8 6958 7198 7439 24I I8I { 07679 79I8 8i58 8398 8637 8877 91I6 9355 9594 9833 239 i82 26 oo007 o3io o548 0787 1025 1263 i50o 1739 1976 2214 238 i83 245I 2688 2925 3162 3399 3636 3873 4109 4346 4582 237 r84 4818 5054 5290 5525 5761 5996 6232 6467 6702 6937 235 I85 7172 7406 7641 7875 8IIo 8344 8578 88I2 9046 9279 234 186 *95I3 9746 9980 *2I3 o446 o679 0912 1144 1377 1609 233 I87 27 1842 2074 2306 2538 2770 3ooi 3233 3464 3696 3927 232 i88 4158 4389 4620 4850 50o8 5311 5542 5772 6002 6232 230 I89 6462 6692 6921 7151 7380 7609 7838 8067 8296 8525 229 Igo * 8754 8982 21II 9439 9667 9895 1*23 o35 o0578 o806 228 191 28 1033 126I I488 715I 1942 2I69 2396 2622 2849 3075 227 192 33oi 3527 3753 3979 4205 4431 4656 4882 5107 5332 226 193 5557 5782 6007 6232 6456 668I 6905 7I30 7354 7578 225 I94 7802 8026 8249 8473 8696 8920 9143 9366 9589 9812 223 195 29 oo35 0257 0480 0702 0925 1147 1369 159I I813 2034 222 196 2256 2478 2699 2920 3141 3363 3584 3804 4025 4246 22I 197 4466 4687 4907 5127 5347 5567 5787 6007 6226 6446 220 198 6665 6884 7104 7323 7542 7761 7979 8198 84I6 8635 2Ig 199 * 8853 9071 9289 9507 9725 9943 OI6I o378 o595 o813 219 200 30 1o3o 1247 1464 i68i 1898 2I14 233I 2547 2764 2980 2I7 201 3196 3412 3628 3844 4059 4275 449' 4706 492I 5i36 216 202 535I 5566 578I 5996 6211 6425 6639 6854 7068 7282 2I5 203 7496 77IO 7924 8137 835i 8564 8778 8991 9204 94I7 213 204 *9630 9843 *o56 0268 0481 o693 o090o6 iii8 33o 1542 212 205 31 754 1966 2177 2389 2600 2812 3023 3234 3445 3656 211 206 3867 4078 4289 4499 4710 4920 513o 5340 5551 5760 210 207 5970 6i8o 6390 6599 6809 7018 7227 7436 7646 7854 209 208 8o63 8272 848I 8689 8898 91o6 9314 9522 9730 9938 208 209 32 0146 o354 o562 0769 0977 1184 I39I I598 I8o5 2012 207 210 2219 2426 2633 2839 3046 3252 3458 3665 3871 4077 206 211 4282 4488 4694 4899 5105 5310 5516 572I 5926 6i3I 205 212 6336 654i 6745 6950 7155 7359 7563 7767 7972 8176 204 213 8380 8583 8787 8991 9194 9398 96O0 9805 0oo8 0211 203 214 33 0414 0617 08I9 1022 1225 1427 163o 1832 2034 2236 202 215 2438 2640 2842 3044 3246 3447 3649 3850 4051 4253 202 216 4454 4655 4856 5057 5257 5458 5658 5859 6o5q 6260 201 217 6460 6660 6860 7060 7260 7459 7659 17858 805 8257 200 218 * 8456 8656 8855 9054 9253 9451 9650 9849 *047 0246 I99 219 34 0444 0642 o84I I39g 1237 I435 I632 183o 2028 2225 I98 N. I 0 1 2 3 4 6 71 8 |9 D. 4 LOGARITHMS OF NUMBERS. TABLE I. N. 0 1 2 3 4 6 6 7 8 9 D. 220 34 2423 2620 2817 3oI4 3212 3409 36o6 3802 3999 4I96 I97 221 4392 4589 4785 498I 5178 5374 5570 5766 5962 61 7 196 222 6353 6549 6744 6939 7135 7330 7525 7720 7915 8iO Ig95 223 83o5 85oo00 8694 8889 9083 9278'9472 9666 9860 *o54 194 224 35 0248 0442 o636 0829 1023 1216 1410 16o3 1796 1989 I93 225 2I83 2375 2568 276I 2954 3147 3339 3532 3724 3916 I93 226 4108 43oi 4493 4685 4876 5068 5260 5452 5643 5834 192 227 6026 62I7 6408 6599 6790 6981 7172 7363 7554 7744 191 228 7935 8125 83i6 8506 8696 8886 9076 9266 9456 9646 I90 229 * 9835 *025 0215 0404 0593 o783 0972 iI6I I350o 539 I19 230 36 1728 9I17 2io5 2294 2482 267I 2859 3048 3236 3424 i88 23I 361I2 3800 3988 4176 4363 4551 4739 4926 5ii3 53o0 188 232 5488 5675 5862 6o49 6236 6423 66io 6796 6983 7169 i 87 233 7356 7542 7729 7915 -8oi 8287 8473 8659 8845 9030 I86 234 *9216 9401 9587 9772 9958 *I43 o328 o5I3 o698 o883 i85 235 37 Io68 1253 1437 1622 I806 IggI 2175 2360 2544 2728 I84 236 2912 3096 3280 3464 3647 383I 40 I 5 4198 4382 4565 I84 237 4748 4932 5115 5298 548I 5664 5846 6029 6212 6394 I83 238 6577 6759 6942 7124 7306 7488 7670 7852 8034 82I6 i82 239 * 8398 8580 876i 8943 9124 9306 9487 9668 9849.03 0 I8i 240 38 0211 o392 0573 0754 0934 1115 I296 147'6 i656 I837 i8i 241 2017 2197 2377 2557 2737 2917 3097 3277 3456 3636 I8o 242 3815 3995 4174 4353 4533 4712 489 5070 5249 5428 179 243 5606 5785 5964 6142 6321 6499 6677 6856 7034 72I2 178 244 7390 7568 7746 7923 8IoI 8279 8456 8634 8811 8989 178 245 * 9gI66 9343 9520 9698 9875 *051 0228 o405 0582 0759 177 240 390935 1112 1288 i464 I64I 1817 1993 2169 2345 2521 176 247 2697 2873 3048 3224 3400 3575 3751 3926 4101 4277 176 248 4452 4627 4802 4977 5152 5326 5501o 5676 5850 6025 175 249 6199 6374 6548 6722 6896 7071 7245 7419 7592 7766 174 250 7940 8114 8287 8461 8634 8808 8981 9154 9328 950o I73 251 *9674 9847 *020 o192 o365 o538 0711 o883 io56 1228 173 252 40 I401 1573 1745 19I7 2089 2261 2433 2605 2777 2949 172 253 3121 3292 3464 3635 3807 3978 4149 4320 4492 4663 171 254 4834 5005 5I76 5346 5517 5688 5858 6029 6199gg 6370 17I 255 6540 6710 688I 7051 7221 7391 7561 773I 7901 8070 170 256 8240 84io 8579 8749 8918 9087 9257 9426 9595 9764 I69 257 *9933 *102 0271 0440 0609 0777 0946 1114 I1283 I45I 169 258 41 1620 1788 1956 2124 2293 2461 2629 2796 2964 3132 I68 259 3300 3467 3635 3803 3970 4137 430o 4472 4639 4806 I67 260 4973 5140 5307 5474 564I 5808 5974 6I4I 6308 6474 I67 26I 664I 6807 6973 7139 7306 7472 7638 7804 7970 8135 I66 262 83oi 8467 8633 8798 8964 91I29 9295 9460 9625 979I i65 263 *9956 *I21 0286 045I o066 078I 0945 1110 1I275 i439 i65 264 42 1604 1768 I933 2097 226i 2426 2590 2754 2918 3082 I64 265 3246 34IO 3574 3737 3901 4065 4228 4392 4555 4718 I64 266 4882 5045 5208 5371 5534 5697 586o 6023 6i86 6349 I63 267 65ii 6674 6836 6999 7I6I 7324 7486 7648 78 I 7973 162 268 8135 8297 8459 8621 8783 8944 9106 9268 9429 9591 162 269 *9752 9914 *o07 0236 0398 0559 0720 o88I 1042 1203 161 270 43 i364 1525 I685 I846 2007 2I67 2328 2488 2649 2809 I6i 271 2969 3i3o 3290 3450 3610 3770 3930 4090 4249 4409 i6o 272 4569 4729 4888 5048 5207 5367 5526 5685 5844 6004 I59 273 6163 6322 648i 6640 6799 6957 7116 7275 7433 7592 I59 274 7751 7909 8067 8226 8384 8542 870I 8859 9017 9175 158 275 *9333 949I 9648 rco6 9964 *122 0279 0437 o594 o752 I58 276 44 0909 io66 1224 i38I I538 i695 I852 2009 2I66 2323 I57 277 2480 2637 2793 295o 3io6 3263 3419 3576 3732 13889j i57 278 4045 |4201 4357 4513 4669 4825 4981 5137 5293 5449 56 279 56o4 S76o S54 2 5604 576o 595 6071 6226 6382 6537 6692 6848 7003 I55 N. 012 3 4 5 6 178 9 i D. TABLE I. LOGARITHMS OF NUMBERS. 5 N. O 1 2 3 4 5 6 7 8 9 D. 280 44 7158 7313 7468 7623 7778 7933 8088 8242 8397 8552 I55 281 * 8706 886I 9015 9170 9324 9478 9633 9787 9941 *095 154 282 45 o0249 o403 0557 0711 o865 i018 1172 1326 1479 I633 I54 283 1786 1940 2093 2247 2400 2553 27o6 2859 3012 3i65 I53 284 33i8 3471 3624 3777 3930 4082 4235 4387 4540 4692 I53 285 4845 4997 515o 5302 5454 56o6 5758 5910 6062 6214 I52 286 6366 65 8 6670 682 6973 7125 7276 7428 7579 7731 152 287 7882 8033 8184 8336 8487 8638 8789 8940 9091 9242 I5I 288 *9392 9543 9694 9845 9995 *I46 0296 0447 o597 0748 151 289 46 o898 o148 1198 I348 1499 I649 I799 1948 2098 2248 i5o 290 2398 2548 2697 2847 2997 3 46 3296 3445 3594 3744 50o 291 3893 4042 4191 434o 4490 4639 4788 4936 5085 5234 149 292 5383 5532 568o 5829 5977 6126 6274 6423 657I 6719 149 293 6868 7016 7164 7312 7460 7608 7756 7904 8052 8200 I48 294 8347 8495 8643 8790 8938 9085 9233 9380 9527 9675 148 295 * 9822 9969 116 0263 0410 0557 0704 o85i 0998 I145 I47 296 47 1292 1438 1585 1732 1878 2025 2171 2318 2464 26io i46 297 2756 2903 3049 3195 334I 3487 3633 377 3925 407 46 298 4216 4362 4508 4653 4799 4944 5090 5235 538i 5526 I46 299 5671 58i6 5962 6107 6252 6397 6542 6687 6832 6976 I45 300 7121 7266 7411I 7555 7700 7844 7989 8133 8278 8422 I45 3oi 8566 8711 8855 8999 9143 9287 9431 9575 9719 9863 144 302 48 0007 015I 0294 o438 o582 0725 o869 1io02 156 1299 144 303 I443 i586 1729 I872 2016 2159 2302 2445 2588 2731 143 304 2874 30o6 3159 3302 3445 3587 3730 3872 4015 4157 143 305 4300 4442 4585 4727 4869 5oii 5153 5295 5437 5579 142 3o6 5721 5863 6005 6147 6289 6430 6572 6714 6855 6997 I42 307 7138 7280 7421 7563 7704 7845 7986 8127 8269 8410 141 308 855I 8692 8833 8974 9114. 9255 9396 9537 9677 9818 I41 309 * 9958 *099 0239 o38o0 520 066I o8o0I 94 Io08i 1222 140 310 49 I362 1502 I642 1782 1922 2062 2201 2341 2481 2621 140 311 2760 290o 304o 3179 3319 3458 3597 3737 3876 40I5 I39 312 4155 4294 4433 4572 47II 4850 4989 5128 5267 54o6 I39 313 5544 5683 5822 5960 6099 6238 6376 65i5 6653 6791 i39 314 6930 7068 7206 7344 7483 7621 7759 7897 8o35 8173 138 315 8311 8448 8586 8724 8862 89999 9137 9275 9412 9550 i38 3i6 * 9687 9824 9962 +099 0236 0374 o51I 0648 0785 0922 I37 317 50 Io5g I196 I333 1470 I607 I744 I880 2017 2154 2291 137 318 2427 2564 2700 2837 2973 3109 3246 3382 35i8 3655 I36 319 379I 3927 4063 4199 4335 4471 4607 4743 4878 50I4 i36 320c'51o50 5286 5421 5557 5693 5828 5964 6099 623 4 6370 136 32I 6505 6640 6776 6911 I 7046 718I 7316 7451 7586 7721 i35 322 7856 7991 8126 8260 8395 8530 8664 8799 8934 9068 135 323 * 9203 9337 9471 9606 9740 9874 0oo9 oi43 0277 0411 134 324 5I o545 o679 o813 0947 o8I 1215 I349 i482 i616 1750 I34 325 i883 2017 2151 2284 2418 2551 2684 2818 2951 3084 I33 326 3218 335i 3484 3617 3750 3883 4oi6 4149 4282 4414 I33 327 4548 468I 4813 4946 5079 52II 5344 5476 5609 5741 I33 328 5874' 6006 6139 6271 6403 6535 6668 6800 6932 7064 i32 329 7196 7328 7460 7592 7724 7855 7987 8119 825i 8382 132 330 8514 8646 8777 8909 9040 917 9303 9434 9566 9697 131 331 * 9828 9959 +090 022I o353 o484 o6i5 0745 o876 1007 i3i 332 52 1138 1269 I400 1 530 I66i 1792 I922 2053 2183 2314 I3i 333 2444 2575 2705 2835 2966 3096 3226 3356 3486 36i6 I30 334 3746 3876 40o6 4136 4266 4396 4526 4656 4785 4915 i3o 335 5045 5174 5304 5434 5563 5693 5822 5951 608I 6210 129 336 6339 6469 6598 6727 6856 6985 7114 7243 7372 75o0 129 337 7630 7759 7888 8oi6 8145 8274 8402 853i 8660 8788 129 338 * 8917 9045 9174 9302 9430 9559 9687 9815 9943 +072 128 339 53 0200 o328 o456 o584 0712 0840 0968 1096 1223 i35i 128 N. 0 1 2 3 4 5 6 7 i 8 9 D. 6 LOGARITHMS OF NUMBERS. TABLE I. N. 0 1 2 3 4 5 6 7 8 9 D. 340 53 1479 1607 1734 1862 IggO990 2117 2245 2372 1 2500 2627 128 34I 2754 2882 30091 336 3264 3391 3518 3645 3772 3899 127 342 4026 4153 4280 4407 4534 466I 4787 4914 504I 5167 I27. 343 5294 5421 5547 5674 58o0 5927 6053 680o 63o6 6432 126 344 6558 6685 68ii 6937 7063 7189 7315 744I 7567 7693 126 345 7819 7945 807I 8197 8322 8448 8574 8699 8825 8951 126 346 *9076 9202 9327 9452 9578 9703 9829 9954 +079 0204 125 347 540329 0455 o580 0705 o83o 0955 io8o 1205 I33o 14541 I25 348 1 579 1704 I829 1953 2078 2203 2327 2452 2576 2701 125 349 2825 2950 3074 3199 3323 3447 357I 3696 3820 3944 24 350 4068 4192 4316 4440 4564 4688 4812 4936 5060 5I83 124 35i 5307 543i 5555 5678 5802 5925 6049 6172 6296 6419 124 352 6543 6666 6789 6913 7036 7159 7282 7405 7529 7652 123 353 7775 7898 8021 8144 8267 8389 8512 8635 8758 888i 123 354 *9o03 9126 9249 937I 9494 96I6 9739 986I1 9984 +io6 123 355 55 0228 o35i 0473 o595 0717 o840 0962 1o84 I206 I328 I22 356 i450 I572 1694 I816 1938 2060 218I 2303 2425 2547 122 35;7 I 2668 2790 2911 3033 3155 3276 3398 3519 364o 3762 I21 358 | 3883 4004 4126 4247 4368 4489 4610 473I 4852 4973 121 359 5094 52I5 5336 5457! 5578 5699 5820 5940 606i 6182 12I 360 6303 6423 6544 6664 6785 6905 7026 7146 7267 7387 120 36i 7507 7627 7748 7868 7988 8108 8228 8349 8469 8589 120 362 8709 8829 8948 9068 9188 9308 9428 9548 9667 9787 120 363 * 9907 o026 046 0265 o385 o504 0624 0743 o863 og982 119g 364 56 iioi 1221 340 I459 1578 1698 18I7 1936 2055 2174 119 365 2293 2412 2531 2650 2769 2887 3006 3125 3244 3362 119 366 348 36oo 3718 3837 395) 4074 4192 431 4429 4548 Ig 367 4666 4784 4903 5021 5139 5257 5376 5494 5612 5730 118 368 5848 5966 6084 6202 6320 6437 6555 6673 6791 6909 1i8 369 7026 7144 7262 7379 7497 7614 7732 7849 7967 8084 ii8 370 8202 8319 8436 8554 8671 8788 8905 9023 9g40 9257 117 371 * 9374 9491 9608 9725 9842 9959 0*o76 oi3 o309 0426 117 372 57 o543 o66o 0776 o893 IOIO 1126 1243 1359 I476 1592 117 373 1709 1825 1942 2058 2174 2291 2407 2523 2639 2755 II6 374 2872 2988 3io4 3220 3336 3452 3568 3684 38oo 3915 ii6 375 4o3I. 4147 4263 4379 4494 46io 4726 484I 4957 5072 116 376 5188 5303 5419 5534 5650 5765 5880 5996 6111 6226 115 377 6341 6457 6572 6687 6802 6917 7032 7147 7262 7377 115 378 7492 7607 7722 7836 7951 8o66 818i 8295 84Io 8525 115 379 8639 8754 868 8983 9097 9212 9326 944I 9555 9669 14 380 *9784 9898 +012 0126 0241 o355 0469 o583 o697 o811 14 38i 58 0925 039 i53 1 267 i38i 1495 6o8 1722 i836 1 950 I14 382 2063 2177 2291 2404 2518 2631 2745 2858 2972 3o85 II4 383 31gg99 3312 3426 3539 3652 3765 3879 3992 4io5 42I8 II3 384 433I 444 4557 4670 4783 4896 5009 5122 5235 5348 ii3 385 546i 5574 5686 5799 5912 6024 1 6I37 6250 6362 6475 II3 386 6587 6700 68I2 6925 7037 7149'7262 7374 7486 7599 112 387 77II 7823 7935 8047 8I6o 8272 8384 8496 8608 8720 112 388 | 8832 |8944 9056 g9167 9279 9391 9503 9615 9726 9838 I12 389 * 9950 *o6I 0173 0284 0396 0507 06190730 0842 o953 112 390 59 1065 1 76 1287 1399 I5IO 162I 1732 i 843 1955 2066 III 391 2I77 2288 2399 2510 2621 2732 12843 2954 3o64 3175 IIi 392 3286 3397 3508 36i8 3729 3840 3950 4o6I 4171 4282 III 393 4393 503 46I4 4724 4834 4945 5055 5i65 5276 5386 IIO 394 5496 5606 5717 5827 5937 6047 16157 6267 1 6377 6487 II0 395 6597 6707 6817 6927 7037 7146 7256 7366 7476 7586 I1o 396 7695 7805 79I4 8024| 834 8243 8353 8462 8572 868 i 110 397 8718 879 00 9009 911I9 9228 9337 9446 9556 9665 9774 Iog 398 - *9883 |9992 +10 0210 Io3i 0428 0537 o646 0755 0864 Og 399 60 0973 1o82 II9I 1299 I408 15I7 1625 I1734 1843 1951 109 N. 0 1 2 3 4 5 6 7 8 9 D. TABLE I. LOGARITHMS OF NUMBERS. N. 0 1 2 3 4 5 6 8 9 D. 400o 602060 269 2277 2386 2494 2603 2711 2819 2928 3036 io08 401 3144 3253 336i 3469 3577 3686 3794 390o2 4oio 4i8 io8 402 4226 4334 4442 455o 4658 4766 4874 4982 5089 5197 108 403 5305 54,3 5521 5628 5736 5844 5951 6059 6166 6274 io8 404 638i 6489 6596 6704 68ii 6919 7026 7133 721 7348 107 405 7455 7562 i669 7777 7884 799 808 8312 849 07 406 8526 8633 8740 8847 8954 9061 9167 9274 938I 9488 107 407 9594 97oi 9808o8 9914 +021 0128 0234 0341I 0447 o54 107 408 6i o66o 0767 0873 0979 io86 1192 1298 14o5 1511 1617 io6 409 1723 1829 1936 2042 2148 2254 2360 2466 2572 2678 io6 410 2784 2890 2996 3102 3207 3313 4i9 3525 363o 3736 io6 ~411 3842 394740o53 459 4264 4370 4475 4581 4686 4792 io6 412 4897 5oo3iIo8 5213 5319 5424 5529 5634 5740 5845 9 o} 413 5950 6o55 6i60o 6265 6370 6476 658i 6686 6790 6895 105 414 7000 7105 7210 7315 7420 7525 7629 7734 7839 7943 105 4'5 8048 8153 8257 8362 8466 8571 8676 878o0 8884 8989 io5 416 *9093 919819302 9406 9511 9615 9719 9824 9928 +032 104 417 62 0136 024/0 o344 0448 0552 o65 0760 84 98 1072 104 o66 76o o86/4 o968 10 418 1176 1280,384 1488 1592 1695 1799 1903 2007 2110 I 104 419 i 2214 231812421 2525 2628 2732 2835 2939 3042 3{ 46 104 420 3249 3353 3456 3559 3663 3766 3869 3973 4076 4179 io3 421 4282 4385 4488 4591 4695 4798 490 5oo0041 5107 5210 io3 422 5312 54,5 55i8 5621 5724 5827 5929 6032 635 6238 io3 423 6340 6443 6546 6648 6751 6853 6956 7058 7161 7263 io3 424 7366 7468 7571 7673 7775 7878 7980 8082 8i85 8287 102 425 8389 8491 8593 8695 8797 8900 9002 9104 9206 9308 102 426 9410 9512 9613 9715 9817 9919 +021 01231 224 0326 102 427 630428 o53o o63i 0733 0835 0936 io38 11391 241 1342 102 428 1444 1545 1647 1748 1849 1951 2052 2153 2255 2356 o101 429 2457 2559 2660 2761 2862 2963 3064 3i65 326 3367 o101 430 3468 3569 3670 3771 3872 3973 4074 4175 4276 4376 100 431 4477 4578 4679 4779 4880 4981 So8i 5182 5283 5383 0oo 432 5484 5584 5685 5785 5886 5986 6087 6187 6287 6388 100 433 6488 6588 6688 6789 6889 6989 7089 789 7290 7390 100 434 74go 7590 7690 7790 7890 7990 8090 8190 8290 9 99 435 8489 8589 8689 8789 8888 8988 9088 1i88 9287 9387 99 436 9486 9586 9686 9785 9885 9984 +o84 oi83 0283 0382 99 437 640481 o58i o68o 0779 0879 0978 1077 1177 1276 1375 99,438 1474 1573 1672 1771 1871 1970 2069 2168 2267 2366 99 439 2465 2563 2662 2761 2860 2959 3o58 3i56 3255 3354 99 440 3453 355i 3650o 3749 3847 3946 4044 4i43 4242 4340 98 441 4439 4537 4636[4734 4832 4931 5029 5127 5226 5324 98 442 5422 5521 5619 5717 5815 5913 6oii 6i0io 6208 630o6 98 443 6404 6502 6600oo 6698 6796 6894 6992 7089 7187 7285 98 444 7383 7481 7579 7676 7774 7872 7969 8067 8i65 8262 98 445 836o 8458. 8555 8653 8750 8848 8945 90o43 9140 9237 97 446 9335 9432 9530o 9627 9724 9821 91)9 +oi6 oII3 021i 97 447 65o0308 o4oS 0502 0599 0696 0793 089o0 o9871 O 84 ii8i 9 448 1278 1375 1472 1 569 i666 1762 1859 I956 2053 2150 97 449 2246 2343 2440 2536 2633 2730 2826 2923 3019 3116 97 45o 3213 3309 3405 3502 3598 3695 3791 3888 3984 4o8o 96 451 4177 42731 4369 4465 4562 4658 4754 4850/4946 5042 96 452 5138 5235 5331 5427 5523 5619 5715 58io 5906 6002 96 453 6098 6194 6290 6386 6482 6577 6673 6769 6864 6960 96 454 7036 7152 7247 7343 7438 7534 7620i7725 7820 7916 96 /7132~~~~~~~~~~~ ~~~~~~~ 72 73436 963 455 8oii 8107 8202 8298 8393 8488 8584 8679 8774 8870 95 456 8965 9060 9155 9250 9346 9441 9536 9631 9726 9821 95 457 * 9916 *011 oio6 0201 0296 0391 0486 058i o676 0771 95 458 660865 0o960 io55'i5o 1245 1339 i43411529 1623 1718 95 459 i8i3 1907 2002 2096 2191 2286 2380 2475 2559 2663 95 N. 0 o 1 3 4 5 6 8- 9 ID. 8 LOGARITHMS OF NUMBERS. TABLE I. N. 0 1 2 3 4 5 6 7 8 9 D. 460 66 2758 2852 2947 3o04 3I35 3230 3324 3418 3512 3607 94 46I 3701 3795 3889 3983 4078 4I72 4266 4360 4454 4548 94 462 4642 4736 4830 4924 50o8 5112 5206 5299 5393 5487 94 463 558I 5675 5769 5862 5956 6050 6143 6237 6331 6424 94 464 65i8 6612 6705 6799 6892 6986 7079 7173 7266 7360 94 465 7453 7546 7640 7733 7826 7920 8o3 81o6 8199gg 8293 93 466 8386 8479 8572 8665 8759 8852 8945 9038 g9131 9224 93 467 * 9317 941 9503 9596 9689 9782 9875 9967 *o6o o153 93 468 67 246 o339 o43 0524 o67 o710o 0802 o895 0988 Io8o 93 469 1173 1265 i358 I45I I543 I636 1728 1821 19g3 2005 93 470 2098 219o 2283 2375 2467 2560 2652 2744 2836 2929 92 471 3021 3113 3205 3297 3390 3482 3574 3666 3758 3850 92 472 3942 4034 4126 4218 431o 4402 4494 4586 4677 4769 92 473 486I 4953 5045 5137 5228 5320 5412 5503 5595 5687 92 474 5778 5870 5962 6053 6145 6236 6328 6419 65ii 6602 92 475 6694 6785 6876 6968 7059 715 7242 7333 7424 7516 91 476 7 0 7698 7789 788 7972 8o63 854 8245 8336 8427 91 477 8518 8609 8700 8791 8882 8973 9064 9155 9246 9337 91 478 *9428 9519 9610 9700 9791 9882 9973 *o63 oi54 0245 9I 479 68o336 0426 o517 o60 7 0698 0789 o879 o0970 o6o 151 9gI 480 1241 I332 1422 15i3 i6o3 I693 1784 I874 1964 2055 90 481 2145 2235 2326 2416 2506 2596 2686 2777 2867 2957 90 482 3047 3I37 3227 3317 3407 3497 3587 3677 3767 3857 90 483 3947 4037 4I27 42I7 4307 4396 4486 4576 4666 4756 90 484 4845 4935 5025 5114 5204 5294 5383 5473 5563 5652 90 485 5742 583 5921 6oio 6ioo 6 89 6279 6368 6458 6547 89 486 6636 6726 68 6.5 6904 6994 7083 7172 7261 735I 7440 89 487 7529 76I8 7o707 77 7975 8o64 8I53 8242 833i 89 488 8420 859 8598 8687 8776 8865 893 9042 9131 92 20 89 489 *9309 9398 9486 9575 9664 9753 9841 9930 oo09 0107 89 490 69 0196 0285 o373 0462 o550 o639 0728 oS16 go905 o993 8 49I Io8I 1170 1258 1347 I435 I524 I612 1700 I789 1877 88 492 I965 2053 2I42 2230 2318 2406 2494 2583 2671 2759 88 493 2847 2935 3023 311I 3I99 3287 3375 3463 355I 3639 88 494 3727 315 393 39399 4078 4166 4254 4342 4430 4517 88 495 4605 4693 478, 4868 4956 5044 513I 5219 5307 5394 88 496 5482 5569 5657 5744 5832 5919 6007 6094 6182 6269 87 497 6356 6444 653i 66i8 6706 6793 6880 6968 7o55 7142 87 498 7229 7317 7404 7491 7578 7665 7752 7839 7926 80o4 87 499 81io 8i88 8275 8362 8449 8535 8622 8709 8796 8883 87 500 870 o9057 9144 923I 93I7 9404 9491 9578 9664 975I 87 501 *9838 9924 *01I 0098 0184 0271 o358 0444 o53I o617 87 502 70 0704 0790 o877 o0963 Io50 II36 1222 I309 1395 I482 86 5o3 i568 I654 174I 1827 1913 I999 2086 2172 2238 2344 86 504 2431 2517 2603 2689 2775 2861 2947 3033 319g 3205 86 505 329I 3377 3463 3549 3635 372 138o7 3895 3979 4065 86 506 4151 4236 4322 4408 4494 4579 4665 4751 4837 4922 86 507 5008 5094. 5179 5265 5350 5436 5522 5607 5693 5778 86 508 5864 5949 6035 6120 6206 6291 6376 6462 6547 6632 85 509 6718 6803 6888 6974 70o59 7144 7229 7315 7400 7485 85 5io 7570 7655 7740 7826 791 7996 808 i 8I66 8251 8336 85 51I 842I 8506 8591 8676 8761 8846 8931 9015 9100ioo 985 85 512 * 9270 9355 9440 9524 9609 9694 9779 9863 9948 *033 85 513 71 o0117 0202 0287 0371 0456 o540 o625 0710 0794 o879 85 514 0963 1048 II32 1217 i3o I385 1470 i1554 1639 1723 84 515 I807 1892 I976 2060 2144 2229 233 12397 2481 2566 84 5I6 2650 2734 28 8 2902 2986 3070 3I54 3238 3323 3407 84 517 3491 3575 3650 3742 3826 39IO'3994 4078 4162 4246 84 518 4330 4414 4497 4581 4665 4749 4833 4916 5ooo 5084 84 5i9 5167 525I 5335 54I8 5502 5586 5669 5753 5836 5920 84 N. 0 1 21 3 4 5 6 7 8 9 D TABLE I. LOGARITHMS OF NUMBERS. 9 N. | 1 2 3 4 5 6 7 8 9 D. 520 7 6oo3 6087 6o70 6254 6337 6421 6504 6588 6671 6754 83 521 6838 6921 7004 7088 717I 7254 7338 7421 7504 7587 83 522 7671 7754 7837 7920 8003 8086 8169 8253 8336 8419 83 523 8502 8585 8668 875i 8834 8917 9000 9083 9165 9248 83 524 * 9331 9414 1 9497 9580 9663 9745 9828 1991 9994.077 83 525 72 0159 0242 0325 0407 0490 0573 o655 0738 0821 0903 83 526 0986 o68 1151 1233 I316 I398 I48I I563 1646 1728 82 527 1 8II I893 1975 2058 2140 2222 2305 2387 2469 2552 82 528 2634 2716 2798 288I 2963 3045 3127 3209 3291 3374 82 529 3456 3538 3620 3702 3784 3866 3948 4030 41I2 4194 82 530 4276 4358 44401 4522 4604 4685 4767 4849 4931 5o13 82 531 5095 5176 5258 5340 5422 5503 5585 5667 5748 583o 82 532 5912 5993 6075 6i56 6238 6320 64o, 6483 6564 6646 82 533 6727 6809 6890 6972 7053 7134 7216 7297 7379 7460 8i 534 7541 7623 7704 7785 7866 7948 8029 810o 8191 8273 8I 535 8354 8435 85i6 8597 8678 8759 8841 8922 9003 9084 8i 536 9165 9246 9327 94o8 9489 9570 9651 9732 9813 9893 8I 537 *9974 *o55 oi36 0217 o298 0378 o459 o5401o 621 0702 8i 538 73 0782 o863 0944 1024 Iio5 Ii86 1266 1347 1428 i5o8 8i 539 I589 I669 1750 i83o II9I 1991 2072 2i52 2233 2313 8I 54o 2394 2474 2555 2635 2715 2796 2876 2956 3037 3117 8o 541 3197 3278 3358 3438 35I8 3598 3679 3759 3839 3919 80 542 3999 4079 460 4240 4320 4400 4480 456o 4640 4720 80 543 4800 4880 4960 5040 5120 5200 5279 5359 5439 5519 80 544 5599 5679 5759 5838 5918 5998 6078 |657 6237 6317 80 545 6397 6476 6556 6635 6715 6795 6874 6954 7034 7113 80 546 7193 7272 7352 7431 7511 7590 7670 7749 7829 798 79 547 7987 8067 18146 8225 8305 8384 8463 8543 8622 8701 79 348 8781 886o 8939 o9018 9097 9177 9256 9333 9414 9493 79 549 *9572 9651 9731 i98i 9889 9968 io47 0126 0205 0284 79 550 74 o363 0442 052I o6oo o678 0757 o836 o915 o994 | 1073 79 551 1152 1230 130q i388 1467 1546 1624 1703 1782 86 79 552 1939 208 20o96 2175 2254 2332 2411 2489 12568 2646 79 553 2723 2804 2882 2961 3039 3118 3196 32753 3353 3431 78 554 3510 3588 3667 3745 3823 3902 3980 40581 436 14215 78 555 4293 437, 4449 4528 4606 4684 4762 4840 4919 4997 78 556 5075 5153 5231 5309 5387 5465 5543 5621 5699 5777 78 557 5855 5933 6011 6089 6167 6245 6323 6401 6479 6556 -8 558 6634 6712 1 67g90 6868 6945 7023 7101 7179 7256 7334'8 559 7412 7489 7567 17645 7722 7800 7878 7955 8o33 8i o 78 56o 8i88 8266 8343 8421 8498 8576 8653 873i 88o8 8885 77 561 8963 9040 91,8 9195 9272 9350 9427 9504 9582 9659 77 562 *9736 9814 989I 9968.o45 0123 0200 0o277 o354 043 77 563 75 o508 o586 o663 0740 0817 o894 0971 1io48 1125 1202 77 564 1279 1356 1433 i510 I587 1664 1741 18i8 1895 1972 77 565 2048 2125 2202 2279 2356 2433 2509 2586 2663 2740 77 566 2816 2893 2970 3047 3123 3200' 3277 33531330 3506 77 567 3583 3660 3736 3813 3889 3966 4042 4119 4195 4272 77 568 4348 4425 450o 4578 4654 4730 4807 4883 49601 536 76 569 5112 5I89 5265 534i 5417 5494 5570 5646 5722 5799 76 570 5875 5951 J6027 6io3 6i8o 6256 6332 640816484 6560 76 57.1 6636 6712 16788 6864 6940 7016 7092 71681721447320 76 572 7396 772 7548 7624 7700 7775 7851 7927 8oo3 8079 76 573 8i55 8230 8306 83821 8458 8533 8609 8685 8761 8836 76 574 8912 8988 9063 9139 9214 9290 9366 944' 9517 9592 76 575 * 9668 9743 9819 9894 9970 *045 0121 o196 0272 o347 75 576 76 0422 0498 0573 0649 0724 0799 0875 0950 1025 110O 75 577 1176 125I 1326 1402 1477 1552 1627 1702 1778 853 75 578 I928 2003 2078 2153 2228 2303 2378 1 2453 12529 2604 75 579 2679 2754 2829 2904 2978 3053 3128 323 327 3353 75 N. 0 1_ 2 3 4 1 5 6 | 8 9 D. 10 LOGARITHMS OF NUMBERS. TABLE I. N. 0 1 2 3 4_ 5 6 7 8 9 D. 58o 763428 35o3 3578 3653 3727 3802 3877 3952 4027 41Io 75 581 4176 4251 4326 4400 4475 4550 4624 4699 4774 4848 75 582 4923 4998 5072 5147 5221 5296 537a 15445 5520 5594 75 583 5669 5743 5818 5892 5966 6o4i 61r5 690! 6264 *6338 74 584 6413 6487 6562 6636 6710 6785 6859 6933 7007 7082 74 585 7156 7230 7304 7379 7453 7527 7601 76751 7749 1 823 74 586 7898 7972 8046 8120 8i94 8268 8342 8416 8490 8564 74 587 8638 8712 8786 886o 8934 9008 9082 9I56' 9230 9303 74 588 *9377 9451 9525 9599 9673 9746 9820 I 9894 9968 +042 74 589 77 015 oi89 0263 o336 04Io o484 o557 o63i 0705 o778 74 590 o852 0926 o999 i073 II46 1220 123 1367 I44o I514 74 591 I 587 I66i 1734 I 808 i88I I955 2028 2102 2175 2248 73 592 2322 2395 2468 2542 2615 2688 2762 2835 2908 298I 73 593 3055 3128 3201 3274 3348 342I 3494 3567 364i7 33713 73 594 3786 386o 3933 4oo6 4079 4152 4225 4298 4371 4444 73 595 4517 4590 4663 4736 4809 4882 4955 5028 51oo 5173 73 596 5246 5319 5392 5465 5538 56io 5683 5756 5829 5902 73 597 5974 6047 6120 6193 6265 6338 641 i6483 6556 6629 73 598 6701 6774 6846 6919 6992 7064 7'37 7209 7282 7354 73 599 7427 7499 7572 7644 77117 7789 7862 7934 8oo6 8079 72 6oo 815i 8224 8296 8368 8441 8513 8583 8658 8730 88021 72 6oi 8874 8947 9019 9091 9 63 9236 9308 9380' 9452 9524 72 602 *9596 9669 9741 9813 9885 9957 029 OIOI 01 oi73 0245 72 6o3 780317 o389 o46I o533 o605 0677 0749 0821 0893 o0965 72 604 1037 I Og Ii81 I253 i324 I396 1468 540 I612 I684 72 605 1755 1827 1899 1971 2042 2114 2186 2258 2329 2401 72 606 2473 2544 2616 2688 2759 283I 2902 2974 3046 3117 72 607 3189 3260 333 2 3403 3475J 3546 3618 3689 37761 383 2 71 60o8 3904 3975 4o46 4I18 4I89 426I 4332 4403 4475 4546 71 609 4617 4689 4760 483I 4902 4974 5045 5II6 5187 5259 71 6Io 533o 540I 5472 5543 56I5 5686 5757 5828 589 5970 71 6ii 6o4i 6iiZ 6i83 6254 6325 6396 6467 6538 6609o 668o 71 612 675I 6822 6893 6964 7035 71o6 7177 7248 7319| 7390 71 613 7460 7531 7602 7673 7744 7815 7885 7956 1 027 8098 71 6I4 8168 8239 83io 838i 845i 8522 8593 8663 8734 8804 71 615 8875 8946 90o16 9087 9157 9228 9299 9369 9440 9510 71 616 * 9581 965i 9722 9792 9863 9933 oo004 0074 0144 0215 70 617 79 O285 o356 0426 o496 0567 0637 0707 0778 o848 o918 70 6i8 0988 1059 1129 1199 1269 i34o 14io 1480 155o I620 70 619 I691 176I I83I 1901 1971 204 211III 2181 2252 2322 70 620 2392 2462 2532 2602 2672 2742 28I2 12882 2952 3022 70 621 3092 3I62 323I 33o1 3371 344i 35iI 358i 3651 3721 70' 622 3790 386o 3930 4000 4070 4139 4209 1.4279 4349 4418 70 623 4488 4558 4627 4697 4767 4836 4906 4976 5o45 51 I5 70 624 5185 5254 5324 5393 5463 5532 5602 5672 5741 5811 70 625 588o 5949 6019 6o88 68 58 6227 6297 6366 6436 65o5 69 626 6574 6644 67I3 6782 6852 6921 690 7060 7129 7198 69 627 7268 7337 7406 7475 7545 7614 7683 7752 7821 7890 69 628 7960 8029 80981 867 8236 83o5 8374 8443 85i3 8582 69 629 8651i 8720 8789 8858 8927 8996 9065 9134 9203 9272 69 63o 934I 9409 9478 9547 96I6 9685 9754 9823 9892 9960 69 63i 80 00oo29 0098 o67 0o236 o3o5 0373 o442 o5II 0580 o648 69 632 0717 0786 o854 0923 0992 io6I 1129 119}8 1266 1335 69 633 1404 1472 1541 160o9 678 1747 I815 1884 1952 2021 69 634 2089 2i58 2226 2295 2363 2432 2500 2568 2637 2705 69 635 2774 2842 2910 2979 3047 3ii6 3184 3252 3321 3389 68 636 3457 352i 35594 3662 3730 3798 3867 3935 4003 4071 68 637 4139 4208 4276 4344 4412 4480 4548 46I6 4685 1 4753 68 638 4821 4889 4957 5025 3093 5i61 5229 5297 5365 5433 68 639 550I 5569 5637 5705 5773 584I 5908 5976 6044 6112 68 N. O 1 t 3 4 6 I 8 9 D. TABLE I. LOGARITHMS OF NUMBERS. 11 N. 0 1 2 3 4 5 6 7 S 9 D.,640 8o 6180 6248 1 6316 16384 645I 6519 6587 6655 6723 6790 68 641 6858 6926 6994 7061 7129 7197 7264 7332 7400 7467 68 642 7535 7603 17670 7738 7806 7873 7941 8oo8 8076 8143 68 643 8211 8279 18346 188414 848, 8549 86i6 8684 875I 8818 67 644 8886 8953 9021 o9088 9156 9223 9290 9358 9425 - 9492 67 645 -9560 9627 9694 9762 9829 9896.9964 03 oo0098 oi65 67 646 81 0233 o3o0 o367 o434 o51o o569 o636 o703 0770o 0837 67 647 0904 0971 1o39 Iio6 1173 1240 1307 I374 I441 i508 67 648 1D75 I642 1709 I776 1843 1910 1977 2044 21II 2178 67 649 2245 2312 2379 2445 2512 2579 2646 27I3 2780 2847 67 650 2913 2980 3047 3I,14 3i8i 3247 33,4 338i 3448 35,4 67 65I 3 58 3648 37I4 378I 3848 3914 398I 4048 41 14 148I 67 652 4248 4314 438I 4447 45i4 458i 4647 4714 4780 4847 67 653 4913 4980 5046 5113 5I79 5246 5312 5378 5445 55II 66,654 5578 5644 5711 5777 5843 5910 5976 6042 6I09 6175 66 655 6241 63o8 6374 6440 6506 6573 6639 6705 677 i 6838 66 656 6904 697o 7036 7102 7169 7235 730I 7367 7433 7499 66 657 7565 763I 7698 7764 7830 7896 7962 8028 8094 816o 66'658 8226 8292 8358 8424 8490 8556 8622 8688 8754 8820 66 659 8885 8951 9017 9083 9149 9215 9281 9346 9412 9478 66 660 * 9544 96IO 9676 974' 9807 9873 9939 o004 0070 oI36 66 66I 82 0201 0267 0333 0399 0464 0530 o595 o66 o0727 0792 66 662 o858 o92 og989 io55 1I120 I86 1251 1317 I382 I448 66 663 i514 I579,645 1710 1775 84 Ig o6 1972 2037 2o103 65 664 2168 2233 2299 2364 2430 2495 2560 2626 2691 2756 65 665 2822 2887 2952 30I8 3083 3I48 3213 3279 3344 3409 65 666 3474 3539 3605 3670 3735 3800 3865 3930 3996 4o6I 65 667 4126 4191 4256 432 4386 445I 45I6 458I 4646 47II 65 668 4776 484I 4906 497I 5o36 5Io, 5i66 523I 5296 5361 65 669 5426 5491 5556 5621 5686 5751 58i5 588o 5945 6oio 65 670 6075 614o 6204 6269 6334 6399 6464 6528 6593 6658 65 671 6723 6787 6852 6917 6981 7046 7I1i 7I75 7240 7305 65 672 7369 7434 7499 |7563 7628 7692 7757 782I 7886 795I 65 673 80o5 8080 8144 8209 8273 8338 8402 8467 853I 8595 64 674 866o 8724 / 8789 8853 8918 8982 49111 9175 9239 64 675 9304 9368 9432 9497 956I 9625 9690 9|75,4 9818 9882 64 676 * 9947 *o0I 0075 0139 0204 0268 0332 o3g6 o46o0 o0525 64 677 83 o589 o653 0717 o78I 0845 o09 0973 I037 1102 Ii66 64 678 1230 1294 I38 1422 |486 15o0 1614 I678 I742 I806 64 679 1870 1934 1998 2062 2126 2189 2253 2317 2381 2445 64 680 2509 2573 2637 2700 2764 2828 2892 2956 3020 3o83 64 68i 3147 321 3275 3338 3402 3466 3530 3593 3657 3721 64 682 3784 3848 3912 3975 4039 4 o3 4166 4230 4294 4357 64 683 4421 4484 4548 4611 4675 4739 4802 4866 4929 4993 64 68,4; 5056 5120 5i83 5247 53io 5373 5437 5500 5564 5627 63 685 569I 5754 5817 588I 5944 6007 607I 6I34 69 626I 63 686 6324 6387 6451 6514 6577 664 67046767 6830 6894 63 687 6957 7o02 7083 71461 7210 7273 7336 7399 7462 7525 63 688 8 7652 771 7778 78 704 7967 o3o 8093 8i56 63 689| 8219 8282 8345 8 4 I 34 8597 8660 8723 8786 63 690 8849 8912 8975 9038 9101oi 9164 9227 9289 9352 9415 63 691 * 9478 9541 9604 9667 9729 9792 9855 991g 9981 +043 63 692 84 1oo6 0169g 0232 0294 0357 0420 0482 o545 o6o8 0671 63 693 0733 0796 85 092I 0984 Io46 109 1 172 1234 1297 63 694 1359 I422 I485 1 547 I6io 1672 1735 1797 I86o 1922 63 695 1985 2047 210 2172 2235 2297 2360 2422 2484 2547 62 696 2609 2672 2734 2796 2859 2921 2983 3046 3io8 3170 62 697 3233 3295 3357 3420 3482 3344 3606 3669 3731 3793 62 698 3855 3918 398o 4042 4104 4166 4229 4291 4353 44I5 62 699 4477 4539 46 i 4664 4726 4788 4850 4912/4974 5036 62 N. 0 1 1 1 3 1 4 5 6 7 8 9 D. 12 LOGARITIIMS OF NUMBERS. TABLE I. N. 0 1 2 3 4 5 6 7 8 9 D. 700 84 5098 5i60 5222 5284 5346 54o8 5470o5532 5594 5656 62 701 5718 5780 5842 5904 5966 6028 60o90 6i51 6213 6275 62 702 6337 6399 6461 6523 6585 6646 6708 6770 6832 6894 62 703 6955 7017 7079 7141 7202 7264 7326 7388 7449 7511 62 704 7573 7634 7696 7758 7819 7881 7943 8004 8066 8128 62 705 8189 8251 8312 8374 8435 8497 8559 8620 8682 8743 62 706 8805 8866 8928 8989 905i 9112 9174 9235 9297 9358 SI 7o0 943 I99481 i9542 9604 9665 9726 9788 9849 9911 9972 6 I 708 85 0033 o095 oi56 0217 0279 o340 40 0462 0524 o585 6 709 0646 0707 0769 o830 o891 0952 10o4 1075 I136 1197 6i 710 1258 i320 I38I 1442 50o3 I564 1625 i686 1747 1809 6i 71I 1870 1931 1992 2053 21 4 2175 2236 2297 2358 2419 6 712 2480 2541 2602 2663 2724 2785 2846 2907 2968 3029 6I 713 3090 3150 3211 3272 3333 3394 3455 35i6 3577 3637 61 714 3698 3759 3820 3881 3941 4002 4063 4124 4185 4245 6I 715 43o6 4367 4428 4488 4549 4610 4670O 4731 4792 4852 6I 716 4913 4974 5034 5095 5i56 5216 5277 5337 5398 5459 6i 717 5519 5580 5640 5701 576I 5822 5882 5943 6003 6064 6i 718 6124 6185 6245 6306 6366 6427 6487 6548 6608 6668 60 719 6729 6789 6850 6910 6970 7o031 7091 7152 7212 7272 60 720 7332 7393 7453 7513 7574 7634 7694 7755 7815 7875 60 72I 7935 7995 8056 8116 8176 8236 8297 8357 8417 8477 60 722 8537 8597 8657 8718 8778 8838 8898 8958 91oi8 9078 60 723 9138 9198 9258 9318 9379 9439 9499 9559 9619 9679 60 724 * 9739 9799 9859 9918 9978 *o38 oo0098 o158 0218 o278 60 725 86 o338 0398 0458 o518 o578 0637 o697 o757 0817 0877 60 726 0o37 0996 o6 iii6. 1176 1236 i295 1355 i415 I475 60 727 1534 1594 1654 1714 1773 i833 1893 1952 2012 2072 60 728 2131 2191 2251 2310 2370 2430 2489 2549 2608 2668 60 729 2728 2787 2847 2906 2966 3025 3o85 3144 3204 3263 6o 730 3323 3382 3442 350o 356i 3620 368o 3739 3799 3858 59 73I 3917 3977 4036 4096 4i55 4214 4274 4333 4392 4452 59 732 04SI 4570 4630 4689 4748 4808 4867 4926 4985 5045 50 733 5Io4 5163 5222 5282 534I 5400 5459 55i9 5578 5637 59 734 5696 5755 5814 5874 5933 5992 6o5I 61106169 6228 59 735 6287 6346 64o5 6465 6524 6583 6642 6701 6760 6819 59 736 6878 6937 696 7055 7114 7173 7232 729I 7350 7409 59 737 7467 7526 7 5 7644 7703 7762 7821 7880 7939 7998 59 738 8056 8115 8174 8233 8292 8350 8409 8468 8527 8586 59 739 8644 8703 8762 8821 8879 8938 8997 9056 9114 9173 59 740 9232 9290 9349 9408 9466 9525 9584 9642 9701 9760 59 741 9818 9877 9935 9994 *o53 0111 0170 0228 0287 0345 59 742 87 0404 0462 0521 o579 o638 o696 o755 o813 o872 og30 58 743 o989 1047 11o6 1164 1223 1281 1339 1 398 1456 155 58 744 1573 163I 1690 1748 I8o6 i865 1923 1981 2040 2098 58 745 2156 2215 2273 2331 2389 2448 2506 2564 2622 2681 58 746 2739 2797 2855 2913 2972 3030 3088 3146 3204 3262 58 747 3321 3379 3437 3495 3553 3611 3669 3727 3785 3844 58 748 3902 3960 4018 4076 4134 4192 4250 4308 4366 4424 58 749 4482 4540 4598 4656 4714 4772 4830 4888 4945 5003 58 750 50o6 5119 5177 5235 5293 5351 5409 5466 5524 5582 58 751 5640 5698 5756 5813 5871 5929 5987 6045 6102 616o 58 752 6218 6276 6333 6391 6449 6507 6564 6622 6680 6737 58 753 6795 6853 6910 6968 7026 7083 7141 7199 7256 7314 58 754 7371 7429 7487 7544 7602 7659 7717 7774,7832 7889 j8 755 7947 8004 8062 8 19 8177 8234 8292 8349 8407 8464 57 756 8522 8579 8637 8694 8752 8809 8866 8924 8981 9039 57 757 9096 9153 9211 9268 9325 9383 9440 9497 9555 9612 57 758 *9669 9726 9784 984I 9898 9956 *oi3 0070 0127 oi85 57 759 88 0242 0299 o356 o43 00471 o528 o585. 0642 0699 o0756 57 N. O 1 2 3 4 5 6 17 8 9 D. TABLE I. LOGARITHMS OF NUMBERS. 13 N. o 1 2 3 4 5 6 7 8 9 D. 760 88084 o871 0928 o985 O1042 1099 ii56 1213 1271 1328 57 761 i385 1442 1499 I556 i6i3 1670 1727 1784 i84i 1898 57 762 1955 2012 2069 2126 2183 2240 2297 2354 2411 2468 57 763 2525 2581 2638 2695 2752 2809 2866 2923 2980 3037 57 764 3093 3io 3207 3264 3321 3377 3434 3491 3548 3605 57 765 366i 3718 3775 3832 3888 3945 4002 4o59 4115 4172 57 766 422 42854342 439912 4569 4625 4682 4739 57 767 479 42841 2 456 462 5J 4682J,i7! 57,67 479,4852 4909 4965 5022 5078 5i35 5192 5248 5305 57 768 536i 54i8 5474 553i 5587 5644 5700 5757 5813 5870 57 769 5926 5983 6039 6096 6152 6209 6265 6321 6378 6434 56 770 6491 6547 6604 666o 6716 6773 6829 6885 6942 6998 56 771 70 54 7111 7167 7223 7280 7336 7392 7449 75o05 7561 56 772 7617 7674 7730 7786 7842 7898 7955 80oi i 8067 8123 56 773 8179 8236 8292 8348 8404 8460 85i6 8573 8629 8685 56 774 8741 8797 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.o030 o0o86 0141 0197 0253 o309 o365 56 777 89 0421 0477 o533 0589 0645 0700 0736 812 o868 0924 56 778 0980 io35 io09 II47 1203 1259 134 1370 1426 i482 56 779 I537 1593 1649 1705 1760 I81i6 872 1928 1983 2039 56 780 2095 2150o 2206 2262 2317 2373 2429 2484 2540 2595 56 781 2651 2707 2762 2818 2873 2929 2985 3040 3096 3131 56 782 3207 3262 33,8 3373 3429 3484 354o 3595 363; 37o6 56 783 3762 3817 3873 3928 3984 4039 4094 4i5o 4205 4261 55 784z 43x6 4371 4427 4482 4538 4593 4648 4704 475914814 55 785 4870 4.925 4980 5o36 50og91 5146 5201 5257 15312 5367 55 786 5423 5478 5533 5588 5644 5699 5754 5809 5864 5920 55 787 5075 6o3o 6o085 6i4o 6195 625i 63o6 636i 6416 6471 55 788 6526 6581 6636 16692 6747 6802 6857 6912 6967 7022 55 789 7077 7132 7187 7242 7297 7352 7407 7462 7517 7572 55 790 7627 7682 7737 7792 7847 7902 7957 8012 8067 8122 55 791 8176 8231 8286 834i 8396 845I 85o6 856i 86i5 8670 55 792 8725 8780 8835 8890 8944 8999 9o54 9I0919164 9218 55 793 9273 9328 9383 9437 9492 9547 9602 9656 9711 9766 55 794 x 9821 9875 9930 9985.039 0094 0149 0203 0258 0312 55 795 900367 o4/22 0476 o531 o586 o64o 0695 0749 o8o41 0859 55 796 0913 0968 1022 1077 1131 1ii86 1240 1 295 1349 1404 55 797;458 1513 1 567 1622 1676 1731 1785 1 840 1894 1948 54 798 2003 2057 2112 2166 2221 2275 2329 2384 2438 2492 54 _~~~~~~~~~~~3g2219 I33 54 799 2547 2601 2655 2710 2764 2818 287 2927 2981 3o36 54 800 3090 3144 3199gg 3253 3307 336i 3416 3470 3524 3578 54 8oi 3633 3687 3741 3795 3849 3904 3958 4012 4066 4120 54 802 4174 4229 4283 4337 4391 4445 4499 4553 4607 4661 54 803 4716 4770 4824 4878 4932 4986 50o40 5094 5148 5202 54 804 5256 53io 5364 54I8 5472 5526 558o 5634 5688 5742 54 8~~~~~~~~~~56 50 53L5685 8o5 5796 585o 5904 5958 60o12 6o66 6I9 6173 6227 6281 54 8o6 6335 6389 6443 6497 655i 6604 6658 6712 6766 6820 54 807 6874 6927 6981 7035 7089 7143 7196 7250 7304 7358 54 8o8 7411 7465 7519 7573 7626 7680 7734 7787 7841 7895 54 809 7949 8002 8o56 88io 8i63 8217 82708324 8378 843i 54 8io 8485 8539 8592 8646 8699 8753 8807 886o 8914 8967 54 8ii 9021 9074 9128 9181 9235 9289 9342 9396 9449 9303 54 812 *e9556 96Io 9663 97i6 I9770 9823 9877 9930 9984'0o37 53 8r3 910091 o0144 0197 025Io3o4 o358 o0411 0464 o5i8 0571 53 814 0624 0678 o73 0784 o838 o891 0944 0998 io5i iio4 53 85 1158 1211I 1264 1317 I37I 1424 I477 i53o 1584 1637 53 8i6 1690 1743 1797 iS8o 1903 1956 2009 2063 2116 2169 53 817 2222 2275 2328 2381 2435 2488 2541 2594 2647 2700 53 8i8 2753 2806 2859 2913 2966 3019 3072 3125 3178 3231 53 819 3284 3337 3390 3443 3496 3549 3602 3655/3708 3761 53 N. 0 1 2 3 4 5 6 7 8 9 D. 14 LOGARITHMS OF NUMBERS. TABLE I. N. 0 1 1 1 2 3 4 5 7 8 7 I D. 820 91 3814 3867 3920 3973 4026 4079 432 4184 4237 4290 53 821 4343 4396 4449 4502 4555 46o8 4660 47I13' 4766 4819 53 822 4872 4925 4977 5030 5083 5136 5i89 524I1 5294 5347' 53 823 5400 5453 55o5 5558 561I 5664 5716 5769 5822 5875 53 824 5927 5980 6033 6o85 6i38 619i 6243 6296 6349 640oi 53 825 6454 6507 6559 6612 6664 6717 6770 6822 6875 6927 53 826 6980 7033 7085 7138 7190 7243 7295 17348 7400 7453 53 827 7506 7558 761I 7663 7716 7768 7820 7873 7925 7978 52 828 8030 8o83 8135 8188 8240 8293 8345 8397 845ol 85o2 52 829 8555 8607 8659 87 12 8764. 88I6 8869 8921 8973 9026 52 830 9078 g913o 9183 9235 9287 9340o 9392 9444 9496 95491 52 83I * 96o0 9653 9706 9758 981o 9862 9914 9967 +019 0071 52 832 92' 0123 0176 0228 0280 0332 0384 0436 0489 o54i 0593 52 833 o645 o697 o78o o853 ogo6 0958 110 1io62 11 14 52 834 i i66 1218 1270 1322 1374 1426 1478 153o 1582 1634 52 835 i686 1738 1790 1842 1894 1946 1998 250 2102 l 2154 5 2 836 2206 2258 2310 2362 2414 2466 25 8 2570 2622 2674 52 837 2725 2777 2829 288I 2933 2985 3037 30891 340 3192 52 838 3244 3296 3348 3399 345I 3503 3555 3607 3658 3710 52 839 3762 38I4 3865 3917 3969 4021 4072 4124 4176 4228 52 840 4279 433, i43S3 4434 4486 4538 4589 4641 4693 4744 52 84I 4796 4848 4899 4951 5003 5054 5io6 5157 5209 5261 52 842 5312 5364 5415 5467 55i8 5570 562I 5673 5725 5776 52 843 5828 5879 593 5982 6034 6085 6I37 6i88 6240 6291 5i 844 6342 6394 6445 6497 6548 66oo00 665i 6702 6754 6805 5i 845 6857 6908 6959 70I I 7062 7114 7I65 7216 7268 73I9 5i 846 7370 7422 7473 7524 7576 7627 7678 7730 7781 7832 5i 847 7883 7935 7986 8037 8088 8i40 8191 8242 8293 8345 5I 848 8396 8447 8498 8549 86oI 8652 8703 8754 88o5 8857 5i 849 8908 8959 9010 9061 9gI2 9163 9215 9266 9317 93 59368 5 85o 9419 9470 9521 9572 9623 9674 9725 9776 9827 9879 5i 85I * 9930 998I +o32 oo83 oi34 oi85 0236 0287 o338 o389 5i 852 93 0440 o491 o542 o592 o643 0694 0745 0796 o847 o89 5I 853 0949 1000 I051 1102 1153 1204 1254 I3o5 I356 1407 5I 854 I458 I509 i56o I6o10 66i 1712 1763 I814 I865 Igi5 5i 855 1966 2017 2068 2118 2I69 2220 2271 2322 2372 2423 5i 856 2474 2524 2575 2626 2677 2727 2778 2829 2879 2930 5i 857 298I 3o3I 3082 3i33 3i83 3234 3285 3335 3386 3437 51 858 3487 3538 3589 3639 3690 3740 3791 384I 3892 3943 5i 859 3993 4044 4094 4145 4195 4246 4296 4347 4397 4448 5I 86o 4498 4549 4599 465o 4700 475 48oi 4852 4902 4953 50o 86I 5003 5054 5io4 5154 5205 5255 5306 5356 54o6 5457 5o 862 5507 5558 56o8 5658 5709 5759 S809 586o 5910 5960 50 863 6oii 6o6i 6i06 6162 6212 6262 6303 6363 6413 6463 5o 8654 65i4 6564 6614 6665 67iS 6765 68i5 6865 6916 6966 5o 865 7016 7066 7117 7167 7217 7267 7317 7367 7418 7468 5o 866 7518 7568 7618 7668 7718 7769 7819 7869 79I9 7969 5o 867 8019 8061 9 806 9 II 869 8219 8269 8320 8370 8420 8470 5o 868 8520 8570o 8620 8670 8720 8770 8820 8870 8920 8970 50 869 9020 9070 9120 9170 9220 9270 9320 9369 94I9 9469 50o 870 95I9 9569 9619 9669 9719 9769 9819 9869 99I8 9968 50 871 94 oo00 oo68 oi oi68 0218 0267 0317 o367 0417 o467 5o 872 o5I6 o566 o616 o666 0716 0765 o8i5 o865 o915 o964 5o 873 1014 Io64 1114 Ii63 1213 1263 1313 1362 1412 1462 50 874 I5II i56i I6ii i660 I7IO i76o 1809o189 85 909 1958 5o 875 2008 2058 2107 2157 2207 2256 2306 2355 2405 2455 50 876 2504 2554 2603 2653 2702 2752 280I 2851 2901 2950 50 877 3000 3049 3099 348 3198 3247 3297 3346 3396 3445 49 878 3495 3544 3593 3643 3692 3742 379I 384 38o90 3939 49 879 3989 4038 4088 4137 4186 4236 42 5 4335 4384 4433 49 N. 0 1 34 5I4 6 718 9 D. TABLE 1. LOGARITHMS OF NUMBERS. 15 N I 0 1 2 314 5 6 7 8 9 D. 88o 94 4483 4532 4581 4631 4680 4729 4779 4828 4877 4927 49 881 4976 5025 5074 5124 5173 5222 5272 532I 5370 5419 49 882 5469 55i8 5567 5616 5665 5715 5764 58I3 5862 5912 49 883 596i 6o1o 6059 6io8 6I57 6207 6256 63o5 6354 6403 49 884 1 6452 6501 655i 66oo 6649 6698 6747 6796 6845 6894 49 885 I 6943 6992 7041 7090 7140 7189 7238 7287 7336 7385 49 886 7434 7483 7532 7581 7630 7679 7728 7777 7826 7875 49 887 7924 7973 8022 8070 8119 68 J8217 8266 8315 8364 49 888 84I3 8462 85ii 856o 8609 8657 8706 8755 8804 8853 49 889 8902 8951 8999 9048 9097 9146 9195 9244 9292 9341 49 890 9390 9439 9488 9536 9585 9634 9683 973I 9780 9829 49 891 *9878 9926 9975 +024 0073 012I 0170 0219 0267 o3i6 49 892 95 o0365 0414 0462 o5i i o560 o6o8 o657 o706 0754 o8o3 49 893 o85 0900 o949 0997 I046 1095 143 1192 1240 1289 49 894 i338 I386 i435 i483 I532 158o0 629 1677 1726 1775 49 895 I823 1872 1920 1969 2017 2066 21,4 2163 2211 2260 48 896 2308 2356 2405 2453 2502 2550 2599 2647 2696 2744 48 897 2792 2841 2889 2938 2986 3034 3083 3i3i 3 318 3228 48 898 3276 3325 3373 3421 3470 3518 3566 36I5 3663 371, 48 899 3760 38o8 3856 3905 3953 4001 4049 4098 4146 4194 48 900 4243 4291 4339 4387 4435 4484 4532 4580 4628 4677 48 901 4725 4773 4821 4869 4918 4966 50 4 5062 5i0 5158 48 902 5207 5255 5303 5351 5399 5447 5495 5543 5592 5640 48 903 5688 5736 5784 5832 588o 5928 5976 6024 6072 6120 48 904 6i68 6216 6265 63I3 636i 6409 6457 65o5 6553 66o0 48 905 6649 6697 6745 6793 6840 6888 6936 6984 7032 7080 48 906 7128 7 76 7224 7272 7320 7368 7416 7464 7512 7559 48 907 7607 7655 7703 775 7799 7847 7894 7942 799o 8o38 48 908 8086 8134 8i8 8229 8277 8325 8373 8421 8468 8516 48 909 8564 8612 8659 8707 8755 88o3 885o 8898 8946 8994 48 910 9041 9089 9137 9185 9232 9280 9328 9375 9423 9471I 48 911 95I8 9566 9614 9661 9709 9757 9804 9852 9900 9947 48 912 *9995 +042 0090 oi38 0185 0233 o280 o328 o376 0423 48 913 96 047I o5i8 o566 o6i3 066i 0709 o756 o8041o i5 0899 48 914 0946 o994 1041 1089 1ii36 i 84 123I 1279 1326 1374 47 915 1421 I469 I5i6 i563 i6ii i658 1706 1753 i8o0I 848 47 9I6 1895 1943 1990 2038 2085 2132 2180 2227 2275 2322 47 917 2369 2417 2464 2511 2559 2606 2653 2701 2748 2795 47 918 2843 2890 2937 2985 3032 3079 3126 3174 322I 3268 47 919 33i6 3363 341 o 3457 3504 3552 3599 3646 3693 374, 47 920 3788 3835 3882 3929 3977 4024 4071 4I i8 4165 4212 47 921 4260 4307 4354 4401 4448 4495 4542 4590 4637 4684 47 922 4731 4778 4825 4872 4919 4966 5o13 50o6 5o8 5155 47 923 5202 5249 5296 5343 5390 5437 5484 553 i 5578 5625 47 924 5672 5719 5766 58i3 586o 5907 5954 6oo00 6048 6095 47 925 6142 6189 6236 6283 6329 6376 6423 6470 6517 6564 47 926 66ii 6658 6705 6752 6799 6845 6 892 6939 6986 7o33 47 927 7080 71 27 7173 7220 7267 7314 7361 74o8 7454 7501 47 928 7548 7595 7642 7688 7735 7782 7829 7875 7922 7969 47 929 80o6 8062 8109 8156 8203 8249 8296 8343 8390 8436 47 930 8483 853o 8576 8623 8670 8716 8763 88io 8856 8903 47 93i 8950 8996 9043 9090 9136 9183 9229 9276 9323 9369 47 932 9416 9463 9509 9556 9602 9649 9695 9742 9789 9835 47 933 * 9882 9928 9975 +021 oo68 o04 oI6I 0207 0254 W00oo 47 934 7 o347 o393 0440 0486 o533 0579 o626 672 0719 0765 4 935 0812 o858 o090o4 951 i 997 I044 1090 I1 37 83 1229 46 936 1276 I322 i 369 1415 I461 50o8 | 554 160o 1 647 1693 46 937 1740 1786 1832 1879 1925 197I1 2018 2064 2110 2157 46 938 2203 2249 2295 2342 2388 2434 248I 2527 1 2573 2619 46 939 2666 27I2 2758 2804 I 285I 2897 2943 2989 3035 3082 46 N. 0 1 2 3 4 5.6 8 9 D. 16 LOGARITHMS OF NUMBERS. TABLE I. N. 0O 1 2 3 4 5 6 7 8 9 D. 940 97 3128 3174 3220 3266 33I3 3359 3405 345i 3497 3543 46 941 3590 3636 3682 3728 3774 3820 3866 3913 3959 4oo 5 46 942 4051I 4097 4143 4189 4235 428I 4327 4374 4420 4466 46 943 4512 4558 4604 4650 4696 4742 4788 4834 4880 4926 46 944 4972 5o18 5o64 5iio 5i56 5202 5248 5294 534o 5386 46 945 5432 5478 5524 5570 56i6 5662 5707 5753 5799 5845 46 946 5891 5937 5983 6029 6075 612 6167 6212 6258 6304 46 947 6350 6396 6442 6488 6533 6579 6625 667I 67I7 6763 46 948 6808 6854 6900 6946 6992 7037 7083 7129 7175 7220 46 949 7266 7312 7358 7403 7449 7495 754I 7586 7632 7678 46 950 7724 7769 7815 7861 7906 7952 7998 8043 8089 8i35 46 951 8i81 8226 8272 8317 8363 8409 8454 85oo008546 8591 46 952 8637 8683 8728 8774 8819 8865 8911 8956 9002 9047 46 953 9093 9138 9184 9230 9275 9321 9366 9412 9457 9503 46 954 9548 9594 9639 9685 9730 9776 9821 9867 9912 9958 46 955 98 0003 0049 0094 o0140 oi85 023 0276 0322 0367 0412 45 956 o458 0503 0549 0594 o64o o685 0730 0776 o 821 0867 45 957 0912 0957 0oo3 1048 IO93 1139 1184 1229 1275 1320 45 958 I366 i4ii i456 i5oi 1547 I592 1637 I683 1728 1773 45 959 181g 1864 g199 1954 2000 2045 2090 2135 2181 2226 45 960 2271 2316 2362 2407 2452 2497 2543 2588 2633 2678 45 961 2723 2769 2814 2859 2904 2949 2994 3o4o 3085 313o 45 962 3175 3220 3265 33io 3356 3401 3446 3491 3536 3581 45 963 3626 3671 3716 376'2 3807 3852 3897 3942 3987 4032 45 964 4077 4122 4167 4212 4257 4302 4347 4392 4437 4482 45 965 4527 4572 4617 4662 4707 4752 4797 4842 4887 4932 45 966 4977 5022 5067 5112 5157 5202 5247 5292 5337 5382 45 967 5426 5471 55i6 556i 5606 565i 5696 574I 5786 583o 45 968 5875 5920 5965 6oio 6055 6ioo 6144 6189 6234 6279 45 969 6324 6369 64I3 6458 6503 6548 6593 6637 6682 6727 45 970 6772 6817 686 69o6 6951 6996 7040 7085 7130 7175 45 97I 7219 7264 7309 7353 7398 7443 7488 7532 7577 7622 45 972 7666 7711 7756 7800 7845 7890 7934 7979 8024 8068 45 973 81i3 8157 8202 8247 8291 8336 838I 8425 8470 8514 45 974 8559 8604 8648 8693 8737 8782 8826 887 8916 8960 45 975 9005 9049 9094 9138 9183 9227 9272 9316 936I 9405 45 976 9450 9494 9539 9583 9628 9672 97I7 9761 9806 9850 44 977 *9895 9939 9983 *028 0072 o117 o161 0206 0250 0294 44 978 99 o339 o383 0428 0472 0516 o56I o605 o65O o694 0738 44 979 0783 0827 087 0916 0960 1004 049 o093 1137 1182 44 980 1226 1270 I315 1359 i4o3 1448 1492 i536 i58o i625 44 981 I 669 1713 1758 I802 1846 189o0 935 I979 2023 2067 44 982 2111 256 2200 2244 2288 2333 2377 2421 2465 2509 44 983 2554 2598 2642 2686 2730 2774 2819 2863 2907 2951 44 984 2995 3039 3o83 3127 3172 3216 326o 33o4 3348 3392 44 985 3436 348o 3524 3568 36i3 3657 3701 3745 3789 3833 44 986 3877 3921 3965 4009 4o53 4097 4141 4i85 4229 4273 44 987 4317 436I 44o5 4449 4493 4537 458 4625 4669 47I3 44 988 4757 48o0 4845 4889 4933 4977 502I 5o65 51o8 5152 44 989 5196 5240 5284 5328 5372 54I6 5460 5504 5547 5591 44 990 5635 5679 5723 5767 5811 5854 5898 5942 5986 6o3o 44 991 6074 6117 6i6i 6205 6249 6293 6337 6380 6424 6468 44 992 65I2 6555 6599 6643 6687 6731 6774 68i8 6862 690o6 44 993 6949 6993 7o37 7080 7124 7168 7212 7255 7299 7343 44 994 7386 7430 7474 7517 756I 7605 7648 7692 7736 7779 44 995 7823 7867 7910 7954 7998 8o4 8085 8129 8172 8216 44 996 8259 8303 8347 8390 8434 8477 852I 1 8564 8608 8652 44 997 8695 8739 8782 8826 8869 8913 8956 go 9o00 9043 9087 44 998 9i31 9174 9218 926I 9305 9348 9392 9435 9479 9J22 44 999 9565 9609 9652 9696 9739 9783'9826 9870 9913 9957 43 N. 0 1 2 3 4 5 |6 8 1 D. TABLE II. LOGARITHMIC SINES AND TANGENTS, FOR EVERY DEGREE AND MINUTE OF THE QUADRANT. If the logarithms of the values in Table III. be each increased by Io, the results will be the values of this table. The logarithmnic Secants and Cosecants are not given. They may be readily obtainedl, as follows:-Subtract the logarithmic Cosine from 20, and the remainder will be the logarithmic Secant; subtract the logarithmic Sine from 20, and the remainder will be the logarithmic Cosecant. 18 LOGARITHMIC SINES, TANGENTS, ETC. TABLE II. 00 1790 / Sine. D. Cosine. D. Tang. D. Cotang. o Inf. Neg. IO-000000 Inf. Neg. Infinite. 60 I 6.463726 501717 000000 oo 6.463726 501717 I3*536274 S5 2 764756 293485 ooooo0 oo 764756 293483 235244 58 3 940847 208231 000000 00 940847 208231 o59153 57 4 7.o65786 16I517 000000 00 7*o65786 I6I5I7 I2.9342I4 56 5 162696 131968 oooooo 0o 162696 131969 837304 55 6 241877 II 575 9'999999 01 241878 111 57 758122 54 7 308824 96653 999999 o1 308825 99653 691175 53 8 366816 85254 999999 oI 366817 82254 633i83 52 9 417968 76263 999999 oI 4I7970 76263 582030 5I 10 463726 68988 999998 oI 463727 68988 536273 50 11 750o5II8 62981 99999998 01 7.505120 62981 12-494880 49 12 542906 57936 999997 oI 542909 57933 457091 48 I3 577668 53641 999997 oI 577672 53642 422328 47 I4 609853 49938 999996 Ol 609857 49939 390I43 46 i5 639816 46714 999996 oi 639820 4671 36oi80 45 i6 667845 4388I 999995 Ol 667849 43882 332151 44 17 694173 41372 999995 01 694179 41373 305821 43 18 718997 39135 999994 I 7190oo3 39136 280997 42 19 742478 37I27 999993 OI 742484 37128 2575I6 4I 20 764754 353I5 999993 oI 764761 35I36 235239 40 21I 7.785943 33672 9.999992 oI 7.78595I 33673 12.214049 39 22 806146 32175 999991 OI 806i55 32176 193845 38 23 825451 30805 999990 oI 825460 308o6 I74540 37 24 843934 29547 999989 02 843944 29549 i56056 36 25 861662 28388 999989 02 861674 28390 I38326 35 26 878695 27317 999988 02 878708 27318 121292 34 27 895085 26323 999987 02 895099 26325 104901 33 28 910879 25399 999986 02 910894 25401 089106 32 29 926119 24538 999985 02 926134 24540 073866 3i 30 940842 23733 999983 02 940858 23735 059142 i30 3I 7.955082 22980 9.999982 02 7.9551o O 22981 I2.o449oo 29 32 968870 22273 999g98 02 968889 22275 o3iiii 28 33 982233 21608 999980 02 982253 21610 OI7747 27 34 995198 2098I 999979 02 995219 20983 004781 26 35 8oo7787 20390 999977 02 80oo7809 20392 11.992191 25 36 020021 19831 999976 02 020044 19833 979956 24 37 o319919 19302 999975 02 o3I945 I305 968o055 23 38 04350I 18801 999973 02 o43527 i88o3 956473 22 39 054781 18325 999972 02 o54809 18327 945191 21 40 065776 7872 99997I 02 o68o6 7874 934194 9 20 4I 81076500 I7441 9.999969 02 8.076531 17444 II.923469 42 086965 17031 999968 02 086997 I7034 913003 I8 43 097183 I6639 999966 02 0972I7 16642 902783 I7 44 107167 I6265 999964 o3 107203 16268 892797 16 45 116926 I5908 999963 o3 II6963 I5910 883037 15 46 126471 i5566 99996I o3 I265Io I5568 873490 14 47 i3581o 15238 999959 03 i3585i I5241 864149 13 48 144953 14924 999958 o3 144996 14927 855004 12 49 153907 14622 999956 o3 I53952 14627 846048 11 5o 162681 I4333 999954 o3 162727 14336 837273 IO 51 8-I71280 14054 9-999952 o3 8.7I1328 14057 11I828672 52 I79713 13786 999950 o3 I79763 13790 820237 53 187985 13529 999948 o3 i88o36 13532 811964 7 54 196102 13280 999946 03 196156 13284 803844 6 55 204070 1304I 999944 o3 204126 i3044 795874 5 56 211895 12810 999942 04 211953 12814 788047 4 57 219581 12587 999940 04 21964 125o90 780359 3 58 227134 12372 999938 04 227195 I2376 772805 2 59 234557 12164 999936 04 23462I 12168 765379 I 60 241855 11963 9999934 04 24192I 11967 758079 0 Cosihie. D. Sine. D. Cotang. D. Tang. I 900 890 TABLE II. LOGARITHMIC SINES, TANGENTS, ETC. 19 10 1780 Sine. D. Cosine. D. Tang. D. Cotang. o 8-24I855 II63 9.999934 04 8-24192I I 967 II*758079 6o I 249033 1I 768 999932 04 249102 11772 750898 5 2 256094 ii58o 999929 04 256I65 II584 743835 58 3 263042 11398 999927 04 263I 5 I 402 736885 57 4 26988I 11221 999925 04 269956 I1225 730044 56 5 276614 110o0 999922 04 276691 I1o54 723309 55 6 283243 io883 999920 04 283323 io887 716677 54 7 289773 ro72I 999918 04 289856 10726 7I0144 53 8 296207 io565 999915I o4 296292 10570 7o3708 52 9 302546 I0413 9999I3 04 302634 I04i8 697366 5i 10 308794 10266 999910 04 3o8884 10270 69I1I6 5o 11 8.314954 10122 9.999907 04 8*3i5046 10126 II684954 49 12 321027 9982 999905 o4 321122 9987 678878 48 I3 327016 9847 999902 04 327114 g985I 672886 47 I4 332924 9714 999899 05 333025 9719 666975 46 I5 338753 9586 999897 o5 338856 9590 66II44 45 i6 344504 9460 999894 05 344610 9465 655390 44 17 350181 9338 999891 o5 350289 9343 6497II 43 i8 355783 92Ig9 999888 05 355895 9224 644Io5 42 19 36i3i5 9103 999885 05 361430 io8 638570 41 20 366777 8990 999882 05 366895 8995 633Io5 40 21 8.372171 8880 9.999879 05 8.372292 8885 II627708 39 22 377499 8772 999876 o5 377622 8777 622378 38 23 382762 8667 999873 o5 382889 8672 6I711 37 24 387962 8564 999870 o5 388092 8570 6II908 36 25 39310I 8464 999867 o5 393234 8470 606766 35 26 398179 8366 999864 o5 398315 837I 60o685 34 27 403 99 8271 99986I o5 403338 8276 596662 33 28 4o8i6i 8177 999858 o5 408304 8I82 59I696 32 29 413068 8o86 999854 05 413213 8091 5 677 3 30 4I7919 7996 99985I o6 418068 8002 58I932 30 3I 8-4227I7 7909 9.999848 o6 8.422869 7914 1157713I 2 32 427462 7823 999844 o6 4276I8 7830 572382 2 33 I 432156 7740 999841 o6 432305 7745 567685 27 34 436800 7657 999838 o6 436962 7663 563038 26 35 441394 7577 999834 o6 441560 7583 558440 25 36 445941 7499 99983I o6 446110 7505 553890 24 37 450440 7422 999827 o6 45o063 7428 549387 23 38 454893 7346 999824 o6 455o07 7352 544930 22 39 4593o0 7273 999820 o6 459481 7279 540519 21 40 463665 7200 9998I6 o6 463849 7206 536I5I 20 41 8-467985 7129 9.9998I3 o6 8-468172 7I35 II-53I828 19 42 472263 7060 999809 o6 472454 7066 527546 43 476498 699I 999805 o6 476693 6998 523307 17 44 480693 6924 999801 o06 480892 6931 5IgI08 16 45 484848 6859 999797 07 485030 6865 5I4950 i5 46 488963 6794 999794 07 489170 68o0 5io83o I4 47 493040 6731 999790 07 493250 6738 506750 i3 48 1497078 |6669 999786 07 497293 6676 502707 12 49 50io80 66o8 999782 0o7 501298 66i5 498702 II 50 505045 6548 999778 07 505267 6555 494733 10 5I 8.508974 6489 9.999774 07 8.509200 6496 11490800 9 52 512867 643I 999769 07 5I3098 6439 486902 8 53 5i6726 6375 999765 07 510696 6382 483039 7 54 52055 6319 99976I 07 520790 6326 479210 6 55 524343 6264 999757 07 524586 6272 475414 5 56 528102 6211 999753 o7 528349 62I8 47I65I 4 57 53I828 6158 999748 07 532080 6i65 467920 3 58 | 535523 6io6 999744 07 535779 6ii3 46422I1 2 59 53986 6055 999740 07 539447 6062 46o553 I 60 542819 6o4 999735 07 543084 602 4569I6 0 4D 07 52836I004 7999835854 Cosine. D. Sine. D.7 Cotang. D. Tang. 910 88~ 491 79 |99|0. 4 4 1 1 7501539 20 LOGARITHMIC SINES, TANGENTS, ETC. TABLE II. 20 177~ r Sine. D. Cosine. D. Tang. D. Cotang. | 0 8.542819 6004 9.999735 07 8 543084 6012 1 I4569I6 6o I 546422 5955 999731 07 54669 I 5962 453309 5 2 549995 5906 999726 07 550268 5914 449732 58 3 553539 5858 999722 o8 553817 5866 446I83 57 4 557054 5811 999717 o8 557336 5819 442664 56 5 560540 5765 9997I3 o8 560828 5773 439172 55 6 563999 5719 999708 o8 564291 5727 435709 54 7 567431 5674 999704 08 567727 5682 432273 53 8 570836 563o 999699 o8 571137 5638 428863 52 9 574214 5587 999694 o8 574520 5595 425480 5i 10 577566 5544 999689 08 577877 5552 422123 5o II 8.580892 5502 9-999685 o8 8.581208 5510 II 418792 4 12 584193 5460 999680 08 584514 5468 415486 48 I3 587469- 5419 999675 o8 587795 5427 412205 47 I4 590721 5379 999670 o8 59105I 5387 408949 46 15 593948 5339 999665 o8 594283 5347 405717 45 i6 5971 52 5300 999660 o8 597492 5308 402508 44 17 600332 5261 999655 o8 600677 5270 399323 43 I8 603489 5223 999650 08 6o3839 5232 396I6I 42 19 606623 5186 999645 09 606978 5194 393022 41 20 609734 5I49 999640 09 6IO094 5I58 389906 40 21 8.612823 5112 9.999635 09 8.6 I3189 5121 I.3868ii 39 22 615891 5076 999629 09 6I6262 5085 383738 38 23 618937 504I 999624 09 619313 5050 380687 37 24 621962 5006 999619 09 622343 5o15 377657 36 25 624965 4972 999614 09 625352 4981 374648 35 26 627948 4938 999608 09 628340 4947 371660 34 27 63o09 11 4904 999603 09 63I3o8 4913 368692 33 28 633854 4871 999597 09 634256 4880 365744 32 29 636776 4839 999592 09 637I84 4848 362816 31 30 639680 48o6 999586 09 640093 4816 359907 30 31 8 642563 4775 9.99958I 09 8 642982 4784 I1I3570i8 29 32 645428 4743 999575 o9 645853 4753 354147 28 33 648274 4712 999570 09 648704 4722 351296 27 34 651102 4682 999564 09 651537 4691 348463 26 35 653911 4652 999558 IO 654352 4661 345648 25 36 656702 4622 999553 IO 657149 4631 342851 24 37 659475 4592 999547 Io 659928 4602 340072 23 38 662230 4563 99954i 10 662689 4573 337311 22 39 664968 4535 999535 10 665433 4544 334567 21 40 667689 45o6 999529 Io 668I60 4526 33I840 20 4I 8.670393 4479 9.999524 IO 8'670870 4488 II.329130 l9 42 673080 445I 999518 lo 673563 446I 326437 I8 43 675751 4424 9995I2 10 676239 4434 323761 17 44 678405 4397 999506 IO 678900 4417 321100 I6 45 681 43 4370 999500 I0 681 544 4380 3 8456 I5 46 683665 4344 999493 10 684172 4354 315828 14 47 686272 4318 999487 10 686784 4328 313216 13 48 688863 4292 99948I Io 68938I 4303 310619 12 49 691438 4267 999475 IO 691963 4277 308037 II 50 693998 4242 999469 10 694529 4252 305471 10 5i 8.696543 4217 9.999463 II 8-6970o8 4228 11302919 52 699073 4192 999456 II 6996I7 4203 300383 8 53 701589 4168 999450 II 702139 4179 29786i 7 54 704090 4144 999443 II 704646 4155 295354 6 55 706577 4121 999437 11 707140 4132 292860 5 56 709049 4097 99943 II 709618 41o8 290382 4 5 711507 4074 999424 11 712083 4085 287917 3 5 713952 4o5I 9994I8 II 714534 4062 285466 2 59 716383 4029 999411 1I 716972 4040 283028 60 718800 4006 999404 II 719396 4017 280604 0 Cosine. D. Sine. j"D. Cotang. D. Tang. I 920 87~ "...~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ TABLE II. LOGARITHMIC SINES, TANGENTS, ETC. 21 30 1760 Sine. D. Cosine. D. Tang. D. Cotang. I o 8.718800 4006 9.999404 II 8.719396 4017 II28o0604 6o 1 721204 3984 999398 II 721806 3995 278194 59 2 723595 3962 99939I II 724204 3974 275796 58 3 725972 3941 999384 11 726588 3952 273412 57 4 728337 3919 999378 Ii 728959 3930 271041 56 5 730688 3898 999371 II 731317 3909 268683 55 6 733027 3877 999364 I2 733663 3889 266337 54 7 735354 3857 999357 12 735996 3868 264004 53 8 737667 3836 999350 I2 738317 3848 261683 52 9 739969 3816 999343 12 740626 3827 259374 51 10 742259 3796 999336 12 742922 3807 257078 50 II 8.744536 3776 9.999329 I2 8.745207 3787 11254793 49 12 746802 3756 999322 12 747479 3768 252521 48 13 749055 3737 999315 I2 749740 3749 250260 47 14 751297 3717 999308 12 751989 3729 248011 46 I5 753528 3698 999301 12 754227 3710 245773 45 I6 755747 3679 999294 12 756453 3692 243547 44 17 757955 366i 999287 12 758668 3673 241332 43 i8 76o05i 3642 999279 12 760872 3655 239128 42 19 762337 3624 999272 12 763065 3636 236935 41 20 764511 3606 999265 12 765246 3618 234754 40 21 8.766675 3588 9.999257 12 8.767417 36o0 11232583 39 22 768828 3570 999250 13 769578 3583 230422 38 23 770970 3553 999242 13 771727 3565 228273 37 24 773101 3535 999235 13 773866 3548 226134 36 25 775223 35r8 999227 I3 775995 353I 224005 35 26 777333 350o 999220 I3 778114 3514 221886 34 27 779434 3484 999212 I3 780222 3497 219778 33 28 781524 3467 999205 I3 782320 3480 217680 32 29 7/83605 345I 999197 I3 784408 3464 215592 3I 30 785675 343I 999189 13j 786486 3447 213514 30 31 8.787736 3418 9.9991 81 13 8788554 343I 11*21446 29 32 789787 3402 999174 i3 790613 34i4 209387 28 33 791828 3386 999166 13 792662 3399 207338 27 34 793859 3370 999158 13 794701 3383 205299 26 35 79588i 3354 999150 13 796731 3368 203269 25 36 797894 3339 999142 I3 798752 3352 201248 24 37 799897 3323 999134 i3 800763 3337 199237 23 38 801892 33o8 999126 13 802765 3322 197235 22 39 803876 3293 999118 I 3 804758 3307 195242 21 40 805852 3278 999110 13 806742 3292 193258 20 4I 8.807819 3263 9.999102 i3 8.808717 3278 II1I9I283 19 42 809777 3249 999094 I4 8Io683 3262 I89317 I 43 811726 3234 999086 14 812641 3248 I87359 17 44 813667 3219 999077 I4 814589 3233 I854II 16 45 8I5599 3203 999069 I4 8I6529 3219 18347I i5 46 817522 319i 999061 14 8I846I 3205 I81539 14 47 819436 3177 999053 I4 820384 319I I796I6 13 48 821343 3i63 999044 14 822298 3177 177702 12 49 823240 3149 999036 14 824205 3i63 175795 11 50 825130 3133 999027 14 826103 350o 173897 10 5I 8.827011 3122 9-9990I9 I4 8.827992 3136 11172008 52 828884 3io8 999010 I4 829874 3123 170126 53 830749 3095 999002 14 831748 3iio 168252 7 54 832607 3082 998993 14 8336i3 3096 166387 6 55 834456 3069 998984 14 835471 3083 I64529 5 56 836297 3056 998976 14 837321 3070 I62679 4 57 838i3o 3043 998967 15 839163 3057 160837 3 58 839956 3030 998958 iS 840998 3045 159002 2 59 841774 3017 998950 15 842825 3032 I57175 I 60 843585 3000 998941 I5 844644 30I9 155356 o,I Cosine. D. Sine. D. Cotang. D. Tang..30. 86~ 22 LOGARITHMIC SINES, TANGENTS, ETC. TABLE II. 40 1750 Sine. D. Cosine. D. Tan. D. Cotang. o 8.843585 3oo5 9.99894I I5 8.844644 3019 II.55356 6o I 845387 2992 998932 I5 846455 3007 I53545 5 2 847183 2980 998923 15 848260 2995 151740 59 3 848971 2967 998914 I5 850057 2982 49943 57 4 850751 2955 998905 15 851846 2970 i48i54 56 5 852525 2943 998896 I5 853628 2958 146372 55 6 854291I 2931 998887 I5 855403 2946 144597 54 7 856049 2919 998878 I5 857171 2935 14282g 53 8 857801 2907 998869 15 858932 2923 4i068 52 9 859546 2896 998860 i5 86o686 2911 139314 5i 10 861283 2884 998851 15 862433 2900 137567 50 II 8.8630I4 2873 gg99884I i5 8.864173 2888 II1I35827 49 12 864738 286I 998832 I5 865906 2877 I34094 48 I3 866455 2850 998823 I6 867632 2866 132368 47 14 868i65 2839 998813 i6 86935I 2854 130649 46 I5 869868 2828 998804 i6 871064 2843 128936 45 I6 871565 28I7 998795 I6 872770 2832 127230 44 I7 873255 2806 998785 16 874469 282I 125531 43 I8 874938 2795 998776 i6 876162 2811 I23838 42 19 8766I5 2786 998766 i6 877849 2800 I22I5I 41 20 878285 2773 998757 i6 879529 2789 12047I 40 2I 8*879949 2763 9.998747 I6 8*88I202 2779 I1118798 39 22 88i6o7 2752 998738 i6 882869 2768 11713I 38 23 883258 2742 998728 i6 884530 2758 1I15470 37 24 884903 2731 998718 I6 886i85 2747 II3815 36 25 886542 272I 998708 i6 887833 2737 11 2167 35 26 888174 2711 998699 I6 889476 2727 110524 34 27 889801 2700 998689 i6 891112 2717 108888 33 28 891421 2690 998679 i6 892742 2707 107258 32 29 893035 2680 998669 17 894366 2697 105634 31 30 894643 2670 998659 I7 895984 2687 104016 30 31 8.896246 2660 9.998649 17 8.897596 2677 I1 I02404 2 32 897842 2651 998639 17 899203 2667 100797 2 33 899432 1 2641 998629 17 900803 2658 099197 27 34 90O17 2631 998619 17 902398 2648 097602 26 35 902596 2622 998609 17 903987 2638 og60oi3 25 36 go4I69 2612 998599 17 905570 2629 o94430 24 37 905736 2603 998589 17 907147 2620 092853 23 38 907297 2593 998578 I7 908719 2610io Og91281 22 39 o908853 2584 998568 17 910285 260I 089715 21 40 910404 2575 998558 17 91 846 2592 088154 20 4I 8-9iic949 2566 9998548 17 8-91340I 2583 i.o86599gg 42 913488 2556 998537 17 914951 2574 085049 I 43 915022 2547 998527 I7 916495 2565 1 o835o5 I7 44 916550 2538 998516 i8 918034 2556 08I966 i6 45 918073 229 998506 i8 919568 2547 080432 I5 46 919591 2520 998495 i8 92IO96 2538 078904 14 47 92I103 2512 998485 i8 922619 2530 077381 i3 48 922610 2503 998474 18 924136 2521 075864 12 49 924112 2494 998464 I8 925649 2512 07435i II 50 925609 2486 998453 i8 927156 2503 072844 10 5I 8-927100.2477 9-998442 I8 i 8.928658 249S 1.I07I342 52 928587 2469 99843I i8 93oi55 2486 o69845 53 930068 2460 998421 i8 93I647 2478 o68353 7 54 931544 2452 998410 I8 933134 2470 o66866 6 55 93301i5 2443 998399 i8 934616 2461 o65384 5 56 934481I 2435 998388 i8 936093 2453 063907 4 57 935942 2427 998377 i8 937565 2445 062435 3 58 937398 2419 998366 i8 939032 2437 060968 2 59 938850 2411 998355 I8 940494 2430 o59506 I 60 940296 2403 998344 I8 941902 242I o58048 o Cosine. D. Sine. D. ( Cotang. D. Tang. 940 850 TABLE II. LOGARITHMIC SINES, TANGENTS, ETC. 23 50 1740 _ Sine. D. Cosine. 1D. I Tang. D. Cotang. 0 8.940296 2403 9.998344 I9 8-941952 2421 II0 58048 60 I 941738 2394 998333 1I 943404 2413 o56596 5q 2 943I74 2387 998322 Ig 944852 2405 055I48 58 3 944606 2379 9983I Ig 946295 2397 o537o5 57 4 946034 237I 998300 9 94774 2390 052266 56 5 947456 2363 998289 19 949168 2382 050832 55 6 948874 2355 998277 Ig 950597 2374 049403 54 7 950287 2348 998266 19 952021 2366 047979 53 8 951696 2340 998255 19 953441 2360 o46559 52 9 953100 2332 998243 I9 954856 2351 045I44 5i 10 954499 2325 998232 19 956267 2344 043733 So II 8.955894 2317 9998220 Ig 8957674 2337 1II-042326 40 12 957284 2310 998209 19 959075 2329 040925 48 13 958670 2302 998I97 I9 960473 2323 o39527 47 I4 960052 2295 998186 Ig 961866 2314 03 I34 46 i: 961429 2288 998174 19 963255 2307 o36745 45 I6 962801 2280 998163 19 964639 2300 o3536i 44 17 964170 2273 998I51 I9 9660Ig 2293 033981 43 I8 965534 2266 998139 20 967394 2286 o032606 42 19 966893 2259 998128 20 968766 2279 03234 41 20 968249 2252 998116 20 970I33 2271 029867 40 21 8.969600 2244 9 998104 20 8'971496 2265 11o028504 3o 22 970947 2238 998092 20 972855 2257 027145 38 23 972289 2231 998080 20 974209 2251 02579I 37 24 973628 2224 998068 20 975560 2244 024440 36 25 974962 2217 998056 20 976906 2237 023094 35 26 976293 2210 998044 20 978248 2230 021752 34 27 977619 2203 998032 20 979586 2223 0204I4 33 28 97894i 2197 998020 20 980921 2217 Ig9079 32 30 981573 2I83 997996 20 983577 2204 016423 30 3I 8'982883 2177 9.997984 20 8'984899 2I97 II1OIoIOl 29 32 984189 2170 997972 20 986217 21I 013783 28 33 985491 2163 997959 20 987532 21 84 012468 27 34 986789 2157 997947 20 988842 2178 01OI58 26 35 988083 2150 997935 21 990I49 2171 00oo85I 25 36 989374 2144 997922 21 991451 2I65 oo8549 24 37 990660 2138 9979 I 21 992750 2158 0072 50 23 38 991943 2131 997897 21 994045 2I 52 oo5955 22 39 993222 2125 997885 21 995337 2146 o004663 21 40 994497 2119 997872 21 996624 2140 003376 20 4I 8.995768 2II2 9.997860 21 8.997908 2134 11002092 1 42 997036 210o6 997847 21 999188 2127 000812 18 43 998299 2100 997835 21 9o000465 2I21 IO-999535 17 44 999560 2094 997822 21 001738 211I 998262 16 45 9.000ooo86 2087 997809 21 003007 2I09 996993 I5 46 002069 2082 997797 21 004272 2103 995728 14 47 oo3318 2076 997784 2I 005534 2097 994466 13 48 004563 2070 997771 21 006792 209 993208 12 49 oo5805 2064 997758 21 008047 92085 953 ii 50 007044 2058 997745 21 009298 2080 990702 IO 5I 9.008278 2052 9.997732 21 9'010546 2074 10o-89454 52 009510 2046 997719 21 011790 2068 988210 8 53 010737 2040 997706 21 oI30o3 2062 986969 7 54 0119g62 2034 997693 22 014268 2056 985732 6 55 013I82 2029 99768o 22 015502 205I 984498 5 56 014400 2023 997667 22 016732 2045 983268 4 57 oi56i3 20o7 997654 22 959 2 040 82041 3 58 016824 2012 997641 22 019183 233 2 59 o01803 2006 997628 22 020403 2028 979597 I 60 019235 2000 997614 22 021620 2023 978380 O' Cosine. D. Sine. 1D. Cotang. D. Tang. 950 840 24 LOGARITHMIC SINES, TANGENTS, ETC. TABLE II. 60 1730 t-. | S. Cosine. D. Tang. D. Cotang. 0o g-019235 2000 9.997614 22 9.021620 2023 I0-978380 60 1 020435 1995 99760o 22 022834 2017 977166 5 2 o21632 1989 997588 22 024044 2011 975956 58 3 022825 I 984 997574 22 02525I 2006 974749 57 4 02401 6 1978 997561 22 026455 2000 973545 56 5 025203 1973 997547 22 027655 1995 972345 55 6 026386 1967 997534 23 028852 1990 971148 54 027567 I962 997520 23 030046 1985 969954 53 028744 1957 997507 23 031237 1979 968763 52 9 029918 195I 997493 23 032425 1974 967575 51 10 03o089 1947 997480 23 o33609 1969 966391 50 II 9.032257 1941 9.997466 23 9.0o34791 I964 I O965209 49 12 033421 1936 997452 23 035069 1958 96403 48 13 034582 1930 997439 23 037144 1953 962856 47 I4 o35741 1925 997425 23 o383i6 1948 961684 46 15 o36896 1920 997411 23 o39485 I943 960515 45 i6 o38048 1915 997397 23 04065i I938 959349 44 37 03 1910 997383 23 o04183 1933 958187 43 I 08 40342 1905 997369 23 042973 1928 957027 42 19 041485 1 99 927355 23 044I3o I923 955870 41 20 042625 1894 997341 23 o45284 1918 954716 4o 21 90o43762 1889 9.997327 24 9.046434 1913 IO0953566 39 2 2 o44895 1884 997313 24 047582 1908 952418 38 23 046026 1879 997299 24 048727 1903 951273 37 24 047154 1875 997285 24 049869 1898 95013I 36 25 048279 1870 997271 24 05o1008 1893 948992 35 26 o049400 865 997257 24 052144 I889 947856 34 27 o50519 I86o 997242 24 053277 I884 946723 33 28 051635 i855 997228 24 054407 1879 945593 32 29 952749 I85o 997214 24 055535 1874 944465 3I 30 o53859 i845 997199 24 o56659 1870 94334I 3o 31 9.054966 1841 9-997185 24 905778I I865 IO 0-942219 2 32 056071 i836 997170 24 058900o 869 94o1100 2 33 057172 i83I 997156 24 o06006 i855 939984 27 34 o58271 1827 997141 24 o6ii30 I851 938870 26 35 059367 1822 997127 24 062240 I846 937760 25 36 o60460 1817 997112 24 063348 1842 936652 24 37 o6i55I 1813 997098 24 064453 I837 935547 23 38 o62639 i8o8 9970o83 25 o65556 I833 934444 22 39 o63724 I804 997068 25 o66655 1828 933345 21 40 064806 1799 997053 25 067752 I824 932248 20 4I 90o65885 1794 9.997039 25 9.0o68846 I819 Io.931154 I 42 066962 1790 997024 25 o69938 I8I5 930062 I8 43 o68036 1786 997009 25 071027 I8IO 928973 17 44 069107 1781 996994 25 072113 i8o6 927887 I6 45 070176 1777 996979 25 073197 1802 926803 15 46 071242 1772 996964 25 074278 1797 925722 14 4 072306 I768 996949 25 075356 1793 924644 13 4 073366 I763 996934 25 076432 1789 923568 12 49 074424 I75g 996919 25 077505 1784 922495 II 50o 075480 1755 996904 25 078576 1780 921424 10 I 9.o76533 175o 9.996889 25 9.079644 1776 10.920356 9 52 077583 1746 996874 25 080710 1772 919290 8 53 078631 1742 996858 25 081773 1767 918227 7 54 079676 1738 996843 25 082833 1763 917167 6 55 o80719 1733 996828 25 08389I I759 916Io9 5 56 o81759 I 729 996812 26 084947 i755 915o53 4 57 082797 1725 996797 26 086000 I751 9I4000 3 58 o83832 1721 996782 26 o87050 1747 912950 2 59 o84864 1717 996766 26 088098 1743 9I 02 I 60 o85894 1713 996751 26 089144 1738 910856 0' Cosine. D. Sine. D. Cotang. D. Tang. I TABLE II. LOGARITHMIC SINES, TANGENTS, ETC. 25 7~ 1720! Sine. D. Cosine. D. Tang. D. Cotang. 0 9-o85894 I713 9.99675I 26 9.089144 1738 I0-gI0856 6o I o86922 1709 996735 26 i ogoI87 I734 909813 5 087947 I704 996720 26 091228 1730 908772 5 3 088970 1700 996704 26 092266'I727 907734 57 4 089990 1696 996688 26 093302 1722 906698 56 5 091008 I692 996673 26 o094336 1719 905664 55 6 092024 1688 996657 26 o95367 17I3 904633 54 7 o093037 1684 996641 26 o096395 1711 903605 53 8 094047 I68o 996625 26 097422 I707 902578 52 9 095o056 1676 996610 26 098446 1703 901554 51 10 096062 I673 996594 26 099468 1699 900532 50 II 9.097065 i668 9.996578 27 9100oo487 I695 10O8995I3 4 12 o098066 i665 996562 27 101504 1691 898496 4z I3 099065 I66I 996546 27 102519 1687 89748I 47 I4 I00062 I657 996530 27 io3532 1684 896468 46 5 oo101056 1653 996514 27 104542 i680 895458 45 I6 102048 I 649 996498 27 io5550 I676 894450 44 17 103037 I645 996482 27 io6556 I672 893444 43 I8 104025 I641 996465 27 107559 1669 892441 42 19 I050io i638 996449 27 108560 i665 891440 4I 20 105992 I634 996433 27 109559 i66i 890441 40 21 9.106973 i63o 9.9964I7 27 9.1i0556 i658 i0-889g44 39 22 10795I I627 996400 27 111551 654 888449 38 23 108927 I623 996384 27 II2543 i65o 887457 37 24 j 109901ggI 6I 996368 27 II3533 I646 886467 36 25 11 0873 1616 99635I 27 114521 1643 885479 35 26 1118 42 I612 996335 27 II5507 1639 884493 34 27 1 I2809 I6o8 9963I8 27 I16491 i636 883509 33 28 1I3774 6o05 996302 28 II7472 1632 882528 32 29 114737 160 996285 28 118452 I629 881548 31 30 1 II5698 1597 I 996269 28 II9429 I625 88057I 3o 31 9 -16656 I594 9.996252 28 9.I20/04 I622 10879596 2 32 117613 1590 996235 28 21377 618 878623 28 33 ii8567 I587 996219 28 122348 i6i5 877652 27 34 1II951 1 i583 996202 28 123317 i6ii 876683 26 35 120469 I580 996185 28 124284 I607 875716 25 36 121417 1576 996168 28 I25249 i604 874751 24 37 1 22362 1573 996I51 28 126211 I61 873789 23 38 123306 1569 996134.28 127172 1597 872828 22 39 124248 i566 996117 28 128130 I594 87I870 21 40 125187 1562 996oo00 28 129087 15gI 870913 20 41 9-I26I25 I559 9.996083 29 9.i3004I I587 Io0869959 1 42 127060 I556 996066 29 130994 I584 869oo6 iS 43 127993 1552 996049 29 131944 1581 868o56 17 44 128925 1549 996032 29 132893 1577 867I107 I6 45 129854 i545 9960I 5 29 I33839 1574 866i61 15 46 130781 1542 995998 29 134784 I57I 8652I6 14 47 131706 1539 995980 29 135726 1567 864274 13 48 132630 1535 995963 29 136667 1564 863333 12 49 13355I I532 995946 29 I37605 I56i 862395 Ii 50 134470 1529 995928 29 138542 i558 861458 10 51i 9.135387 1525 9.995911 29 9-I39476 555 IO860524 9 52 i36303 1522 995894 29 140409 55i 859591 8 53 137216 1519 995876 29 I4134o0 548 85866o 7 54 138128 I5i6 995859 29 I42269 I545 857731 6 55 139037 I512 995841 29 I43196 I542 856804 5 56 139944 I509 995823 29 I4412I I539 855879 4 57 140850 i5o6 995806 29~ I45044 I535 854956 3 58 141754 i5o3 995788 29 145966 1532 854034 2 59 142655 1500 995771 29 146885 1529 853115 I 60 1 I43555 1496 995753 29 147803 1526 852197 0 Cosine. D. Sine. D. Cotang. D I. Tang. 970 - 820 2 26 LOGARITHMIC SINES, TANGENTS, ETC. TABLE II. 8~ 171~ t Sine. D. Cosin. D. Tang. D. Cotang. 0 9. 43555 1496 9-995753 30 9 I47803 1526 10*852197 60 I44453 I493 995735 30 I48718 I523 851282 5 2 145349 1490 995717 30 149632 I520 850368 58 3 146243 1487 995699 30 150544 1517 849456 57 4 147 136 I484 995681 30 15 I 454 I 514 848546 56 5 148026 I48 I 995664 3o 152363 i5 i 847637 55 6 148915 I478 995646 30 r53269 I508 84673I 54 149802 1475 995628 3o0 54174 I505 845826 53 8 50686 1472 995610 30 155077 I502 844923 52 9 151569 1469 995591 30 155q78 499 844022 51 10 152451 1466 995573 30 156877 1496 843123 50 II 9-.53330 1463 9-99555'5 30 9157775 1493 10-842225 49 12 154208 1460 995537 30 I58671 1%4 841329 48 I3 i55o83 1457 995519 3o0 59565 I487 840435 47 I4 I55957 1454 995501 31 160457 1484 839543 46 i5 i5683o I45I 995482 31 I61347 48 &38653 45 i6 I57700 1448 995464 3I I62236 I479 837764 44 17 I58569 I445 995446 3I I63123 1476 836877 43 i8 I59435 1442 995427 3I 164008 1473 835992 42 19 i6030i 1439 995409j 31 64892 1470 835io8 41 20 161164 i436 995390 31 165774 1467 834226 40 21 9 I62025 I433 9.995372 31 9.I66654 I464 io833346 39 22 I62885 I430 995353 31 167532 146i 832468 38 23 163743 I427 995334 3i 168409 I458 831591 37 24 164600 1424 995316 31 169284 I455 830716 36 25 I65454 1422 995297 3I I70o57 I453 829843 35 26 166307 1419 995278 31 17I029 I450 82897I 34 27 I67I59 1416 995260 3i I7I899 I447 82810i 33 28 i68oo8 I413 995241 32 172767 I444 827233 32 29 168856 14I0 995222 32 I73634 1442 826366 3I 30 169702 1407 995203 32'74499 1439 825501 30 3I 9.170547 I405 9.995184 32 9.I75362 I436 s10824638 2 32 I71389 1402 995165 32 176224 433 823776 28 33 172230 1399 995146 32 177084 I43I 822916 27 34 173070 1396 995127 32 177942 1428 822058 26 35 173908 1394 995108 32 178799 1425 821201 25 36 174744 I391 995089 32 179655 1423 820345 24 37 175578 388 995070 32 8o80508 420 819492 23 38 1764II i386 995051I 32 i8I360 I417 8i8640 22 39 177242 I383 995032 32 I82211 I145 817789 21 40 I78072 i380 9950I3 32 I83059 I412 816941 20 41 9-178900 1377 9-994993 32 9 -83907 I409 10-8I6093 19 42 179726 I374 994974 32 I84752 I407 815248 18 43 I8055J 1372 994955 32 185597 I404 814403 17 44 I81374 I369 994935 32 I86439 1402 8I356I I 45 I82I96 i366 994916 33 187280 1399 812720 15 46 I83oi6 I364 994896 33 I88120 1396 8ii880 14 47 183834 I36I 994877 33 188958 I393 811042 i3 48 184651 1359 994857 33 189794 1391 8Io206 12 49 I85466 I356 994838 33 o90629 1389 80937 I I 50 I86280 i353 994818 33 191462 i386 808538 io 5I 9.187092 I35I 9.994798 33 9-I92294 I384 I0-807706 9 52 187903 i348 994779 33 I93124 38i 806876 8 53 I88712 i346 994759 33 I93953 1379 806047 7 54 189519 1343 994739 33 I94780 1376 805220 6 55 190325 1341 994720 33 195606 1374 804394 5 56 191130 I338 994700 33 196430 137I 803570 4 57 I91933 I336 994680 33 I97253 1369 802747 3 58 192734 1333 994660 33 198074 i366 801926 2 59 I93534 I33o 994640 33 198894 1364 80o110o6 I 60 194332 1328 994620 33 1997 I3 i36i 800287 0 t Cosine. D. Sine. D. Cotang. I D. Tang. T 980 -10 TABLE II. LOGARITHMIC SINES, TANGENTS, ETC. 27 90go 170 Sine. D. Cose... D. Tang. D. Cotang. _ O 9.-94332 1328 9.994620 33 9.I997I3 I361 10-800287 60 I 195129 1326 994600 33 200529 1359 79947I S 2 195925 1323 994580 33 201345 I356 798655 3 1967I9 1321 994560 34 202159 1354 79784I 57 4 I975II I3i8 994540 34 202971 1352 797029 56 5 198302 I316 994519 34 203782 1349 796218 55 6 199091 I313 994499 34 204592 I347 795408 54 7 i99879 13ii 994479 34 205400 1345 794600oo 53 8 200666 i3o8 994459 34 206207 1342 793793 52 9 20145I I3o6 994438 34 207013 I34o 792987 51 10 202234 1304 994418 34 207817 I338 792I83 50 II 9-2030I7 i30o 9-994398 34 9-2086I9 1335 IO-79I38I 4 12 203797 1299 994377 34 209420 1333 790580 48 13 204577 1296 994357 34 210220 1331 789780 47 14 205354 1294 994336 34 2108 1328 788982 46 I5 206131 1292 994316 34 211815 326 78885 45 i6 206906 1289 994295 34 212611 1324 787389 44 17 207679 1287 994274 35 213405 1321 786595 43 i8 0o8452 I285 994254 35 214I98 1319 785802 42 19 209222 1282 994233 35 214989 1317 78501 41 20 209992 1280 994212 35 215780 1315 784220 40 21 9-210760 I278 9.99419I 35 9-2I6568 13I2 IO-783432 39 22 211526 1275 994171 35 217356 I3io 782644 38 23 212291 I1273 994150 35 218142 I308 781858 37 24 213055 1271 994129 35 218926 I305 781074 36 25 213818 1268 994Io0 35 219710 I3o3 780290 35 26 214579 I266 994087 35 220492 I30o 7795o8 34 27 215338 I264 994066 35 221272 1299 778728 33 2 216097 1261 994045 35 222052 1297 777948 32 29 216854 1259 994024 35 222830 1294 777170 3i 30 217609 1257 994003 35 223607 1292 776393 30 3I 9.218363 1255 9.993982 35 9-224382 1290 10-775618 2 32 219116 1253 993960 35 225156 1288 774844 28 33 2I9868 1250 993939 35 225929 I286 77407I1 27 34 2206I8 1248 993918 35 226700 1284 773300 26 35 22I367 1246 993897 36 227471 128I 772529 25 36 222115 1244 993875 36 228239 1 279 77176I 24 37 222861 1242 993854 36 229007 127 77 0993 23 38 223606 I239 993832 36 229773 1275 770227 22 39 224349 1237 993811 36 230539 1273 76946I 21 40 225092 1235 993789 36 23I302 1271 768698 20 4I 9.225833 1233 9.993768 36 9.232065 1269 10-767935 I9 42 226573 1231 993746 36 232826 1267 767174 I8 43 2273II 1228 993725 36 233586 1265 766414 17 44 228048 1226 993703 36 234345 1262 765655 i6 45 228784 I224 993681 36 235103 1260 764897 15 46 2295I8 1222 993660 36 235859 1258 764141 14 47 230252 1220 993638 36 236614 1256 763386 13 48 230984 1218 9936I6 36 237368 I254 762632 12 49 2317I5 I216 993594 37 238120 1 252 76188o0 i 50 232444 1214 993572 37 238872 1250 761128 10 5I 9.233I72 1212 9.993550 37 9.239$22 1248 10.760378 52 233899 1209 993528 37 240371 1246 759629 8 53 234625 1207 993506 37 241118 1244 758882 7 54 235349 1205 993484 37 241865 1242 758135 6 55 236073 I203 993462 37 242610 1240 757390 5 56 236795 I20I 993440 37 243354 1238 756646 4 57 237515 1199 993418 37 244097 1236 755903 3 58 238235 1197 993396 37 244839 1234 755i6I 2 59 238953 II95 993374 37 245579 1232 754421 I 60 239670 1193 993351 37 246319 1230 753681 0 | Cosine. D. - Sine. D. Cotang. - D. Tan-. I 28 LOGARITHMIC SINES, TANGENTS, ETC. TABLE II. 100 1690 t | Sine. D. Cosine. D. Tang. D. Cotang. I 0 9.239670 1193 9'99335I 37 9-2463I9 1230 Io0753681 60 1 240386 II9I 993329 37 247057 1228 752943 59 2 24110I 1189 993307 37 247794 1226 752206 58 3 241814 II87 993284 37 248530 I224 751470 57 4 242526 ii85 993262 37 249264 1222 750736 56 5 243 237 II83 993240 37 249998 1220 750002 55 6 243947 1181 993 2 7 38 250730 12i8 749270 54 7 244656 1179 993195 38 251461 1217 748539 53 8 245363 II77 993172 38 252I91 12I5 747809 52 9 246069 1175 993 49 38 252920 12I3 747080 5I 10 246775 1173 993127 38 253648 12II 746352 50 II 9'247478 II71 9.993104 38 9.254374 20o9 10745626 49 12 24818I II69 993o08 38 255Ioo00 207 744900 48 I3 248883 II67 993059 38 255824 1205 744176 47 I4 249583 ii65 993036 38 256547 203 743453 46 I5 250282 II63 993013 38 257269 I20I 74273I 45 i6 250980 II6i 992990 38 257990 1200 7420IO 44 17 25I677 II59 992967 38 258710 II98 741290 43 i8 252373 II58 992944 38 259429 1196 74057I 42 19 253067 ii56 992921 38 260146 I94 739854 41 20 25376I I154 992898 38 260863 1192 739137 40 21 9-254453 1152 9.992875 38 9-26I578 1190 IO-738422 39 22 255144 I i5o0 992852 38 262292 189 737708 38 23 255834 I 48 992829 39 263005 I87 736995 37 24 256523 ii46 992806 39 263717 I 85 736283 36 25 257211 1144 992783 39 264428 I83 735572 35 26 257898 I1142 992759 39 265138 I8I 734862 34 2 258583 I1141 992736 39 265847 I79 734153 33 259268 1139 992713 39 266555 I78 733445 32 29 259951 I37 992690 39 267261 1176 732739 31 30 260633 I1I35 992666 39 267967 1174 732033 30 31 9.26I3i4 II33 9.992643 39 9-26867I II72 10-731329 29 32 261994 II3I 9Q2619 39 2693j5 II70 730625 28 33 262673 II3o 992596 39 270077 II69 729923 27 34 26335I 1128 99~572 39 270779 1167 729221 26 35 264027 1126 992549 39 271479 II65 72852I 25 36 264703 1124 992525 39 272178 1I64 727822 24 37 265377 1122 99250I 39 272876 1162 727124 23 38 26605 I1120 992478 40 273573 ii60 726427 22 39 266723 1119 992454 40 274269 II58 725731 2I 40 267395 1117 992430 40 274964 I57 725036 20 4I 9 268065 Iii5 9.992406 40 9-275658 Io55 10*724342 I9 42 268734 II3 992382 40 27635i 153 723649 I8 43 269402 II I 992359 40 277043 II5I 722957 17 44 270069 M IIO 992335 40 277734 ii5o 722266 I6 45 270735 iio8 9923ii 40 278424 I 48 721576 I5 46 271400 II06 992287 40 279I3 1147 720887 I4 47 272064 II05 992263 40 27980I II45 720I99 I3 48 272726 io03 992239 40 280488 1i43 7I9512 12 49 273388 1101 992214 40 281174 1141 7I8826 I 50 274049 I099 992190 40 28I858 1140 7I8142 10 5I 9.274708 1098 9.992166 40 9-282542 ui38 10.7i7458 9 52 275367 Io96 992142 40 283225 I36 716775 53 276025 1094 992118 4I 283907 II35 716093 7 54 27668I 1092 992093 41 284588 1133 715412 6 55 277337 09 I 992069 4I 285268 113I 714732 5 56 277991 I089 992044 4I 285947 I30o 7i4053 4 57 278645 1087 992020 41 286624 II28 7I3376 3 58 279297 io86 99I996 4I 287301 1126 712699 2 59 279948 io84 991971 4I 287977 1125 712023 I 60 280599 I082 991947 4I 288652 1123 711348 0 I Cosine. D. Sine. 3D. Cotang. "D. Tang.. 100~ 790 TABLE II. LOGARITHMIC SINES, TANGENTS, ETC. 29 110 16680' Sine. D. Cosine. D. Tang. D. Cotang. I o 9 280599 1082 9'991947 41 9.288652 1123 10o7I 348 60 281248 I081 991922 41 289326 1122 7Io674 5 2 281897 1079 99I 89g7 41 289999 1120 7Ioo0 58 3 282544 1077 99I873 41 29067I II8 709329 57 4 25831go I076 99I848 41 291342 1117 70865 56 5 288336 i 1074 991823 41 292013 15 707987 55 6 284480 - 1072 991799 4I 292682 1114 707318 54 7 285124 1071 991774 42 293350 I112 706650 53 8 285766 1069 991749 42 294017 I 705983 52 9 286408 IO67 991724 42 294684 II09 705 i6 5i 10 287048 io66 991699 42 295349 1107 70465I 50 I[ 9'287688 IO64 9.991674 42 9.296013 11o6 10.703987 49 12 288326 o163 99g649 42 296677 1104 703323 48 13 288964 io6I 991624 42 297339 iio3 702661 47 4 289600oo o59 991 599 42 298001 1101 701999 46 15 290236 Io58 991574 42 298662 I100 701338 45 i6 290870 i056 991549 42 299322 IO98 700678 44 17 291504 io54 991524 42 299980 1096 700020 43 i8 292137 Io53 991498 42 3oo638 109o 699362 42 19 292768 Io5I 99I473 42 301295 1093 698705 41 20 293399 Io50 99I448 42 30o95I 1092 698049 40 21 I 9*294029 Io48 9.991422 42 9.302607 Io9o 10*697393 3 22 294658 1o46 991397 42 30326I I089 696739 38 23 295286 Io45 991372 43 303914 1087 696086 37 24 295913 I043 99I346 43 304567 io86 695433 36 25 296539 1042 g99321 43 305218 1084 694782 35 26 297I64 1040 991295 43 305869 io83 694131 34 2;7 297788 Io39 991270 43 3o65i9 Io81 69348 33 2 298412 1037 991244 43 3071 68 08 692832 32 29 299034 io36 99I218 43 307816 1078 692I84 3i 30 299655 io34 991193 43 308463 Io77 69I537 3o 3I 9.300276 I032 9.99II67 43 9.309I09 I075 IO-6gO89I 2 32 300895 io3I 991141 43 309754 1074 690246 2 33 3oi514 1029 99II15 43 310399 1073 689601 27 34 302132 1028 99I090 43 3 042 107I 688958 26 35 302748 1026 991064 43 31i685 1070 6883i5 25 36 303364 1025 gg991038 43 312327 io68 687673 24 37 303979 1023 991012 43 312968 1067 687032 23 38 304593 1022 990986 43 3i3608 io65 686392 22 39 305207 I020 990960 43 3I4247 Io64 685753 21 40 305819 1019 990934 44 314885 1062 685ii5 20 41 9.306430 1017 9.990908 44 9.3I5523 io6i I0.684477 19 42 307041 1io6 990882 44 316159 Io6o 683841 I8 43 307650 io14 990855 44 316795 io58 683205 17 44 308259 IoI3 990829 44 317430 I057 682570 i6 45 308867 lOI 990803 44 318o64 io55 681936 i5 46 309474 10 I 990777 44 318697 io54 68i303 14 47 3ioo8o ioo8 990750 44 319330 io53 680670 13 48 3Io685 10o7 990724 44 31996I Io5i 68oo39 I2 49 311I289 I005 990697 44 320592 Io5o 679408 II 5 31893 100oo4 990671 44 321222 1048 678778 io 51 9.312495 ioo3 9.990645 44 9-32185I 1047 10-678149 9 52 313097 1001 9906I8 44 322479 1045 677521 8 53 313698 000 99o591 44 323o106 1044 676894 7 54 314297 998 990565 44 323733 1043 676267 6 55 31 4897 997 g90538 44 324358 1041 675642 5 56 315495 996 9905II 45 324983 io4o 675017 4 57 316092 994 990485 45 325607 1039 674393 3 58 3I6689 993 990458 45 326231 Io 37 673769 2 59 317284 J991 gg99o43I 45 326853 io36 673147 I 6o 3I7879 990 990404 45 327475 io35 672525 0 Cosine. D. Sine. D. Cotang. D. Tang. _j~f 30 LOGARITHMIC SINES, TANGENTS, ETC. TABLE II. 120 1667~ t I Sine. D. Cosine. D. Tang. D. Cotan. o 9-317879 990 9.990404 45 9.327475 Io35 Io.672525 6o I 318473 988 990378 45 328095 io33 671905 5 2 319066 987 99035I 45 328715 Io32 671285 58 3 3i9658 986 990324 45 329334 Io3o 670666 57 4 320249 984 990297 45 329953 Io029 670047 56 5 320840 983 990270 45 330570 I028 669430 55 6 321430 982 990243 45 331187 I026 6688I3 54 322019 980 990215 45 33i8o3 1o25 668197 53 322607 979 99oi0188 45 332418 I124 667582 52 9 323i94 977 99016I 45 333033 1023 666967 5i Io 323780 976 990I34 45 333646 i021 666354 50 II 9.324366 975 9.9900o 7 46 9.334259 1020 Io166574I 49 I2 324950 973 990079 46 33487I 1019 665129 48 I3 325534 972 990052 46 335482 IOI7 664518 47 I4 326117 970 990025 46 336093 IoI6 663907 46 I5 326700 969 989997 46 336702 ioi5 663298 45 i6 32728I 968 989970 46 3373II IOI3 662689 44 17 327862 966 989942 46 337919 1012 662081 43 I8 328442 965 9899I5 46 338527 iolI 661473 42 19 329021 964 989887 46 339133 Ioio 660867 41 20 329599 962 989860 46 339739 I008 660261 40 21 9.330176 961 9.989832 46 9.340344 I007 IO0659656 3 22 330753 960 989804 46 340948 ioo6 659052 38 23 331329 958 989777 46 34I552 1oo004 658448 37 24 331903 957 989749 47 342155 oo003 657845 36 25 332478 956 989721 47 342757 1002 657243 35 26 333o5i 954 989693 47 343358 1ooo 656642 34 27 333624 953 989665 47 343958 999 656042 33 28 334195 952 989637 47 344558 99 655442 32 29 334767 950 9896Io 47 345157 997 654843 31 30 335337 949 989582 47 345755 996 654245 3o 3i 9.335906 948 9-989553 47 9.346353 994 IO0653647 29 32 336475 946 989525 47 346949 993 653o5I 28 33 337043 945 989497 47 347545 992 652455 27 34 3376o10 944 989469 47 348I4I 99I 651859 26 35 338176 943 989441 47 348735 990 65I262 25 36 338742 94' 9894i3 47 349329 988 65067I 24 37 339307 940 989385 47 349922 987 650078 23 38 339871 939 989356 47 35o034 986 649486 22 39 340434 937 989328 47 35iio6 985 648894 ~ 2I 40 340996 936 989300 47 351697 983 6483o3 20 4I 9.341558 935 9-989271 47 9.352287 982 Io-6477I3 1 42 342119 934 989243 47 352876 98I 647124 I8 43 342679 932 9892 4 47 353465 980 646535 17 44 343239 93I 989I86 47 354053 979 645947 i6 45 343797 930 989157 47 354640 977 645360 I5 46 344355 929 989I28 48 355227 976 644773 14 47 344912 927 989i00 48 3558i3 975 644187 I3 48 345469 926 989071 48 356398 974 643602 12 49 346024 925 98904Z2 48 356982 973 6430I8 ii 50 346579 924 989014 48 357566 971 642434 Io 5I 9-347I34 922 9-988985 48 9-358I49 970 Io.64I85I 52 347687 921 988956 48 35873i 969 64I269 8 53 348240 920 988927 48 3593i3 968 640687 7 54 1 348792 919 988 98 48 359893 967 640107 6 55 349343 917 988869 48 360474 966 630526 5 56 349893 916 988840 48 36io53 965 638947 4 57 350443 9I5 98881I 49 361632 963 638368 3 58 350992 914 988782 49 3622IO 962 637790 2 59 35i54o0 93 988753 49 362787 961 637213 I 60 352088 9I1 988724 49 363364 960 636636 o / Cosine. D. Sine. D. Cotang. D. Tang. t TABLE II. LOGARITHMIC SINES, TANGENTS, ETC. 31 13~ 1660' Sine. D. Cosine. D. Tang. D. | Cotang. 0 9-352088 911 9.988724 49 9.363364 960 - Io.636636 60 I 352635 910 988695 49 363940 959 636060 51 2 3538I 8i 909 988666 49 3645i5 958 635485 58 3 353726 908 988636 49 365090oo 957 63491 0 57 4 35427 I 907 988607 49 365664 955 634336 56 5 3548I5 905 988578 49 366237 954 633763 55 6 355358 904 988548 49 3668o0 953 633190 54 7 355oi 90o3 9885I9 49 367382 952 6326I8 53 8 356443 902 988489 49 367953 951 632047 52 9 356984 01 988460 49 368524 950 631476 5i 10 357524 899 988430 49 369094 949 630906 50 ii 9.358064 898 9.98840, 49 9.369663 948 10 630337 40 12 358603 897 988371 49 370232 946 629768 48 i3 359141 896 988342 49 370799 945 629201 47 I4 359678 895 988312 50q 371367 944 628633 46 I5 3602I5 893 988282 50o 371933 943 628067 45 i6 360752 892 988252 50o 372499 942 627501 44 I 361287 89I 988223 50o 373064 941I 626936 43 I8 36I822 890 988I93 50 373629 940 62637I 42 I9 362356 889 988I63 50o 37493 939 625807 4' 20 362889 8 988133 50o 374756 938 625244 40 21 9.363422 887 9.988I03 50 9.3753I9 937 I0 62468i 39 22 363954 885 988073 50 37588i 935 624119 38 23 364485 884 988043 50 376442 934 623558 37 24 365oi6 883 988013 50o 377003 933 622997 36 25 365546 882 987983 50 377563 932 622437 35 26 366075 88i 987953 50 378I22 931 62I878 34 27 366604 880 987922 50o 378681 930 62I3I9 33 28 367131 879 987892 50o 379239 929 62076I 32 29 367659 877 987862 50 379797 928 620203 3i 3o 368 I85 876 987832 51 38o354 927 619646 30 3i 9-3687I 875 9.98780I 5i 9'3809I0 926 I0-619090 29 32 369236 874 98777I 5I 381466 925 6I8534 28 33 36976 873 987740 5i 382020 924 617980 27 34 370285 872 987710 5I 382575 923 617425 26 35 370808 87 I 987679 51 383129 922 6I687 25 36 371330 870 987649 5i 383682 921 6,63i8 24 37 371852 869 987618 5I 384234 920 6I5766 23 38 372373 867 987588 5i 384786 919 615214 22 39 372894 866 987557 5i 385337 9I8 614663 21 40 373414 865 987526 5i 385888 917 614I12 20 4I 9.373933 864 9.987496 5i 9.386438 9Io5 1o6I3562 19 42 374452 863 987465 5i 386987 914 6i30oi3 i8 43 374970 862 987434 51 387536 913 612464 17 44 375487 86i 987403 52 388084 912 6I 916 i6 45 376003 86o 987372 52 38863i 91I 611369 i5 46 3765I9 859 98734I 52 389I78 910 6I0822 I4 47 377035 858 9873IO 52 389724 90 6io276 i3 48 377549 857 987279 52 390270 90 60973 12 49 378063 856 987248 52 39o8i5 907 6o0985 ii 50 378577 854 987217 52 391360 906 6o8640 10 5I 9.379089 853 9 987186 52 9 39190o3 90o5 Io'6o8o97 52 379601 852 987155 52 392447 904 607553 53 38o0I3 85I 987I24 52 392989 90o3 60701I 7 54 380624 850 987092 52 393 53I 902 606469 6 55 38ii34 849 987061 52 394073 901 6o05927 5 56 38I643 848 987030 52 394614 900 605386 4 5 382152 847 986998 52 39554 899 604846 3 5 38266I 846 986967 52 395694 898 6043o6 2 59 383i68 845 986936 52 396233 897 603767 I 60 383675 844 986904 52 396771 896 603229 0 I Cosine. D. [ Sine. D. Cotang. D. | Tang. I 103o 76~0 32 LOGARITHMIC SINES, TANGENTS, ETC. TrABLE II. 140 1650 Sine. D. Cosine. D. Tang. D. Cotan,. O 9.383675 844 9-986904 52 I 9.39677I 896 io0603229 6o I 384182 843 986873 53 397309 896 602691 5 2 384687 842 986841 53 397846 895 602154 5 3 385192 84I 986809 53 398383 894 601617 57 4 385697 840 986778 53 398919 893 6oio8i1 56 5 386201 839 986746 53 399455 892 600545 55 6 386704 838 986714 53 399990 89I 6ooo io 54 7 387207 837 986683 53 4oo024 80o 599476 53 8 387709 836 98665I 53 40oio58 88 598942 52 9 388210 835 986619 53 401591 888 59840 5i 10 388711 834 986587 53 402124 887 597876 50 II 9.3892II 833 9.986555 53 9.402656 886 o.597344 4 12 3897 1 1 832 986523 53 403 I 87 885 5968i3 48 I3 390210 83i 986491 53 403718 884 596282 47 14 390708 830 986459 53 404249 883 59575I 46 15 39I206 828 986427 53 404778 882 595222 45 I6 391703 827 986395 53 405308 88i 594692 44 I7 39219 826 986363 54 405836 880 594i64 43 I8 39269 825 986331 54 406364 879 593636 42 19 3931 824 986299 54 406892 878 5931io8 41 20 393685 823 986266 54 407419 877 59258I 40 21 9.394179 822 9.986234 54 9.407945 876 10592055 3 22 394673 821 986202 54 408471 875 591529 38 23 395166 820 986169 54 408996 874 59i004 37 24 395658 8I9 986I37 54 40952I 874 590479 36 25 396I50 8I8 986104 54 4I00oo45 873 589955 35 26 39664I 8I7 986072 54 4Io569 872 58943I 34 27 397132 817 986039 54 411092 871 588908 33 28 397621 8i6 986007 54 4ii6i5 870 588385 32 29 398111 815 985974 54 412137 869 587863 31 30 398600 8i4 985942 54 412658 868 587342 30 3I 9-399088 8I3 9.985909 55 9.4I3179 867 0-58682 2 32 399575| 812 985876 |55| 413699 866 58630oI 28 33 400062 8ii 985843 55 414219 865 58578I 27 3,4 4o05492 8io 9858ii 55 414738 864 585262 26 35 401033 809 985778 55 4I5257 864 54743 25 36 401520 808 985745 55 4i5775 863 584225 24 37 402005 807 985712 55 4I6293 862 583707 23 38 402489 806 985679 55 4I68io 86I 583I9O 22 39 402972 805 985646 55 417326 86o 582674 21 40 403455 804 985613 55 417842 859 582158 20 4I 9-403938 803 9-985580 55 9.418358 858 IO-581642 1 42 404420 802 985547 55 418873 857 581I27 I8 43 404901 801 985514 55 419387 856 5806I3 17 44 405382 8oo00 98548o 55 419901 855 5800oo i6 45 405862 799 985447 55 4204I5 855 579585 1i5 46 4o634I 798 985414 56 420927 854 579073 14 47 406820 797 985381 56 421440 853 578560 3 48 407299 796 985347 56 421952 852 578048 12 49 407777 795 9853I4 56 422463 85I 577537 11 50 408254 794 985280 56 422974 850 577026 IO 5I 9.40873I 794 9.985247 56 9-423484 849 Io-576I6 52 409207 793 9852T3 56 423993 848 576007 53 409682 792 985I80 56 424503 848 575497 7 54 4oI57 |791 985146 56 425oII 847 574989 6 55 410632 790 985II3 56 425519 846 57448, 5 56 411io6 789 985079 56 426027 845 573973 4 57 4Ii579 788 985045 56 426534 844 573466 3 58 412052 787 9850II 56 427041 843 572951 2 59 412524 786 984978 56 427547 843 572453 60 4I2996 785 984944 56 428o52 842 57I948 0 Cosine. D. Sine. D. Cotang. D. Tang. | 1040 - 750 TABLE II. LOGARITHMIC SINES, TANGENTS, ETC. 33 150 1640 I Sine. D. Cosine. D. Tang. D. Cotang. I 0 9.412996 785 9.984944 57 9.428052 842 10o 571948 6o I 4I3467 784 984910 57 428558 84i 571442 59 2 413938 783 984876 57 429062 840 570938 58 3 414408 783 984842 57 429566 839 570434 57 4 414878 782 984808 57 430070 838 569930 56 5 4I5347 78I 984774 57 430573 838 569427 55 6 4158i5 780 984740 57 431075 837 568925 54 7 416283 779 984706 57 431577 836 568423 53 8 41675i 778 984672 57 432079 835 56792I 52 9 4172I7 777 984638 57 432580 834 567420 5I Io 4I7684 776 984603 57 433080 833 566920 50 11 9I.48150 775 9.984569 57 9.433580 832 Io0566420 49 I2 4186I5 774 984535 57 434080 832 565920 48 13 4I9079 773 984500 57 434579 83i 56542I 47 i4 4I9544 773 984466 57 435078 830 564922 46 i5 420007 772 984432 58 435576 829 564424 45 i6 420470 77I 984397 58 436073 828 563927 44 17 420933 770 984363 58 436570 828 563430 43 i8 4213q5 769 984328 58 437067 827 562933 42 I9 421857 768 984294 58 437563 826 562437 41 20 422318 767 984239 58 438059 825 56I941 40 2I 9-422778 767 9.-984224 58 9.438554 824 Io.56I446 39 22 423238 766 984190 58 439048 823 560952 38 23 423697 765 984155 58 439543 823 560457 37 24 4241 56 764 984120 58 440036 822 559964 36 25 424615 763 984085 58 440529 821 559471 35 26 425073 762 984050 58 44I022 820 558978 34 425530 76I 9840I5 58 4415I4 8x9 558486 33 2 425987 760 98398I 58 442006 8iq 557994 32 29 426443 760 983946 58 442497 8I 557503 3I 30 426899 759 9839II 58 442988 817 557012 3o 31 9.427354 758 9.983875 58 9.443479 8i6 Io.55652I 29 32 427809 757 983840 59 443968 8i6 556032 28 33 428263 756 98380o5 59 444458 815 555542 27 34 4287I7 755 983770 59 444947 814 555053 26 35 429170 754 983735 59 445435 8I3 554565 25 36 429623 753 983700 59 445923 812 554077 24 37 430075 752 983664 59 4464II 812 553589 23 38 430527 752 983629 09 446898 81ii 553102 22 39 430978 75I 983594 59 447384 810 5526I6 2I 40 431429 750 983558 59 447870 809 552130 20 4I 9.431879 749 9.983523 59 9'448356 809 10.55I644 19 42 432329 749 983487 59 448841 808 55ii59 18 43 432778 748 983452 59 449326 807 550674 17 44 433226 747 9834I6 59 449810 806 550190 I6 45 433675 746 98338I 59 450294 806 549706 I5 46 434122 745 983345 59 450777 805 549223 I4 47 434569 744 983309 59 45126o 804 548740 13 48 4350I6 744 983273 60 45I743 803 548257 I2 49 435462 743 983238 60 452225 802 547775 II 50 435908 742 983202 60 452706 802 547294 io 5I 9.436353 741 9.983166 60 9.453I87 8oi 10o54683 9 52 436798 740 983130 60 453668 800 546332 53 437242 740 983094 60 454i4z8 799 545852 54 437686 739 983058 60 454628 799 545372 55 438129 738 983022 60 455107 798 544893 5 56 438572 737 982986 60 455586 797 5444i4 4 57 4390oi4 736 982950 60 456064 796 543936 3 58 439456 736 982914 60 456542 796 543458 2 59 439897 735 982878 60 457oi9 795 542981 I 60 440338 734 982842 60 457496 794 542504 0 Cosine. D. Sine. I D. Cotang. D. Tang. 105~ 40 2* 34 LOGARITHMIC SINES, TANGENTS, ETC. TABLE II. 160 1630 r Sine. D. Cosine. D. | Tang. D. Cotang. t o 9*440338 734 9.982842 6o 9.457496 794 10-542504 6o I 440778 733 982805 60 457973 793 542027 5 2 441218 732 982769 6i 458449 793 54I55I 5 3 441658 73I 982733 6I 458923 792 541075 57 4 442096 731 982696 6I 459400 791 540600 56 5 442535 730 982660 6I 459875 790 540125 55 6 442973 729 982624 6I 460349 790 53965I 54 4434io 72g 982587 6I 460823 789 539177 53 443847 727 98255I 61 46I297 78 538703 52 9 444284 727 982514 6i 461770 788 538230 5i 10 444720 726 982477 6i 462242 787 537758 50 I1 9.445155 725 9.982441 6i 9.4627I5 786 Io0537285 4 12 445590 724 982404 6i 463186 785 536814 48 13 446025 723 982367 6I 463658 785 536342 47 I4 446459 723 982331 6I 464128 784 535872 46 I5 446893 722 982294 6I 464599 783 535401 45 i6 447326 72I 982257 6i 465069 783 534931 44 17 447759 720 982220 62 465539 782 53446I 43 i8 448i9I 720 982183 62 466008 78I 533992 42 19 448623 719 982146 62 466477 780 533523 41 20 449054 7I8 982109 62 466945 780 533055 40 21 9-449485 717 9-982072 62 9-4674I3 77 I0o.532587 3 22 449915 716 982035 62 467880 77 532120 38 23 450345 716 981998 62 468347 778 53I653 37 24 450775 715 98196I 62 468814 777 53ii86 36 25 451204 714 981924 62 469280 776 530720 35 26 451632 713 98I886 62 469746 775 530254 34 27 452060 713 981849 62 470211 775 529789 33 28 452488 712 981812 62 470676 774 529324 32 29 452915 711 98I774 62 471141 773 528859 3I 30 453342 7I1 981737 62 47I605 773 528395 3o 3i 9.453768 710 9-981700 63 9.472069 772 10.527931 29 32 454194 709 981662 63 472532 771 527468 28 33 454619 708 98I625 63 472995 771 527005 27 34 455044 707 981587 63 473457 770 526543 26 35 455469 707 981549 63 473919 769 52608I 25 36 455893 706 981512 63 47438I 769 525619 24 37 456316 705 981474 63 474842 76 525158 23 38 456739 704 981436 63 475303 767 524697 22 39 457162 704 981399 63 475763 767 524237 21 40 457584 703 98136I 63 476223 766 523777 20 4I 9-458006 702 9.981323 63 9.476683 765 10.5233I7 1 42 458427 701 981285 63 477142 765 522858 i 43 458848 701 981247 63 477601 764 522399 17 44 459268 700 981209 63 478059 763 52194I i6 45 459688 699 981171 63 478517 763 521483 I5 46 46oio8 69 98 9833 64 478975 762 521025 I4 47 460527 698 981095 64 479432 76I 520568 13 48 460946 697 981057 64 479889 761 520111 12 49 461364 696 981019 64 480345 760 519655 ii 50o 461782 695 980981 64 480801 759 519199 10 51 9. 462199 695 9.980942 64 9.481257 759 10*518743 9 52 4626I6 694 980904 64 481712 758 518288 53 463032 693 980866 64 482167 757 517833 7 54 463448 693 980827 64 482621 757 517379 6 55 463864 692 980789 64 483075 756 516925 5 56 464279 691 980750 64 483529 755 516471 4 57 464694 690 980712 64 483982 755 5I6oi8 3 58 4651o8 690 980673 64 484435 754 5i5565 2 59 465522 689 980635 64 484887 753 5 5ii3 I 60 465935 688 980596 64 485339 753 51466i 0 f Cosine. D. Sine. D. Cotang. 1D. Tang. 7 1060 3 TABLE II. LOGARITHMIC SINES, TANGENTS, ETC. 35 170 1 620 _ Sine. D. Cosine. D. Tang. D. Cotang.' 0 9-465935 688 9.980596 64 9.485339 755 io051466I 60 I 466348 688 980558 64 48579I 752 514209 59 2 46676I 687 98o0519 65 486242 751 513758 58 3 467173 686 980480 65 486693 75I 5i33o0 57 4 467585 685 980442 65 487I43 750 5I2857 56 5 467996 685 980403 65 487593 749 512407 55 6 468407 684 980364 65 488043 749 511957 54 7 468817 683 980325 65 488492 748 5ii5o8 53 8 469227 683 980286 65 488041 747 5II059 52 9 469637 682 980247 65 489390 747 5io6Io 5i Io 470046 68i 980208 65 489838 746 510162 50 II 9.470455 680 9.980oi69 65 9.490286 746 Io050974I 49 12 470863 680 980130 65 490733 745 509267 48 I3 471271 679 98009 I 65 491180 744 508820 47 I4 471679 678 980052 65 49I627 744 508373 46 I5 472086 678 9800oo2 65 492073 743 507927 45 I6 472492 677 979973 65 492519 743 507481 44 17 472898 676 979934 66 492965 742 507035 43 i8 473304 676 979895 66 493410 74, 506590 42 I9 473710 675 979855 66 493854 740 506I46 41 20 474115 674 9798I6 66 494299 740 505701 40 21 9.4745i9 674 9-979776 66 9 494743 740 Io0505257 3 22 474923 673 979737 66 495186 739 504814 38 23 475327 672 979697 66 495630 738 504370 37 24 475730 672 979658 66 496073 737 503927 36 25 476133 671 979618 66 496515 737 503485 35 26 476536 670 979579 66 496957 736 5o3o43 34 27 476938 669 979539 66 497399 736 50260o 33 28 477340 669 979499 66 49784I 735 502159 32 29 47774I 66 979459 66 498282 734 50I718 31 30 478142 667 979420 66 498722 734 50o1278 30 3I 9.478542 2 667 9-979380 66 9.499I63 733 Io.5oo837 29 32 478942 666 979340 66 499603 733 500397 28 33.479342 665 979300 67 500042 732 499958 27 34 47974I 665 979260 67 50048I 731 4995I9 26 35 480140 664 979220 67 500920 731 499080 25 36 480539 663 979180 67 501359 730 498641 24 37 480937 663 979I40 67 501797 730 498203 23 38 48I334 662 979100 67 502235 729 497765 22 39 48I73i 66i 979059 67 502672 728 497328 21 40 482128 66I 9790I9 67 503I09 728 49689I 20 41 9.482525 660 9.978979 67 9.503546 727 10.496454 19 42 482921 659 978939 67 503982 727 4960I8 i8 43 483316 659 978898 67 504418 726 495582 17 44 4837I2 658 9788 67 504854 725 495I46 i6 45 484I07 657 978817 67 505289 725 4947II 15 46 484501 657 978777 67 505724 724 494276 14 47 484895 656 978737 67 506159 724 49384 I3 40 485289 655 978696 68 506593 723 493407 12 49 485682 655 978655 68 507027 722 492973 ii 50 486075 654 978615 68 507460 7-22 492540 10 5I 9-486467 653 9'978574 68 9.507893 72I 10-492IO7 52 486860 653 978533 68 508326 721 491674 8 53 48725I 652 978493 68 508759 720 491241 7 54 487643 65i 978452 68 509191 719 490809 6 55 488034 65I 978411 68 509622 719 490378 5 56 488424 65o 978370 68 5oo0054 718 489946 4 488814 65o 978329 68 5Io485 7I8 4895 5 3 58 | 489204 64 978288 68 5io9i6 77 489o84 2 59 489593 64 978247 68 5ii346 716 488654 I 6o 489982 648 | 978206 68 511776 7 6 488224 0 I Cosine. D. S ine. D. Cotang. D. T ang. 1070 o12 _ _o 36 LOGARITHMIC SINES, TANGENTS, ETC. TABLE II. 180 1610 Sine. D. Cosine. D. Tang. D. Cotang. / o 9.489982 648 9.978206 68 9 511776 716 10-488224 60 I 49037I 648 978165 68 512206 7I6 487794 50 2 490759 647 978124 68 512635 715 487365 58 3 49II47 646 978083 69 5I3o64 714 486936 57 4 491535 646 978042 69 513493 714 486507 56 5 49I922 645 97800I 69 5I392I 713 486079 55 6 492308 644 977959 69 514349 713 48565i 54 492695 644 977~18 69 5I4777 712 485223 53 493o08I 643 977 77 69 515204 712 484796 52 9 493466 642 977835 69 5563i 711II 484369 5i I0 493851 642 977794 69 516057 710 483943 50 II 9-494236 64I 9.977752 69 9 5I6484 710 Io.4835i6 4 12 494621 64i 9777II 69 516910 709 483090 4 13 495005 64o 977669 69 517335 709 482665 47 14 495388 639 977628 69 51776I 708 482239 46 I5 495772 639 977586 69 5i8186 708 481814 45 i6 496154 638 977544 7o 518610 707 48139o 44 I7 496537 637 977503 70 519034 706 480966 43 I8 496919 637 97746I 70 519458 706 480542 42 19 497301 636 977419 70 519882 705 480oI8 41 20 497682 636 977377 70 520305 705 479695 40 21 9.498064 635 9.977335 70 9.520728 704 10479272 3 22 498444 634 977293 70 52115I 703 478849 38 23 498825 634 977251 70 521573 703 478427 37 24 499204 633 977209 70 521995 703 478oo5 36 25 499584 632 977I67 70 522417 702 477583 35 26 499963 632 977125 70 522838 702 477162 34 27 500342 63i 977083 70 523259 701 47674I 33 28 500721 63i 977041 70 523680 o01 476320 32 29 501099 630 976999 70 524100 700 475900 3I 30 501476 629 976957 70 524520 699 475480 3o 3I 9.50I854 629 9-976914 70 9.524940 699 I0.475060 29 32 502231 628 976872 7I 52535 69 47464 28 33 502607 628 976830 71 525778 698 474222 27 34 502984 627 976787 71 526197 697 473803 26 35 503360 626 976745 71 5266I5 697 473385 25 36 503735 626 976702 71 527033 696 472967 24 37 504II0 625 976660 7I 52745I 696 472549 23 38 504485 625 976617 71 527868 695 472132 22 39 504860 624 976574 71 528285 695 471715 21 40 505234 623 976532 71 528702 694 471298 20 41 9.505608 623 9.976489 7I 9.529I19 693 10.47088I 10 42 50598i 622 976446 71 529535 693 470465 18 43 506354 622 976404 71 529951 693 470049 I7 44 506727 621 97636I 71 530366 692 469634 16 45 507099 620 976318 71 530781 69I 469219 15 46 507471 620 976275 7I 531196 69I 468804 14 47 507843 6I9 976232 72 5316. 690 468389 13 48 508214 619 976189 72 532025 690 467975 12 49 508585 6I8 976I46 72 532439 689 46756I II 50 508956 618 976103 72 532853 689 467I47 10 5I 9.509326 617 9.976060 72 9.533266 688 10.466734 52 509696 616 976017 72 533679 688 46632I 8 53 5Ioo65 616 975974 72 534092 687 465908 7 54 5I0434 6i5 975930 72 534504 687 465496 6 55 5io8o3 6I5 975887 72 534916 686 465084 5 56 511172 614 975844 72 535328 686 464672 4 57 51I540 6I3 975800 72 535739 685 46426I 3 58 511907 613 975757 72 536i5o 685 463850 2 59 512275 612 975714 72 53656i 684 463439 I 60 5I2642 612 975670 72 536972 684 463028 0 / Cosine. D. Sine. D. Cotang. D. Tang. t 108~ 71o TABLE II. LOGARITHMIC SINES, TANGENTS, ETC. 37 190 1600 I Sine. D. Cosine. D. Tang. D. Cotang.' o 9.512642 612 9.975670 73 9.536972 684 Io.-463028 6o I 5I3oo 6I 975627 73 537382 683 4626I8 5 2 5I3375 6II 975583 73 537792 683 462208 5 3 51374I 6io 975539 73 538202 682 46I798 57 4 514107 609 975496 73 5386ii 682 461389 5 5 514472 609 975452 73 539020 68I 460980 55 6 54837 60 975408 73 539429 68i 46057I 54 7 5-.5202 608 975365 73 539837 680 460163 53 8 5I5566 607 975321 73 540245 680 459755 52 9 S5I93o 607 975277 73 540653 679 459347 51 10 516294 6o6 975233 73 54Io06 679 458939 5o Ii 9.5I6657 605 9.975I89 73 9.54I468 678 1o.458532 44 12 517020 605 975145 73 541875 678 458I25 48 13 517382 604 97510I 73 542281 677 457719 47 14 517745 604 975057 73 542688 677 4573I2 46 I5 518IO7 603 975oI3 73 543o94 676 456906 45 i6 5i8468 603 974969 74 543499 676 4569oI 44 17 518829 602 974925 74 543965 675 456095 43 i8 5I 9I 6oi 97488o 74 544310 675 455690 42 19 51955I 6oi 974836 74 544715 674 455285 41 20 5199ii 6o00 974792 74 545II9 674 45488I 40 21 9-52027I 600 9.974748 74 9.545524 673 Io-454476 39 22 52063I 599 974703 74 545928 673 454072 38 23 520990 599 974659 74 54633I 672 453669 37 24 521349 598 974614 74 546735 672 45326 36 25 521707 598 974570 74 547138 671I 452862 35 26 522066 597 974525 74 547540 671 452460 34 27 522424 596 97448I 74 547943 670 452057 33 28 52278I 596 974436 74 548345 670 45I655 32 29 523138 595 97439I 74 548747 669 451253 31 30 523495 595 974347 75 549149 669 45o85I 30 JI 9.523852 594 9.974302 75 9.54955o 668 Io.450450 29 32 524208 594 974257 75 54995I 668 450049 28 33 524564. 593 974212 75 550352 667 449648 27 34 524920 593 974I67 75 550752 667 449248 26 35 525275 592 974122 75 551153 666 448847 25 36 525630 59I 974077 75 55I552 666 448448 24 37 525984 591 974032 75 551952 665 448048 23 38 526339 590 973987 75 55235I 665 447649 22 39 526693 590 973942 75 552750 665 447250 21 40 527046 589 973897 75 553I49 664 446851 20 41 9-5274400 589 9-973852 75 9.553548 664 I0-446452 19 42 527753 588 973807 75 553946 663 446054 I8 43 528105 588 97376I 75 554344 663 445656 I7 44 528458 587 973716 76 554741 662 445259 i6 45 528810 587 97367I 76 555i39 662 444861 I5 46 529161 586 973625 76 555536 66i 444464 14 47 5295I3 586 97358o 76 555933 66i 444067 13 48 529864 585 973535 76 556329 660 443671 I2 49 530215 585 973489 76 556725 660 443275 II 50 530565 584 973444 76 55712I 659 442879 io 5I 9.53o09I5 584 9.973398 76 9.' 5575I7 659 io.442483 52 531265 583 973352 76 557913 659 442087 53 53i614 582 973307 76 55830o8 658 44I692 7 54 531963 582 97326I 76 558703 658 441297 6 55 532312 58i 9732I5 76 559097 657 440903 5 56 53266I 58I 973I69 76 559491 657 440509 4 57 533009 580 973I24 76 559885 656 4401i5 3 58 533357 580 973078 76 560279 656 439721 2 59 533704 579 973032 77 560673 655 439327 I 6o 534052 578 972986 77 56io66 655 438934 O I I Cosine. D. Sine. D. Cotang. I D. Tang.! 1090 170o 38 LOGARITHMIC SINES, TANGENTS, ETC. TABLE II. 200 1590! Sine. D. Cosine. D. Tang. D. Cotang. t 0 9.534052 578 9.972986 77 9.56IO66 655 Io.438934 6o I 534399 577 972940 77 561459 654 43854I 5S 2 534745 577 972894 77 56i851 654 438149 58 3 535092 577 972848 77 562244 653 437756 57 4 535438 576 972802 77 562636 653 437364 56 5 535783 576 972755 77 563028 653 436972 55 6 536129 575 972709 77 5634I9 652 43658I 54 7 536474 574 972663 77 56381I 652 436189 53 8 5368i8 574 9726I7 77 564202 65i 435798 52 9 537163 573 972570 77 564593 65i 435407 5i 10 537507 573 972524 77. 564983 650.4350I7 50 II 9.53785I 572 9.972478 77 9.565373 650 Io0434627 4 12 538i94 572 972431 78 565763 649 434237 48 I3 538538 571 972385 78 566153 649 433847 47 14 53888o 571 972338 78 566542 649 433458 46 i5 539223 570 97229I 78 566932 648 433068 45 i6 539565 570 972245 78 567320 648 432680 44 57 39907 569 972198 78 56770 647 432291 43 i 540249 569 9721 1 78 568o98 647 43 902 42 I9 540590 568 972105 78 568486 646 43 154 4I 20 54093i 568 972058 78 568873 646 431127 40 21I 9.541272 567 9.97201I 78 9.56926I 645 Io.430739 30 22 541613 567 971964 78 569648 645 430352 38 23 54I953 566 971917 78 570035 645 429965 37 24 542293 566 971 87 78 570422 644 429578 36 25 542632 565 971823 78 570809 644 429I91 35 26 542971 565 971776 78 57II9 643 428805 34 27 5433io 564 97I729 79 57I58I 643 428419 33 2 543649 564 97I682 79 57I967 642 428033 32 29 543987 563 971635 79 572352 642 427648 3i 30 544325 563 971588 79 572738 642 427262 30 3i 9.544663 562 9-971540 79 9.573I23 64I I0.426877 22 32 545000 562 971493 79 573507 64i 426493 2 33 545338 561 97,446 79 573892 640 426108 27 34 545674 56i 971398 79 574276 640 425724 26 35 5460oi 560 971351 79 574660 639 425340 25 36 546347 560 971303 79 575044 639 424956 24 37 546683 559 971256 79 575427 639 424573 23 38 54701o9 559 971208 79 575810 638 424190 22 39 547354 558 97116 79 576193 638 423807 21 40 547689 558 971113 79 576576 637 423424 20 41 9.548024 557 9.971066 80 9.576959 637 10.423041 I1 42 548359 557 97II018 80 57734I 636 422659 I 8 43 548693 556 970970 80 577723 636 422277 17 44 549027 556 970922 80 578104 636 421896 I6 45 549360 555 970874 80 578486 635 421514 15 46 549693 555 970827 80 578867 635 421 33 14 47 550026 554 970779 80 579248 634 420752 13 4 55o359 554 970731 80 579629 634 420371 12 49 550692 553 970683 80 580009 634 41999, II 50 551024 553 970635 80 580389 633 419611 10 5I 9-551356 J 552 9-970586 80 9-580769 633 10-419231 52 551687 552 970538 80 581149 632 4i885I 8 53 5520o8 552 970490 80 581528 632 418472 7 54 552349 551 970442 80 58I907 632 418093 6 55 552680 55I 970394 80 582286 63i 417714 5 56 553oio 550 970345 8I 582665 63i 417335 4 5 55334I 550 970297 8I 583044 630 416956 3 553670 549 970249 8I 583422 630 4i6578 2 59 554000 549 970200 81 583800 629 416200 I 60 554329 548 970152 8I 584177 629 415823 0' Cosine. D. Sine. D. Cotang. D. Tang. t 1100 690 TABLE II. LOGARITHMIC SINES, TANGENTS, ETC. 39 21~ 158~ t Sine. D. Cosine. D. Tang. D. Cotang. r o 9.554329 548 9.970152 8i 9.584I77 629 IO.415823 6o I 554658 548 970103 8I 584555 62 4I5445 59 2 554987 547 970055 8I 584932 628 4i5o68 58 3 5553I5 547 970006 8I 585309 628 4I4691 5 4 555643 546 969957 8i 585686 627 4i43I4 56 5 555971 546 969909 8i 586062 627 413938 55 6 556299 545 969860 8i 586439 627 413561 54 556626 545 9698ii 8i 586815 626 413I85 53 556953 544 969762 8i 587190 626 412810 52 9 557280 544 969714 8i 587566 625 412434 5I Io 557606 543 969665 8i 58794t 625 4I2059 50 II 9. 557932 543 9.969616 82 9.588316 625 io.4ii684 49 12 558258 543 969567 82 588691 624 4II309 48 I3 558583 542 9695I8 82 589066 624 410934 47 I4 558909 542 969469 82 589440 623 4io560 46 15 559234 54i 969420 82 589814 623 41oi86 45 i6 559558 54I 969370 82 590188 623 409812 44 I7 559883 540 969321 82 590562 622 409438 43 i8 560207 540 969272 82 590935 622 409065 42 19 56o53i 539 969223 82 591308 622 408692 4I 20 560855 539 969I73 82 59I68i 621 408319 40 21 9.56ii78 538 9.969124 82 9.592054 62I 10407946 39 22 56i5oi 538 969075 82 592426 620 407974 38 23 56I824 537 969025 82 592799 620 407201 37 24 562146 537 968976 82 593171 6I9 406829 36 25 562468 536 968926 83 593542 6I 406458 35 26 562790 536 968877 83 593914 61 406086 34 27 5631I2 536 968827 83 594285 6i8 405715 33 28 563433 535 968777 83 594656 6I8 405344 32 29 563755 535 96872 83 595027 617 404973 3I 30 564075 534 968678 83 595398 617 404602 3o 3I 9.564396 534 9.968628 83 9.595768 6I7 IO. 404232 2 32 5647i6 533 968578 83 596i38 6I6 403862 28 33 565036 533 968528 83 596508 6i6 403492 27 34 565356 532 968479 83 596878 6i6 403122 26 35 565676 532 968429 83 597247 6i5 402753 25 36 565995 53i 968379 83 5976I6 6i5 402384 24 37 566314 53i 968329 83 597985 6i5 4020o5 23 38 566632 53I 968278 83 598354 614 401646 22 39 56695i 530 968228 84 598722 614 401278 2I 40 567269 530 968178 84 599o9I 6i3 400909 20 41 9.567587 529 9.968128 84 9.599459 6I 3 o.40054I 19 42 567904 529 968078 84 599827 6I3 400173 18 43 568222 528 968027 84 600194 612 399806 17 44 568539 528 967977 84 600562 612 399438 16 45 568856 528 967927 84 600929 6ii 39907I 15 46 569I72 527 967876 84 601296 6ii 398704 14 47 569488 527 967826 84 6oi663 6ii 398337 13 48 569804 526 967775 84 602029 6Io 397971I 12 49 570I20 526 967725 84 602395 6io 397605 II 50 570435 525 967674 84 60276I 6io 397239 io 51 9.570751 525 9.967624 84 9.603127 609 io.396873 9 52 57IO66 524 967573 84 603493 609 396507 53 57I380 524 967522 85 603858 609 396I42 7 54 57I695 523 967471 85 604223 608 395777 6 55 572009 523 967421 85 604588 608 395412 5 56 572323 523 967370 85 604953 607 395047 4 57 572636 522 967319 85 605317 607 394683 3 58 572950 522 967268 85 605682 607 394318 2 59 573263 521 967 2I7 85 606046 606 393 54 I 6o 573575 52 967166 85 60640 6o6 393590 o Cosine. D. Sine. D. Cotang. D. Tang. r 1110 680 40 LOGARITHMIC SINES, TANGENTS, ETC. TABLE II. 220 157o I Sine. D. Cosine. D. Tang. D. Cotang.! 0 9-573575 521 9.967I66 85 9.606410 606 Io0393590 6o 1 573888 520 967115 85 606773 6o6 393227 5 2 574200 520 967064 85 607137 605 392863 58 3 5745I2 519 9670I3 85 607500 605 392500 57 4 574824 519 96696I 85 607863 604 392I37 56 5 575136 5I9 966910 85 608225 604 391775 55 6 575447 5I8 966859 85 608588 604 3914I2 54 575758 5i8 966808 85 608950 603 39I050 53 8 576069 5I7 966756 86 609312 603 390688 52 9 576379 517 966705 86 609674 603 390326 51 Io 576689 5i6 966653 86 6i0036 602 389964 50 11 9.576999 5i6 9.966602 86 9.6I0397 602 io0389603 49 I2 5773o 5i6 966550 86 610759 602 38924I 48 I3 5776I8 515 966499 86 6II120 6oi 388880 47 14 577927 515 966447 86 611480 601 388520 46 15 578236 514 966395 86 6i i84I 60I 388I59 45 i6 578545 5I4 966344 86 612201 600 387799 44 I7 578853 5I3 966292 86 61256I 6oo 387439 43 I8 579162 513 966240 86 61292I 600 387079 42 19 579470 513 966188 86 61328I 599 386719 4I 20 579777 512 966I36 86 6I364I 599 386359 4o 21 9.580085 5I2 9.966085 87 9.6I4000 598 io0386000 3 22 580392 5I 966033 87 6143S 598 38564I 38 23 580699 5II 96598I 87 6147I8 598 385282 37 24 58oo005 51 965928 87 615077 597 384923 36 25 5813I2 5Io 965876 87 6I5435 597 384565 35 26 5816I8 5Io 965824 87 615793 597 384207 34 27 581924 509 965772 87 6i615I 596 383849 33 2 582229 509 965720 87 6i6509 596 38349I 32 29 582535 509 965668 87 616867 596 38313 3I 30 582840 508 9656i5 87 617224 595 382776 3o 3I 9.583i45 508 9965563 87 9'617582 595 I0*382418 29 32 583449 507 9655I I 87 617939 595 382061 28 33 583754 507 965458 87 618293 594 381705 27 34 584058 506 9654o6 87 618652 594 381348 26 35 584361 506 965353 88 619008 594 380992 25 36 584665 506 96530o 88 619364 593 38o636 24 37 584968 505 965248 88 619720 593 380280 23 38 585272 505 965195 88 620076 593 379?24 22 39 585574 504 965143 88 620432 592 379568 21 40 585877 504 965090 88 620787 592 379213 20 4 9q.586I79 50o3 9965037 88 9.62II42 592 1o-378858 I9 42 586482 503 964984 88 621497 59I 378503 18 43 586783 503 96493I 88 621852 59 378148 17 44 587o85 502 964879 88 622207 59o 377793 i6 45 587386 502 96482 88 62256I 59o 377439 I5 46 587688 50 964773 88 6229I5 59o 377085 14 47 587989 5oi 964720 88 623269 589 376731 i3 48 588289 5oi 964666 89 623623 589 376377 12 49 588590 500 964613 89 623976 589 376024 11 50 588890 o00 964560 89 624330 588 375670 10 5I 9.589190 499 9.964507 89 9.624683 588 io037537 52 589489 499 964454 89 6250o36 588 374964 53 589789 499 964400 89 625388 587 374612 7 54 5900oo88 498 964347 89 625741 587 374259 6 55 590387 498 964294 89 626093 587 373907 5 56 590686 497 964240 89 626445 586 373555 4 57 590984 497 964187 89 626797 586 373203 3 58 59I282 497 964I33 89 627I49 586 37285I 2 59 59i58o 496 964080 89 62750oI 58 372499 I 60 59i878 496 964026 89 627852 585 37248 o Cosine. D. Sine. D. Cotang. D. Tang. TABLE II. LOGARITHMIC SINES, TANGENTS, ETC. 41 230 1560 I Sine. D. Cosine. D. Tang. D. Cotang. I o 9.591878 496 9.964026 89 9-627852 585 Io.372148 6o I 592I76 495 963972 89 628203 585 371797 5 2 592473 495 9639j9 89 628554 585 371446 5 3 592770 495 963865 9go 628905 584 371095 57 4 593067 494 9638Ii 9o 629255 584 370745 56 5 593363 494 963757 90 629606 583 370394 55 6 593659 493 963704 9o 629956 583 370044 54 7 593955 493 963650 9go 630306 583 369694 53 8 594251 493 963596 9go 630656 583 369344 52 9 594547 492 963542 9go 63ioo5 582 368995 5I 10 594842 492 963488 90 63i355 582 368645 50 Ii 9.595137 491 9.963434 9o 4.631704 582 ~10368296 44 I2 595432 49I 963379 90 632053 58i 367947 48 I3 595727 491 963322 90 632402 58I 367598 47 14 59602I 490 963271 90 632750 58i 367250 46 i5 596315 490 963217 90 633099 580 366901 45 i6 596609 489 963163 9go 633447 580 366553 44 17 596903 489 963108 91 633795 580 366205 43 I8 597196 489 963054 91 634143 579 365857 42 I9 5974 0 488 962999 91 634490 579 36551o 4I 20 597783 488 962945 91 634838 579 365162 40 21 9-598075 487 9.962890 91 9.635I85 578 io.3648I5 3 22 598368 487 962836 9I 635532 578 364468 38 23 598660 487 962781 91 635879 578 364121 37 24 598952 486 962727 91 636226 577 363774 36 25 599244 486 962672 91 636572 577 363428 35 26 599536 485 962617 91 636919 577 363o8I 34 27 599827 485 962562 91 637265 577 362735 33 28 6ooii8 485 962508 91 637611 576 362389 32 29 600409 484 962453 91 637956 576 362044 31 30 600700 484 962398 92 638302 576 361698 30 31 9-60099o 484 9.962343 92 9.638647 575 io.36I353 29 32 601280 483 962288 92 638992 575 36ioo8 28 33 60o570 483 962233 92 639337 575 360663 27 34 60o86o 482 962178 92 639682 574 36o318 26 35 60215o 482 962123 92 640027 574 359973 25 36 602439 482 962067 92 64037I 574 359629 24 37 602728 481 962012 92 640716 573 359284 23 38 603017 481 961957 92 641060 573 358940 22 39 6o33o5 481 961902 92 641404 573 358596 21 40 603594 480 961846 92 641747 572 358253 20 4I 9'603882 480 9.961791 92 9.642o91 572 Io035709 19 42 604170 479 961735 92 642434 572 357566 18 43 604457 479 961680 92 642777 572 357223 17 44 604745 479 961624 93 643120 571 356880 16 45 605032 47 96i569 93 643463 571 356537 I5 46 605319 478 961513 93 643806 571 356194 I4 47 605606 478 961458 93 644148 570 355852 I3 48 605892 477 961402 93 644490 570 3555io 12 49 606179 477 961346 93 644832 570 355i68 ii 50 606465 476 961290 93 645174 569 354826 10 5I 9.60675I 476 9.961235 93 9.645516 569 io0354484 52 607036 476 96I179 93 645857 569 354143 53 607322 475 961123 93 646199 569 3538oI 7 54 607607 475 96I067 93 646540 568 35346o 6 55 607892 474 9610I 93 64688I 568 353119 5 56 608177 474 960955 93 647222 568 352778 4 57 60846I 474 960899 93 647562 567 352438 3 58 608745 473 960843 94 647903 567 352097 2 59 609029 473 960786 94 648243 567 351757 I 60 609313 473 960730 94 648583 566 35I417 0 / Cosine. D. Sine. D1. Cotang. D. Tang. I 1130 66~ 42 LOGARITHMIC SINES, TANGENTS, ETC. TABLE II. 240 15&0' Sine. D. Cosine. |D. Tang. D. ICotang. t o 9 609313 473 9'96073' 94 9.648583 566 Io.3514I7 6o I 609597 472 960674 94 6489.23 566 351077 5 2 609880 472 960618 94 649263 566 350737 58 3 6ioi664 472 960561 94 649602 566 350398 57 4 610447 47I 960505 94 649942 565 350058 56 5 6I0729 471 960448 94 65028I 565 349719 55 6 6IIOI2 470 960392 94 65o620 565 349380 54 7 6I I294 470 960335 94 650959 564 34904I 53 8 611576 470 960279 94 651297 564 348703 52 9 611858 469 960222 94 65i636 564 348364 51 10 612140 469 96dI65 94 651974 563 348026 50 II 9.612421 469 9.960Io0 95 9.652312 563 IO.347688 44 12 612702 468 960052 95 652650 563 347350 4 I3 612983 468 959995 95 652988 563 347012 47 I4 613264 467 959938 95 653326 562 346674 46 I5 613545 467 959882 95 653663 562 346337 45 I6 613825 467 959825 95 654000 562 346000 44 17 6i14o05 466 959768 95 654337 56I 345663 43 I8 614385 466 9597II 95 654674 56i 345326 42 19 6I4665 466 959654 935 6550oi 56I 344989 4I 20 614944 465 959596 95 655348 56i 344652 4O 21 9.615223 465 9'959539 95 9.655684 560 IO-3443I6 3 22 6I5502 465 959482 95 656020 560 343980 38 23 615781 464 959425 95 656356 560 343644 37 24 6i6060 464 959368 95 656692 559 343308 36 25 6i6338 464 959310 96 657028 559 342972 35 26 6I66i6 463 959253 96 657364 559 342636 34 27 616894 463 959195 96 657699 559 342301 33 28 617172 462 959138 96 658034 558 341966 32 29 617450 462 959080 96 658369 558 34I631 31 30 617727 462 959023 96 658704 558 34t296 30 31 961I8004 46 9.958965 96 9.659039 558 lo34096I 2 32 6I828I 461 9589o8 96 659373 557 340627 28 33 6I8558 46i 95885o 96 659708 557 340292 27 34 618834 46o 958792 96 660042 557 339958 26 35 619110 460 958734 66 660376 557 339624 25 36 619386 460 958677 96 660710 556 339290 24 37 6I9662 459 9586I9 96 66I043 556 338957 23 3 6I9938 459 95856i 96 661377 556 338623 22 39 620213 459 958503 97 66I710 555 338290 21 40 620488 458 958445 97 662043 555 337957 2C 4I 9.620763 458 9.958387 97 9.662376 555 ro0337624 1 42 621038 457 958329 97 662709 554 337291 I1 43 621313 457 958271 97 663042 554 336958 I7 44 621587 457 958213 97 663375 554 336625 i6 45 621861 456 958154 97 663707 554 336293 i5 46 622135 456 958096 97 664039 553 33596i 14 47 622409 456 958038 97 664371 553 335629 13 48 622682 455 957979 97 664703 553 335297 12 49 622956 455 957921 97 665035 553 334965 ii 50 623229 455 957863 97 665366 552 334634 10 5I 9.623502 454 9.957804 97 9.665698 552 IO-334302 52 623774 454 957746 98 666029 552 333971 53 624047 454 957687 98 666360 551 333640 7 54 6243I9 453 957628 98 66669I 551 333309 6 55 624591 453 957570 98 667021 55i 332979 5 56 624863 453 957511 98 667352' 551 332648 4 57 625135 452 957452 98 667682 550 332318 3 58 625406 452 957393 98 668oi3 550 331987 2 59 625677 452 957335 98 668343 550 33I657 I 60 625948 45I 957276 98 668673 550 331327 0 t Cosine. D. Sine. D. Cotang. D. Tang. _ 114~ 65- TAB-LE II. LOGARITHMIC SINES,'TANGENTS, ETC. 43 25~ 1540 I Sine. D. Cosine. D. Tang. D. Cotan. I o 9.625948 451 9.957276 98 9.668673 550 1Io331327 60 I 626219 45I 957217 98 669002 549 330998 5q 2 626490 451 957158 98 669332 549 33o668 58 3 626760 450 957099 98 66966I 549 330339 57 4 627030 450 957040 98 66999I 548 330009 56 5 627300 45o 95698I 98 670320 548 329680 55 6 627570 449 95692I 99 670649 548 32935I 54 7 627840 449 956862 99 670977 548 329023 53 8 628109 449 9568o3 99 671306 547 328694 52 9 628378 448 956744 99 67l635 547 1 328365 5i Io 628647 448 956684 99 67I963 547 328037 50 II 9-628916 457 9.956625 99 9.672291 547 10.327709 4 I2 629185 447 956566 99 672619 546 327381 48 I3 629453 447 956506 99 672947 546 327053 47 I4 629721 446 956447 99 673274 546 326726 46 15 629989 446 956387 99 673602 546 326398 45 i6 630257 446 956327 99 673929 545 32607, 44 I7 630524 446 956268 99 674257 545 325743 43 I8 630792 445 956208 100 674584 545 325416 42 19 631059 445 956148 100 674911 544 325089 4I 20 631326 445 956089 100 675237 544 324763 40 21 9.631593 444 9.956029 100 9.675564 544 Io0324436 3q 22 63I859 444 955969 100 675890 544 324110 38 23 632123 444 955909 100 676217 543 323783 37 24 632392 443 955849 100 676543 543 323457 36 25 632658 443 955789 100 676869 543 323131 35 26 632923 443 955729 100 677194 543 322806 34 27 633189 442 955669 100 677520 542 322480 33 28 633454 442 955609 100 677846 542 322154 32 29 633719 442 955548 I00 678171 542 321829 31 30 633984 441 955488 100 678496 542 321504 30 31 9.634249 441 9'955428 I OI 9.678821 54I I0-321 79 29 32 634514 440 955368 101 679146 54I 320854 28 33 634778 440 955307 IoI 679471 541 320529 27 34 635042 440 955247 101 679795 54I 320205 26 35 6353o6 439 955186 101 680120 540 3I9880 25 36 635570 439 955126 101 680444 540 319556 24 37 635834 439 955065 101 680768 540 3I 232 23 38 636097 438 955005 101 681092 540 3 890o8 22 39 636360 438 954944 101 681416 539 318584 21 40 636623 438 954883 101 681740 539 318260 20 4I 9.636886 437 9.954823 101 9.682063 539 o03I7937 19 42 637I48 437 954762 IoI 682387 53q 317613 i8 43 6374II 437 954701 IO1 682710 538 317290 17 44 637673 437 954640 101 683033 538 316967 i6 45 637935 436 954579 101 683356 538 316644 i5 46 6381.97 436 954518 102 683679 538 316321 4 47 638458 436 954457 102 684001 537 315999 I3 48 638720 435 954396 102 684324 537 315676 I2 49 63898I 435 954335 I02 684646 537 315354 II 50 639242 435 954274 102 684968 537 3150o32 I1 51 9.639503 434 9-954213 e2 9685290 536 1o.314710 9 52 639764 434 954152 102 685612 536 3i4388 8 53 640024 434 954090 102 685934 536 314066 7 54 640284 433 954029 102 686255 536 313745 6 55 640544 1 433 953968 102 686577 535 313423 5 56 640804 433 953906 102 686898 535 1 313102 4 57 641064 432 953845 102 687219 535 312781 83 58 641324 432 953783 102 687540 535 312460 2 59 64I583 432 953722 Io3 68786I 534 312139 I 60 641842 43i 95366o io3 688I82 534 311818 0 Cosine. I D. Sine. D. Cotang. D. Tang. / 115~ 4~ 44 LOGARITHMIC SINES, TANGENTS, ETC. TABLE II. 260 1530' Sine. D. Cosine. D. Tang. D. Cotang. 0 9-.64I842 43i 9.953660 103 9.688I82 534 Io-3I818 60 I 642IOI 43 953599 103 688502 534 31 498 5 2 642360 43I 953537 io3 688823 534 3111I77 58 3 642618 430 953475 io3 689143 533 310857 57 4 642877 430 953413 1o3 689463 533 310537 56 5 643135 430 953352 1io3 689783 533 3102I7 55 6 643393 430 953290 io3 690103 533 309897 54 643650 429 953228 Io3 690423 533 309577 53 643908 429 953166 io3 690742 532 309258 52 9 644165 429 953104 io3 6910io62 532 308938 5I Io 644423 428 953042 o103 691381 532 3I869 50 II 9.644680 428 9.952980 104 9.691700 53I Io-3083oo 4 12 644936 428 952918 104 692019 531 307981 4 I3 645193 427 952855 I04 692338 53i 307662 47 I4 645450 427 952793 104 692656 53I 307344 46 I5 645706 427 95273I I04 692975 53I 307025 45 I6 645962 426 952669 104 693293 530 306707 44 I7 646218 426 952606 104 693612 530 306388 43 i8 646474 426 952544 104 693930 530 306070 42 19 646729 425 952481 104 694248 530 305752 41 20 646984 425 952419 104 694566 529 3o5434 40 21 9.647240 425 9.952356 104 9.694883 529 Io0305II7 39 22 647494 424 952294 104 695201 529 304799 38 23 647749 424 95223I 104 695518 529 304482 37 24 648004 424 952168 Io5 695836 529 304164 36 25 648258 424 95210o6 105 696153 528 303847 35 26 648512 423 952043 io5 696470 528 303530 34 648766 423 95I980 io5 696787 528 3032I3 33 2 649020 423 951917 1o5 697103 528 302897 29 649274 422 95I854 io5 697420 527 3025o80 3 30 649527 422 951791 io5 697736 527 302264 30 3I 9.64978I 422 9.951728 10o5 9.698053 527 io030I947 2 32 65oo0034 422 951665 io5 698369 527 30o63I 2 33 650287 421 951602 105 698685 526 3oi315 27 34 650539 421 951539 Io5 699001 526 300999 26 35 650792 421 951476 io5 699316 526 3oo684 25 36 651044 420 951412 Io5 699632 526 300368 24 37 651297 420 95I349 Io6 699947 526 300053 123 38 651549 420 951286 Io6 700263 525 299737 22 39 6518oo 4I9 951222 106 700578 525 299422 21 40 652052 419 95II59 io6 700893 525 299107 20 4I 9.652304 419 9.951096 IO6 9.70I208 524 IO298792 19 42 652555 418 951032 106 701523 524 298477 I8 43 652806 418 950968 10o6 70o837 524 298163 17 44 653057 4i8 950905 io6 702152 524 297848 i6 45 653308 4I8 950841 io6 702466 524 297534 15 46 653558 417 950778 io6 702781 523 297219 14 47 653808 417 950714 io6 703095 523 296905 i3 48 654059 417 95o650 io6 703409 523 296591 12 49 654309 4i6 950586 Io6 703722 523 296278 iI 50 654558 4i6 950522 107 704036 522 295964 10 5i 9.654808 4I6 9.950458 107 9-704350 522 10o29565o0 52 655058 4i6 950394 107 704663 522 295337 8 53 655307 4I5 950330 107 704976 522 295024 7 54 655556 415 950266 107 705290 522 294710 6 55 655805 4I5 950202 107 705603 521 294397 5 56 656054 414 950138 107 705916 521 294084 4 57 656302 414 950074 I07 706228 521 293772 3 58 65655I 414 950010ooo 107 706541 521 293459 2 59 656799 4I3 949945 107 706854 521 293146 I 60 657047 413 949881 107 707I66 520 292834 0 Cosinc. D. Sine. D. Cotang. D. Tang. 1160 630 TABLE II. LOGARITHMIC SINES, TANGENTS, ETC. 45 270 1520 I Sine. D. | Cosine. D.| Tang. | D. Cotang. / 0 9 657047 4I3 9.94988I 107 9.707'66 520 10-292834 60 1 657295 413 949816 107 707478 520 292522 5 2 6575)42 412 949752 107 707790 520 2922I0 58 3 6577c0 412 949688 io8 708102 520 291898 57 4 658037 412 949623 108 708414 519 29I586 56 5 658284 412 949558 108 708726 5I9 291274 55 6 65853I 411 949494 108 709037 519 290963 54 7 658778 4I 949429 108 709349 519 290651 53 8 659025 411 949364 0i8 709660 519 290340 52 9 65927I 410 949300,o8 709971 518 290029 5I I0 659517 410 949235 o108 710282 518 289718 50 I 9.659763 4I0 9.94917o io8 9.710593 518 Io0289407 49 12 6600oo 409 949105 io8 710904 518 289096 48 13 660253 409 949040 io8 7II215 518 288785 47 14 66o50o 409 948975 io8 711525 517 288475 46 i5 66o746 409 948910 I o8 711836 517 288164 45 6 660991 408 948845 10o8 712146 517 287854 44 17 661236 408 948780 109 712456 517 287544 43 I8 66I48I 408 948715 1o9 712766 5i6 287234 42 I9 661726 407 948650 Iog09 7306 516 286924 41 20 661970 407 948584 o09 713386 5I6 286614 40 21 9.662214 407 9.948519 109 9.713696 5i6 10.286304 3q 22 662459 407 948454 109 714005 5i6 285995 38 23 662703 4o6 948388 109 714314 515 285686 37 24 662946 406 948323 I og09 7 14624 5I5 285376 36 25 663190 406 948257 I o09 74933 5I5 285067 35 26 663433 405 948192 I09 7I5242 515 284758 34 27 663677 405 948126 og09 75551 514 284449 33 28 663920 405 948060 og09 71586 514 2841,40 32 29 664163 4o5 947995 I10 716168 514 283832 31 30 664406 404 947929 O0 716477 5I4 283523 30 31 9.664648 404 9-947863 o0 9*7I6785 5I4 10283215 29 32 66489I 404 947797 110 717093 513 282907 28 33 665133 403 94771 110 717401 513 282599 27 34 665375 4o3 947665 Io 717709 513 28229 26 35 665617 4o03 947600' 0 7I8017 513 28983 25 36 665859 402 947533 II0 718325 5i3 28I675 24 37 666ioo 402 947467 11 0 78633 512 28I367 23 38 666342 402 947401 I10 718940 512 28I060 22 39 666583 402 947335 I10 719248 512 280752 21 40 666824 401 947269 110 7i9555 512 280445 20 41 94667065 401 9-947203 110 9-719862 512 10-280I38 I9 42 667305 401 947136 III 720I69 511 27983I I8 43 667546 401 947070 III 720476 5ii 279524 17 44 667786 400 947004 II1 720783 51 279217 i6 45 668027 400 946937 III 72 I89 511 27891 15 46 668267 400 946871 I II 72396 511 278604 14 47 6685o6 399 946804 III 721702 5io 278298 I3 48 668746 399 946738 III 722009 51io 277991 12 49 668986 399 946671 III 7223I5 510io 277685 II 50 669225 399 946604 III 722621 5I0 277379 10 5i 9-669464 398 9-946538 III 9-722927 5io 10 I0-277073 9 52 669703 398 94647I III 723232 509 276768 8 53 669942 398 946404 III 723538 509 276462 7 54 670I8I 397 946337 III 723844 509 276I56 6 55 6704I9 397 946270 112 724149 509 27585I 5 56 670658 397 946203 112 724454 50o 275546 4 57 670896 397 946I36 I12 724760 5o8 275240 3 5 67I3 396 946069 112 725065 508 274935 2 59 671372 396 946002 112 725370 58 274630 I 60 67I609 396 945935 112.725674 508 274326 0 Cosine. D. Sine. D1. Cotang. D. Tang. I 1170 620 4:6 LOGARITHMIC SINES, TANGENTS, ETC. TABLE II. 280 1510 I Sine. D. Cosine. D. Tang. D. Cotang. 0 9.67i609 396| 9.945935 112 9.725674 5o8 10o274326 6o I 67i847 395 945868 112 725979 508 27402I 59 2 672084 395 945800 112 726284 507 273716 5 3 672321 395 945733 11I2 726588 507 273412 57 4 672558 395 945666 112 726892 507 2731o8 56 5 672795 394 945598 112 727197 507 272803 55 6 673032 394 94553I 112 727501 507 272499 54 7 673268 394 945464 I3 727805 5o6 272195 53 8 673505 394 945396 113 728109 506 27I891 52 9 673741 393 945328 113 7284I2 5o6 271588 5I 0O 673977 393 945261 I3 728716 5o06 271284 50 II 9.67423 393 9-945I93 I13 9-729020 5o6 IO.270980 4 12 674448 392 945125 I 3 729323 505 270677 4 13 674684 392 945058 113 729626 505 270374 47 I4 674919 392 944990 I3 729929 505 270071 46 I5 675I55 392 9454922 I13 730233 505 269767 45 i6 675390 39I 944854 II 730535 55 269465 44 17 675624 391 944786 113 730838 504 260262 43 i8 675859 39I 944718 I3 731141 504 268859 42 I9 676094 391 944650 113 731444 504 268556 4I 20 676328 390 944582 114 731746 504 268254 40 21 9.676562 390 9-944514 114 9-732048 504 IO-267952 3 22 676796 390 944446 I14 73235I 503 267649 38 23 677030 390 944377 114 732653 5o3 267347 37 24 677264 389 944309 II4 732955 503 267045 36 25 677498 389 944241 114 733257 503 266743 35 26 67773I 389 944172 114 733558 503 266442 34 27 677964 388 944I04 114 733860 502 266140 33 28 678197 388 944036 114 734162 502 265838 32 29 678430 388 943967 114 734463 502 265537 31 30 678663 388 943899 114 734764 502 265236 30 31 9I 678895 387 9.943830 114 9.735066 502 I0-264934 29 32 679128 387 94376I 114 735367 502 264633 28 33 679360 387 943693 115 735668 50oi 264332 27 34 679592 387 943624 II5 735969 501 26403I 26 35 679824 386 943555 115 736269 50o 26373I 25 36 68oo0056 386 943486 115 736570 50 263430 24 37 680288 386 943417 115 736870 50I 263130 23 38 680519 385 943348 115 737171 500 262829 22 39 680750 385 943279 115 737471 500 262529 21 40 680982 385 9432I0 115 73777I 500 262229 20 4I 9-68123I 385 9.943I4I 115 9-73807I 500 IO-26I1929 1 42 681443 384 943072 115 738371 500 2691629 I8 43 681I674 384 943003 115 738671 499 261329 17 44 681905 384 942934 115 738971 499 261029 i6 45 68235 384 942864 I 5 739271 499 260729 15 46 682365 383 942795 II6 739570 499 26430 14 47 682595 383 942726 1I6 739870 499 260130 13 48 682825 383 942656 II6 740169 499 25983i I2 49 683o55 383 942587 ii6 740468 498 259532 II 50 683284 382 942517 Iu6 740767 498 259233 I0 5i 9-6835I4 382 9-942448 II6 9-741066 498 10-258934 52 683743 382 942378 i6 741365 498 258635 53 683972 382 942308 Iu6 741664 498 258336 7 54 684201 38i 942239 ii6 741962 497 258038 6 55 68443o 38i 942169 I6 74226I 497 257739 5 56 J684658 38i 942099 I 6 742559 497 25744 4 57 684887 38o 942029 I 6 742858 497 257I42 3 58 685iI5 38o 941959 ii6 743156 497 256844 2 59 685343 38o 941889 117 743454 497 256546 I 60 68557I 380 941819 I57 743752 496 256248 0 Cosine. D. Sine. D. Cotang. D. Tan. 1180 610,3,, 6793.o TABLE II. LOGARITHMIC SINES, TANGENTS, ETC. 47 290 1500, Sine. D. Cosine. D. Tang. D. Cotan g. 0o 9-68557I 380 9.94I8I9 117 9.743752 496 10-256248 6o I 685799 379 941749 117 744050 496 2559 j 5 2 686027 379 941679 I17 744348 496 255652 3 686254 379 941609 II7 744645 496 255355 57 4 686482 379 94I539 117 744943 496 255057 56 5 686709 378 941469 I17 745240 496 254760 55 6 686936 378 94I398 117 745538 495 254462 54 687163 378 941328 II7 745835 495 254165 53 8 687389 378 94I258 IiI7 746132 495 253868 52 9 687616 377 941187 117 746429 495 25357I 5I 10 687843 377 941117 117 746726 495 253274 50 II 9.688069 377 9'94I046 II8 9.747023 494 10'252977 49 12 688295 377 940975 118 747319 494 25268I 48 13 688521 376 940905 II8 747616 494 252384 47 I4 688747 376 940834 II8 7479I3 494 252087 46 15 688972 376 940763 /I8 748209 494 251791 45 I6 689198 376 940693 118 748505 493 251495 44 17 689423 375 940622 II8 74880I 493 251199 43 I8 689648 375 94055i II8 749097 493 250903 42 I9 689873 375 940480 II8 749393 493 250607 41 20 690098 375 940409 ii8 749689 493 250311 40 21 9.690323 374 9-940338 II8 9-749985 493 10.250015 39 22 69o0548 374 940267 118 750281 492 2497I9 38 23 690772 374 940196 I 8 750576 492 249424 37 24 690996 374 940125 I19 750872 492 249I28 36 25 691220 373 940054 II9 751167 492 248833 35 26 691444 373 939982 II9 751462 492 248538 34 27 691668 373 93991 I 9 751757 492 248243 33 2 691892 373 939840 119 752052 49I 247948 32 29 692I15 372 939768 I19 752347 491 247653 3i 30 692339 372 939697 119 752642 491 247358 30 3i 9.692562 372 9'939625 119 9-752937 49[ I0-247063 29 32 692785 371 939554 119 753231 49I 246769 28 33 693008 371 939482 I119 753526 49' 246474 27 34 693231 371 9394I0 119 753820 490 246180 26 35 693453 371 939339 119 75415 490 245885 25 36 693676 370 939267 120 754409 490 245591 24 37 693898 370 939195 120 754703 490 245297 23 38 694120 370 939123 120 754997 490 245003 22 39 694342 370 939052 120 75529I 490 244709 21 40 694564 369 938980 120 755585 489 244415 20 41 9.694786 369 9.938908 I20 9.755878 489 10*244122 19 42 695007 369 938836 I20 756172 489 243828 18 43 695229 369 938763 120 756465 489 243535 17 44 69545o 368 93869I 120 756759 489 243241 i6 45 69567I 368 938619 120 757052 489 242948 15 46 695892 368 938547 I20 757345 488 242655 I4 47 69611ii3 368 938475 120 757638 488 242362 13 48 696334 367 938402 121 75793i 488 242069 12 49 696554 367 938330 121 758224 488 241776 ii 50 696775 367 938258 I21 758517 488 241483 io 5I 9,696995 367 9.938I85 I2I 9-7588Io 488 I241190 9 52 697215 366 938113 121 759102 487 240898 53 697435 366 938040 I21 759395 487 240605 7 54 697654 366 937967 21 759687 487 240313 6 55 697874 366 937895 121 759979 487 240021 5 56 698094 365 937822 121 760272 487 239728 4 57 698313 365 937749 121 76o0564 487 239436 3 58 698532 365 937676 121 76o856 486 239144 2 59 69825I 365 937604 121 76II48 J486 238852 I 600 6o 698970 364 |93753I 12 761439 486 238561 0 Cosine. D. Sine. D. Cotang. D. Tang. | 119~ 60~ 48 LOGARITHMIC SINES, TANGENTS, ETC. TABLE II. 300 1490 / Sine. D. Cosine. D. Tang. D. Cotang. o 9.698970 364 9.93753i 121 9.76I439 486 10.23856i 6o I 699 189 364 937458 2 2 76I73I 486 238269 5 2 699407 364 937385 122 762023 486 237977 58 3 699626 364 937312 I22 762314 486 237686 57 4 699844 363 937238 122 762606 485 237394 56 5 700062 363 937165 122 762897 485 237103 55 6 700280 363 937092 122 763188 485 236812 54 700498 363 937019 122 763479 485 23652I 53 700716 363 936946 122 763770 485 236230 52 9 700933 362 936872 122 764061 485 235939 5I 10 701151 362 936799 122 764352 484 235648 50 II 9.701368 362 9.936725 I22 9.764643 484 Io.235357 4 12 701585 362 936652 I23 764933 484 235067 48 I3 701802 36I 936578 I23 765224 484 234776 47 I4 702019 36I 936505 123 765514 484 234486 46 15 702236 36i 936431 123 765805 484 234195 45 i6 702452 36I 936357 123 766095 484 233905 44 I7 702669 360 936284 I23 766385 483 233615 43 i8 702885 360 936210 123 766675 483 233325 42 19 703o101 360 936I36 123 766965 483 233035 4I 20 703317 360 936062 I23 767255 483 232745 40 21 9-703533 359 9-935988 i23 9-767545 483 IO232455 3 22 703749 359 935914 I23 767834 483 232166 38 23 703964 359 935840 123 768124 482 231876 37 24 704179 359 935766 124 768414 482 231586 36 25 704395 359 935692 124 768703 482 231297 35 26 704610 358 935618 124 768992 482 231008 34 27 704825 358 935543 124 76928I 482 230719 33 28 705040 358 935469 124 769571 482 230429 32 29 705254 358 935395 124 769860 48i 230140 3I 30 705469 357 935320 124 770148 481 229852 30 31 9.705683 357 9.935246 124 9.770437 48 I O.229563 29 32 705898 357 935171 124 770726 48I 229274 28 33 706112 357 935097 124 771015 48i 228985 27 34 706326 356 935022 124 771303 48I 228697 26 35 706539 356 934948 124 771592 48I 228408 25 36 706753 356 934873 I24 771880 480 228120 24 37 706967 356 934798 125 772168 480 227832 23 38 707180 355 934723 I25 772457 480 227543 22 39 707393 355 934649 I25 772745 480 227255 21 40 707606 355 934574 125 773033 480 226967 20 4I 9.7078I9 355 9-934499 I25 9-773321 48o 10.226679 I 42 708032 354 934424 125 773608 479 226392 I 43 708245 354 934349 125 773896 479 226Io4 17 44 708458 354 934274 125 774184 479 225816 16 45 708670 354 934199 125 77447I 479 225529 I5 46 708882 353 934123 125 774759 479 225241 14 47 709094 353 934048 125 775046 479 224954 i3 4 709306 353 933973 125 775333 479 224667 12 49 709518 353 933898 126 775621 47 224379 II 50 709730 353 933822 126 775908 478 224092 10 5I 9.709941 352 9.933747 I26 9-776I 95 478 10.223805 52 710153 352 933671 126 776482 478 2235i8 8 53 710364 352 933596 126 776768 478 223232 7 54 710575 352 933520 126 777055 478 222945 6 55 710786 35I 933445 126 777342 478 222658 5 56 710997 35I 933369 126 777628 477 222372 4 5 71I208 35I 933293 I26 7779I5 477 222085 3 58 711419 351 933217 126 778201 477 221799 2 59 711629 350 933141 126 778488 477 221512 I 60 711839 350 933066 126' 778774 477 221226 o j Cosine. D. Sine. D. Cotang. D. Tang. I 1200 590 TABLE II. LOGARITHMIC SINES, TANGENTS, ETC. 49 810~ 1480 t Sine. D. Cosine. D. Tang. D. Cotang. I 0 97iI839 350 9.933066 126 9.778774 477 10 22I226 60 I 712050 350 932990 127 779060 477 220940 59 2 712260 35o 932914 127 779346 476 220654 58 3 712469 349 932838 127 779632 476 220368 57 4 712679 349 932762 127 779918 476 220082 56 5 712889 349 932685 I27 780203 476 219797 55 6 713098 349 932609 127 780489 476 219511 54 713308 349 932533 127 780775 476 219225 53 7 I 3 51 7 3 4V 932457 127 781060 476 218940 52 9 713726 348 932380 127 781346 475 218654 5i 10 713935 348 932304 127 781631 475 2I8369 50 II 9-714144 348 9.932228,127 9.781916 475 10o2I8084 49 I2 714352 347 932151 127 782201 475 217799 48 13 714561 347 932075 128 782486 475 217514 47 I4 714769 347 931998 128 78277I 475 217229 46 15 714978 347 93192I 128 783056 475 216944 45 i6 715186 347 931845 128 783341 475 216659 44 17 715394 346 931768 128 783626 474 216374 43 i8 715602 346 931691 128 783910 474 216090 42 19 715809 346 931614 128 784195 474 215805 41 20 716017 346 931537 128 784479 474 215521 40 21 9.716224 345 9.931460 128 9.784764 474 10.215236 3 22 716432 345 931383 128 785048 474 214952 38 23 7i6639 345 931306 128 785332 473 214668 37 24 716846 345 931229 129 785616 473 214384 36 25 7I7053 345 931152 129 785900 473 214100 35 26 717259 344 93I075 I29 786184 473 213816 34 27 7I7466 344 930998 129 786468 473 213532 33 28 717673 344 930921 129 786752 473 213248 32 29 717879 344 930843 129 787036 473 212964 3I 30 718085 343 930766 129 787319 472 21268I 3o 3I 9.71829I 343 9.930688 129 9.787603 472 10-212397 29 32 718497 343 9306II 129 787880 472 212114 28 33 718703 343 930533 129 788170 472 211830 27 34 718909 343 930456 129 788453 472 211547 26 35 719114 342 930378 I29 788736 472 211264 25 36 719320 342 930300 i3o 789019 472 210981 24 37 719525 342 930223 i3o 789302 47' 210o698 23 38 719730 342 93o0145 i30 789585 471 210415 22 39 719935 34I 930067 i3o 789868 47' 210132 2I 40 7201i40 341 929989 i30 790oi5I 471 209849 20 41 9*7 20345 34I 9.-9299II 13o 9.790434 471 1O 209566 19 42 -720549 34I 929833 i30 790716 47' 209284 I8 43 720754 340 929755 i3o 790999 471 209001 17 44 720958 340 929677 i30 79128I 471 208719 i6 45 721162 340 929599 i3o 791563 470 208437 i5 46 721366 340 929521 i30 791846 470 208154 I4 47 721570 340 929442 i3o 792128 470 207872 I3 48 721774 339 929364 i3I 792410 470 207590 12 49 721978 339 929286 i3i 792692 470 207308 II 50 722181 I 339 929207 13I 792974 470 207026 Io 51 9.722385 339 9.929129 131 9.793256 470 10.206744 52 722588 33 92o9050 131 793538 469 206462 53 722791 338 928972 I31 793819 469 206181 7 54 722994 338 928893 131 79410I 469 205899 6 55 723197 338 928815 I3I 794383 469 20o5617 5 56 723400 338 928736 i3i 794664 469 205336 4 57 723603 337 928657 I3I 794946 469 205054 3 58 723805 337 928578 13i 795227 469 204773 2 59 724007 337 928499 i3i, 795508 468 204492 i 60 724210 337 928420 i31 795789 468 204211 0 Cosine. D. Sine. D. Cotang. D. Tang. I 1210 5o 60 LOGARITHMIC SINES, TANGENTS, ETC. TABLE II. 320 1470 Sine. D. Cosine. D. Tang. D. I Cotang. o 9.724210 337 9.928420 132 9.795789 468 10.204211 6o I 724412 337 92'8342 I32 79607C 468 203930 59 2 724614 33.6 928263 I32 796351 468 203649 58 3 724816 336 928183 132 796632 468 203368 57 4 725017 336 928104 132 796913 468 203087 56 5 725219 336 928025 132 797194 468 202806 55 6 725420 335 927946 132 797474 468 202526 54 7 725622 335 927867 I32 797755 468 202245 53 8 725823 335 927787 132 798036 467 201964 52 9 726024 335 927708 I32 798316 467 201684 5I. I0 726225 335 927629 132 798596 467 20I404 50 11 9-726426 334 9-927549 I32 9.798877 467 I0.201123 4 12 726626 334 927470 133 799157 467 200843 48 13 726827 334 927390 133 799437 467 200563 47 I4 7'27027 334 927310 133 799717 467 200283 46 I5 727228 334 927231 133 799997 466 200003 45 i6 727428 333 927151 133 800277 466 I99723 44 17 727628 333 92707I 133 800557 466 I99443 43 I8 727828 333 926991 133 800836 466 Ig9964 42 19 728027 333 926911 133 801116 466 I98884 41 20 728227 333 926831 133 801396 466 198604 40 21 9-728427 332 9-926751 133 9.801675 466 IO.-98325 3 22 728626 332 926671 I33 801955 466 198045 23 728825 332 926591 i33 802234 465 197766 37 24 729024 332 926511 134 802513 465 197487 36 25 729223 33I 926431 I34 802792 465 I97208 35 26 729422 331 926351 134 803072 465 I96928 34 27 72962I 33I 926270 I34 80335i 465 I96649 33 28 729820 33i 926190 i34 803630 465 196370 32 29 73001oo8 330 926110 134 803909 465 196091 31 30 730217 330 926029 134 804187 465 1958I3 30 31 9.730415 330 9'925949 I34 9'804466 464 o. I95534 2 32 730613 330 925868 134 804745 464 195255 28 33 730811 330 925788 134 805023 464 I94977 27 34 73oo100 329 92707 134 805302 464 194698 26 35 731206 329 925626 134 805580 464 194420 25 36 731404 329 925545 135 805859 464 I9414I 24 37 73I602 329 925465 I35 806I37 464 193863 23 38 731799 329 925384 35 8064I5 463 i93585 22 39 73I996 328 925303 I35 806693 463 193307 2! 40 732193 328 925222 135 80697 463 I93029 20 4I 9-73239o 328 9-925141 135 9.807249 463 I0-. I92751 I1 42 732587 328 925060 I35 807527 463 I92473 I8 43 732784 328 924979 135 807805 463 i92195 17 44 732980 327 924897 135 808083 463 I91917 I6 45 733177 327 924816 135 80836i 463 191639 15' 46 733373 327 924735 i36 808638 462 191362 14 4 733563 327 924654 I36 808916 462 191084 13 4 733765 327 924572 I36 809193 462 190807 I2 49 73396I 326 924491 I36 809471 462 190529 II 50 734157 326 924409 i36 809748 462 190252 10 51 9-734353 326 9-924328 136 9.810025 462 I0o.89975 52 734549 326 924246 i36 810302 462 I89698 8 53 734744 325 924164 I36 8io580 462 189420 7 54 734939 325 924083 i36 810857 462 189143 6 55 735135 325 924001 136 8III34 46I I88866 5 56 735330 325 9239I9 I36 8II4Io 46i i885go 4 57 735525 325 923837 I36 811687 46I I883i3 3 58 735719 324 923755 137 811964 46I i88036 2 59 735914 324 923673 I37.8I2241 46i I87759 I 60 7361o9 324 923591 137 8125I7 46I I87483 0 t Cosine. D. Sine. D. Cotang. D. Tang. z 122 570 TABLE II. LOGARITHMIC SINES, TANGENTS, ETC. 51 330 1460 _ Sine. D. Cosine. D. Tang. D. Cotang. 0 9.736109 324 9.92359I 137 9.812517 46I IO I87483 60 1 736303 324 923509 137 8I2794 46[ 187206 5 2 736498 324 923427 137 813070 46 186930 58 3 736692 323 923345 137 813347 460 i86653 57 4 736886 323 923263 I37 813623 460 186377 56 5 737080 323 92318I 137 813899 460 186101 55 6 737274 323 923098 137 814176 460 185824 54 7 737467 323 923016 137 814452 460 185548 53 8 737661 322 922933 137 814728 460 185272 52 9 737855 322 92285I I37 815004 460 184996 51 Io 738048 322 922768 I38 815280 460 184720 50 II 9.738241 322 9.922686 I38 9.8i5555 459 Io0I84445 4 12 738434 322 922603 I38 81583i 459 184169 48 I3 738627 32I 922520 I38 816107 459 I83893 47 I4 738820 321 922438 138 816382 459 i836i8 46 i5 739013 321 922355 138 8I6658 459 183342 45 i6 739206 321 922272 i38 816933 459 I83067 44 17 739398 321 922189 i38 817209 459 182791 43 18 739590 320 922106 i38 817484 459 182516 42 19 739783 320 922023 138 817759 459 182241 41 20 739975 320 921940 I38 8i8035 458 181965 40 21 9.740I67 320 9.92I857 I39 9.8I83IO 458 10i18I690 3 22 740359 320 921774 139 8i8585 458 I8I415 38 23 740550 319 921691 I39 88860o 458 I 8 I 40 37 24 740742 319 921607 139 819135 458 i80865 36 25 740934 319 921524 I39 819410 458 180590 35 26 741125 319 921441 139 819684 458 i80316 34 27 741316 3i9 921357 139 819959 458 18004I 33 28 741508 318 92I274 139 820234 458 179766 32 29 741699 3i8 921190 139 820508 457 179492 31 30 741889 3i8 921107 i39 820783 457 179217 30 3I 9.742080 3i8 9.921023 I39 9.821057 457 IO.I78943 29 32 742271 3i8 920939 140 821332 457 I78668 28 33 742462 317 920856 140 821606 457 178394 27 34 742652 3I7 920772 140 821880 457 178120 26 35 742842 317 920688 I40 822154 457 J77846 25 36 743033 317 920604 140 822429 457 17757I 24 37 743223 317 920520 I40 822703 457 I77297 23 38 743413 3i6 920436 140 822977 456 177023 22 39 743602 3i6 920352 140 823251 456 176749 21 40 743792 316 920268 140 823524 456 176476 20 41 9.743982 3i6 9.920184 I40 9.823798 456 10IO76202 19 42 744171 316 920099 140 824072 456 175928 I8 43 74436I 3I5 92001I 140 824345 456 175655 17 44 744550 315 919931 141 824619 456 17538I I6 45 744739 315 919846 I41 824893 456 I75107 I5 46 744928 315 919762 141 825166 456 174834 14 47 745117 315 9I9677 141 825439 455 I7456I 13 48 745306 314 919593 141 825713 455 174287 12 49 745494 314 919508 I4I 825986 455 174014 11 50 745683 314 919424 141 826259 455 173741 10 5I 9'745871 314 9.919339 141 9.826532 455 10.173468 0 52 746060 314 919254 14I 826805 455 173195 53 746248 313 9.19169 14 827078 455 172922 7 54 746436 3I3 919085 14I 827351 455 172649 6 55 746624 313 919000 141 827624 455 172376 5 56 7468I2 3I3 918915 I42 827897 454 172103 4 57 746999 313 918830 142 828170 454 171830 3 747187 312 918745 142 828442 454 I71558 2 59 747374 312 918659 142 828715 454 I71285 I 60 747562 312 918574 142 828987 454 17I013 O t Cosine. D. Sine. D. Cotang. D. Tang. 1230 56~ 52 LOGARITHMIC SINES, TANGENTS, ETC. TABLE II. 340 1450 / Sine. D. Cosine. D. Tang. D. Cotang. 0 9.747562 312 9.918574 I42 9'828987 454 IO-I710I3 6o I 747749 312 918489 142 829260 454 170740 59 2 747936 312 918404 142 829532 454 170468 58 3 748123 311 918318 142 829805 454 170195 57 4 748310 311 9.18233 142 830077 454 169923 56 5 748497 311 918147 142 830349 453 I69651 55 6 748683 311 918062 I42 830621 453 I69379 54 7 748870 3ii 917 76 143 830893 453 I69107 53 8 749056 3io 917891 I43 83ii65 453 I68835 52 9 749243 3io 917805 143 831437 453 i68563 51 Io10 749429 3io 917719 I43 831709 453 I68291 50 II 9-7496I5 310 9-917634 I43 9.83198I 453 I0'I680I9 4c 12 749801 3o10 917548 I43 832253 453 I67747 48 13 749987 309 917462 I43 832525 453 167475 47 I4 750172 309 917376 143 832796 453 I67204 46 I5 750358 309 917290 143 833068 452 166932 45 i6 750543 309 917204 143 833339 452 I66661 44 I7 750729 309 917118 144 833611 452 166389 43 I 750914 30o 917032 144 833882 452 i66I 8 42 19 751099 308 9I6946 I44 834154 452 165846 4I 20 751284 308 916859 I44 834425 452 165575 40 21 9. 751469 308 9-9I6773 144 9. 834696 452 Io.I65304 3 22 75I654 308 916687 144 834967 452 I65o33 38 23 75I839 308 916600 144 835238 452 164762 37 24 752023 307 916514 144 835509 452 I6449I 36 25 752208 307 916427 I44 835780 45I 164220 35 26 752392 307 9I6341 I44 83605I 451 163949 34 27 752576 307 916254 144 836322 45I 163678 33 28 752760 307 916167 I45 836593 45I 163407 32 29 752944 306 916081 145 836864 451 i63I36 3i 30 753128 306 915994 I45 837134 45I 162866 30 31 9.753312 306 9-915907 I45 9.837405 451 Io.I62595 29 32 753495 306 915820 145 837675 451 162325 28 33 753679 306 915733 145 837946 451 I62054 27 34 753862 305 915646 145 838216 45I I61784 26 35 754046 305 915559 145 838487 450 I615I3 25 36 754229 305 915472 I45 838757 450 I61243 24 37 754412 305 915385 I45 839027 450 160973 23 3 754595 305 915297 I45 839297 450 160703 22 39 754778 304 9152I0 145 839568 450 I60432 21 40 754960 304 915123 I46 839838 450 160162 20 41 9.755143 304 9.915035 I46 9.840108 450 10o159892 19 42 755326 304 914948 146 840378 450 I59622 1 43 755508 304 914860 I46 840648 450 159352 17 44 755690 304 914773 I46 840917 449 159083 16 45 755872 303 914685 I46 841187 449 i588I3 I5 46 756054 303 914598 I46 841457 449 I58543 14 47 756236 303 914510 146 841727 449 I58273 I3 48 756418 303 914422 146 841996 449 158004 12 49 756600 303 914334 I46 842266 449 57734 II 50 756782 302 914246 147 842535 449 I57465 Io 51 9.756963 302 9914158'47 9.842805 449 10I57I95 9 52 757144 302 914070 I47 843074 449 156926 53 757326 302 913982 I47 843343 449 156057 7 54 757507 302 913894 147 843612 449 i56388 6 55 757688 3oi 913806 147 843882 448 I56ii8 5 56 757869 3oi 913718 147 844151 448 I55849 4 758050o 3o0I 913630 I47 844420 448 I55580 3 758230 3oi 9 354I I47 844689 448 I553ii 2 59 75841I 3oi 9I3453 I47 844958 448 i55042 I 60 75859i 3oi 913365'47 845227 448 154773 o I Cosine. D. Sine. D. Cotang. D. Tang.' 1240 660~~~~~~~~~~~~5 TABLE II. LOGARITHMIC SINES, TANGENTS, ETC. 53 8650 1440 t Sine. D. Cosine. D. Tang. D. Cotang.! 0 9.75859I 301 9.9I3365'47 9.845227 448 IO.154773 6o I 758772 300 913276 I47 845496 448 I54504 5 2 758952 300 913187 148 845764 448 154236 58 3 759132 300 913099 i48 846033 448 I53967 57 4 7593I2 300 913010 148 846302 448 I53698 56 5 759492 300 912922 I48 846570 447 153430 55 6 759672 299 912833 I48 846839 447 I5316I 54 7 759852 299 912744 i48 847108 447 I52892 53 8 760031 299 912655 i48 847376 447 152624 52 9 760211 299 912566 I48 847644 447 152356 51 10 760390 299 912477 I48 847913 447 I52087 50 II 9.760569 298 9-912388 I48 9.84818I 447 IO1I5I819 49 12 760748 298 912299 149 848449 447 151551 48 13 760927 298 9122I0 149 848717 447 i51283 47 I4 76Iio6 298 9I212I I49 848986 447 151014 46 15 761285 298 912031 149 849254 447 50746 45 i6 761464 298 9119 42 149 849522 447 150478 44 I7 761642 297 911853 I49 849790 446 I5021O 43 i8 761821 297 911763 149 850057 446 149943 42 19 76I999 297 911674 149 850325 446 149675 41 20 762177 297 9II584 149 850593 446 I49407 40 21 9.762356 297 991 I495 I49 9.85086i 446 Io.*Ii49 39 22 762534 296 911405 I49 851129 446 14887I 38 23 762712 296 9II3i5 50o 85i396 446 148604 37 24 762889 296 911226 50o 85i664 446 I48336 36 25 763067 296 911136 I5O 85193I 446 I48069 35 26 763245 296 9I1046 50o 852199 446 147801 34 27 763422 296 910956 I50 852466 446 147534 33 28 763600 295 910866 i50 852733 445 I47267 32 29 763777 295 9I1076 I50 853ooi 445 146999 31 30 763954 295 9Io686 50o 853268 445 146732 3o 31 9 764131 295 9.9io0596 I50 9.853535 445 Io.46465 29 32 764308 295 910o506 150 853802 445 146198 28 33 764485 294 910415 i50 854069 445 145931 27 34 764662 294 910325 151 854336 445 145664 26 35 764838 294 910235 i5I 854603 445 I45397 25 36 7650I5 294 9IOI44 I5I 854870 445 I4530o 24 37 76519I 294 910054 151 855137 445 144863 23 38 765367 294 909963 151 855404 445 144596 22 39 765544 293 909873 15i 85567I 444 144329 21 40 765720 293 909782 151 855938 444 144062 20 41 9.765896 293 9.909691 151 9.856204 444 I0.143796 19 42 766072 293 909601 151 856471 444 143529 I8 43 766247 293 909510 I51 856737 444 143263 I 7 44 766423 293 909419 i51 857004 444 I42996 16 45 766598 292 909328 I52 857270 444 142730 15 46 766774 292 909237 152 857537 444 142463 I4 47 766949 292 909146 152 857803 444 142197 I3 48 767124 292 909055 i52 858069 444 I41931 I2 49 767300 292 908964 I52 858336 444 141664 II 50 767475 291 908873 152 858602 443 I4I398 io 5I 9.767649 291 9-90878I 152 9.858868 443 IO.I4132 52 767824 291 908690 152 859134 443 140866 53 767999 291 908599 152 859400 443 140600 7 54 768173 291 908507.152 859666 443 I4o334 6 55 %768348 290 908416 i53 859932 443 400oo68 5 56 768522 290 908324 i53 860198 443 139802 4 57 768697 290 908233 I53 860464 443 139536 3 58 76887i 290 908141 I53 860730 443 139270 2 59 769045 290 908049 i53 860995 443 139005 I 60 769219 290 907958 I53 861261 443 138739 0 T Cosine. D. Sine. D. Cotang. D. Tang. 125~ o. 54~ 54 LOGARITHMIC SINES, TANGENTS, ETC. TABLE II. 360 1430 t Sine. D. Cosine. D. Tang. D. Cotang. / 0 9-7692I9 290 9.907958 53 9.861261 443 Io.i38739 6o I 769393 289 07866 53 861527 443 138473 5 2 769566 289 907774 i53 861792 442 I38208 58 3 769740 289 907682 i53 862058 442 137942 57 4 769913 289 907590 I53 862323 442 137677 56 5 770087 289 907498 153 862589 442 1374 11 55 6 770260 288 907406 153 862854 442 I37146 54 7 770433 288 907314 154 863119 442 I3688I 53 8 770606 1 288 907222 I54 863385 442 i366i5 52 9 770779 288 907129 I54 863650 442 13635o 5I I0 770952 288 907037 144 863915 442 i36085 50 II 9.77II25 288 9.906945 I54 9.864I80 442 I0.I35820 4 12 771298 287 906852 I54 864445 442 135555 I3 771470 287 906760 I54 864710 442 135290 47 14 771643 287 906667 i54 864975 441 I35025 46 15 771815 287 906575 I54 865240 441 134760 45 i6 771987 287 906482 i54 8655o05 441 134495 44 I7 772159 287 906389 I55 865770 44I 134230 43 I8 772331 286 906296 155 866o35 441 133965 42 19 772503 286 906204 55 8663oo00 44I 133700 41 20 772675 286 906111 I55 866564 44I 133436 40 2I 9.772847 286 9-9060I8 155 9.866829 441 I0-I33I7I 30 22 7730I8 286 905925 I55 867094 44I I32906 38 23 773190 286 905832 o 55 867358 441 I32642 37 24 77336I 285 905739 155 867623 441 132377 36 25 773533 285 905645 155 867887 44I 132113 35 26 773704 285 905552 155 868I52 440 131848 34 27 773875 285 905459 I55 8684I6 440 I31584 33 28 774046 285 905366 I56 86868o 440 131320 32 29 774217 285 905272 i56 868945 440 131o55 3i 30 774388 284 905179 i56 869209 440 I3079I 30 3I 9.774558 284 9o905085 I56 9869473 440 I0*I30527 2 32 774729 284 904992 i56 869737 440 130263 28 33 774899 284 904898 i56 870001 440 129999 27 34 775070 284 904804 i56 870265 440 129735 26 35 775240 284 9047II i56 870529 440 I29471 25 36 775410 283 904617 i56 870793 440 129207 24 37 775580 283 904523 I56 87I057 440 128943 23 38 775750 283 904429 I57 87 32I 440 I28679 22 39 775920 283 904335 157 871585 440 128415 21 40 776090 283 904241 157 871849 439 I28151 20 41 9.776259 283 9 904147 157 9.872II12 439 10-127888 19 42 776429 282 904053 I57 872376 439 127624 18 43 776598 282 903959 157 872640 439 127360 17 44 776768 282 903864 157 872903 439 I27097 i6 45 776937 282 903770 I57 873167 439 126833 i5 46 777106 282 903676 157 873430 439 I26570 I4 47 777275 281 90358i 167 873694 439 126306 I3 48 777444 281 903487 157 873957 439 I26043 12 49 777613 281 903392 I58 874220 439 I25780 11 50 777781 28I 903298 i58 874484 439 1255I6 10 5I 9'777950 281 9-903203 I58 9.874747 439 10 I25253 52 778II9 28I 903108 i58 8750Io 439 124990 8 53 778287 280 9030I4 I58 875273 438 I24727 7 54 778455 280 902919 I58 875537 438 I24463 6 55 778624 280 902824 58 8758oo00 438 124200 5 56 778792 280 902729 I58 876063 438 123937 4 57 778960 280 902634 i58 876326 438 123674 3 58 779128 280 902539 I59 876589 438 123411 2 59 779295 279 902444 I59 876852 438 123148 I 60 779463 279 902349 159 877114 438 122886 0 I Cosine. D. Sine. D. Cotang. D. Tang. 1260 530 TABLE II. LOGARITHMIC SINES, TANGENTS, ETC. 65 370 142~ 1 Sine. i D. Cosine. D. Tang. D. Cotang. _ o 9.779463 279 9.902349 159 9.877 14 438 10 22886 60 I 779631 279 902253 159 877377 438 122623 5S 2 779798 279 902158 159 877640 438 122360 58 3 779966 279 902063 159 877903 438 122097 57 4 780I33 279 90oi67 159 878165 438 121835 56 5 780300 278 901872 159 878428 438 121572 55 6 780467 278 901776 159 878691 438 I21309 54 780634 278 90I68I I59 878953 437 121047 53 780801 278 901585 159 879216 437 120784 52 9 780968 278 901490 159 879478 437 120522 51 10 781134 278 901394 i60 879741 437 120259 50 I 9.78I30O 277 9.90I298 I60 9.880003 437 10119997 49 12 78I468 277 901202 I60 880265 437 II9735 48 I3 78I634 277 90ii06 I60 880528 437 119472 47 I4 78I800 277 901010 I60 880790 437 1192IO 46 15 781966 277 900914 I60 881052 437 118948 45 i6 782132 277 9008 8 i60 88I134 437 118686 44 I 7 782298 276 900722 I60 881577 437 118423 43 782464 276 900626 16o 881839 437 ii8i6i 42 19 782630 276 900529 i6o 882IOI 437 I17899 41 20 782796 276 900433 161 882363 436 117637 40 21 9-78296I 276 9.900337 I6i 9.882625 436 IO.-II7375 3 22 783127 276 900240 I6i 882887 436 117113 38 23 783292 275 900144 i61 883148 436 116852 37 24 783458 275 900047 161 883410 436 ii659o 36 25 783623 275 899951 161 883672 436 116328 35 26 783788 275 899854 161 883934 436 1r6o66 34 27 783953 275 899757 i6i 884196 436 I 5804 33 28 784118 275 899660 161 884457 436 115543 32 29 784282 274 899564 i61 8847I9 436 1I528I 31 30 784447 274 899467 162 884980 436 115020 30 3I 9.784612 274 9.-899370 I62 9.885242 436 10.114758 29 32 784776 274 899273 162 885504 436 114496 28 33 784941 274 899176 162 885765 436 114235 27 34 7851o5 274 899078 162 886026 436 113974 26 35 785269 273 898981 I62 886288 436 113712 25 36 785433 273 898884 162 P86549 435 iI345I 24 37 785597 273 898787 162 8868ii 435 113189 23 38 78576I 273 898689 i62 887072 435 112928 22 39 785925 273 898592 I62 887333 435 112667 21 40 786089 273 898494 i63 887594 435 112406 20 41 9-786252 272 9-898397 i63 9.887855 435 IO112145 I9 42 786416 272 898299 i63. 888116 435 111884 I 43 786579 272 898202 i63 888378 435 111622 17 44 786742 272 898104 i63 888639 435 111361 16 45 786906 272 898006 i63 888900 435 IIIo00 I5 46 787069 272 897908 I63 889161 435 I1o839 14 47 787232 271 8978Io I63 889421 435 I1o579 13 4 787395 271 897712 i63 889682 435 IIo3I8 12 49 787557 271 897614 I63 889943 435 110057 II 50 787720 271 897516 i63 890204 434 109796 IO 5I 9.787883 271 9.897418 I64 9.890465 434 IO-I09535 52 788045 271 897320 I64 890725 434 109275 53 788208 271 897222 i64 890986 434 109014 7 54 788370 270 897123 I64 891247 434 108753 6 55 788532 270 897025 I64 89i507 434 108493 5 56 788694 270 896926 i64 891768 434 108232 4 57 788856 270 896828 I64 892028 434 Io07972 3 58 789018 270 896729 I64 892289 434 107711 2 59 789180 270 89663I I64 892549 434 o0745I I 60 789342 269 896532 I64 892810 434 107190 o t Cosine. D. Sine. D. Cotang. D. Tang. 127 o 52~ 66 LOGARITHMIC SINES, TANGENTS, ETC. TABLE II. 380 141~ Sine. D. Cosine. D. Tang. D. Cotang. 0 9.789342 269 9.896532 I64 9.8928IO 434 10I07Ig1 6o I 789504 269 896433 165 893070 434 106930 5 2 789665 269 896335 I65 89333i 434 106669 58 3 789827 269 896236 i65 893591 434 106409 57 4 789988 269 896137 i65 893851 434 106149 56 5 790149 269 896038 I65 89411 I 434 105889 55 6 790310 268 895939 165 894372 434 105628 54 7 790471 268 895840 i65 894632 433 105368 53 8 790632 268 895741 165 894892 433 Io5io8 52 9 790793 268 895641 i65 895152 433 104848 51 IO 790954 268 895542 i65 895412 433 104588 5o II 9.79III5 268 9.895443 I66 9.895672 433 IOIo4328 4 12 791275 267 895343 I66 895932 433 10o4o68 48 13 791436 267 895244 i66 896192 433 io38o8 47 14 791596 267 895145 i66 896452 433 103548 46 15 791757 267 895045 i66 896712 433 o103288 45 i6 791917 267 894945 i66 896971 433 Io3029 44 I7 792077 267 894846 i66 89723i 433 102769 43 I8 792237 266 894746 I66 897491 433 102509 42 19 792397 266 894646 i66 897751 433:02249 41 20 792557 266 894546 i66 898010 433 10 I990 40 21 9.792716 266 9-894446 I67 9.898270 433 IO-.I0173o 3 22 792876 266 894346 I67 898530 433 IOI470 38 23 793035 266 894246 I67 898789 433 101211 37 24 793195 265 894146 I67 899049 432 10oo951 36 25 793354 265 894046 167 899308 432 Ioo692 35 26 793514 265 893946 167 899568 432 00oo432 34 27 793673 265 893846 167 899827 432 oo100173 33 28 793832 265 89374.5 i67 900087 432 o99913 32 29 793991 265 893645 167 900346 432 099654 31 30 794I50 264 893544 I67 900605 432 099395 30 3i 9'794308 264 9.893444 I68 9.900864 432 I0.099136 2 32 794467 264 893343 I68 901124 432 098876 28 33 794626 264 893243 I68 901383 432 0986I7 27 34 794784 264 893142 I68 901oi642 432 098358 26 35 794942 264 893041 i68 901901 432 og98099 25 36 79 101 264 892940 i68 902160 432 097840 24 37 795259 263 892839 i68 902420 432 097580 23 38 795417 263 892739 I68 902679 432 09732I 22 39 795575 263 892638 i68 902938 432 og097062 21 40 795733 263 1 892536 I68 903197 43I 096803 20 4I 9.79589I 263 9.892435 I69 9.903456 43I Io.o96544 1 42 796049 263 892334 I69 903714 43i o96286 8 43 796206 263 892233 I69 903973 43i o96027 17 44 796364 262 892132 I69 904232 43i o95768 i6 45 796521I 262 892o30 169 904491 43I o9550o9 i5 46 796679 262 891929 i69 904750 43i 095250 I4 47 796836 262 891827 I69 905008 43I 094992 I3 48 796993 262 891726 I69 905267 43 I 094733 1I2 49 7971 o 26i 89I624 I69 905526 43I 094474 II 50 797307 26I 891523 170 | 905785 43i 094215 io 5I 9-797464 26I 9.891421 170 9-906043 43I Io.o93957 9 52 797621I 26i 8913I9 17o 906302 43 09I o3698 8 53 797777 26I 891217 I70 906560 431 093440 7 54 797934 26I 89III5 I70 g9068g19 43i 09318I 6 55 798og9I 26i 891OI3 17o 907077 43 092923 5 56 798247 26I 8909I1 170 907336 43I 092664 4 57 798403 260 890809 170 907594 43I 092406 3 58 798560 260 890707 17o0 907853 43I o92147 2 59 798716 260 89o6o5 I70 90811 I 430 091 I889 I 6o 798872 260 89o5o3 170 908369 43o 091631 0 t Cosine. D. Sine. D. I Cotang. D. Tang. 128~ 610 TABLE II. LOGARITHMIC SINES, TANGENTS, ETC. 57 390 1400 Sine. | D. Cosine. D. Tang. D. Cotang. 0 9.798872 260 9-89o503 I 170 9908369 430 0o.o91631 6o I 799028 260 890400 I7I 908628 430 091372 59 2 799 84 260 890298 171 908886 430 o9111I 4 5 3 799339 259 890195 171 909144 43o 090856 57 4 799495 259 890093 171 909402 43o 090598 56 5 79965I 259 889990 I7I 909660 43o ogo340 55 6 799806 259 889888 171 9099I8 430 0o90082 54 7 799962 259 889785 171 910177 430 089823 53 8 800117 259 889682 171 910435 430 o89565 52 9 800272 258 889579 171 9I0693 430 o89307 5i IO 800427 258 889477 171 9095I 43 0890go49 50 II 9.800582 258 9.889374 172 9'911209 430 10.08879I 40 12 800737 258 889271 172 9II467 430 o88533 48 13 800o892 258 889168 172 911725 43o o088275 47 I4 801047 258 889064 172 911982 43o 0880o8 46 i 5 80120 258 888961 172 9I291240 430 087760 45 I6 801356 257 888858 172 912498 43o 087502 44 I7 80oi5i 257 888755 172 912756 430 087244 43 i8 801665 257 88865i 172 9I3014 429 086986 42 19 801819 257 888548 172 913271 429 o86729 41 20 801973 257 888444 173 913529 429 o86471 4o 21 9.802128 257 9.888341 173 9'913787 429 I0-086213 3q 22 802282 256 888237 173 914044 429 o85956 38 23 802436 256 888,34 173 9I4302 429 085698 37 24 802589 256 888030 173 914560 429 o85440 36 25 802743 256 887926 I73 914817 429 o85183 35 26 802897 256 887822 173 915075 429 084925 34 27 803050 256 887718 173 915332 429 084668 33 28 803204 256 887614 173 915590 429 o844io 32 29 803357 255 887510 173 915847 429 084153 31 30 803511 255 887406 174 916104 429 o83896 30 31 9.803664 255 9.887302'74 99i6362 429 o.o83638 29 32 803817.255 887198 174 916619 429 o83381 28 33 8o3970 255 887093 174 916877 429 o83123 27 34 804123 255 886989'74 9[7134 429 082866 26 35 804276 254 886885 I74 917391 429 082609 25 36 804428 254 886780 174 917648 429 082352 24 37 80458I 254 886676 I74 917906 429 082094 23 38 804734 254 886571 174 918163 428 08I837 22 39 8o48/6 254 886466 174 918420 428 o8i580 21 40 8o5oS9 254 886362 I75 918677 428 08I323 20 4I 9 805Ig9 254 9-.886257 175 9-gI8934 428 io o8io66 I 42 805343 253 886152 175 919191 428 o8080og I 43 805495 253 886047 175 919448 428 oSo552 17 44 805647 253 885942 175 919705 428 080295 i6 45 805799 253 885837 175 919962 428 o8oo38 i5 46 805951 253 885732 175 920219 428 07978I I 4 47 8o6Io3 253 885627 175 920476 428 079524 I3 48 806254 253 885522 175 920733 428 079267 12 49 806406 252 885416 175 920990 428 079010 II 50 806557 252 885311 I76 921247 428 078753 Io 5 9.8806709 252 9.885205 176 9-921503 428 100o78497 9 52 806860 252 885ioo 176 921760 428 078240 8 53 807011 252 884994 176 922017 428 077983 7 54 807163 252 884889 176 922274 428 077726 6 55 80731 4 252 884783 176 922530 428 077470 5 56 807465 251 884677 176 922787 428 077213 4 57 8o7615 251 884572 176 923044 428 076956 3 58 807766 251 884466 176 923300 428 076700 2 59 807917 25I 884360 176 923557 427 o76443 I 60 808067 251 884254 77 923814 427 076186 0 Cosne. D. Sine. D. Cotang. D. Tang. 1290~ 50 3* 58 LOGARITHMIC SINES, TANGENTS, ETC. TABLE II. 400 139~ Sine. D. Cosine. 1 D. Tang. D. Cotang. 0 9.808067 25I 9.8842a54 177 9 —9238I4 427 10-076I86 6o I 808218 251 884148 177 924070 427 075930 5 2 808368 251 884042 177 9,24327 427 075673 58 3 808519 250 883936 177 924583 427 075417 57 4 8o8669 250 883829 177 924840 427 075r60 56 5 808819. 250 883723 177 925096 427 074904 55 6 808969 250 883617 I77 925352 427 074648 54 809119 250 8835io 177 925609 427 07439,I 53 8 809269 250 883404 177 925865 427 074135 52 9 809419 249 883297 178 926122- 427 073878 51 10 809569 249 883191 1,78 926378 427 073622 50 11 9 8097 18 249 94883084 178 9.926634 427 Io1o73366 4 12 809868 249 882977 178 926890 427 0731I1 48 I3 8I0017 249 882871 1.78 927I47 427 072853 47 I4 810167 249 882764 178 927403 427 072597 46 I5 810o3.6 248 882'657 178" 927659 427 072-341 45 i-6 810465 248 882550 178 927915 427 072085 44 I7 8I0614 248, 882443 1,78 928171 427 071829 43 I8 810763 248 882336 179 928427 427 071573 42 19 810912 248 882229 179 928684 427 071316 41 20 811061 248 882121 179. 928940 427 07 ro60 40 21 9-8112IO 248 9.882014 179 9.929I96 427 10070804 39 22 8ii358 247 88I907 179' 929452 427 070548 38 23 811507 247 881799 179. 929708 427 070292 37 2'4 8ii655 247 881692 1.79 029964 426 070036 36 25 8 11804 247 881584 179 930220 426 o69780 35 26 811952 2.47 881477 1.79 930475 426 069525 34 812100 247 88I369 179 93073I 426 069269 33 2 812248 247 88126I I80 930987 426 o69oi3 32 29 812396 246 881153 I8o 931243 426 o68757 31 3o. 812544 246 88 o46 I80 931499 426 o6850o 30 3I 9.812692 246 9.880938 8o, 9.931755 426 I0.068245 2 32' 81284c 246 88830 o 80 932010 426 o67990 28 33 812988 246 880722 I80 93 2266 42'6 067734 27 34 813135 246 88o063 i,8o 932522T 426 o67478 26 35 813283 246 880505 18o 932778 426 o672-22 25 36 813430 245 880397 I8o 933033 426 066967 24 37 8I3578 245 880289 i8i 933289 426 o6671I 23 3 813725 245 88oi80 i8i. 933545 426 o66455 22 39 813872 245 880072' 181 933800 426 066200 21 40 814019 245 879963 181 934056 426 o65944 20 41 9.814166 245 9.879855 i81 9.9343II 426 io.o6568q 19 42 814313 245 879746 1I81 934567 426 065433 I8 43 814460 244 879637 18i 934822 426 o65178 17 44 814607 244 879529 181 935078 426 064922 I6, 45 814753 244 879420 I81 935333 426 064667 15 46 814900 244 8793I I 181 935589 426 o6441 14 47 815046 244 8792o2 182 935844 426 064i56 13 48 815193 244 879093 182 936100 426 o63goo 12 49 815339. 244 878984 I82 936355 426 063645 ii 50 8I5485 243 878875 I82 936611 426 063389 10 51 9.8563I 243 9'878766 182' 9.936866 425 10.o63134 52 815778 243 878656 182 93,7121 425 062879 53 8I5924; 243 878547 182 937377 4a5 062623' 7 54 81606g 243 878438 I82 937632 425 062368 6 55 8I6215 243 878328 182 937887 425 062II3 5 56 81636i 243 878219. i83 938142 425 o6i858 4 57 816507 242 878109 i83 938398 425 061602 3 58 816652 242 877999 183 938653 425 061347 2 59. 816798 242 877890 183 938908 425 061092 I 60 816943 242 877780 i83 939163 425 060837 o f Cosine. D. Sine. D. Cotang. D. Tang. I 130~ 490, ~ ~ ~ ~ ~ ~ ~ ~ ~ ~... TABLE II. LOGARITHMIC SINES, TANGENTS, ETC. 59 410 1380 Sine. D. Cosine. D. Tang. D. Cotang. I o 9.8I6943 242 9.877780 I83 9.939163 425 io.0o60837 6o I 817088 242 877670 i83 939418 425 060582 5 2 817233 242 877560 183 939673 425 060327 58 3 817379 242 877450 I83 939928 425 o060072 57 4 817524 241 877340 183 940183 425 059817 56 5 817668 24I 877230 I84 940439 425 059561 55 6 817813 241 877120 184 940694 425 o59306 54 7 817958 241 877010 184 940949 425 o05go5 53 8 8181o3 24I 876899 I84 941204 425 058796 52 9 818247 24I 876789 I84 941459 425 o5854I 5i IO 8I8392 241 876678 184 941713 425 058287 50 II 9.818536 240 9.876568 i84 9.941968 425 IO.o58o32 49 12 8i868i 240 876457 184 942223 425 057777 48 I3 818825 240 876347 I84 942478 425 057522 47 I4 818969 240 876236 i85 942733 425 057267 46 I5 819gI3 240 876125 i85 942988 425 057012 45 i6 819257 240 876014 i85 943243 425 056757 44 I7 81940 240 875904 i85 943498 425 056502 43 i8 8I9545 239 875793 I85 943752 425 056248 42 19 819689 239 875682 i85 944007 425 055993 41 20 819832 -239 875571 185 944262 425 055738 40 21 9.8I9976 239 9.875459 I85 9.944517 425 io.o55483 39 22 820120 239 875348 i85 944771 424 055229 38 23 820263 239 875237 I85 945026 424 054974 37 24 820406 239 875126 i86 945281 424 054719 36 25 820550 238 875014 186 945535 424 054463 35 26 820693 238 874903 186 945790 424 054210 34 27 820836 238' 874791 I86 946045 424 053955 33 28 820979 238 874680 i86 946299 424 05370o 32 29 821122 238 874568 i86 946554 424 053446 31 30 821265 238 874456 t86 946808 424 053192 30 31 9.821407 238 9.874344 i86 9.947063 424 Io0052937 29 32 821550 238 874232 187 947318 424 052682 28 33 821693 237 874121 187 947572 424 052428 27 34 821835 237 874009 I87 947827 424 o52173 26 35 821977 237 873896 187 948081 424 051919 25 36 822120 237 873784 187 948335 424 o5i665 24 37 822262 237 873672 187 948590 424 051410 23 38 822404 237 873560 I87 948844 424 o51156 22 39 822546 237 873448 187 949099 424 050901 21 40 822688 236 873335 I87 949353 424 050647 20 4I 9.822830 236 9.873223 I87 9.949608 424 Io.o503 2 19 42 822972 236 873110 I88 949862 424 o5oi38 I 43 823114 236 872998 i88 950116 424 049884 17 44 823255 236 872885 I88 950371 424 049629 I6 45 823397 236 872772 188 950625 424 049375 15 46 823539 236 872659 i88 950879 424 049121 I4 47 823680 235 872547 I88 951133 424 048867 i3 48 823821 235 872434 i88 951388 424 o48612 12 49 823963 235 872321 i88 951642 424 048358 II 50 824104 235 872208 i88 951896 424 048Io4 I0 5I 9.824245 235 9.872095 I89 9.952150 424 10.047850 52 824386 235 87198I I89 952405 424 047595 8 53 824527 235 871868 189 952659 424 047341 7 54 824668 234 87I755 I89 952913 424 047087 6 55 824808 234 871641 189 953167 423 046833 5 56f 824949 234 871528 189 953421 423 046579 4 57 825090 234 871414 189 953675 423 046325 3 58 825230 234 87130o I89 953929 423 04607I 2 59 825371 234 871187 I89 954I83 423 o45817 I 6o 825511 234 871073 190 954437 423 045563 0 _ Cosine. D. Sine. D. Cotang. D. Tang.' 1310 480 18i_..... 60 LOGARITHMIC SINES, TANGENTS, ETC. TABLE II. 420 1370 Sine. D. Cosine. D. Tang. D. Cotang. o 9.825511 234 9.871073 190 9.954437 423 10-045563 6o I 82565I 233 870960 19Igo 954691 423 045309 5 2 825791 233 870846 1go9 954946 423 o45054 58 3 825931 233 870732 Igo 955200 423 o448o0 57 4 826071 233 870618 190 955454 423 044546 56 5 826211 233 870504 190 955708 423 044292 55 6 826351 233 870390 190 955961 423 044039 54 7 826491 233 870276 190 956215 423 o43785 53 8 826631 233 870161 190 956469 423 o4353 52 9 826770 232 870047 191 956723 423 043277 5i 10 826910 232 869933 191 956977 423 043023 50 II 9-827049 232 9.8698I8 9IgI 9957231 423 I0-042769 49 12 827189 232 869704 191 957485 423 042515 48 13 827328 232 869589 191 957739 423 04226I 47 14 827467 232 869474 191 957993 423 042007 46 I5 827606 232 869360 191 958247 423 041753 45 I6 827745 232 869245 191 9585o00 423 04i500 44 17 827884 231 869130 191 958754 423 041246 43 18 828023 231 8690g5 192 959008 423 040992 42 19 828162 23I 868900 192 959262 423 040738 4I 20 8283oi 23i 868785 192 959516 423 040484 4o 21 9.828439 231 9.868670 192 9.959769 423 IO04023I 39 22 828578 231 868555 192 960023 423 039977 38 23 828716 231 868440 I92 960277 423 039723 37 24 828855 230 868324 192 960530 423 039470 36 25 828993 230 868209 192 960784 423 o39216 35 26 829131 230 868093 192 96io38 423 038962 34 27 829269 230 867978 193 96I292 423 o38708 33 2 829407 230 867862 I93 961545 423 038455 32 29 829545 230 867747 193 961799 423 038201 31 30 829683 230 867631 193 962052 423 o37948 30 3i 9'829821I 229 9-8675I5 193 9.962306 423 IO.o37694 29 32 829959 229 867399 I93 962560 423 037440 28 33 830097 229 867283 I93 962813 423 o37187 27 34 830234 229 867167 193 963067 423 036933 26 35 830372 229 867051 193 963320 423 o36680 25 36 830509 229 866935 194 963574 423 036426 24 37 830646 229 866819 I94 963828 423 036172 23 38 830784 229 866703 194 96408 I 423 o35919 22 39 830921 228 866586'94 964335 423 o35665 21 40 83io58 228 866470 I94 964588 422 035412 20 4I 9.83II95 228 9.866353 194 9.964842 422 IO.o35I58 1 42 831332 228 866237 194 965095 422 o3490o5 43 831469 228 866120 194 965349 422 o34651 17 44 83i606 228 866004 195 965602 422 o34398 i6 45 831742 228 865887 195 965855 422 o34145 15 46 831879 228 865770 I95 966109 422 033891 14 47 832015 227 865653 195 966362 422 o33638 13 48 832152 227 865536 195 966616 422 033384 12 49 832288 227 865419 195 966869 422 o3313i II 50 832425 227 865302 195 967123 422 032877 10 5I 9.83256I 227 9'865185 195 9.967376 422 Io.o032624 52 832697 227 865068 195 967629 422 032371 53 832833 227 864950 195 967883 422 032II7 7 54 832969 226 864833 196 968136 422 o31864 6 55 833o105 226 8647I6 196 968389 422 o3161i 5 56 833241 226 864598 196 968643 422 031357 4 57 833377 226 86448I 196 968896 422 03o104 3 58 833512 226 864363 196 969149 422 o3o85i 2 59 833648 226 864245 196 969403 422 o30597 I 60 833783 226 864127 196 969656 422 030344 o Cosine. D. Sine. D. Cotang. D. Tang. / 1320 470 TABLE II. LOGARITHMIC SINES, TANGENTS, ETC. 61 430 136~o _ Sine. D. Cosine. D. Tang. D. Cotang. 0 9.833783 226 9.864127 I96 9.969656 422 2io-o30344 6o I 833919 225 [ 864010 I 96 969909 422 o3o091 5 2 834054 225 863892 197 970I62 422 029838 58 3 834189 225 863774 I97 970416 422 029584 57 4 834325 225 863656 197 970669 422 029331 56 5 834460 223 863538 197 970922 422 029078 55 6 834595 225 863419 I97 971175 422 01 28825 54 7 83z4730 225 863301 I97 971429 422 02857I 53 8 834865 225 863183 197 971682 422 0283I8 52 9 834999 224 863064 197 97I935 422 o28065. 5i 10 835134 224 862946 198 972188 422 027812 50 I 9.835269 224 9.862827 198 9'97244I 422 10027559 44 12 835403 224 862709 198 972695 422 027305 4 13 835538 224 862590 I98 972948 422 027052 47 14 835672 224 862471 198 973201 422 026799 46 15 835807 224 862353 198 973454 422 026546 45 16 835941 224 862234 198 973707 422 026293 44 17 8360o75 223 862115 198 973960 422 026040 43 I8 836209 223 861996 198 97421 422 025787 42 19 836343 223 861877 198 974466 422 025534 41 20 836477 223 861758 199 974720 422 025280 40 2I 9.836611 * 223 9.861638 199 9.974973 422 I0.025027 39 22 836745 223 861519 199 975226 422 024774 38 23 836878 223 861400 199 975479 422 024521 37 24 837012 222 861280 199 975732 422 024268 36 25 837146 222 861161 199 975985 422 024015 35 26 837279 222 86o041 199 976238 422 023762 34 27 837412 222 860922 199 976491 422 023509 33 28 837546 222 860802 199 976744 422 023256 32 29 837679 222 860682 200 976997 422 023003 31 30 837812 222 860562 200 977250 422 022750 30 31 9.837945 222 9.860442 200 9.977503 422 10022497 29 32 838078 221 860322 200 977756 422 022244 28 33 838211II 221 860202 200 978009 422 021991 27 34 838344 22I 860082 200 978262 422 021738 26 35 838477 221 859962 200 978515 422 021485 25 36 8386o10 221 859842 200 978768 422 021232 24 37 838742 221 859721 201 979021 422 020979 23 38 838875 221 859601 201 979274 422 020726 22 39 839007 22I 859480 201 979527 422 020473 21 40 839140 220 859360 201 979780 422 020220 20 41 9.839272 220 9.859239 20I 9.980033 422 IO.oI9967 19 42 839404 220 8591i19 201 980286 422 019714 I8 43 839536 220 858998 201 98o0538 422 oi9462 I7 44 839668 220 858877 201 980791 42I 019209 16 45 839800 220 858756 202 981044 421 oI8956 I5 46 839932 220 858635 202 981297 42I OI8703 14 47 840064 219 858514 202 981550 421 018450 I3 48 840,96 219 858393 202 981803 421 018I97 12 49 840328 219 858272 202 982056 421 017944 II 50 840459 219 85815i 202 982309 42I 017691 10 5I 9.8405gI 219 9-858029 202 9.982562 421 IOO17438 52 840722 219 857908 202 982814 421 017186 53 840854 219 857786 202 983067 421 016933 7 54 840985 219 857665 203 983320 42I oi668o 6 55 841II6 2I8 857543 203 983573 421 OI6427 5 56 841247 218 857422 203 983826 421 016174 4 57 841378 2I8 857300 203 984079 421 015921 3 58 841509 218 857178 203 984332 421 oi5668 2 59 84i64o 218 857056 203 984584 421 oi54i6 I 60 841771 218 856934 203 984837 421 oi5i63 O " Cosine. D. Sine. D. Cotang. 1D. Tang. I 1330 46 _~~~~~~...., 62 LOGARITHMIC SINES, TANGENTS, ETC. TABLE IL 440 1350 I Sine. D. Cosine. D. Tang. D. Cotang. 0 9-84I771 218 9.85634 203 9.984837 421 io-oI5I63 6o I 841902 218 856812 203 985090 421 oI49I0 5 2 842033 218 85669c 204 985343 421 014657 3 842163 2I7 856568 204 985596 421 014404 57 4 842294 217 856446 204 985848 421 OI4152 56 5 842424 217 856323 1 204 986101 421 013899 55 6 842555 217 856201 204 986354 42I 013646 54 7 842685 217 856078 204 986607 42I 013393 53 8 842815 217 855956 204 986860 421 oi3i40 52 9 842946 217 855833 204 987112 42I 012888 51 Io 843076 217 855711 205 987365 421 012635 50 II 9.843206 216 9.855588 205 9*9876I8 42I 10012382 4 12 843336 216 855465 205 987871 421 012129 48 13 843466 2 6 855342 205 988123 421 011877 47 I4 843595 216 855219 205 988376 421 OII624 46 15 843725 216 855096 205 988629 421 01137I 45 I6 843855 216 854973 205 988882 421 oIIIi8 44 I7 843984 216 854850 205 989134 42I oio866 43 I8 844114 215 854727 206 989387 421 oio6I3 42 19 844243 215 854603 2o6 989640 421 oio360 41 20 844372 215 854480 206 989893 421 010107 40 21 9.844502 215 9.854356 206 9.990I45 42I 0o.oo9855 39 22 844631 215 854233 206 990398 421 o009602 38 23 844760 215 854o09 2056 99065I 42I 0o9349 37 24 844889 215 853986 206 990903 421 009097 36 25 845018 215 853862 206 991156 421 008844 35 26 845I47 215 853738 206 991409 421 oo8591 34 27 845276 214 853614 207 991662 421 oo8338 33 2 845405 214 853490 207 991914 421 oo8086 32 29 845533 214 853366 207 992167 421 007833 3i 30 845662 214 853242 207 992420 421 007580 30 3I 9.845790 214 9.853118 207 9.992672 42I IO0oo7328 2 32 845919 214 852994 207 992925 42 I 007075 28 33 846047 2I4 852869 207 993178 42I oo6822 27 34 846I75 2I4 852742 207 99343I 421 oo6569 26 35 846304 214 852620 207 993683 421 oo6317 25 36 846432 213 852496 208 993936 421 oo6064 24 37 846560 213 852371 208 994189 421 0oo58I 23 38 846688 213 852247 208 994441 421 oo5559 22 39 8468I6 213 852122 208 994694 421 oo5306 21 40 846944 213 851997 208 994947 421 oo005053 20 41 9.847071 2I3 9.85I872 208 9.995199 42I I-0od480I 19 42 847199 213 851747 208 995452 42I 004548 I 43 847327 213 851622 208 995705 421 o04295 17 44 847454 212 851497 209 995957 421 004043 i6 45 847582 212 851372 209 996210 421 o03790 15 46 847709 212 851246 209 996463 421 oo3537 I4 47 847836 212 851121 209 996715 421 oo3285 I3 48 847964 212 850996 209 996968 421 003032 12 49 848091 212 850870 209 997221 421 002779 II 50 848218 212 850745 209 997473 421 002527 IO 51 9.848345 212 9'850619 209 9'997726 421 0I.002274 52 848472 211 850493 210 997979 421 002021 53 848599 211 850368 210 998231 42I 00oo769 7 54 848726 211 850242 210 998484 421 oo001516 6 55 848852 211 850oi6 210 998737 421 oo001263 5 56 848979 211 849990 210 998989 421 ooo000 4 57 8491o6 211 849864 210o 999242 421 000758 3 5 849232 21I 849738 210 999495 421 ooo000505 2 59 849359 211 849611 210 999747 421 000253 I 60 849485 211 849485 2IO 10O000000 42I I 000000 O Cosine. D. Sine. D. Cotang. D. Tang. TABLE III., OF N ATURAL SINES AND TANGENTS; TO EVERY DEGREE AND MINUTE OF THE QUADRANT. IF the given angle is less than 450, look for the degrees and the title of the column, at the toap of the page; and for the minutes on the left. But if the angle is between 450 and go90, look for the degrees and the title of the column, at the bottom; and for the minutes on the right. The Secants and Cosecants, which are not inserted in this table, may be easily supplied. If I be divided by the cosine of an arc, the quotient will be the secant of that arc. And if I be divided by the sine, the quotient will be the cosecant. The values of the Sines and Cosines are less than a unit, and are given in decimals, although the decimal point is not printed. So also, the tangents of arcs less than 450, and cotangents of arcs greater than 450, are less than a unit and are expressed in decimals with the decimal point omitted. 64 NATURAL SINES AND COSINES. TABLE III. I I 00' 1o 20 30 40 Sine. lCosine. Sine. Cosine. Sine. Cosine. Sine. Cosine. Sine. Cosine. o ooooo000 Unit. 01745 99985 o3490 99939 05234 99863 06976 99756 60 I 00029 Unit. 01774 99984 o3519 99938 05263 9986I 07005 99754 59 2 ooo58 Unit. oi8o3 99984 o3548 99937 05292 99860 07034 99752 58 3 ooo87 Unit. oI832 99983 03577 99936 05321 99858 07063 99750 57 4 ooii6 Unit. OI862 99983 o3606 99935 o5350 99857 07092 99748 56 5 00145 Unit. oi8g9 99982 o3635 99934 o5379 99855 07I21 99746 55 6 00175 Unit. 01920 99982 03664 99933 o54o08 99854 07I50 99744 54 7 00204 Unit. OI949 9998I o3693 99932 L 05437 99852 07179 99742 53 8 00233 Unit. 01978 99980 03723 99931 05466 99851 07208 99740 52 9 00262 Unit. 02007 9998o o3752 99930 05495 99849 07237 99738 5 10 00291 Unit. 02036 99979 o378I 99929 o5524 99847 07266 99736 50 ii 00320 99999 02065 99979 038io 99927 o5553 99846 07295199734 49 12 00349 99999 02094 99978 03839 99926 o5582 99844 07324 9973I 48 I3 00378 99999 02123 99977 o3868 99925 o56I 99842 07353 99729 47 I4 00407 99999 02152 99977 03897 99924 o5640 9984i 07382 99727 46 I5 00436 99999 02181 99976 03926 99923 o5669 99839 0741I 99725 45 i6 00465 99999 02211 99976 03955 99922 05698 99838 07440 99723 44 17 00495 99999 02240 99975 o3984 99921 05727 99836 07469 99721 43 I 00524 99999 02269 99974 o40o3 999i9 05756 99834 07498 99719 42 I9 oo553 99998 02298 99974 04042 999I8 05785 99833 07527 99716 4I 20 00oo582 99998 02327 99973 04071 99917 o58I4 9983I 07556 99714 40 21 006ii 99998 02356 99972 0410 9991g6 o5844 99829 07585 99712 39 22 oo640 99998 02385 99972 04I29 99915 05873 99827 07614 997IO 38 23 oo00669 99998 02414 99971 04I59 999I3 o5902 99826 07643 99708 37 24 o0698 99998 02443 99970 04i88 99912 o593I 99824 07672 99705 36 25 00727 99997 02472 99969 04217 99911 o5960 99822 07701 99703 35 26 00756 99997 0250I 99969 04246 999Io o5989 99821 07730 9970I 34 27 00785 99997 02530 99968 o4275 99909 o6oi8 99819 07759 99699 33 2 00814 99997 02560 99967 04304 99907 o6047 99817 07788 99696 32 29 oo844 99996 02589 99966 04333 99906 o6076 998I5 07817 99694 3t 30 00873 99996 02618 99966 04362 99905 o6io5 998I3 07846 99692 30 3 oo00902 99996 02647 99965 04391 99904 o6i34 99812 07875 99689 29 32 oo93I 99996 02676 99964 04420 99902 o6i63 99810 07904 99687 28 33 oog60o 99995 02705 99963 04449 9990I o6I92 99808 07933 99685 27 34 oog898 99995 02734 99963 04478 999o 0622I 99806 07962 99683 26 35 01018 99995 02763 99962 o4507 99898 o6250 99804 07991 99680 25 36 01047 99990 02792 99961 04536 99897 o6279 99803 o8020 99678 24 3 o501076 99994 02821 99960 04565 99896 o63o0 9980I o804 99676 23 3 oio5 99994 02850 99959 04594 99894 o6337 99799 o8o78 99673 22 39 oiI34 99994 02879 99959 04623 99893 o6366 99797 o8Io7 9967I 21 40 oii64 99993 029o8 99958 04653 99892 o6395 99795 o8i36 99668 20 4I oII93 99993 02938 99957 04682 99890 o6424 99793 08I65 99666 I9 42 01222 99993 02967 99956 04711 99889 06453 99792 08194 99664 I8 43 01251 99992 02996 99955 04740 99888 o6482 99790 08223 99661 I7 44 01280 99992 o3025 99954 04769 99886 o65i 99788 o8252 99659 16 45 oI309 99991 o3054 99953 o4798 99885 o6540 99786 08281 99657 5 46 oi338 9999 o3o83 99952 04827 99883 o6569 99784 o8310 99654 14 47 01367 99991 o31I2 99952 04856 99882 o6598 99782 o8339 99652 13 48 o01396 9999o o314i 9995i 04885 99881 o6627 99780 o8368 99649 12 49 01425 9999 0o3170 99950 049I4 99879 o6656 99778 08397 99647 I 50 01454 99989 03199 99949 04943 99878 o6685 99776 08426 99644 1 5I oI483 99989 o3228 99948 04972 99876 06714 99774 o8455 99642 52 o1513 99989 03257 99947 o500oo 99875 o6743 99772 08484 99639 53 01542 99988 o3286 99946 o503o 99873 06773 99770 08513 99637 7 54 OI571 99988 o33I6 99945 o5059 99872 o6802 99768 o8542 99635 6 55 01600 99987 o3345 99944 o508 99870 o683 99766 o8571 99632 5 56 OI62g 99987 03374 99943 05117 99869 o686o 99764 o8600oo 99630 4 57 oi658 99986 o3403 99942 05146 99867 o6889 99762 08629 99627 3 58 oi687 99986 o3432 99941 o5175 99866 o6g918 99760 o8658 99625 2 59 O716 99985 o3461 99940 0520o5 99864 o6947 99758 08687 99622 I 6o 01745 99985 o349o 99939 o5234 99863 o6976 99756 087I6 9g6Ig o Cosine. Sine. Cosine. Sine. Cosine. Sine. Cosine. Sine. Cosine. Sine. 890 880 87~ 860 850 TABLE III. NATURAL SINES AND COSINES. 65 _ 50 0 6, 80 _ 90go Sine Cosine. Sine. iCosine. Sine. Cosine. Sine. Cosine. Sine. Cosine. 0 o8716 996i 9 io453 99452 12I87 99255 I39I7 99027 i 5643 98769 60 o08745 99617 Io482 99449 12216 99251 13946 99023 I5672 98764 5 2 08774 99614 105oI 99446 12245 99248 13975 99019 15701 98760 5 3 o88o3 99612 I0540 99443 I2274 99244 I4004 990OI I5730 98755 57 4 o883 99609 Io569 99440 12302 99240 I4o33 99011 15758 987 5 56 5 o8860 99607 I0597 99437 1233I 99237 I4o6i 99006 I5787 98746 55 6 o888 99604 I0o626 99434 12360 99233 14090 99002 158i6 98741 54 7 08gi 99602 io655 9943I I2389 99230 I4I-9 98998 5845 98737 53 8 08947 99599 10684 99428 12418 99226 I4148 98994 i5873 98732 52 9 08976 99596 I07I3 99424 I12447 99222 14I77 98990 I 590a 98728 51 I 09005 99594 I0742 99421 12476 9921 I4205 98986 I5931 98723 50 I 09ogo34 9959 I0o77I 99418 I2504 992I5 14234 98982 I5959 98718 49 12 090go63 99588 Io8oo00 9945 I2533 9921 I 4263 98978 I5988 98714 48 i3 o9092 99586 10829 99412 12562 99208 I4292 98973 I60o7 98709 47 E4 OgI21 99583 io858 99409 259gI 99204 I4320 98969 i6046 98704 46 5 09150 99580 10887 99406 I 2620 99200 I4349 98965 i6074 98700 45 I6 ogI79 99578 IogI6 99402 I 2649 99197 I4378 9896I i6Io3 98695 44 I7 og09208 99575 I0945 99399 12678 993 I 4407 98957 I6I32 98690 43 i8 09237 99572 10973 99396 I2706 991 9 I4436 98953 16160 98686 42 19 09266 99570 11002 99393 I2735 99186 I4464 98948 16I89 9868I 4I 20 09295 99567 iio3x 99390 I 2764 99182 14493 98944 I6218 98676 40 2 o9324 99564 iio6o 99386 I 2793 99I78 14522 98940 I6246 9867 I 3 22 09og353 99562 IIo89 99383 12822 99175 I455I 98936 I6275 98667 3 23 o09382 99559 IIIi 99380 12851 99171 14580 9893I i6304 98662 37 24 0941I 99556 1I147 99377 12880 99167 I46o8 98927 i6333 98657 36 25 09440 99553 II176 99374 12908 99I63 14637 98923 i636i 98652 35 26 o0946 9955I 11205 99370 12937 99160 14666 989I9 16390 98648 34 27 o9498 99548 11234 99367 12966 99156 14695 98914 I6419 98643 33 28 o09527 99545 11263 99364 12995 99152 I4723 989Io 16447 98638 32 29 09556 99542 I 1291 99360 I3024 99148 I4752 98906 i6476 98633 31 30 og585 99540 11320 99357 i3o53 99I44 I478I 98902 i65o5 98629 30 31 og9614 99537 II349 99354 i3o8i 9914 I 48Io 98897 I6533 98624 29 32 o09642 99534 II378 9935i i3iio 991 37 I4838 98893 i6362 98619 2 33 o0967I 99531I II407 99347 I3I39 99I33 14867 98889 I6591 98614 27 34 09700 99528 I 436 99344 I3I68 99129 14896 98884 I6620 98609 26 35 097 29 99526 I465 9934 I 3197 99I25 I4925 98880 16648 98604 25 36 09758 99523 II494 99337 13226 99122 I 4954 98876 I6677 98600 24 37 09787 99520 11523 99334 I3254 991I8 I4982 98871 16706 98595 23 38 o9816 99517 II552 99331 I3283 99II4 150o1 98867 I6734 98590 22 39 o9845 99514 II58o 99327 1332 9911ggI I5040 98863 I6763 98585 21 40 09874 9951i I1609 69324 I334 99Io6 50o69 98858 I6792 98580 20 4I oggo3 99508 II638 99320 13370 99gI02 50o97 98854 16820 98575 19 42 o9932 99506 I 667 99317 399 99098 I5126 98849 6849 98570o I 8 43 o99g6I 99503 II696 99314 I3427 99094 ISI55 98845 I6878 98565 I 7 44 09o90 99500 11725 9931o I3456 99o9g I5184 98841 I6906 9856 I 6 45 10019 99497 II754 99307 I3485 99087 I52I2 98836 i6935 98556 i5 46 00oo48 99494 11783 99303 13514 99083 I524I 98832 16964 9855 I 14 47 10077 9949I 11812 99300 I3543 99079 15270o 98827 I6992 98546 13 48 IOIo6 99488 I 84o 99297 i3572 99075 I5299 98823 17021 98541 I2 49 ioi35 99485 11869 99293 i36oo 9907I I5327 9888 7o050o 98536 I 50 Ioi64 99482 1 I 898 99290 13629 99067 i5356 98814 17078 9853 I i 5I 1092 99479 11927 99286 I3658 99063 I5385 98809 17107 98526 9 52 10221 99476 1I956 99283 I3687 99059 i54I4 98805 17136 98521 53 10250 99473 11985 99279 I37I6 99055 15442 98800 17I64 985i6 7 54 10279 99470 12014 99276 I3744 99o5I i547I 98796 I7I93 985I 6 55 io3o8 99467 12043 99272 13773 99047 55oo 987917 2 2 2 98506 5 56 10337 99464 I207I 99269 I3802 99043 I5529 98787 17250 9850I 4 57 io366 9946I 12100 99265 i383I 99o39 i5557 98782 17279 98496 3 58 Io395 99458 12I29 99262 I386o 99035 i5586 98778 I7308 9849I 2 59 10424 99455 12I58 99258 i3889 99031 56I5 98773 17336 98486 60o 1453 99452 12I87 99255 I39I7 99027 i5643 98769 I7365 98481 0 |Cosine. Sine. Cosine. Sine. Cosine. Sine. Cosine. Sine. Cosine. Sine. 840 830 820 810 80~ 66 NATURAL SINES AND COSINES. TABLE III. 100 110 120 13~ 140 Sine. Cosine. Sine. ICosine. Sine. Cosine. Sine. Cosine. Sine. Cosine. o 17365 9848I I908I 98I63 2079I 978I5 22495 97437 24,92 197030 6o I I17393 98476 19109 98157 20820 Q7809 22523 97430 24220 97023 59 12 7422 98471 g9138 98152 20848 97803 22552 97424 24249 97015 58 3 1745I 98466 19I67 98146 20877 97797 22580 97417 24277 97008 57 4 17479 9846I I9195 98140 20905 97791 22608 974II 24305 97001 56 5 17508 98455 19224 98135 20933 97784 22637 97404 24333 96994 55 6 17537 98450 19252 98129 20962 97778 22665 97398 24362 96987 54 7 17565 98445 1928I 98124 20990 97772 22693 97391 24390 96980 53 8 17594 98440 19309 98118 21019 97766 22722 97384 24418 96973 52 9 17623 98435 19338 98I12 21047 97760 22750 97378 24446 96966 5I 10 1765 I 98430 19366 98107 21076 97754 22778 97371 24474 96959 50 11 17680 98425 i9395 98101 21104 97748 22807 97365 24503 96952 49 I2 17708 98420 19423 98096 21132 97742 22835 97358 24531 96945 48 I3 27737 98414 19452 98090 21161 97735 22863 97351 24559 96937 47 14 I7766 98409 19481 98084 21189 97729 22892 97345 24587 96930 46 I5 I7794 98404 19509 98079 21218 97723 22920 97338 24615 96923 45 i6 17823 98399 19538 98073 21246 97717 22948 97331 24644 96916 44 17 I7852 98394 i9566 98067 21275 977I11 22977 97325 24672 96909 43 18 I7880 98389 I9595 98061 21303 97705 23005 97318 24700 96902 42 19 17909 98383 I9623 98056 2133I 97698 23033 973II 24728 96894 41 20 17937 98378 I9652 98050 21360 97692 23062 97304 24756 96887 40 21 17966 98373 I9680 98044 21388 97686 23090 97298 24784 96880 39 22 I7995 98368 I9709 98039 21417 97680 23118 97291 248J3 96873 38 23 18023 98362 I9737 98033 21445 97673 23146 97284 24841 96866 37 24 I8052 98357 19766 98027 21474 97667 23175 97278 24869 96858 36 25 18o8I 98352 19794 9802I 21502 97661 23203 97271 24897 96851 35 26 18IO1 98347 9Ig823 9806 21530 97655 2323I 97264 24925 96844 31 27 I8i38 g834I i985I 980oi 21559 97648 23260 97257 24954 96837 33 2 181 66 98336 I9880 98004 21587 97642 23288 9725I 24982 96829 32 29 18195 98331 99gg8 97998 21616 97636 23316 97244 25010 96822 3i 3o 18224 98325 19937 97992 21644 97630 23345 97237 25038 96815 3o 3I 18252 98320 I9965 97987 21672 97623 23373 97230 25066 96807 29 32 18281 98315 I9994 97981 21701 97617 23401 97223 25094 96800 28 33 18309 983IO 20022 97975 2I729 976II 23429 97217 25I22 96793 27 34 i8338 98304 2005I 97969 21758 97604 23458 97210 2515I 96786 26 35 18367 98299 20079 97963 21786 97598 23486 97203 25179 96778 25 36 18395 98294 20Io8 97958 218I4 97592 23514 9716 25207 96771 24 37 I8424 98288 20I36 97952 21843 97585 23542 97Ig9 25235 96764 23 38 8452 98283 20165 97946 2187I 97579 23571 97I82 25263 96756 22 39 I848I 98277 20193 97940 2I899 97573 23599 97I76 25291 96749 2I 40 I850g 98272 20222 97934 21928 97566 23627 97I69 25320 96742 20 41 I8538 98267 20250 97928 21956 97560 23656 97I62 25348 96734 19 42 I8567 98261 20279 97922 2I985 97553 23684 97155 25376 96727 I8 43 18595 98255 20307 979I6 22013 97547 237I2 97148 25404 967I9 17 44 I8624 98250 20336 97910 22041 97541I 23740 97I4I 25432 96712 I6 45 18652 98245 20364 97905 22070 97534 23769 97I34 25460 96705 15 46 I868i 98240 20393 97899 22098 97528 23797 97127 25488 96697 14 47 I8710 98234 20421 97893 22126 9752I 23825 97120 25516 96690 13 4 I8738 98229 20450 97887 22I55 975I5 23853 97II3 25545 96682 12 49 18767| 98223 20478 97881 22183 97508 23882 97106 25573 96675 II 50 I8795 98218 20507 97875 22212 97502 23910 97100 2560I 96667 10 51 I8824 98212 20535 97869 22240 97496 23938 97093 25629 96660 52 I8852 98207 20563 97863 22268 97489 23966 97086 25657 96653 53 i888i 98201 20592 97857 22297 97483 23995 97079 25685 96645 54 18910 98196 20620 9785I 22325 97476 24023 97072 25713 96638 6 55 18938 98190 20649 97845 22353 97470 24051 97065 25741 96630 5 56 I8967 98i85 20677 97839 22382 97463 24079 97058 25769 96623 4 57 i8995 98179 20706 97833 22410 97457 24108 9705I 25798 966I5 3 58 19024 98174 20734 97827 22438 97450 24136 97044 25826 96608 2 59 19052 98168 20763 97821 22467 97444 24I64 97037 25854 96600 60 1o908I 9863 20791 97815 22495 97437 24192 97030 25882 06593 0 Cosine. Sine. lCosine. Sine. Cosine. Sine. Cosine. Sine. Cosine. Sine. 1790 780 7 70 760 750 TABLE III. NATURAL SINES AND COSINES. 67 150 160 170 180 190 Sine. ICosine. Sine. |Cosine. Sine. Cosine. Sine. Cosine. Sine. Cosine. o 25882 96593 27564 96I26 29237 95630 30902 95I06 32557 94552 6o I 25910 96585 27592 96118 29265 95622 30929 95097 32584 94542 59 2 25938 96578 27620 961 0 29293 i 956I3 30957 95088 32612 94533 58 3 25966 96570 27648 96I02 29321 95605 30985 95079 32639 94523 57 4 25994 96562 27676 96094 29348 95596 31012 95070 32667 945I4 56 5 26022 96555 27704 96086 29376 95588 31040 9506I 32694 94504 55 6 26050 96547 2773 96078 29404 95579 3io68 95052 32722 94495 54 7 26079 96540 27759 96070 29432 9557 3o1095 95043 32749 94485 53 8 26107 96532 27787 96062 29460 95562 3I123 95033 32777 94476 52 9 26135 96524 27815 96054 29487 95554 3II5I 95024 32804 94466 5I 10 26I63 965I7 27843 96046 2955 95545 3II78 950I5 32832 94457 50 11 26191 96509 27871 96037 29543 95536 31206 95006 32859 94447 49 12 26219 96502 27899 96029 2957 95528 31233 94997 32887 94438 48 I3 26247 96494 27927 96021 29599 95519 3I26I 94988 32914 94428 47 14 26275 96486 27955 960 3 29626 9551 31289 94979 32942 94418 46 I5 26303 96479 27983 96005 29654 95502 3i3i6 94970 32969 94409 45 I6 2633 I96471 28011 95997 29682 95493 3344 94961I 32997 94399 44 I7 26359 96463 280o39 95989 297Io 95485 31372 94952 33024 9439o 43 81 26387 96456 280o67 9598 29737 95476 3I399 94943 33o5i 94380 42 19 264I5 96448 28095 95972 29765 95467 31427 94933 33079 94370 41 20 26443 96440 28123 95964 29793 95459 31454 94924 33io6 9436, 40 21 26471i 96433 28I50 95956 29821 95450 31482 949i5 33134 9435I 39 22 26500 96425 28I78 95948 29849 9544I 3I5Io 94906 33I6I 94342 38 23 26528 964I7 28206 95940 29876 95433 3I537 94897 33189 94332 37 24 26556 9641o 28234 9593I 29904 95424 3I565 94888 33216 94322 36 25 26584 96402 28262 95923 29932 95415 31593 94878 33244 94313 35 26 26612 96394 28290 959 5 29960 95407 3I620 94869 33271 94303 34 27 26640 96386 283I8 95907 29987 95398 3I648 94860 33298 94293 33 28 26668 96379 28346 95898 3oo005 95389 3675 94851 33326 94284 32 29 26696 9637I 28374 95890o 30043 95380 3703 94842 33353 94274 3i 30 26724 96363 28402 95882 3007 I 95372 3I730 94832 3338i 94264 30 3 I26752 96355 28429 95874 30098 95363 31758 94823 334o8 94254 29 32 26780 96347 28457 95865 30126 95354 31786 94814 33436 94245 28 33 26808 96340 28485 95857 3oi54 95345 3i8i3 94805 33463 94235 27 34 26836 96332 28513 95849 30182 95337 3184I 94795 33490 94225 26 35 26864 96324 2854, 9584I 30209 95328 3I868 94786 33518 942I5 25 36 26892 963 6 28569 95832 30237 95319 3I896 94777 33545 94206 24 37 26920 96308 28597 95824 30265 953io 31923 94768 33573 94 96 23 38 26948 96301 28625 958i6 30292 953o0I 3I95 94758 336oo 94I86 22 39 26976 96293 28652 95807 30320 95293 31979 94749 33627 94176 21 40 27004 96285 28680 95799 30348 95284 32006 94740 33655 94167 20 41 27032 96277 28708 9579I 30376 95275 32034 94730 33682 94 57 19 42 27060 96269 28736 95782 304o3 95266 3206I 94721 337I0 94147 18 43 27088 96261 28764 95774 3o43I 95257 32089 94712 33737 94I37 I7 44 271I6 96253 28792 95766 30459 95248 32II6 94702 33764 94127 I6 45 27144 96246 28820 95757 30486 95240 32144 94693 33792 94II8 I5 46 27172 96238 28847 95749 3o5I4 95231 3217I 94684 338I9 94Io8 I4 47 27200 96230 28875 9574o 30542 95222 32I99 94674 33846 94098 13 48 27228 96222 28903 95732 3o057o 9523 32227 94665 33874 94088 2 49 27256 96214 28931 95724 30597 95204 32254 94656 3390oi 94078 i 50 27284 96206 28959 95715 30625 95I95 32282 94646 33929 94068 io 5I 273I2 96I98 28987 95707 30653 95I86 32309 94637 33956 94058 9 52 27340 96190 29015 95698 30680 95I77 32337 94627 33983 94049 8 53 27368 96182 29042 95690 30708 95168 32364 94618 34011 94039 7 54 27396 96I74 29070 9561i 30736 95159 32392 94609 34038 94029 6 55 27424 96166 29098 95673 30763 95i50 324I9 94599 34065 94019 5 56 27452 9618 29126 95664 30791 95I42 32447 94590 34093 94009 4 57 27480 96I50 29I54 95656 308I9 95I33 32474 94580 34I20 93999 3 58 27508 96142 29I82 95647 3o846 95124 32502 9457I 34147 93989 2 59 27536 96134 29209 95639 30874 95ii5 32529 9456I 34I75 93979 I 60 27564 96I26 29237 9563o 30902 95Io6 32557 94552 34202 93969 0 Cosine. Sine. Cosine. Sine. Cosine. Sine. Cosine. Sine. Cosine. Sine. 740 730 172 710 700 68 NATURAL SINES AND COSINES. TABLE III. 200 210 220 230 240 Sine. Cosine. Sine. Cosine. Sine. Cosine. Sine. Cosine. Sine. Cosine. o 34202 93969 35837 93358 37461 927I8 39073 92050 40674 9I355| 60 I 34229 93959 35864 93348 37488 92707 391oo 92039 40700 91343 59 2 34257 93949 35891 93337 37515 92697 39127 9202 40727 9133i 58 3 34284 93939 35918 93327 37542 92686 39153 920I6 40753 91319 57 4 343 I 93929 35945 93316 37569 92675 39180 92005 40780 9I307 56 5 34339 93919 35973 93306 37595 92664 39207 91994 4o8o6 91295 55 6 34366 93909 36ooo 93295 37622 92653 39234 91982 4o833 91283 54 7 34393 93899 36027 93285 37649 92642 39260 91971 4o86o 91272 53 8 34421 93889 36054 93274 37676 9263I 39287 91959 40886 91260 52 9 34448 93879 36o08 93264 377o3 92620 39314 91948 40013 91248 51 I 34475 93869 36Io8 93253 3773o 92609 39341 91936 4o939 91236 50o I I 345o3 93859 36135 93243 37757 92598 39367 91925 40966 91224 49 I2 3453o 93849 36162 93232 37784 92587 39394 91914 40992 91212 48 I3 34557 93839 3690o 93222 378I I 92576 3942I 91902 4I01 9gI200 47 14 34584 93829 36217 93211 37838 92565 39448 g91891 41045 9gI88 46 15 34612 93819 36244 9320I 37865 92554 39474 91879 41072 91176 45 i6 34639 93809 36271 93190 37892 92543 39501 91868 410 98 91164 44 17 34666 93799 36298 93I80 37919 92532 39528 91856 41125 91152 43 i8 34694 93789 36325 93169 37946 92521 39555 91845 41151 91140 42 19 34721 93779 36352 93159 37973 925o10 39581 9g833 41178 91128 41 20 34748 93769 36379 93148 37999 92499 39608 91822 41204 91116 40 2I 34775 93759 36406 93I37 38026 92488 39635 91810 41231 9II04 39 22 34803 93748 36434 93127 38053 92477 3966I 91799 41257 91092 38 23 34830 93738 3646i 93116 38080 92466 39688 91787 41284 91 o80 37 24 34857 93728 36488 93o106 38107 92455 397 5 91775 413io 91068 36 25 34884 93718 36515 93o95 38134 92444 39741 91764 41337 9io56 35 26 34912 93708 36542 93084 3816i 92432 39768 91752 4i363 91044 34 2 34939 93698 36569 93074 38I88 9242I 39795 9I74I 4I390 91032 33 2 34966 93688 36596 93063 382I5 9241O 39822 9I729 4I4I6 91020 32 29 34993 93677 36623 93052 38241 92399 39848 9178 4,443 9100oo8 3i 30 35021 93667 3665o 93042 38268 92388 39875 91706 41469 90996 30 31 35048 93657 36677 93031 38295'92377 39902 g9164 4I496 90984 29 32 35075 93647 36704 93020 38322 92366 39928 9683 41522 90972 28 33 35I02 93637 3673I 930IO 38349 92355 39955 91671 41549 90960 27 34 35i30 93626 36758 92999 38376 92343 39982 91660 41575 90948 26 35 35157 936I6 36785 92988 38403 92332 4ooo8 9I648 41602 90936 25 36 35I84 936o6 36812 92978 38430 92321 40035 91636 41628 90924 24 37 352II 93596 36839 92967 38456 923IO 40062 91625 4I655 90o11 23 38 35239 93585 36867 92956 38483 92299 40088 91613 4168I 90899 22 39 35266 93575 36894 92945 385io 92287 4oIi5 91601 41707 90887 21 40 35293 93565 36921 92935 38537 92276 40oi4 9159o 4I734 90875 20 41 35320 93555 36948 92924 38564 92265 40168 91578 41760 90863 19 42 35347 93544 36975 92913 3859i 92254 40oi95 91566 41787 908851 i 43 35375 93534 37002 92902 386I7 92243 40221 9I555 4I813 90839 17 44 35402 93524 37029 92892 38644 9223I 40248 91543 41840 90826 I6 45 35429 93514 37056 92881 38671 92220 40275 9I53I 4I866 90814 i5 46 35456 935o3 37083 92870 38698 92209 4030I 9I5 41892 90802 14 47 35484 93493 37110 92859 38725 92198 40328 91508 41919 90790 I3 48 355II 93483 37137 92849 38752 92186 40355 9I496 4I945 90778 12 49 35538 93472 37I64 92838 38778 92175 4038I 9i484 4I972 90766 1 50 35565 93462 37191 92827 38805 92164 40408 91472 41998 90753 10 5I 35592 93452 372I8 928I6 38832 92152 40434 91461 42024 90741 52 35619 93441 37245 92805 38859 92141 4o46I 91449 42051 90729 53 35647 93431 37272 92794 38886 92130 40488 91437 42077 90717 7 54 35674 93420 37299 92784 38912 92119 40514 91425 42104 90704 6 55 35701 934,0 37326 92773 38939 92107 4054I 91414 42130 90692 5 56 35728 93400 37353 92762 38966 92096 40567 91402 42156 90680 4 57 35755 93389 37380 9275I 38993 92085 40594 91390 42183 90668 3 58 35782 93379 37407 92740 39020 92073 4062I 91378 422091 0o655 2 59 358io 93368 37434 92729 39046 92062 40647 91366 42235 90643 I 60 35837 93358 3746I 92718t 39073 92050 40674 91355 42262 90631 o.... I -- _ _ _ _ 1_1____1__ Cosine. Sine. Cosine.Sine. Cosine. Sine. Cosine. S Sine. Cosine. Sine. 690 680 67~0 66 65~ TABLE III. NATURAL SINES AND COSINES. 69 250 260 270 280 290,I - — I Sine. Cosine. Sine. Cosine. Sine. Cosine. Sine. Cosine. Sine. Cosine. 0 42262 9063I 43837 89879 45399 89IOI 46947 88295 4848I 87462 6o I 42288 9o6i81 43863 89867 45425 89087 46973 88281 48506 87448 5q 2 423I5 90606 43889 89854 4545I 89074 46999 88267 48532 1 87434 58 3 4234I 90594 439I6 89841 45477 89061 47024 88254 48557 87420 57 4 42367 90582 43942 89828 455o3 89o48 47050 88240 48583 87406 56 5 42394 90569 43968 89816 45529 89o35 47076 88226 48608 8739I 55 6 42420! 9o557 43994 89803 45554 8902, 47Ioi 88213 48634 87377 54 7 42/44/6 90545 44020 89790 45580 89008 47127 88199 48659 87363 53 8 42473 90532.44046 89777 45606 88995 47I53 88I85 48684 87349 52 9 42499 1 90520 44072 89764 45632 8898, 47178 88I72 487Io 8733 5I Io 42525 90507 44098 89752 45658 88968 47204 88I58 48735 8732I 5o II 42552 90495 44124 89739 45684 88955 47229 88,44 4876I 87306 49 I2 42578 90483 44I5I 89726 457Io 88942 47255 88I30 48786 87292 48 I3 426041 90470 44I77 89713 45736 88928 4728i 88I 7 488Ii| 87278 47 I4 4263I 9o458 44203 89700 45762 889I5 47306 88io3 48837 87264 46 I5 42657 90446 44229 89687 45787 88902 47332 88089 48862 87250 45 I6 42683 90433 44255 89674 458I3 88888 47358 88075 48888 87235 44 17 42709 9042I 4428I 89662 45839 88875 47383 88062 489I3 8722I 43 i8 42736 90408 44307 89649 45865 88862 47409 88048 48938 87207 42 I9 42762 9o396 44333 89636 4589I 88848 47434 88o34 48964 87I93 41 20 42788. 90383 4435g 89623 459I7 88835 47460 88020 48989 87I78 40 2I 42815 90371 44385 896Io 45942 88822 47486 88oo6 490I4 87I64 39 22 4284I 90358 444,I 89597 45968 888o8 475II 87993 49040 87150 38 23 42867 90346 44437 89584 45994 88795 47537 87979 49065 87I36 37 24 42894 90334 44464 8957I 46020 88782 47562 87965 49090 8712I 36 25 42920 9032, 44490 89558 46046 88768 47588 87951 49,I6 87Io7 35 26 42946 90309 445i6 89545 46072 88755 476I4 87937 49I4I 87093 34 27 42972 90296 44542 89532 46097 88741 47639 87923 49I66 87079 33 28 42999 90284 44568 89519 46123 88728 47665 87909 49192 87064 32 29 43025 9027I 44594 89506 46i49 887I15 47690 87896 49217 87050 31 30 43o5i 90259 4462o 89493 46175 88701 477I6 87882 49242 87036 30 3I 43077 90246 44646 89480 4620I 88688 4774I 87868 49268 87021 29 32 43I0o4 90233 44672 89467 46226 88674 47767 87854 49293 87007 28 33 43i3o 9022, 44698 89454 46252 8866i 47793 87840 493I8 86993 2 34 43I56 90208 44724 8944I 46278 88647 478I8 87826 49344 86978 26 35 43182 gOI90196 44750 89428 46304 88634 47844 878I2 49369 86964 25 36 43209 o9083 44776 894I5 4633o 88620 47869 87798 49394 86949 24 37 43235 90I7, 44802 89402 46355 88607 47895 87784 494I 86935 23 38 43261 9o158 44828 89389 4638I 88593 47920 87770 4944 86921 22 39 43287 o9046 44854 89376 46407 8858o 47946 87756 49470 86906 21 40 43313 o903-3 44880 89363 46433 88566 4797I 87743 49495 86892 20 4I 4334o0 90oI20 44906 89350 46458 88553 47997 87729 4952I 86878 I 42 43366 o90o8 44932 89337 46484 88539 48022 87715 49546 86863 I8 43 43392 i 90095 44958 89324 465I0o 88526 48048 87701 4957I 86849 17 44 43418 90082 44984 89311 46536 88512 48073 87687 49596 86834 I6 45 43445 90070 450oI 89298 4656I 88499 48099 87673 49622 86820 5 46 43471 90057 45o36 89285 46587 88485 48I24 87659 49647 868o5 14 47 43497 90045 45062 89272 46613 88472 48I5o 87645 49672 8679I I3 48 43523 1 90032 45o88 89259 46639 88458 48175 8763I 49697 86777 I2 49 43549 90019 45II4 89245 46664 88445 4820I 876I7 49723 86762 II 50 43571 90007 4540 89232 46690 88431 48226 87603 49748 86748 1 5I 43602 19994 45166 89219 46716 88417 48252 87589 49773 86733 q 52 43628 8998 45I92 89206 46742 88404 48277 87575 49798 86719 8 53 43654 89968 452I8 89I93 46767 88390 483o03 87561 49824 86704 7 54 43680 89956 45243 89180 46793 88377 48328 87546 49849 86690 6 55 43706 89943 45269 89I67 468I9 88363 48354 87532 49874 86675 5 56 43733 89930 45295 89I53 46844 88349 48379 875I8 49899 8666I 4 5 43759 1 89918 4532I 89140 46870 88336 484o5 87504 49924 86646 3 43782 89905 45347 89127 46896 88322 48430 87490 49950 86632 2 59 438II 89892 45373 89II4 4692 88308 48456 87476 49975 86617 I 60 43837 89879 45399 890oI 46947 88295 4848I| 87462 500o00ooo 86603 o Cosine. I Sine. Cosine. Sine. Cosine. Sine. Cosine. Sine. Cosine. Sine. 640 630 62~ 610 600 70 NATURAL SINES AND COSINES. TABLE III. 800 310 320 330 340 Sine. Cosine. Sine. Cosine. Sine. Cosine. Sine. Cosine. Sine. Cosine 0 50000oooo 866o3 i150o4 85717 02992 848o5 54464 83867 55919 82904 60 I 5o0025 86588 51529 85702 530l7 84789 54488 8385i 55943 82887! 5 2 5oo0050 86573 5i554 85687 53041 84774 54513 83835 55968 8287I 5 3 50076 86559 51579 85672 53066 84759 54537 83819 55992 82855 57 4 5oio1 86544 5i6o04 85657 53091 84743 54561 8380o4 56oi016 82839 56 5,5o26 86530 ~1628 85642 53115 84728 5586 83788 56o040: 82822 55 6 50oi5 865i5 51653 85627 53i4o 84712 546io 83772 56064 82806 54 7 50176 8650oi 5i678 85612 53164 84697 54635 83756 56088 82790 53 8 5020O 86486 51703 85597 531891846811 54659 83740 56112 82773 52 9 50227 86471 5I728 85582 532,4 84666 54683 83724 56136 82757 51 oI 5o252 86457 51753 85567 53238 84650 54708 83708 56s6o 82741 So s1 50277 8644/2 5I778 8555i 53263 84635 54732 83692 56S84 82724 4 1 50302 86427 518o3 85536 53288 84619 54756 83676 56208 82708 4 1I3 50327 86413 51828 85521 53312 84604 54781i 8366o 56232 82692 47 14 50352 86398 51852 85506 53337 84588 54805 83645 56256 82675 46 i5 50377 86384 51877 85491 53361i 84573 54829 83629 56280 82659 45 i6 50403 86369 51902 85476 53386 84557 54854 836i3 563o5 82643 44 17 50428 86354 51927 85461 534Ii 84542 54878 83597 56329 82626 43 18 5o453 86340 51952 85446 53435 84526 54902 83581 56353 8261o 42 19 50478 86325 51977 8543I 5346o 845ii 54927. 83565 56377 82593 4i 20 5o5o3 863io 52002 85416 53484 84495 54951 83549 564oi 82577 40 21 50528 86295 52026 854oi 535o09 84480 54975 83533 56425 8256i1 39 22 5o553 86281 52051 85385 53534 84464 54999 83517 56449 82544 38 23 50578 86266 52076 85370 53558 84448 55024 835oi 56473 82528 3 24 5o6o3 86251 52io.1 85355 53583 84433 55048 83485 56497 82511i 36 25 50628 86237 52126 8534o 53607 84417 55072 83469 56521 82495 35 26 5o654 86222 52151 85325 53632 84402 55097 83453 56545 82478 34 27 50679 86207 52175 853io 53656 84386 55121 83437 56569 82462 33 28 50704 86192 522oo00 85294 5368i 84370 54514 83421 56593 82446 32 29 50729 86178 52225 85279 53705 84355 55169 83405 56617 82429 3i 3o 50754 86163 52250 85264 53730 84339 55194 83389 56641 8241 3o 3i 50779 86148 52275 85249 53754 84324 55218 83373 56665 82396 2 32 50804 186i33 52299 85234 53779 843o8 55242 83356 56689 82380 2 33 5o829 86II9 52324 85218 53804 84292 55266 8334o 56713 82363 27 34 50854 86104 52349 85203 53828 84277 55291 83324 56736 82347 26 35 50879 86089 52374 85188 53853 84261 55315 833o8 56760 82330 25 36 5090o4 86074 52399 85173 53877 84245 55339 83292 56784 82314 24 37 50929 86o59 52423 85157 53902 84230 55363 83276 5680o8 82297 23 38 50954 8604o 52448 85142 53926 84214 55388 83260 56832 82281 22 39 50979 86o3o 52473 85127 53951 84198 55412 83244 56856 822641 21 40 j5oo4 86o15 52498 85112 53975 84182, 55436 83228 5688o 82248 20 41 51029 86oo0o 52522 85096 54o000oo 84167 55460 83212 56904 82231 19 42 5IO54 85985 52547 85o8I 54024 8415I 55484 83195 56928 82214 18 43 51079 85970 52572 85o66 54o49 84135 55569 83179 56952 82198 17 44 5I o4 85956 52597 85o5I 54073 84120 55533 83163 56976 82181 s6 45 51129 85941 52621 85o35 54097 84104 55557 83147 57000 821651 15 46 51s54 85926 52646 85020 54122 84088 5558I 83s3i 57024 82148 14 47 5I179 85911 52671 85oo005 5446 84072 55605 83ii5 57047 82132 13 48 51204 85896 52696 84989 54171 84057 5563o 83098 57071 82115 12 49 51229 85881 52720 84974 54195 84041 55654 83o82 57095 82098 sI 50 5I254 85866 52745 84959 5422o 84025 55678 83o66 57119 82082 10 5i 51279 85851 52770 84943 54244 84oo009 55702 83o5o 57143 820651 52 53o4 85836 52794 84928 54269 83994 55726 83034 57167 82048 53 51329 85821 528I9 84913 54293 83978 5575o 830oi7 57191 82032 7 54 5354 858o6 52.844 84897 54317 83962 55775 83oos 57215 82o15 6 55 51379 85792 52869 84882 54342 83946 55799 82985 57238 81999 5 56 51404 85777 52893 84866 54366 83930 55823 82969 57262 81982 4 57 51429 85762 529s8 8485I 54391 83915 55847 82953 57286 8965j 3 58 51454 85747 52943 84836 544i5 83899 55871 82936 57310 81949 2 59 51479 85732 52967184820 5444o 83883 55895 8292o 57334 81932 5 60 5I504 85717 52992 84805 54464 83867 55919 82904 57358 81915 o Cosine. Sine. Cosine. Sine. Cosine. Sine. Cosine. Sine. Cosine. Sine. 590 580 0 560 550 590 580 51o IJ 560 ~ 550 TABLE III. NATURAL SINES AND COSINES. 350 360 370 380 390 Sine. Cosine. Sine. Cosine. Sine. Cosine. Sine. Cosine. Sine. Cosine. o 57358 81915 58779 80902 6o182 79864 6i566 7880! 62932 77715 60 I 57381 81899 58802 8o885 60205 79846 61589 78783 62955 77696 S 2 57405 81882 58826 80867 60228 79829 61612 78765 62977 77678 5 3 57429 81865 58849 8o85o 60251 798Ii 6i635 78747 6300oo 77660 57 4 57453 81848 58873 8o833 60274 79793 6i658 78729 63022 77641 56 5 57477 81832 58896 8o8i6 60298 79776 6168i 78711 63045 77623 55'6 57501 81815 58920 80799 60321 79758 61704 78694 63o68 77605 54 57524 81798 58943 8o782 60344 79741 61726 78676 63090 77586 53 57548 81782 58967 80765 60367 79723 6I749 78658 63ii3 77568 52 9 57572 81765 58990 80748 60390 79706 61772 78640 6335 77550 5i lo 57596 81748 59o014 80730 6o414 79688 61795 78622 63158 77531 50 576I9 81731 59037 80713 60437 79671 6i8i8 78604 63i8o 77513 4 i2 57643 81714 59061 80696 60460 79653 61841 78586 63203 77494 4 13 57667 81698 59084 80679 60483 79635 61864 78568 63225 77476 47 54 S769i 8i68I 59I108 80662 60506 70618 61887 78550 63248 77458 46 IS 57715 81664 5913i 80644 60529 796oo00 61909 78532 63271 77439 45 i6 57738 81647 59154 80627 60553 79583 61932 78514 63293 77421 44 17 57762 8163i 59i78 8o6io 60576 79565 61955 78496 63316 77402 43 i8 57786 81614 592o1 80593 6o599 79547 61978 78478 63338 77384 42 19 578Io 8i597 59225 80576 60622 79530 6200oo1 78460 63.36i 77366 4i 20 57833 8i58o 59248 8o558 60645 79512 62024 78442 63383 77347 40 21 578S7 8i563 59272 8o54i 60668 79494 62046 78424 63406 77329 39 22 5788i 81546 59295 80524 60691 79477 62069 78405 63428 77310 38 23 57904 81530 59318 8o5o7 60714 79459 62092 78387 63451 77292 37 24 57928 8i51 3 59342 80489 60738 79441 62115 78369 63473 77273 36 25 57952 81496 59365 80472 60761 79424 62138 78351 63496 77255 35 26 57976 81479 59389 80455 60784 79406 62160o 78333 635i8 77236 34 27 57999 81462 59412 80438 60807 79388 6a2183 78315 63540 77218 33 28 58023 81445 59436 80420 60830 7937i 62206 78297 63563 77199 32 29 58047 81428 59459 80403 60853 79353 62229 78279 63585 7718i 3i 30 5807o 81412 59482 80386 60876 79335 62251 78261 63608 77162 30 31 58o94 8i395 595o6 8o368 60899 79331 8 62274 78243 63630 77144 29 32 58118 81378 59529 8o35i 60922 79300 62297 78225 63653 77125 28 33 58141 81361 59552 80334 6o945 79282 62320 78206 63675 77107 27 34 58165 81344 59576 8o316 60968 79264 62342 78188 63698 77088 26 35 58189 81327 59599 80299 60991 79247 62365 78170 63720 77070 25 36 58212 8i3Io 59622 80282 6ioi5 79229 62388 78152 63742 77051 24 3 58236 81293 59646 80264 6io38 792II 624II 78134 63765 77033 23 3 5826o 81276 59669 80247 6io6i 79193 62433 78116 63787 77014 22 39 58283 81259 59693 80230 61084 79176 62456 78o98 638o10 76996 21 40 58307 81242 59716 80212 61107 79158 62479 78079 63832 76977 20 4I 58330 81225 S9739 8oi95 6ii3o 79140 62502 178061 63854 76959 i 42 58354 81208 59763 8078 6ii53 79122 62524 78043 63877 76940 I1 43 58378 81191 59786 8oi6o 61176 79105 62547 78025 63899 76921 17 44 58401i 81174 59809 8o0143 61199 79087 62570 78007 63922 76903 i6 45 58425 8ii57 59832 80o125 61222 79069 62592 77988 63944 76884 I5 46 58449 81140 59856 8010oio8 61245 79051 62615 77970 63966 76866 14 47 58472 81I123 59879 80091 61268 79033 62638 77952 63989 76847 I3 48 58496 8iio06 59902 80073 61291 79o016 62660 77934 64011 76828 12 49 585i9 81089 59926 8oo56 61314 78998 62683 77916 64033 76810 1i 50 58543 81072 59949 8oo0038 61337 78980 62706 77897 64056 76791 10 5i 58567 8io055 59972 80021 61360 78962 62728 77879 64078 76772 52 58590 8o1038 59995 8oo0003 6i383 78944 62751 77861 64100 76754 53 586i4 8102I 60019 79986 61406 78926 62774 77843 64123 76735 7 54 58637 81004 60042 79968 61429 78908 62796 77824 64145 76717 6 55 5866i 80987 6oo0065 7995I1 61451 78891 628I9 77806 64167 76698 5 56 58684 80970 60089 79934 61474 78873 62842 77788 64190 76679 4 57 58708 80953 601I2 79916 61497 78855 62864 77769 64212 76661 3 58 58731 80936 6oi0135 79899 61520 78837 62887 77751 64234 76642 2 59 58755 80919 6oi0158 79881 6i543 78819 62909 77733 64256 76623 I 60 58779 80902 60182 79864 6i566 78801 62932 77715 64279 76604 0 Cosine. Sine. Cosine. Sine. Cosine. I Sine. Cosine. Sine. Cosine. Sine. 540 530 520 10 500 540 530 520 51~ 500 72 NATURAL SINES AND COSINES. TABLE III. 400 410 420 430 440 Sine. Cosine. Sine. Cosine. Sine. Cosine. Sine. Cosine. Sine. Cosine. o 64279 76604 656o6 75471 66913 743I4 68200 73135 69466 7I934 60 1 643oi 76586 65628 75452 66935 74295 68221 73I16 69487 71914 5? 2 64323 76567 65650 75433 66956 74276 68242 73096 69508 71894 5 3 64346 76548 65672 75414 66978 74256 68264 73076 69529 71873 57 4 64368 76530 65694 75395 66999 74237 68285 73056 69549 71853 56 5 64390 76511 65716 75375 67021 74217 68306 73036 69570 71833 55 6 64412 76492 65738 75356 67043 74198 68327 73016 69591 71813 54 7 64435 76473 65759 75337 67064 74I78 68349 72996 696I2 71792 53 8 64457 76455 65781 75318 67086 74159 68370 72976 69633 71772 52 9 64479 76436 65803 75299 67107 74139 6839I 72957 69654 7I752 5i io 64501 76417 65825 75280 67129 74120 68412 72937 69675 71732 50 Ii 64524 76398 65847 75261 67151 74100 68434 72917 69696 71711 49 12 64546 76380 65869 7524I 67I72 74080 68455 72897 69717 7I691 48 I3 64568 7636I 65891 75222 67I94 7406I 68476 72877 69737 7167I 47 I4 64590 76342 65913 75203 67215 7404I 68497 72857 69758 71650 46 5 64612 76323 65935 75184 67237 74022 685i8 72837 69779 7I630 45 i6 64635 76304 65956 75165 67258 74002 68539 72817 69800 71610 44 I7 64657 76286 65978 75146 67280 73983 6856I 72797 69821 7I590 43 I 8 64679 76267 66000 75126 6730I 73963 68582 72777 69842 71569 42 19 64701 76248 66022 75I07 67323 73944 68603 72757 69862 71549 4 20 64723 76229 66044 75088 67344 73924 68624 72737 69883 71529 40 21 64746 762IO 66o66 75069 67366 73904 68645 72717 69904 71508 39 22 64768 76192 66088 75050 67387 73885 68666 72697 69925 71488 38 23 64790 76173 66109 75030 67409 73865 68688 72677 69946 71468 37 24 64812 76154 6613I 75011 67430 73846 68709 72657 69966 71447 36 25 64834 76135 66i53 74992 67452 73826 68730 72637 69987 7I427 35 26 64856 76II6 66175 74973 67473 73806 68751 72617 70008 71407 34 27 64878 76097 66197 74953 67495 73787 68772 72597 70029 7I386 33 28 6490o1 76078 662I8 74934 675I6 73767 68793 72577 70049 71366 32 29 64923 76059 66240 74915 67538 73747 68814 72557 70070 71345 3I 30 64945 7604I 66262 74896 67559 73728 68835 72537 70091 71325 30 3I 64967 76022 66284. 74876 67580 73708 68857 72517 70112 71305 29 32 64989 76003 66306 74857 67602 73688 68878 72497 70132 7I284 28 33 650oi 75984 66327 74838 67623 73669 68899 72477 70153 71264 27 34 65033 75965 66349 74818 67645 73649 68920 72457 70174 7I243 26 35 65055 75946 66371 74799 67666 73629 6894I 72437 70195 71223 25 36 65077 1 5927 66393 74780 67688 73610 68962 724I7 70215 71203 24 37 65ioo 1 75908 66414 74760 67709 73590 68983 72397 70236 71182 23 38 65122 75889 66436 7474I 67730 73570 69004 72377 70257 7II62 22 39 65144 75870 66458 74722 67752 7355I 69025 72357 70277 71141 21 40 65i66 7585I 66480 74703 67773 73531 69046 72337 70298 71121 20 41 65i88 75832 665oi 74683 67795 73511 69067 72317 70319 71100 19 42 65210 75813 66523 74664 678I6 73491 69088 72297 70339 7IO80 I 43 65232 75794 66545 74644 67837 73472 69Io9 72277 70360 71059 17 44 65254 75775 66566 74625 67859 73452 69130 72257 7038I 71039 i6 45 65276 75756 66588 74606 67880 73432 69I5I 72236 70401 71019 15 46 65298 75738 66610 74586 67901 73413 69172 72216 70422 70998 I4 47 653201 75719 66632 74567 67923 73393 69193 72196 70443 70978 13 48 65342 75700 66653 74548 67944 73373 69214 72176 70463 70957 12 49 65364 75680o 66675 74528 67965 73353 69235 72156 70484 70937 11 50 65386 7566I 66697 74509 67987 73333 69256 72136 70505 70916 IO 5I 65408 75642 667I8 74489 68008 73314 69277 72II6 70525 70896 52 65430 75623 66740 74470 68029 73294 69298 72095 70546 70875 8 53 65452 75604 66762 7445I 68o5i 73274 69319 72075 70567 70855 7 54 65474 75585 66783 7443I 68072 73254 69340 72055 70587 70834 6 55 65496 75566 66805 74412 68093 73234 6936I 72035 70608 70813 5 56 655i8 75547 66827 74392 681I5 732I5 69382 72o05 70628 70793 4 57 65540 75528 66848 74373 68I36 73195 69403 71995 70649 70772 3 58 65562 75509 66870 74353 68157 73i75 69424 71974 70670 70752 2 59 65584 75490 6689I 74334 68179 73155 69445 71954 706go 7073I I 60 65606 75471 66913 1743I4 68200 73135 69466 71934 707II 707II 0 Cosine. Sine. Cosine. Sine. Cosine. Sine. Cosine. Sine. Cosine. Sine. 490 480 4~0 460 450 TABLE III. NATURAL TANGENTS AND COTANGENTS. 73 Tangent. Cotant,. Tangent. Cotang. Cotag Tangent. Cotangg. 00. ___~ Tangent.Ctag o ooooo Infinite. OI746 57.2900 03492 28-6363 0524I 19-08II 6o I 00029 3437-75 Oi775 56.3506 o3521 28.3994 05270 I8*9755 59 2 ooo58 1718.87 01804 55-4415 o3550 28.1664 05299 I8.87II 58 3 ooo87 I I45 92 oi833 54-5613 o3579 27 - 9372 o5328 I8-7678 57 4 OOII6 859-436 01862 53 - 7086 o3609 27 7117 05357 i 8.6656 56 5 OO145 687 549 01891 52.882 I o3638 27'4899 o5387 I 8.5645 55 6 00175 572.957 01920 52 0 o807 o3667 27.2715 o54i6 I 8.4645 54 7 00204 491i- i06 OI949 51 -3032 o3696 27 -o566 o5445 i8.3655 53 8 oo00233 429'7I8 01978 50.5485 o3725 26'8450 o5474 I8-2677 52 9 oo00262 38I.971 02007 49.8 57 o3754 26.6367 o5503 18. I708 5i IO 0029I 343-774 02036 49- I039 o3783 26-4316 o5533 18-0750 50 II oo320 312*52I 02066 48.*4I 2 03812 26 - 2296 o5562 I7*9802 49 12 oo349 286-478 o2095 47.7395 o3842 26.-o307 o559i 17.8863 48 I 3 oo00378 264-441 02 I 24 47.o853 o3871 25 - 8348 o5620 17. 7934 47 14 00oo407 245.552 02153 46.4489 o39oo 25.6418 o5649 I7.7o015 46 I 5 oo436 229.I82 02I82 45.8294 o3929 25.45J17 o5678 I7.6Io6 45 i6 oo465 214-858 02211 45.226iI o3958 25.2644 o57o8 17.5205 44 17 00oo495 202-219 02240 44.6386 o3987 25.o798 o5737 17.4314 43 i 8 00oo524 I90.-984 o226 44.o66I o4oi16 24. - 8978 o5766 17'3432 42 19 oo553 i8o. 32 0229g 43.5o8I o4046 24-7185 o5795 17.2558 4I 20 oo582 171. 885 02328 42.964i o4o75 24.-5418 o5824 17.-693 4o 21 oo6ii I63.700 02357 42.4335 0o4Io4 24.3675 o5854 I7.o837 39 22 oo64o I56.259 o2386 41.9I58 o4i33 24.-1957 o5883 I6.9990o 38 23 oo669 149g465 02415 41.4Io6 04i62 24.o263 o5912 I6 95o0 37 24 oo698 I43.237 o2444 40o.9174 04i9gI 23'8593 0594I I6.8319I 36 25 00oo727 I37.507 o2473 40o.4358 04220 23.6945 o597o i6. 7496 35 26 oo756 I32.219 o2502 39. 655 o4250 23-5321 o5999 I6.668i 34 27 oo785 I127'32I 0253I 39. 5059 o4279 23'3718 o6029 I6. 5874 33 28 oo814 122774 02560 39. o568 o43o8 23.2137 o6o58 i 6.50o7 5 32 29 oo844 II8.54o 02589 38'6177 o4337 230o577 o6087 I6.4283 3i 30 oo873 II4.589 02619 38.I885 o4366 22-9038 o6ii6 16.3499 3o 3I 00902 IIO-.892 02648 37.7686 04395 22'7519 o6i45 I6.2722 29 32 oog093 IO7.426 o2677 37.3579 04424 22.6020 06i75 I6-1952 28 33 oog0960 IO4. I7 o2706 36.9560 o4454 22-4541 o6204 I6.-119O 27 34 oo0989 Io01.IO7 02735 36-5627 o4483 22.3081 o6233 I6.o435 26 35 oio 8 98.2179 o2764 36.I776 o45I2 22.I640 06262 I5.9687 25 36 OI047 95.4895 o2793 35.8oo6 04541 22.02I7 06291 15.8945 24 37 o1076 92.9085 02822 35.4313 o457o 21.88;3 o632 15.8211 23 38 OIi05 90-.4633 0285i 35. o695 o4599 217426 o635o 15-7483 22 39 01135 88.- I436 0288I 34.71l5 04628 21.6056 o6379 I5.6762 21 40 oii64 85. 9398 02910 34.3678 o4658 21.4704 o6408 15.6048 20 4 OII193 83. 8435 02939 34.0273 o4687 21.3369 o6437 I5-534o 19 42 01222 81.8470 o2968 33'6935 o47I6 21.2049 o6467 15.4638 I8 43 o125I 79-9434 02997 33.3662 o4745 21-0747 o6496 i5-3943 I7 44 OI280 78 -I263 o3026 330o452 04774 20-9460 o6525 I5.3254 i6 45 oi309 76.39oo o3o55 32.7303 o48o3 20-.8I88 o6554 15-.257I i5 46 oi338 74.7292 o3o84 32.4213 o4832 20-6932 o6584 I5.I893 I4 47 oi367 73- 39 o3i1I4 32*.I81 o4862 20- 5691 o66i3 I5.I1222 I3 48 oi396 7I-6I5I o3i43 31.82o5 0489I 20.4465 06642 15-.o557 12 o49 OI425 7o.i533 03172 3I.5284 04920 20-3253 06671 14.9898 II 5o oi455 68-75oi 0320I 3I.2416 o4949 20.20o56 o6700 14.9244 Io 5i 01484 67-40i9 o3230o 3o.9599 o4978 20-o872 o6730 I4-8596 9 52 oI5I13 66.-io55 o3259 3o.6833 o5oo7 19-9702 o6759 14.7954 8 53 O1542 64.858o o3288 3o.4II6 o5o37 Ig-9.8546 o6788 14-7317 7 54 o157I 63-6567 03317 3o. I446 o5o66 19.7403 o068I7 14.6685 6 55 oi6oo 62.4992 o3346 29.8823 o5o95 i9g-6273 o6847 I4.6o59 5 56 1I62 6i.3829 o3376 29.6245 o5124 ig-9.5I56 o6876 I4.5438 4 57 o0I65 6o-3058 o34o5 29.3711 o5i53 I9.4o05I o69o5 14.4823 3 58 oI687 59-2659 o3434 29-1220 o5182 19-2959 o6934 I4-4212 2 59 oi0716 58-2612 o3463 28.877I o05212 I9.-879 06963 14.3607 I 6o 0I746 57.2900 o3492 28.6363 o5241 Ig-9.0811 o6993 I4.3007 0o Cotang. Tangent. Cotang. Tangent. Cotang. Tangent. Cotang. Tangent. 89t~~ s( 880 87~0 86~ i~~~~~~~~~~~ 74 NATURAL TANGENTS AND COTANGENTS.. TABLE III. _ 40 50 6~ 10 Tangent. Cotang. Tangent. Cotang. Tangent.- Cotang. Tangent. Cotang. 0 o6993 I4-3007 o8749 ili.43oI i05iO 9.5I436 12278 8.I4435 60 1 07022 14-24II o8778 I1I39I9 Io540 9.4878I 12,308 8.1248I 5 2 0705I I4 I821 o8807 II 3540 10569 9.46I4I 12338 8.io536 58 3 07080 14-1235 08837 II.363 10599 9.43515 12367 8.o8600 57 4 07110 4. o655 o8866 11.2789 10628 9 40904 12397 8.06674 56 5 07 I4.0oo79 08895 11 2417 o1657 938307 I2,426 8.04756 55 6 O76g 3 9507 o8g92 5 I 2048 I o687 9'35724 12456 8 o2848 54 7 07197 I3 38940 08954 II I68I 10716 9.33154 12485 800oo948 53 8 07227 I3'8378 08983 ii.i3i6 10746 9.30599 12515 7.99058 52 9 07256 13 782i O09013 I.og54 10775 9-28058 12544 7'971I76 5I IO 07285 I3'7267 09042 11.59g4 Io805 9*25530 I2574 7.95302 50 11 07314 I3.67I1 09071 II.o237 io834 9 230I6 12603 7.93438 49 12 07344 I3.6I74 o09101 IO9882 o10863 9-205I6 12633 7.91582 48 I3 o7373 I3.5634 09130 Io.9529 10893 9Ii8028 I2662 7.89734 47 I4 07402 I3.5098 o9159 Io.9I'78 10922 9-I5554 I2692 7.87895 46 I5 07431 I3 *4566 O9189 O.8 82;9 10952 9. 30o3 12722 7.86o64 45 1,6 07461 I3 4039 09218 io-8483 109g8 9 I0646 12751 7.84242 44 I17 07490 I3.35I5 09247 Io:.8I39 IIOII 9.08211 12781 7.82428 43 r8 07519 I3.2996 09277 o. 7797 11040 9 - o5789 12810 7.80622 42 19 07548 I3.248o 09306 10:7457 Io070 9.03379g 12840 7-78825 41 20 07578 I3.i969 o9335 Io0-71I9 9 I1o09 o983 I2869 7.77035 40 21 07607 I3.146 o09365 Io-.6783 i I I 28 898598 12899 7*75254 39 22 07636 13.og958 o9394 IO.6450 ii158 8.96227 I2929 7.-73480 38 23 07665 I3.o458 o9423 IO.6ii8 11187 8.93867 I2958 7.71715 37 24 07695 1r29962, 09453 10.5789 11217 8-9152o 12988 7.69957 36 25 07724 1'2.9469 o9482 I.5462: 112,46 889185 13017 7 -68208 35 26 07753 12.8981 o9511 io.5I36 11276 8.86862 13047 7-66466 34 27 07782 12.8496 o9541 I O48I3I ii3o5 8.8455I 13076 7.64732 33 28 07812 12-80I4 09570 10.449I 11335 8.82252 13I06 7.63005 32 29 o784i 12-7536 o09600 iO-4I72 i364 8.79964 i3i36 7.61287 3i 30 o7870 I2.7062 09629 io-3854 11394 8.77689 13165 7.59575 30 31 07899 I2-.659I o9658 io-3538 11423 8.75425 13195 7.57872 29 32 07929 12.6I24 09688 IO-3224 11452 8.73.172 13224 7-56I76 28 33 07958 I2.5660 09717 O12913 11482 8.7g931 I3254 7.54487 27 34 07987 12.5I99 09746 O12602 11511i 8.68701 13284 7.528o6 26 35 o8o07 1I24742 09776 IO2294 1154i 8-66482 13313 7.5II32 25 36 08046 12.4288 09805 10.I988 11570 8-64275 i3343 7-49465 24 37 o8075 12-3838 09834 io.i683 ii6oo 8.62078 13372 7.47806 23 38 08104 12.3390 o9864 IO.138I r1629 8.59893 I3402 7-46-154 22' 39 08134 12.2946 09893 io-io8o 1'659 8-57718 13432 7.44509 21 40 o8163 12.2505 o9923 10-078O0 1688 8.55555 03461 7-42871 20 4I 08I92 I2.2067 09952 I-0o483 11718 8.53402 i3491 7-41240 I9 42 0822I I2.1632 09981 IO-OI87 11747 8-51259 1352I 7.396I6 18 43 08251 I2.I20I 10IOOII 9.98930 rI777 8-49128 03550 7.37999 17 44 08280 12'0772 10040 9.96007 I80o6 8.47007 13580 7.36389 i6 45 08o309 120346 10069 99310I I1I836 8-44896 I3609 7.34786 I5 46 08339 i II9923 Ioo0 9.90211 ri865 8.42795 13639 7.33I90 14 47 o8368 I1.9504 IOI28 9.87338 11895 8'40705 13669 7-31600 I3 48 08397 11.9087 o0158 9.84482 11924 8-38625 13698 7.30018 I2 49 08427 II.8673 10187 9.8I64I |1954 8.36555 13728 7.28442 11 50 08456 II-82*62 10216 9g788I7 11983 8-34496 r3758 7.26873 10 5I 08485 11.7853 10246 9.76009 120I3 8.32446 r3787 7.253IO 9 52 o8514 i 11.7448 10275 9-732I7 12042 8.30406 13817 723754 8 53 o8544 I1.7045 10o305 970441 12072 8'28376 13846 7.22204 7 54 o8573 11.6645 io334 9-6768 I210I 8-26355 13876 7.2066I 6 56 0 8602 II.6248 o10363 9.64935 1213 8-24345 13906 7-19I12 5 56 08632 II-5853 10393 9.62205 12I60 8-22344 I3935 7-I7594 4 57 o866I II-546I 10422: 9-50490' 12190 8-20352 I3965 7-I607I 3 58 o8690 II-5072 10452 9-5679I 12219 8-I8370 13995 71I4553 2 59 o8720 1 II4685 Io48, 9.54106 I2249 8-6398 14024 7.I3042 I 60 08749 II-43oi 10510 9.51436 12278 8.I4435 i4o54 7.11537 0 - _____________________2_9o 11 352 13965 i I _I Cotang. Tangent. Cotang. Tangent. Tangent. Cotang. Tangent. 850 840 830 820 0 TABLE III. NATURAL TANGENTS AND COTANGENTS. 76 1 8____ 90 1. 110 Tangent. Cotang. Ta. Tangent. Cotang. Tangent. Cotang. O 14054 7-II537 i5838 6.3,375 17633 5.67128 19438 5.-4455 60 I I4o84 7- o0038 I5868 6-30I89 17663 5.66i65 I9468 5. 3658 59 2 I4ii3 7.-o8546 I5898 6- 2900oo7 17693 5.65205 19498 5.12862 58 3 I4I43 7.0o7059 5928 6.27829 17723 5.64248 19529 5I 2069 57 4 14173 70o5579 I5958 6.26655 17753 5.63295 I9559 5.II279 56 5 I4202 7. 04105 I5988 6.254861 17783 5.62344 19589 0.10490 55 6 I4232 7.02637 16017 6.2432, 17813 5-61397 19619 5-09704 54 I4262 7.OII74 16047 6 123160 I7843 5.60452 19649 5 -892 I 53 I429I 6.99718 I6077 6-22003 17873 5.595II 19680 5-08139 52 9 I4321 6. 98268 16107 6-2085I 17903 5.58573 197IO 5'07360 5i Io i435I 6.96823 16137 6-I9703 17933 5.57638 19740 5-.06584 50 II I438I 6.95385 I6I67 6-i8559 I7963 5.56706 i9770 5.o5809 4 12 144o0 6,93952 I6196 6.17419 17993 5.55777 I98oi 5.05037 4 13 14440 6.92525 I6226 6-I6283 18023 5*5485I 1983I 5o04267 47 I4 14470 6-.IIo4 I6256 6.I5I5I i8053 5.53927 1986I 5.03499 46 I5 I4499 6 89688 I6286 6.I4023 i8083 5.53007 I9891 5-o2734 45 i6 14529 6-88278 I63I6 6-I2899 i8ii3 5.520o90 19921 5O9g7I 44 I7 14559 6.86874 I6346 6. I779 I8I43 5 5I I76 19952 5.01210 43 I8 14588 6.85475 I6376 6.-:0664 I8173 5.50264 I9982 5.oo0045I 42 19 I46i8 6.84082 I6405 6.o9552 I8203 5.49356 20012 4.99695 4I 20 14648 6.82694 I6435 6-o08444 I8233 5-4845I 20042 4-98940 40 21 I4678 6.8I312 I6465 6-07340 I8263 5.47548 20073 4-98I88 3 22 14707 6.79936 I6495 6.06240 I8293 5.46648 20103 4-97438 38 23 14737 6-78564 i6525 6.05143 18323 5.45751 20133 4.96690 37 24 14767 6.77199 i6555 6.0405I I8353 5-44857 20164 4-95945 36 25 14796 6-75838 i6585 6-02962 i8383 5-43966 20194 4.9520I 35 26 1486 6-.4483 I66i5 6.0o878 18414 5-.43077 20224 4-94460 34 27 14856 6.73I33 I6645 6.00797 I8444 5'42192 20254 4-93721 33 28 14886 6-7I789 I6674 5-99720 18474 5-4I309 20285 4-92984 32 29 14915 6.70450 I6704 5.98646 i8504 5-40429 20315 4-92249 31 30 14945 6-69116 i6734 5.97576 i8534 5-39552 20345 94-gI5I6 30 3i 14975 6.67787 I6764 5.965io i8564 5-38677 20376 4-90785 29 32 i5005 6.66463 16794 5.95448 18594 5-37805 20406 4-90056 2 33 i5034 6.65144 i6824 5-9439o 18624 5-36936 20436 4.89330 27 34 I5o64 6.6383I i6854 5.93335 18654 5-.36070 20466 4.88605 26 35 I5094 6.62523 I6884 5-92283 18684 5.35206 20497 4-87882 25 36 15124 6.612I9 i6914 5-91235 18714 5.34345 20527 4.87162 24 37 15153 6-5992I i6944 5.90191 18745 5-33487 20557 4.86444 23 38 i5i83 6.58627 I6974 5.8915i 8775 5-32631 20588 4.85727 22 39 1;I5213 6.57339 I7004 5-88II4 18805 5.3I778 206I8 4.850I3 21 40 15243 6.56055 17033 5.87080 i8835 5.30928 20648 484300 20 4I I5272 6.54777 I7063.586o5i I8865 5.30080 20679 4.83590 19 42 I5302 6.535o3 17093 5.85024 18895 5-29235 20709 4.82882 |I 43 I5332 6.52234 17123 5-84ooI 18925 5-28393 20739 4.82175 17 44 I5362 6.50970 17153 5-82982 18955 5-27553 20770 4-8I47I I6 45 I539I 6-49710 17I83 5.8I966 I8986 5.26715 20800oo 480769 15 46 15421 6-.48456 17213 5.80953 19016 5-25880 20830 4.80068 14 47 I545I 6.47206 17243 5.79944 19046 5-25048 20861 4-79370 13 48 1548I 6-4596I i7273 5-78938 I9076 5-24218 20891 4.78673 1I 49 55,, 6-44720 17303 5-77936 19106 5-23391 20921 4- 77978 II 5o I554o 6.43484 17333 5-76937 19136 5-22566 20952 477286 1o 5I 15570 6-42253 17363 5-7594,I 9166 5-2I744 20982 4-76595 52 i56oo00 6-41026 17393 5-74949 19I97 5-20925 21013 4-75906 8 53 I563o 6-3q804I I7423 5-.73960 I9227 5-20107 2IO43 4-752I9 7 54 566 638587 I7453 5-72974 1I9257 5-I9293 21073 4-745341 6 55 15689 6'37374 17483 5-71992 19287 5.18480 2IIo 4-7385I 5 56 15719 6.36I65 17513 5-7IOI3 19317 5-17671 2Ii34 4.73I70 4 57 I5749 6-3496I 17543 5-70037 i9347 5-i6863 2164 472490 3 58 I5779 6.3376I 17573 5-69o64 I9378 5-i6058 2II95 4-7I8I3 2 59 158o9 |6'32566 17603 5.68094 19408 5-I5256 21225 4'7I37 1 6o0 5838 63I375 17633 5-67I28 19438 5-I4455 21256 4-70463 0 Cotang. Tangent. Cotang. Tangent. Cotang.! Tangent. Cotang. Tangent. 81~ 80Ta net 79~ 7 810 80 Igo 780,t~~~~~~~~ 7907 16 NATURAL TANGENTS AND COTANGENTS. TABLE III. 120 13~ 140 150 ITangent. Cotang. Tangent. C:otang. Tangent. Cotang. Tangent. Cotang. 0 21256 4- 70463 23087 4-33I48 24933 4.010o78 26795 3.73205 6o I 21286 4-6979I 23Ii7 4-32573 24964 400oo582 26826 3 - 7277 59 2 21316 4.69121 23148 4.3200I 24995 4000ooo86 26857 3 72338 58 3 2 I 347 4-.68452 23179 4.3I430 25026 3 99592 26888 3'71907 57 4 21377 4-67786 23209 4-30860 25056 3 99099 26920 3.71476 56 5 2I408 4-67121 23240 4-30291 25087 3.98607 2695I 3.71046 55 6 21438 4.66458 23271 4.29724 25ii8 3-98117 26982 3-70616 54 7 2I469 4-65797 2330I 4291I59 25149 3'97627 270I3 3 70I88 53 8 21499 4-65I 8 23332 428595 25180 3.97139 27044 3. 6976 52 9 21529 4.64480 23363 4-.28032 25211 3.9665I 27076 3.69335 5i IO 21560 4.63825 23393 4-2747I 25242 3.96165 27107 3.68909 50 II 21590 4-63I71 23424 4.269I1 25273 3.95680 2I38 368485 49 I2 2I62I 4.625I8 23455 4-26352 25304 3-95I96 27I69 3-68o6I 48 i3 2165i 4.6i868 23485 4-25795 25335 3-94713 27201 3-67638 47 I4 2682 4.61219 235I6 4-25239 25366 3-.94232 27232 38 67217 46 15 21712 4.60572 23547 4-24685 25397 3-9375I 27263 3-66796 45 I6 21743 4-59927 23578 4.24I32 25428 3.9327I 27294 3.66376 44 I7 21773 4-59283 23608 4-23580 25459 3.92793 27326 3-65957 43 i8 21804 4-58641 23639 4-23030 25490 3-923I6 27357 3-65538 42 Ig9 2I834 4-58ooI 23670 4-22481?552I 3.91839 27388 3.6512I 4I 20 21864 4.57363 23700 4-21933 25552 3'91364 274I9 3'64705 40 21 218.95 4-56726 2373I 4-21387 25583 3-90890 2745I 3-64289 3 22 21925 4.56091 23762 4-20842 25614 3'904I7 27482 363874 3 23 21956 4-55458 23793 4-20298 25645 3-89945 27513 3-6346I 37 24 21986 4.54826 23823 4- 9756 25676 3-89474 27545 3-63048 36 25 22017 4.54196 23854 4-19215 25707 3-89004 27576 3-62636 35 26 22047 4.53568 23885 4-I8675 25738 3.88536 27607 3-62224 34 27 22078 4.5294I 239I6 4 I8I37 25769 3.88068 27638 3-6I8I4 33 28 22108 4.523I6 23946 4.17600oo 25800 3-8760I 27670 3-6I405 32 29 22139 4.5i693 23977 4-I7064 2583I 3-87I36 27701 3-60996 31 30 22169 4 5I07I 24008 4-I6530 25862 3-8667I 27732 3-60588 30 31 22200 4.5045I 24039 4-I5997 25893 3-86208 27764 3 6oi8i 29 32 22231 4.49832 24069 4.I5465 25924 3-85745 27795 3-59775 28 33 22261 4-492I5 24100 4-I4934 25955 3-85284 27826 3*59370 27 34 22292 4.48600 24131 4-'4405 25986 3-84824 27858 358966 26 35 22322 4.47986 24i62 4-I3877 260I7 3-84364 27889 3.58562 25 36 22353 4-47374 24193 4-.3350 26048 3-83906 27920 3-58I6o 24 37 22383 4.46764 24223 4-I2825 26079 3.83449 27952 3-57758 23 38 22414 4.46I55 24254 4-I2301 26110 3-82992 27983 3-57357 22 39 22444 4-45548 24285 4-11778 26141 3.82537 28015 3.56957 21 40 22475 4.44942 243i6 4-II256 26172 3.82083 28046 3-56557 20 41 22505 4.44338 24347 4-.o736 26203 3-8I63o 28077 356159 1 42 22536 4.43735 24377 4-Io026 26235 3-8II77 28iog 3-55761 i8 43 22567 4.43134 24408 4-o96q9 26266 3-80726 28140 3 55364 I7 44 22597 4-42534 24439 4-og982 26297 380276 28172 3-.54968 i6 45 22628 4-4I936 24470 4o08666 26328 3-79827 28203 3-54573 i5 46 22658 4-4I340 24501i 408152 26359 3-79378 28234 3-54I79 14 47 22689 4-40745 24532 4-o7639 26390 3.78931 28266 3.53785 13 48 22719 4-40I52 24562 4-07127 26421 3-78485 28297 3.53393 12 49 22750 4.39560 24593 4-066I6 26452 3.78040 28329 3-5300oo I 50 2278I 4.38969 24624 4-06I07 26483 3-77595 28360 3-52609 IO 5I 228II 4.38381 24655 4-o5599 265I5 3-77152 2839I 3-522I9 9 52 22842 4.37793 24686 4-o5092 26546 3-76709 28423 3-51829 8 53 22872 4.37207 24717 40o4586 26577 3.76268 28454 3-5I441 7 54 22903 4.36623 24747 4.o4o8I 26608 3.75828 28486 3-5Io53 6 55 22934 4.36040 24778 4o03578 26639 3.75388 28517 3-50666 5 56 22964 4.35459 24809 4-o3075 26670 3-74950 28549 350279 4 57 22995 4-34879 24840 4.02574 2670I 3-74512 28580 3-49894 3 58 23026 4.34300oo 2487I 4-o2074 26733 3.74075 28612 3-495o9l 2 59 23056 4.33723 24902 4.oI576 26764 3.73640 28643 3-49125 1 60 23087 4.33I48 24933 4.oo078 26795 3732o05 28675 3.48741 0 r Cotang. Tangent. Cotang. ent. Cotang. Tangent. Cotang. | Tangent. I 770 760 750 740 E~~~~~~~~~5 TABLE IIL NATURAL TANGENTS AND COTANGENTS. 77 1 160 1~ 18~ 19~ fangent. Cotang. Tangent. Cotang. Tangent. Cotang. Tangent. Cotang. 0 28675 3-48741 30573 3.27085 32492 3.07768 34433 2'90421 6o 28706 3-48359 30605 3.26745 32524 3-07464 34465 290,0I47 5 2 28738 3-47977 30637 3-26406 32556 3.07I6o 34498 2.89873 58 3 28769 3-47596 30669 3-26067 32588 3-o6857 34530 2-89600 57 4 28800 3-47216 30700 3-25729 3262I 3.o654 34563 2.89327 56 5 28832 3.46837 30732 3*25392 32653 3.06252 34596 2 89055 55 6 28864 3.46458 30764 3*25o05 32685 3.o5950 34628 2.88783 54 7 28895 3*46080 30796 3*247I9 32717 3.o5649 34661 2.88511 53 8 28927 3.45703 30828 3-24383 32749 3-o5349 34693 2.88240 52 9 2895 3*45327 30860 3*24049 32782 3*o5049 34726 2*87970 5i 10 28990 3.4495I 308g9 3-23714 328I4 3o04749 34758 2.87700 50 II 2902I 3.44576 30923 3 2338I 32846 3.o445o 34791 2-87430 49 12 29053 3-44202 30955 3-23048 32878 3.04I52 34824 2.8716I 48 I3 29084 3-43829 30987 3-227I5 32911 30-o3854 34856 2.86892 47 14 29116 3.43456 31019 3-22384 32943 3*o3556 34889 2.86624 46 I5 29147 3*43o84 3io05 3*22053 32975 3/03260 34922 2.86356 45 i6 29179 3*427I3 3io83 3-2I722 33007 30o263 34954 2.86089 44 I7 29210 3*42343 3iii5 3*2I392 3304o 3o02667 34987 2.85822 43 I8 29242 34I1973 3I147 3*20 63 33072 3.02372 35019 2.85555 42 19 29274 3-41604 31178 320734 33104 3-o2077 35052 2.85289 41 20 29305 3*4I236 31210 3-20406 3336 301oI783 35o85 2.85023 4o 21 29337 3-40869 3I242 3-20079 33i69 3O01489 35II7 2.84758 39 22 29368 3-40502 31274 3-I9752 3320I 3-OII96 35i5o 2-84494 38 23 29400 3.40I36 3i306 3.19426 33233 30oogo3 35i83 2*84229 37 24 29432 3.3977I 3I338 3-IgIoo 33266 3oo6II 35216 2.83965 36 25 29463 3.39406 3I370 3*I8775 33298 30oo31 35248 2.83702 35 26 29495 3.39042 3402 3845 3333o 300028 35281 2.83439 34 27 29526 3*38679 3I434 318I27 33363 2.99738 35314 2*83I76 33 2 29558 3.38317 31466 31I7804 33395 2.99447 35346 2*829I4 32 29 29590 3-37953 3I498 3-I748I 33427 2'99I58 35379 2-82653 31 30 29621 3.37594 3i53o 3.17159 33460 2.98868 35412 2-8239I 30 31 29653 3-37234 31562 3-I6838 33492 2-98580 35445 2-82130 29 32 29685 3.36875 31594 3-I6517 33524 2-98292 35477 2.81870 28 33 297I6 3-365i6 3i626 3-I6I97 33557 2-98004 355iO 2.81610 27 34 29748 3.36i58 3i658 3-.5877 33589 2'97717 35543 2-8I350 26 35 29780 3-358oo00 31690g 3-i5558 33621 2.97430 35576 2.81091g 25 36 29811 3.35443 31722 3-I5240 33654 2-97I44 356o8 2.80833 24 37 29843 3.35087 31754 3-I4922 33686 2-96858 3564I 2.80374 23 38 29875 3-34732 3I786 3-I4605 33718 296573 35674 2.803i6 22 39 29906 3.34377 3i8i8 3.14288 33751 2.96288 35707 2-80059 21 40 29938 3.34023 3i850 3-I3972 33783 2.96004 35740 2-79802 20 4I 29970 3.33670 31882 3-i3656 338i6 2.9572I 35772 2-79545 19 42 3ooo0 3.333I7 31914 3I -334i 33848 2.95437 35805 2-79289 I8 43 3oo33 3.32965 3I946 3-13027 3388I 2-95I55 35838 2-79033 I7 44 3oo65 3.32614 31978 3127I3 33913 2-94872 35871 2-78778 i6 45 30097 3.32264 320IO 3.I2400 33945 2-94590 35904 2.78523 i5 46 30128 3-3I914 32042 3-I2087 33978 2-94309 35937 2.78269 I4 47 3oi6o 3.3i565 32074 3.11775 34010 2-94028 35969 2.7SO84 13 48 30I92 3.31216 32106 3-II464 34043 2-93748 36002 2-7776I 12 49 30224 3.30868 32139 3-III53 34075 2.93468 36o35 2.77507 II 50o 30255 3.30521 32171 3I.10842 34io8 2-93I89 36068 2-77254 IO 5I 30287 3-30I74 32203 3-I0532 34I40 2 929O10 36IOI 2-77002 9 52 30319 3.29829 32235 3I0O223 34173 2-92632 36I34 276750 53 3o35i 3.29483 32267 3-09914 34205 2-92354 36I67 2.76498 7 54 30382 3|29139 32299 3-og606 34238 2.92076 369gg 2.76247 6 55 304I4 3.28 795 3233 3.09298 34270 2.91799 36232 2-75996 5 56 30446 3-28452 32363 3-0899I 34303 2-9I523 36265 2-73746 4 30478 13.289og 32396 3:o8685 34335 2-91246 36298 2-75496 3 30509 i3-27767 32428 3-o8379 34368 2-9097I 3633i 2-7246 2 59 30541 3.27426 32460 3.08073 34400 2-90696 36364 2.74997 I 60 30573 3.27085 32492 3-07768 34433 2-9042I 36397 2-74748 0 Cotang. Tangent. Cotang. Tangent. Cotang. Tangent. Cotang. Tangeiit. 730 12 i10 70 78 NATURAL TANGENTS AND COTANGENTS. TABLE III. 200 2120 220 230 l Tangent. Cotang. Taingent.J Cotang. Tangent. Cotang. Tangent. Cotang. o 36397 2-74748 38386 260509 40403 2.47509 42447 2.35585 60 36430 2 74499 38420 2 60283 40436 2 42482 2 35395 52 36463 2-7425I 1 3 8454 3 260oo57 40470 2.47095 42516 2.35205 5 3 36496 2 7400~4f 38487 2 59831 40504 2 46888 42551 2 350I5 57 4 36529 2 73756 I 38520 2-59606 40538 2 -46682 42585 2 34825 56 5 36562 2-73509 38553 2.5938I 40572 2 46476 42619 2.34636 55 6 36595 2.73263 38587 2.59156 40606 2.46270 42654 2.34447 54 7 36628 2.73017 38620 2.58932 40640 2 46065 42688 2.34258 53 8 3666i 2-7277I 38654 2.58708I 40674 2.45860 42722 2.34069 52 9 36694 2 * 72526 38687 2 58484 40707 2.45655 42757 2.3388I 5I Io 36727 2-7228I 38721I 2'58261 40741 2-4545I 4279I 2 33693 50 ii 36760 2-72036 38754 2.58038 40775 2.45246 42826 | 2 33505 4z 1 2 36793 2 71792 38787 2.57851 40809 2 45043 42860 2.33317 48 I3 36826 2.71548 38821 2.57593 40843 2-44839 42894 2-33I30 47 I4 36859 2.7I305 38854 2 57371 40877 2 44636 42929 2.32943 46 5 36892 2. 7I 62 38888 257I50 40911 2 44433 42963 2.32756 45 I6 36925 2 70819 3892I 2.56928 40945 2 44230 42998 2 32570 44 17 36958 2'70577 38955 2 56707 40979 2.44027 43032 232383 43 i8 36991 2.70335 38988 2 56487 41013 2.43825 43067 2397 42 19 37024 270094 39022 2.56266 41047 2.43623 43101 2.32012 41 20 37057 2.69853 39o55 2.56046 4108i 2.43422 43136 2-31826 40 2I 37090 2'69612 39089 2.55827 41115 2-43220 43170 2.3I64I 39 22 37124 269371 39122 2.55608 41149 2.430I9 43205 2-31456 38 23 37157 2.6913I 39156 2.55389 41183 2.42819| 43239 2.3127I 37 24 37190 2-68892 39190 2.55170 412I7 2-42618 1 43274 2.31086 36 25 37223 2.68653 39223 2.54952 41251 2-424I8 43308 2.30902 35 26 37256 2.68414 39257 2.54734 41285 2-42218 43343 2.30718 34 27 37289 2.68175 39290 254516 413I9 2.42019 43378 2.30534 33 28 37322 2.67937 39324 2-54299 41353 2.418I9 43412 2.3035I 32 29 37355 2-67700 39357 2.54082 41387 2-4I620 43447 2.30I67 3i 30 37388 2-67462 39391 2.53865 41421 2-4142I 43481 2-29984 3o 3i 37422 2.67225 39425 2.53648 4I455 2.41223 435i6 2-29801 29 32 37455 2.66989 39458 2.53432 /41490 2-41025 43550 2-296I9 28 33 37488 2.66-752 39492 2.53217 4I524 2.40827 43585 2.29437 27 34 37521 2.66516 39526 2.-5300oo 41558 2.-40629 43620 229254 26 35 37554 2-6628i 39559 2.52786 /4I592 240432 43654 229073 25 36 37588 2.66046 39593 2.52571 41626 2-40235 43689 2-28891 24 37 37621 2-65811 39626 2.52357 41660 2.40038 43724 2.28710 23 38 37654 2.65576 39660 2.52I42 4694 2.39841 43758 228528 22 39 37687 2.65342 39694 2.51929 41728 2.39645 43793 2.28348 21 40 37720 2-65109 39727 2.517I5 41763 239449 43828 228I67 20 4I 37754 2.64875 3976I 2-5I502 41797 2.39253 43862 2-27987 I9 42 37787 2'64642 39795 2-5I289 4i83I 2-39o58 143897 2.278o6 IS 43 37820 2.64410 39829 25I1076 4I865 2.38862 43932 2-27626 17 44 37853 2.64177 39862 2.5o864 41899 2.38668 43966 2.27447 6 45 37887 2.63945 39896 2.50652 4I933 2.38473 4400o 2-27267 I5 46 37920 2.637I4 39930 2.50440 41968 2.38279 44o36 2.27088 14 47 37953 2.63483 39963 2.50229 42002 2.38084 44071 2.26909 13 48 37986 2.63252 39997 2.-500I8 42036 2.3789I 441o5 2-26730 12 49 j 38020 2*6302I 4003I 2.49807 42070 2.37697 44140 2-26552 II 50o 38053 2.62791 40065 249597 42105 2.37504 44175 2.26374 1 5I 38086 2.6256I 40098 2-49386 42139 2-37311 442I0 226I96 9 52 38120 2.62332 40132 2-49177 42173 2.37II8 44244 2.26018 8 53 38i53 2-62103 40I66 2-48967 42207 2.36925 44279 2-25840 7 54 38i86 2.6I874 40200 2.48758 42242 2.36733 443I4 2.25663 6 55 38220 2-6I646 40234 2.48549 42276 2.36541 44349 2.25486 5 56 38253 2.614I8 40267 2-48340 423IO 2.36349 44384 2-25309 4 57 38286 2.6190| 4o3oi 2-48I32 42345 2.36158 44418 2.2532 3 58 38320 2.60963 40335 2.47924 42379 2.35967 44453 12.24956 2 59 38353 2.60736 40369 2'47716 42413 2.35776 44488 2.24780 60o 38386 2.6G509 40403 2947509 42447 2.35585 44523 2o24604 0 Cotang. Tangent. Cotang. Tangent. Cotang. Tangent. Cotang. Tangent. I -680'7_ 660 690 68 670 66~ TABLE IIL NATURAL TANGENTS AND COTANGENTS. 79 24~ 250 260 270 Tangent. Cotang. Tangent. Cotang. Tangent. Cotang. Tangent. Cotang. o 44523 2.24604 46631 2.-I445I 48773 2.o5030 50953 I 9626I 6o I 44558 2.24428 46666 2.I4288 48809 2o04879 50989 I 96120 59 2 44593 2 24252 46702 2 I4125 48840 2.04728 51026 1.95979 58 3 44627 2.24077 46737 2.3963 4888I 2.04577 5io63 1.95838 57 4 44662 2 23902 46772 2.380o 48917 2. 04426 5io99 I.95698 56 5 44697 2.23727 46808 2-I3639 48953 2.04276 51136 I.95557 55 6 44732 2 23553 46843 2 - I3477 48989 2.04125 51173 1.95417 54 7 44767 2*23378 46879 2.I3316 49026 2.03975 51209 1.95277 53 8 44802 2.23204 46914 2.-3154 49062 2.03825 51246 I.95I37 52 9 44837 2.23030 46950 2.12993 49098 2*03675 51283 I 94997 51 Io 44872 2.22857 46985 2.12832 49134 2.o3526 51319 I 94858 5o II 44907 2.22683 47021 2I2671 49I 70 2.03376 5I356 I 94718 49 I2 44942 2.225IO 47056 2.I25II 49206 2.03227 5I393 I*94579 48 I3 44977 2 22337 47092 2.12350 49242 20o3078 5I43o0 194440 47 14 45012 2.22I64 47128 2.I2190 49278 2.02929 51467 1'9430I 46 I5 45047 2.21992 47163 2-I2030 49315 2.02780 51503 I-94I62 45 i6 45082 2-.28I9 47199 2.-I871 49351 2.0263I 5I540 1'94023 44 17 45117 2.2I647 47234 2-1171I 49387 2.02483 51577 I.93885 43 I8 45152 2-2I475 47270 21II552 49423 2.02335 5I614 I'93746 42 I9 45187 2.2I304 47305 2.-i392 49450 2.02187 5165I 193608 41 20 45222 2-21I32 47341 2.11233 49495 2.-203g 5I688 -.93470 40 21 45257 2.2096I 47377 2-II075 49532 2-OI89g 51724 I-93332 39 22 45292 2.20790 474I2 2-I0916 49568 2.-OI743 5I761I I93195 38 23 45327 2.20619 47448 2'I0758 49604 2OI596 5I798 I'93057 37 24 45362 2.20449 47483 2.I0600 49640 2.OI449 5i835 I 92920 36 25 45397 2.20278 47519 2.10442 49677 2.OI302 51872 1.92782 35 26 45432 2-20108 47555 2.I0284 49713 2-0155 5Igog I.92645 34 27 45467 2-I9938 47590 2.IOI26 49749 2.OI008 51946 1.92508 33 28 45502 2.19769 47626 2.09969 49786 2.00862 51983 -.9237I 32 29 45537 2-I9599 47662 2.098II 49822 2.007I5 52020 1-92235 3I 30 45573 2.19430 47698 2.09654 49858 2.oo569 52057 1.92098 30 3I 45608 2-I926I 47733 2-o09498 49894 2.00oo423 52094 I'91962 29 32 45643 2.-9092 47769 2-09341 4993I 2-00277 52131 I 9I826 28 33 45678 2I18923 47805 2-09184 49967 2-oo03I 52168 Ig91690 27 34 45713 2-I8755 47840 2.og9028 50004.99986 52205 I.91554 26 35 45748 2-I8587 47876 2.-o8872 50040.99841 52242 I'9148 25 36 45784 2-I84Ig 47912 2.087I6 50076 1.99695 52279 1.91282 24 37 45819 2.18251 47948 2.o8560 50oi3 I.99550 523I6 1.91147 23 38 45854 2 -8084 47984 2.o8405 5o014g i994o6 52353 19gI012 22 39 45889 2-I79I6 48oi9 2.o8250 5oi85 1.9926I 52390 -90876 2I 40 45924 2-17749 48o55 2o08094 50222 1.99II6 52427 I.90741 20 41 45960 2.17582 48091 2.o79g3 50258 |I98972 52464 I 90607 19 42 45995 2.I74I6 48I27 2.07785 50295 I 98828 5250I I 90472 I8 43 4603o 2.I7249 48I63 2.07630 5o331 I.98684 52538 1.90337 I7 44 46065 2.I7083 48198 2'07476 5o368 1.98540 52575 1.90203 i6 45 46IOI 2.16917 48234 2.0732I 50404 I.98396 526I3 1.90069 I5 46 46136 2-.6751 48270 2.07I67 5o44 I.98253 52650 1.89935 I4 47 46I71 2.I6585 48306 2.070I4 50477 1.98110 52687 |I8980I 13 48 46206 2.-6420 48342 2.o6860 50514 I.97966 52724 I89667 12 49 46242 2.-6255 48378 2.-o6706 50550.97823 52761 i-89533 ii 50 46277 2-I6090 48414 2.06553 5o587 1.97680 52798 1.89400 10 5I 46312 2.15925 48450 2.o6400 50623 1-97538 52836.-89266 52 46348 2.-5760 48486 2.06247 50660 1-97395 52873 I-89I33 53 46383 2.-5596 4852I 2.06094 50696 I'97253 5291o I'89000 7 54 464I8 2-I5432 48557 2.05942 50733 I-9711I 52947 I-88867 6 55 46454 2.-5268 48593 2.05790 50769 I'96969 52984 I.88734 5 56 46489 2-5Io04 48629 2.05637 5o806 I 96827 53022 I.88602 4 57 46525 2-I4940 48665 2.05485 50843 I-96685 53059 -.88469 3 58 46560 2-I4777 48701 2.05333 50879 1.96544 53096 I.88337 2 59 46595 2-I4614 48737 2.05182 50o96 I.96402 53I34 1.882o5 i 60 46631 2-I445I 48773 2.0o5030 50953 1.9626I 53I7I I-88073 o Cotan Tangent. Cotang. Tangent. Cotang. Tangent. Cotang. Tangent. 65 64 630 620 I 80 NATURAL TANGENTS AND COTANGENTS. TABLE IIL 280 290 300 310 Tangent. Cotang. Tangent. Cotang. Tangent. Cotang. Tangent. Cotang. o 53171 I 88073 5543i i.80405 57735 I 73205 60086 I 66428 60 I 53208 1.87941 55469 1.8028I 57774 1.73089 60126 1.663i8 5q 2 53246 1.87809 55507 i.80o58 57813 1.72973 60o65 I.66209 58 3 53283 1.87677 55545 1.80034 5785I 1.72857 60205 1.66099 57 4 53320 1.87546 55583 1.799II 57890 1.7274I 60245 1.65990 56 5 53358 1.874I5 55621 1.79788 57929 I'72625 60284 1I6588I 55 6 53395 I187283 55659 1I79665 57968 I172509 60324 1.65772 54 7 53432 1.87152 55697 179542 58007 1.72393 60364 I.65663 53 8 53470 1'87021 55736 I1794I9 58046 1.72278 604o3 1.65554 52 9 53507 I'8689I 55774 1.79296 58085 1.72163 60443 1.65445 5I io 53545 1.86760 55812 I.79174 58124 1.72047 60483 1.65337 50 11 53582 I.86630 55850 1.79051 58162 I 7I932 60522 1.65228 49 12 53620 1'86499 55888 1.78929 58201 1-71I17 60562 I 65I20 48 I3 53657 I*86369 55926 1.788o07 58240 1.71702 60602 I.650oi 47 I4 53694 I'86239 55964 1.78685 58279 1-7I588 60642 1.64903 46 I5 53732 I 86o09 56oo3 1.78563 583i8 1.71473 6o68I 1.64795 45 I6 53769 1.85979 5604i I*7844I 58357 1'71358 60721 1.64687 44 I7 53807 I 8585o 56079 1'783I1 58396 I'71244 60761 I'64579 43 I8 53844 1'85720 56II7 I178198 58435 1.71129 6080I 164471 42 19 53882 1.8559I 56I56 1.78077 58474 I.71015 6084I 1.64363 4I 20 53920 1.85462 56194 1.77955 585I3 I 7090I 6088I I.64256 40 2I 53957 I.85333 56232 I 77834 58552 I 70787 60921 1.64148 39 22 53995 1.85204 56270 1'77713 5859I 1'70673 60960 1.6404I 38 23 54032 1.85075 56309 I177592 5863 170560 6ooo000 1.63934 37 24 54070 1'84946 56347 I 7747I 58670 I170446 61040 I 63826 36 25 54107 1.84818 56385 I'7735I 58709 1.70332 6io80 1.637I9 35 26 54145 1.84689 56424 1.77230 58748 I 70219 61120 i 63612 34 27 54i83 I.8456I 56462 1I771io 58787 1.701o6 6Ii6o0 I635o5 33 28 54220 I.84433 56500 1.76990 58826 1.69992 61200 I163398 32 29 54258 I.843o5 56539 1.76869 58865 1.69879 61240 I 63292 3i 30 54296 I.84177 56577 I176749 58904 1.69766 61280 i 63 85 30 31 54333 1.84049 566i6 1.76630 58944 1.69653 6I320 1.63079 29 32 54371 I.83922 56654 I.765io 58983 1.6954I 6I36o I.62972 28 33 54409 1.83794 56693.76390o 59022 1.69428 61400 1.62866 27 34 54446 1.83667 56731 I 76271 5906I 1.69316 6I440 1.62760 26 35 54484 I.8354o 56769 1.7615i 59o101 1.69203 61480 1.62654 25 36 54522 I.834I3 56808 1.76032 59140 1.6909I 61520 I162548 24 37 54560 I.83286 56846 1759I3 59I79 1.68979 6I56i 162442 23 38 54597 I.83I59 56885 1 75794 592I8 i.68866 6i6oi I.62336 22 39 54635 I.83o33 56923 I 75675 59258 1.68754 6i64I 1.62230 21 40 54673 1.82906 56962 I 75556 59297 i 68643 6I681 1.62125 20 41 5471 I I82780 57000 1 75437 59336 i 6853i 61721 16209 I9 42 54748 1.82654 57039 1 753I9 59376 I 684i 61676I I6I914 8 43 54786 1.82528 57078 I 75200 594i5 i 68308 6i8oi 1.618o8 17 44 54824 1.82402 57116 1.75082 59454 I.68I96 61842 I.61703 i6 45 54862 1'82276 57I55 I 74964 59494 I 68085 61882 1.6I598 I5 46 54900 I.8250o 57i93 I'74846 59533 i 67974 6I922 1'6I493 14 47 54938 1.82025 57232 1.74728 59573 1.67863 61962 I.6I388 13 48 54975 1.81899 57271 1.746io 59612 I.67752 62003 I.61283 12 49 55oi3 1.8I774 57309 1.74492 5965I I.6764I 62043 I.6II79 1I 50 5505I 1.81649 57348 1.74375 5969I 1.67530 62083 1.6I074 10 5i 55089 1.8I524 57386 1.74257 59730 1.67419 62124 I*60970 9 52 55127 I-8I399 57425 I-74140 59770 1.67309 62164 I-6o865 8 53 55i65 1.81274 57464 1.74022 59809 1.67198 62204 I-6076I 7 54 55203 I-8II5o 57503 1-73905 59849 -.67088 62245 1.60657 6 55 55241 I-8I025 5754I I-73788 59888 1.66978 62285 I-6o553 5 56 55279 -8o901 57580 I-7367I 59928 1.66867 62325 1.60449 4 57 5537 1-80777 576I9 1.73555 59967 I-66757 62366 I-60345 3 58 55355 I-8o653 57657 I-73438 60007 I-66647 62406 1.6024I 2 59 55393 I.80529 57696 I-73321 60046 I-66538 62446 1-60137 I 60 5543u I8o0403 57735 1.73205 60086 1-66428 62487 I-60033j o Cotang. Tanuent. Cotang. Tangent. Cotan/. Tangent. Cotang. Tangent. 610 600 590 i58 _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 8 _ _ _ TABLE III. NATURAL TANGENTS AND COTANGENTS. 81 320 330 340 350 Tangent. Cotang. Tangent.I Cotang. Tangent. Cotang. Tanent. Cotan. o 62487 I 6oo33 64941 i 53986 67451 1 48256 70021 1.428I5 6o I 62527 I 59930 64982 I.53888 67493 1.48I63 70064 I.42726 5 2 62568 1.59826 65023 I'5379I 67536 1I48070 70I07 1'42638 58 3 62608 I.59723 65065 1.53693 67578 I*47977 70I5I 142550 57 4 62649 1.59620 65io6 1.53595 67620 1.47885 70194 1*42462 56 5 62689 1.595I17 65148 1.-53497 67663 1.47792 70238 1.42374 55 6 62730 1.594I4 65189 I.534oo 67705 1.47699 7028I 142286 54 7 62770 1.593II 65231 1.53302 67748 1.47607 70325 1.42I98 53 8 62811 159208 65272 1.53205 67790 1.47514 70368 1.42IIO 52 9 62852 1.59I05 653I4 1.53107 67,832 1.47422 704I2 1.42022 51 10 62892 1.59002 65355 I.530Io 67875 1.47330 70455 1.41934 50 ii 62933 58900oo 65397 I'52913 67917 1.47238 70499 1.41847 49 12 62973 I'58797 65438 1.52816 67960 I147146 70542 1'41759 48 t3 63oi4 I.58695 65480 1.527I9 68002 1.47053 70586 1.4I672 47 I4 63055 1.58593 6552I 1.52622 68045 1.46962 70629 1-4I584 46 15 63095 I'58490o 65563 1.52525 68088 1.46870 70673 1'41497 45 I6 63i36 I.58388 65604 1.52429 68i30 1.46778 707I7 1.41409 44 I7 63177 1.58286 65646 I 52332 68173 I.46686 70760 1.4I322 43 i8 63217 I 58I84 65688 1.52235 68215 1.46595 70804 1I41235 42 19 63258 I.58o83 65729 1.52139 68258 I.46503 70848 14II48 41 20 63299 1I5798I 65771 1.52043 6830I I.464II 7089I 1.4IO6I 40 2I 63340 I 57879 658I3 I.51946 68343 I146320 70935 I 40974 3 22 63380 1.57778 65854 I 5I850 68386 1.46229 70979 140887 38 23 6342I 1.57676 65896 1.5I754 68429 I 46I37 7I023 1.40800 37 24 63462 1.57575 65938 I.5I658 68471 I146046 7IO66 I.40714 36 25 63503 1.57474 65980 1.5I562 68514 1.45955 7IIIO I.40627 35 26 63544 1'57372 66021 I.5I466 98557 1.45864 7II54 I.40540 34 27 63584 1.57271 66063 1.5I370 68600 1.45773 71198 I.40454 33 28 63625 I-57I70 66I05 1.51275 68642 1.45682 71242 I.40367 32 29 63666 I*57069 66147 1.51179 68685 1.45592 7I285 1 4028I 31 30 63707 1.56969 66189 I.5Io84 68728 1.4550I 71329 1.40I95 30 31 63748 I.56868 66230 I.50988 68771 1.454I0 71373 1I40I09 29 32 63789 1.56767 66272 I'50893 68814 I-45320 71417 I'40022 28 33 63830 1.56667 66314 1.50797 68857 1.45229 71461 I139936 27 34 6387I I.56566 66356 I150702 68900 I*45I39 7I505 I*39850 26 35 63912 I.56466 66398 1.50607 68942 1I45049 71549 1.39764 25 36 63953 I.56366 66440 1.50512 68985 1.44958 7I593 I 39679 24 37 63994 1.56265 66482 1.50417 69028 1.44868 7 637 I 39593 23 38 64035 1.56I65 66524 1.50322 6907I 1.44778 71681 139507 22 39 64076 I.56065 66566 1.50228 69114 II44688 71725 I*39421 21 40 64117 1'55966 66608 I.5oi33 69157 I.44598 71769 1.39336 20 4I 64158 i.55866 66650 500oo38 69200 I.44508 71813 1-39250 19 42 64199 1.55766 66692 I'49944 69243 I 444I8 7i857 i 39 65 I8 43 64240 I 55666 66734 1'49849 69286 1.44329 7190I 1.39079 17 44 6428I I.55567 66776 1.49755 69329 1.44239 71946 1.38994 I6 45 64322 I.55467 668i8 1.4966I 69372 1'44149 71990 1.38909og 46 64363 I.55368 66860 I.49566 69416 1.44060 72034 1.38824 I4 47 64404 I'55269 66902 I 49472 69459 1I43970 72078 1.38738 13 48 64446 1.55I7o 66944 1.49378 69502 1.43881 72I22 i.38653 12 49 64487 I'5507I 66986 1I49284 69545 1*43792 72166 I*38568 ii 50 64528 1.54972 67028 I'49190 69588 1.43703 72211 I.38484 10 5i 64569 I-54873 6707I 1I49097 6963I 1.436I4 72255 I*38399 52 646I0 I'54774 6713 1.49003 69675 1.43525 72299 I.383I4 53 64652 I*54675 67155 I*4gqo0 69718 I.43436 72344 I'3822q 7 54 64693 I.54576 67I97 1'48816 69761 I.43347 72388 I.3845 6 55 64734 I.54478 67239 1.48722 69804 1.43258 72432 I.38o6o 5 56 64775 I.54379 67282 1.48629 69847 1I43I69 72477 I'37976 4 57 64817 1.5428I 67324 1.48536 6989 I.43o80 7252I I.3789I 3 58 64858 I*54i83 67366 1*48442 69934 1.42992 72565 1.37807 2 59 64899 I.54o85 67409 1.48349 69977 1.42903 72610 1.37722 I 60 6494I I'53986 6745I 1.48256 70021 1.42815 72654 1.37638 o Cotang. Tangent. Cotang. Tangent. Cotang. Tangent. Cotang. Tangent. 570 56~ 550 540 4* 82 NATURAL TANGENTS AND COTANGENTS. TABLE IIL t 360 370 380 390 Tangent. Cotang. Tangent. Cotang. Tangent. Cotang. Tangent. Cotang. 0 72654 I.37638 75355 i*32704 78129 1.27994 80978 I.23490 o 6o, I 72699 1.37554 7540I I-32624 78173 1-27917 81027 1.234161 5 2 72743 I*37470 75447 I132544 78222 1'27841 81075 1 23343 58 3 72788 1 37386 75492 I*32464 7826 I127764 81123 I 23270 57 4 72832 I*37302 75538 i-32384 78316 I 1.27688 8I I23I96 56 5 72877 I-37218 75584 1.32304 78363 I-276II 81220 1.23123 55 6 7292I I*37I34 75629 I'32224 78410 1127535 81268 I 2300' 54 7 72966 1-37050 75675 I-32144 78457 1-27458 8i3/ 6 1-22977 53 8 73oio010 1.36967 75721 1.32064 78504 I*27382 81364 22904 52 9 73055 1-36883 75767 I-31984 78551 1-27306 81413 1-228311 5I 10 73100 I*36800 75812 I'31904 78598 1'27230 81461 1.22758 50 11 73144 I136716 75858 I.31825 78645 1.2753 81510io 122685 49 12 73189 i 36633 75904 I 31745 78692 1.27077 81558 i.226I2 48 13 73234 1-36549 75950 1-31666 78739 I,27001 81606 1.22539 47 14 73278 I-36466 75996 I-31586 78786 I-26925 8i655 1-22467 46 15 73323 I36383 76042 1'31507 78834 1 26849 81703 1'22394 45 I6 73368 I*36300 76088 I.3I427 78881 1.26774 81752 1.22321 44 17 73413 1.36217 76134 I.3I348 I 78928 1.26698 8i800 122249 43 i8 73457 I-36I33 76180 1.31269 78975 1.26622 8I849 I-22176 42 19 73502 I-36o05 76226 1.3II90 79022 1.26546 81898 1.22104 41 20 73547 I.35968 76272 -31110I 79070 1-26471 81946 1.22031 40 21 73592 I-35g85 76318 I-3io3i 79117 1.26395 81995 1.2I959 3 22 73637 1I35802 76364 1.30952 79164 I1263I9 82044 1*2I886 38 23 73681 1.357I9 76410 1.30873 79212 126244 82092 1-2I814 37 24 73726 1-35637 76456 I-30795 79259 1-26169 82141 1I21742 36 25 73771 I 35554 76502 I.30716 79306 1*26093 82190 1216570 35 26 73816 1 35472 76548 1.30637 79354 1.260I8 82238 1-21598 34 2 73861 1'35389 76594 i 3o558 79401 1 25943 82287 1.21526 33 20 73906 -35307 76640 I-30480 79449 I.25 67 82336 1.21454 32 29 7395I 1-35224 76686 1.3040, 79496 I-25792 82385 1-2I382 3 30 73996 I 35142 76733 I-30323 79544 1.25717 82434 1 I-21310 3o 3I 7404I I-35o6o 76779 1.30244 7959I I-25642 82483 -21238 29 32 74086 I 34978 76825 I-30o66 79639 1i25567 82531 I21166 28 33 7413I.I34896 76871 1.30087 79686 1-25492 82580 1-21094 27 34 74176 i.34814 76918 I-30009 79734 1 254I7 82629 I-2I023 26 35 74221 I34732 76964 1.29931 79781 I25343 82678 I.2095I 25 36 74267 I-34650 77010 1.29853 79829 125268 82727 i.208791 24 37 743I2 I-34568 77057 1.29775 79877 1-25193 82776 1-20808 23 38 74357 1.34487 77103 I-29696 79924 I-25II8 82825 1.20736 22 39 74402 I34405 77149 1.29618 79972 I 25044 82874 1.20665 21 40 74447 1.34323 77196 1.29541 80020 1I-24969 82923 1.20593 20 41 74492 I-34242 77242 1.29463 80067 1.24895 82972 120522 1I 42 74538 I-34i16o 77289 1.29385 8015 1-24820 83022 1-2045I I8 43 74583 1-34~079 77335 |-29307 28063 |1-24746 83071 1.20379 17 44 74628.33998 77382 129229 80211 1.24672 83120 1I-2308 i6 45 74674 I-3396 77428 1.29152 80258 1-24597 83169 1-20237 15 46 74719 i 33835 77475 I-29074 80306 1'24523 83218 1-20I66 14 47 74764 1.33754 77521 I1.28997 80354 1-24449 83268 1.20095 13 4 74810 I 33673 77568 1.2899 80402 JI24375 83317 1.20024 12 49 74855 1.33592 77615 I28842 8o45o I-24301 83366 ii9953 11 50 74900 1 3351 7766I 1.28764 80498 1.24227 834I5 1-19882 10 51 74946.33430 77708 1.28687 80546 1.24I53 83465 I-I98I1 9 52 74991 I 33349 77754 1.28610 8o0594 1-24079 835I4 119740 8 53 75037 I33268 77801 I 1.28533 80642 I.24005 83564 1I9669 7 54 75082 1.33I87 77848 1.28456 80690 I.2393i 836i3 I'-1999 6 55 75128 I33107 77895 1.28379 80738 1.23858 83662 1-9528 5 56 75173 1-33026 77941 1-28302 80786 1.23784 83712 I1- 9457| 4 5 75219 32946 77988 1.28225 80834 1.237Io 83761 i 193871 3 5 75264 I3265 78035 1i.2814811 80882 1.23637 838ii i 1I9316 2 59 7531o 1.32785 78082 1-28071 80930 1.23563 8386o I-19246 I 60 75355 1I32704 78129 1.27994 80978 11.23490 8391o i1917i1 0 Cotang. Tangent, Cotang. Tangent. Cotang. Tangent. Cotag. Tangent. I 530 520 1 510 00 TABLE II1. NATURAL TANGENTS AND COTANGENTS. 83 400 410 420 430 Tangent. Cotang. Tangent. Cotang. Tangent. Cotang. Tangent. Cotang. o 83910 I-I9I75 86929 I-I5037 90040 I-IIo6I 93252 Io07237 6o I 83960 I.I9Io5 86980 1II4969 900oo93 Io996 93306 Io7174 59 2 84009 I 1.1935 87031 II4902 90I46 119 o3I 93360 Io07II2 58 3 84059 118964 87082 i.i*4834 90OI9 1.IO867 934i5 I'07049 57 4 8410 I.I8894 87I33.II4767 9025I I.I0802 93469 1.06987 56 5 84158 I.I8824 87184 1.14699 90304 IO10737 93524 I.06925 55 6 84208 I.I8754 87236 1.I4632 90357.IIO672 93578 I.O6862 54 7 84258 I-I8684 87287 I'14565 904IO 1.10o607 93633 I.o68oo00 53 8 84307 I I8614 87338 I I4498 90463 1I0o543 93688 I.06738 52 9 84357 I I8544 87389 I I443o 90516 IO10478 93742 I.O6676 51 Io 84407 I 18474 8744I I 14363 90569 1I10414 93797 i*o6603 50 ii 84457 I I8404 87492 I I4296 90621 Io10349 93852 I*o655I 49 12 84507 I I8334 87543 I1I4229 90674 1I10285 93906 10o6489 48 I3 84556 I I8264 87595 I I4I62 90727 I1I0220 9396I 10o6427 47 I4 84606 I I8194 87646 I I4095 9078I IxIOI56 940I6 I.o6365 46 I5 84656 I I8125 87698 I I4028 90834 1-10I09 94071 i.o6303 45 I6 84706 I*i8055 87749 1.1396I 90887 I*I0027 94125 I.0624I 44 17 84756 I 17986 87801 I I3894 90940 I*o9963 94180 1 06179 43 I8 84806 I I7916 87852 I I3828 90993 I0o9899 94235 I.O617 42 19 84856 1.17846 87904 I'I376I 91046 Io09834 94290 I o6056 41 20 84906 I'17777 87955.I13694 91099 1'09770 94345 I0o5994 40 21 8'4956 I.17708 88007 II13627 91153 1.09706 Q4400 I.O5932 3 22 85006 I*I7638 88059.I*356i 91206 Io09642 94455 I.o5870 38 23 85057 II.17569 88110 I*I3494 91259 Io09578 94510 I1o5809 37 24 85107 I 1.175 88I62 I'13428 91313 1'O95I4 94565 I.057471 36 25 85I57 II17430 88214 I*I336I 91366 109og450 94620 I.o5685 35 26 85207 I-I736I 88265 1.3295 91419 1.09386 94676 1.05624 34 27 85257 II17292 883I7 I 13228 91473 1I09322 9473I I*o5562 33 28 85307 I1.17223 88369 I 13162 91526 I'09258 94786 i.o55oi 32 29 85358 I.I7I54 88421 I.3096 91580 1.09g95 94841 i.o5439 3I 30 85408 1.I7085 88473 I I3029 91633 1.09I3I 94896 1.05378 30 31 85458 1.17016 88524 1.12963 91687 Ioo9067 94952 1.05317 29 32 85509 1.I6947 88576 I I2897 91740 I 0ooo3 95007 1.05255 28 33 85559 I.I6878 88628 I 283 I 91794 I 0 940 95062 Io05I94 27 34 85609 1.16809 88680 I I2765 91847 I 08876 95118 I.05I33 26 35 85660 1.1674I 88732 1*I2699 91901 I 088I3 95173 1I05072 25 36 85710 I.16672 88784 I 12633 91955 1'08749 95229 Io50oIO 24 37 8576I I.i6603 88836 I I2567 92008 I o8686 95284 I104949 23 38 858ii I-16535 88888 I-12501 92062 I-08622 95340 1.04888 22 39 85862 II6466 88940 1-I2435 92116 I-08559 95395 I-04827 21 40 85912 1.16398 88992 |112369 92I70 I o8496 95451 104766 20 4I 85963 I1I6329 89045 I 12303 92223 I 08432 95506 1I04705 I9 42 86014 I.I626I 89097 I I2238 92277 I 08369 95562 1'04644 18 43 86064 1.16192 89149 I I2172 02331 I.o8306 95618 1.04583 17 44 86ii5 1.16I24 89201 I I2I06 92385 1 08243 95673 Io04522 I6' 45 86i66 I.I6o56 89253 I1-2041 92439 I-08179 95729 1.o446I 15 46 86216 1.I5987 89306 I-II975 92493 Io08II6 95785 1.04401 I4 47 86267 1I15919 89358 I I-Ii9 92547 I-o8o53 9584I I.o0434o I3 48 863I8 I.I585I 89410 I I 44 92601 1o07990 95897 1.04279 12 49 86368 i.i5783 89463 I-11778 92655 I-07927 95952 I-04218 II 50 86419 I*I57i5 8951I I*17I3 927-0 107864 96008 Io04I58 I 5I 86470 I-I5647 89567 1.11648 92763 1-0780I 96064 I-04097 52 8652I i.i5579 89620 III582 92817 1-07738 96120 i 04036 53 86572 i-I55II 89672 I115I7 92872 1.07676 96176 I-03976 7 54 86623 I-I5443 89725 11II452 92926 I-076I3 96232 -0395 6 55 86674 i.-5375 89777 |.-I387 92980 I-07550 96288 i0o3855 5 56 86725 i.i5308 89830 i II32I 93034 I'07487 96344 I 03794 4 57 86776 I-I5240 89883 I-II256 93088 I-07425 96400 I-03734 3 58 86827 I-I5I72 89935 1.III9I 93143 1.-07362 96457 I1o3674 2 59 86878 I1I5i04 89988 I 1II26 93I97 1.07299 965i3 i.o36i3 I 60 86929 1.15037 90040 I iI06I 93252 I-07237 96569 I-03553 0 Cotang. Tangent. Cotang. Tangent. Cotang. i Tangent. Cotang. Tangent. 490 480 4~0 - 460 84 NATURAL TANGENTS AND COTANGENTS. TABLE IIL 440 I 440 Tangent. Cotang. Tangent. Cotang. o 96569 I.o3553 6o 3I 98327 1.01702 29 I 96625 I 03493 32 98384 I.OI642 28 2 9668I I0o3433 5 33 98441 I.oi583 27 3 96738 1.o3372 57 34 98499 I.oI524 26 4 96794 I.o3312 56 35 98556 I.o1465 25 5 96850 1.o3252 55 36 98613 I.oi406 24 6 96907 1.o3192 54 37 9867i I.oi347 23 7 96963 1.03132 53 38 98728 1.01288 22 8 97020 I.o3072 52 39 98786 1.01229 21 9 97076 1.o30o2 5i 40 98843 I oI17o 20 Io 97133 1.02952 50 41 98901 IOI12 19 II 97189 I.02892 49 42 98958 I oio53 i8 12 97246 1.02832 48 43 99016 oo00994 17 13 97302 1.02772 47 44 99073 I.oo935 i6 14 97359 1.-027I3 46 45 9913I I-00876 15 15 974I6 1o02653 45 46 99189 I oo8r8 14 I6 97472 1.o2593 44 47 99247 00oo759 13 I7 97529 I 02533 43 48 99304 1.00oo701 12 I8 97586 1.02474 42 49 99362 I00642 II I9 97643 1o02414 41 50 99420 I.oo583 Io 20 97700 I-02355 40 51 99478 1.oo00525 9 21 97756 1.02295 39 52 99536 1.00467 22 97813 I.02236 38 53 99594 I1oo408 7 23 97870 I.02I76 37 54 99652 i oo35o 6 24 97927 1.02Ii7 36 55 99710 I.oo00291 5 25 97984 1.02057 35 56 99768 1-00233 4 26 98041 1O01998 34 57 99826 I00oo175 3 27 98098 I oi01939 33 58 99884 I.ooii6 2 28 98155 I o1879 32 59 99942 10ooo58 I 29 98213 1.I01820 31 6o Unit. Unit. o 30 98270 1.oI761 30 Cotang. Tangent. Cotang. Tangent. 450 450 TABLE OF CONSTANTS. Base of Napier's system of logarithms.................. s = 2.718281828459 Mod. of common syst. of logarithms =.... com. log. e = A = o0.4342944819o3 Ratio of circumference to diameter of a circle =........... r- = 3. I41592653590 log. r = 0.49714987269.4 72 = 9-86960440Io89............./ 7r = I.772453850906 Are of same length as radius =........... 1800 = Io8oo' -- = 648000" -- 1800. — = 570.2957795130...........................log. = I 7 58122632409 io8oo' -- = 3437'.7467707849.........................lo. = 3.536273882793 648000" -. = 2o6264".8o6247o964,..................... log. = 5.314425133176 Tropical year = 365d. 5h. 48m. 47s..588 = 365d..2422I7456, log. = 2.562581o Sidereal year.= 365d. 6h. 9m. IOS..742 = 365d..256374332, log. = 2.5625978 24h. sol. t.=24h. 3m. 56s..555335 sid. t.=24h. X oo00273791, log. I.002=0oo00I 874 24h. sid. t.=24h. — (3m.55s..90944) sol. t.=24h. X 0.9972696, log. o.997=9.99881 26 British imperial gallon = 277.274 cubic inches............... log. = 2.4429091 Length of sec. pend., in inches, at London, 39. 3929; Paris, 39. 1285; New York, 39.1285. French metre = 3.2808992 Englishfeet = 39.3707904 inches. i cubic inch of water (bar. 30 inches, Fahr. therm. 620) = 252.458 Troy grains. D. APPLETON ~' CO., PUBLISHERS. MATHEMATICAL WORKS. BY GIEO. IT PERKINS, LL. D. ELEMENTS OF ALGEBRA. 12mo. sheep. Price 7'5 Cents Designed as an introduction to the author's "Treatise on Algebra." TREATISE ON ALGEBRA. 8vo. sheep, 420 pp. Price, $1 50. Adapted to the wants of advanced schools and colleges. It will be found to contain a full and complete development of all the various subjects usually taught in our colleges, including a demonstration and application of the THEOREM OF STURM. The present edition has been carefully revised and considerably enlarged. One entire chapter on the subject of Continued Fractions, treated in a general manner, has been added. The subject of RECURRING SERIES has been rewritten and simplified, and many other important changes have been made. ELEMENTS OF GEOMETRY, wrrH PRACTICAL APPLICATIONS. 12mo Sheep. Price $1. The following from the author's preface will give the teacher an idea of the design of the work: "We have found from experience in teaching, that, as a general thing, beginners in the study of Geometry consider it as a dry, unin teresting science. They have but little difficulty in following the demonstration, and arriving at a full conviction of its truth; but they ask, what if the proposition is true? what use can be made of it? " Now, to meet these difficulties we have, in the body of the work, added in a smaller sized type, such remarks, suggestions, and practical applications as we have found to interest the pupil." PLANE TRIGONOMETRY, AND ITS APPLICATION TO MENSURATION AND LAND SURVEYING, ACCOMPANIED WITH ALL THE NECESSARY LOGARITHMIO AND TRIGONOMETRIC TABLES. 8vo. sheep. Price $1 50. The work will be found as simple and practical as the nature of the different subjects will admit, and differs in many respects from other similar works. The chapter upon the division of land is new and highly interesting to all lovers of the science. Great care has been taken to have the tables accurate; they are from the old style of figures, which are less fatiguing to the eye than those in common use Qua D. APPLETON' CO., PUBLlbtIERS. RECOMMENDATIONS OF PERKINS' MATHEMATICAL SERIES, From G. B. DOCHAPRTY, Pro0f of Math., Free Academy, New York. "Ihave examined with more than ordinary interest a'Treatise on Algebra' by George R. Perkins, L.L. D., and also a' Treatise on Plane Trigonometry,' by the same author. I consider them both works of great merit. The'Treatise on Algebra, as a text-book, has no superior. Its arrangement is excellent; the rules explicit, the explanations lucid, and the style in which it is written is well adapted to lead the tyro rapIdly to a skilful knowledge of the higher analysis. "The Treatise on Plane Trigonometry is just such a manual as the;,upil requires It contains every thing necessary for a correct understanding of the principles of the science, and its practical application to the division of land. It will give me great pleasure at all times to recommend either of the above-named works to the teachers and students of my acquaintance." lFrom Professors JACKSON and POTTER, Of Anion College. "I have examined Prof. Perkins' Elements of Algebra. It is a work in which the peculiar merits of the French and English systems are combined; the practical and theoretical being made to illustrate each other. It is consequently better adapted to elementary instruction in our seminaries than any foreign work I have seen. Indeed, t is equally fitted for the common school and the college, as the elementary principles are exhibited sufficiently in detail and with admirable clearness, and the higher parts of the science are fully and ably discussed." From B. F. JOSLIN, Prof. of Math., and Nat. Philosophy, in the City of Boston. "The treatise on Algebra by George R. Perkins, is evidently the work of an experienced and judicious teacher. The illustrations possess in a good degree the merits of simplicity and conciseness, whilst the prominence given to important theorems and rules, facilitates both recollection and reference." From B. W. W. REDFIELD, Asst. Prof. of Malthematics, iV. Y. University. "I have examined, with much satisfaction, Perkins' Mathematical Series, and believe the treatise is inferior to none in point of clearness, accuracy and logical development. For his Elementary work on Algebra, the author deserves the thanks of teachers. He seems to have been the first to discover that. even in the education of children, a course of conclusive reasoning may with profit supersede the old system of dogmatical dictation." From J. ZEINER, Prof of Jfathematics, College of St. 31ichael and All Angels " It contains in a condensed form all the principles of Plane Trigonometry, that wfi5 most likely be needed in its application to the practical purposes of life. They ar* explained and demonstrated in a plain and concise manner, and the various examples and problems solved and unsolved are well calculated to make the student familiar with those principles, and expert in the application of them. The practical bearing of the book will, no doubt, make it a useful manual to the surveyor and engineer, particularly to those who have no opportunity to consult more extensive works on that subject, and wish to obtain the most necessary knowledge of it in a short time." From Wa. J. ROLFE, recent Principal of Day Academy, Wrentham, lfass. "Ihave put two classes through'Perkins' Trigonometry,' and do not hesitate to place it very far above any similar work that I have ever taught. I do not believe there is a book before the public, which contains half as much practical matter on the subject of surveying. I have taken every opportunity to recommend it to my fellow teachers, and have no doubt that some of them will adopt it."