ELEMENTS OF GEOMETRY AND TRIGONOMETRY, FROM THE WORKS Of M&. L.EG E,)X=NDRE. ADAPTED TO THE COURSE OF M3ATIIEMATICAL INSTRUCTION IN TIIE UNITED STATES, BY CITARLES DAVIES, LL.D., AJTtEOR OF ARITHIMETIC, ALGEBRA, PRAC~TIC.,L MATHIEAIATICS FOR PRACTICAL MEM, ELEMENTS OF DESCRIPTIVE AND OF ANALYTICAL GEOMIETRY, ELEMENTS OF DIFFERENTIAL AND INTEGRAL CALCULUS, AND SIADES, SHADOWS, AND PERSPECTITE. NEW YORK: BARNES & BURR, PUBLISHERS, 51 & 53 JOIHN ST. CHICAGO: GEORGE, SHERWOOD, 1lS LAKE ST. CINCINNATI: RICKEY AND CARROLL. ST. LOUIS: KEITIt AND WOODS. 1863. D ab s' 1burst of at#cniatailc talies' Pr3mary fritmnletic AND abIec3l3oolt-Designed for Beginners; containing the elementary tables of Addition, Subtraction, Multiplication, Division, and Denominate Numbers; with a large number oi easy and practical questions, both mental and written. IDabits' ff trst Aesons in rvitlrmetic-Combining the Oral Method with the Method of Teaching the Combinations of Figures by Sight. tgabirs' Entellectual Oritjmctifc —An Analysis of the Science of Numbers, with especial reference to Mental Training and Development. Vabi rs' en X i I c)ool Artif)mnctic-Analytical and Practical. Rev to Diablcs' Whe) 5crool 2rttitretic. abemes';raimmar of 2ritbmetic-An Analysis of the Language of Numbers and the Science of Figures. Daics' NeW Untibersittp ritftnctic —Embracing the Science of Numbers, and their Applications according to the most Improved Methods of Analysis and Cancellation. Bet? to labtes' Ncin EUnibcrsitp &tritmnetic. Uiabies' ~lementtarp lAtcbra-Embracing the First Principles of the Science.?ep to 3abits' ZSltmentarp iSlgcbra. abies,' ElItnuntargn foametrP AND FrigonometrTi-With Applications in Mensuration. Babies' 4Oractical Xattnematics —With Drawing and Mensuration applied to the Mechanic Arts. a(alcfs' [tffbesrsftp IgeIbtra-Embracing a Logical Development of the Science, with graded examples. -3abfes' Bourbon's algebra-Including Sturm's and Horner's Theorems, and practical examples. Dables' AZetenbre's Crometzr alnb Frfgonometrv —Revised and adapted to the course of Mathematical Instruction in the United States. Vabfts' Eemtentls of cutrbeifng AND TIabflattln-Containing descriptions of the Instruments and necessary Tables. Dabaes' rnalIltTtal &eonmetril-Embracing the Equations of the Point, the Straight Line, the Conic Sections, and Surfaces.of the first and second order. lDatbes' TifetrentiaI AND lzntegral talcutus. Babiis' V3scriptibe Geometrp — With its application to Spherical Trigonometry, Spherical Projections, and Warped Surfaces. Babifes' %Daales, ItaOtWS, bAND uerspetfibe. Babies' RLogic anb Wtilitp of XtatIlmatics-With the best methods of In3'4ruction Explained and Illustrated. labies' anb!ectt's Xattematical DIctfonarp anlr Qtvclopcbfa of _zattbe, watft;al Centc — Comprising Definitions of all the terms employed in Mathematics-an Analysis of each Branch, and of the whole, as forming a single Science. ETNTERED, according to Act of Congress, in the year one thousand eight hundred ana sixty-two, by CHARLES DAVIES, in the Clerk's Office of the District Court of tho United States for the Southern District of New York. WntAm DESYsE, STERomYPER AND Exacrr:oFYPEm, 183 William Street, New York. PR E FACE. OF the various Treatises on Elementary Geometry which have appeared during the present century, that of M. LEGIENDRE stands preeminent. Its peculiar merits have won for it not only a European reputation, but have also caused it to be selected as the basis of many of the best works on the subject that have been published in this country. In the original Treatise of LEGENDRE, the propositions are not enunciated in general terms, but by means of the diagrams employed in their demonstration. This departure firom the method of EUCLrID is much to be regretted. The propositions of Geometry are general truths, and ought to be stated in general terms, without reference to particular diagrams. In the following work, each proposition is first enunciated in general terms, and afterwards, with reference to a particular figure, that figure being taken to represent any one of the class to which it belongs. By this arrangement, the difficulty exper;enced by beginners in comprehending abstract truths, is lessened, without in any manner impairing the generality of the truths evolved. The term solid, used not only by LEGENDRE, but by many other authors, to denote a limited portion of space, seems calculated to introduce the foreign idea of matter iv PREFACE. into a science, which deals only with the abstract pro. perties and relations of figured space. The term volsume, has been introduced in its place, under the belief that it corresponds more exactly to the idea intended. Many other departures have been made from the original text, the value and utility of which have been made manifest in the practical tests to which the work has been subjected. In the present Edition, numerous changes have been made, both in the Geometry and in the Trigonometry. The definitions have been carefully revised-the demonstrations have been harmonized, and, in many instances, abbreviatedthe principal object being to simplify the subject as much as possible, without departing from the general plan. These changes are due to Professor Peck, of the Department of Pure Mathematics and Astronomy in Columbia College. For his aid, in giving to the work its present permanent form, I tender him my grateful acknowledgements. CHARLES DAVIES. COLUMBIA COLLEGE, Nzw YoPui, April, 1862. CONTENTS. GEOMETRY. PAGL ANTBODTCTION,............................................ 9 BOOEK L Definitions,........................, 13 Propositiong,................................................ 20 BOOK II. Ratios and Proportions,..................................., 50 BOOK III. The Circle, and the Measurement of Angles,........................ 59 Problems relating to the First and Third Books,............. 82 B00K IV. Proportions of Figures-Measurement of Areas,................ 93 Problems relating to the Fourth Book,............................ 129 BOOK V. Regular Polygons-Measurement of the Circle,,.. 136 BOOK VI. Planes, and Polyedral Angles,.......................,... 157 BOOK VII. Polyedrons,..............178...........a 178 vi CONTENTS. BOOK VIII. PAG'. Cylinder, Cone, and Sphere,............................... 210 BOOK IX. Spherical Geometry,..........,............................. 235 PLANE TRIGONOMETRY. INTRODUCTION. Definition of Logarithms,..................................... 3 Rules for Chlaracteristics,........................................ 4 General Principles............................................ 5 Table of Logarithms,.............................7.......... 7 Manner of Using the Table,...................................... 8 Multiplication by Logarithms,........................................ 11 Division by Logarithms,......................................... 12 Arithmetical Complement......................................... 13 Raising to Powers by Logarithms,..................... 15 Extraction of Roots by Logarithms,.............................. 16 PLANE TRIGONO.METRY. Plane Trigonometry Defined,.................................... 17 Functions of the Arc,........................................... 18-22 Table of Natural Sines,.......................................... 22 Table of Logarithmic Sines,..................................., 22 Use of the Table,.............................................. 23-27 S lution of Right-angled Triangles................................ 27-35 Solution of Oblique-angled Triangles,............................ 36-47 Prololems of Applicationu,.................................. 48 ANALYTICAL TRIGONOMETRY. Analytical Trigonometry Defined,............................. 51 Definitions and General Principles,............................... 51-54 Rules for Signs of the Functions,.......,........ C............ 54 CONTENTS. vib PAG.L Limiting value of Circular Functions.......................... 55 Relations of Circular Functions,................................ 57-59 Functions of Negative Arcs,.................................... 60-62 Particular values of Certain Functions,.......................... 63 Formulas of Relation between Functions and Arcs,................ 64-66 Functiong of Double and Half Arcs,............................ 67 Additional Formulas,........................................... 68-70 Method of Computing a Table of Natural Sines.................. 71 SPHERICAL TRIGONOMETRY. Spherical Trigonometry Defined,................................. 73 General Principles.............................................. 73 Formulas for Right-angled Triangles,............................ 74-76 Napier's Circular Parts.......................................... 77 Solution of Right-angled Spherical Triangles...................... 80-83 Quadraatal Triangles............................................. 84 Formulas for Oblique-angled Triangles............................. 85-92 Solution of Oblique-angled Triangles............................. 92-104 MENSURATION. Mensuration Defined,.............................................. 105 The Area of a Parallelogram....................................... 106 The Area of a Triangle,.......................................... 106 Formula for the Sine of Half an Angle,.1........................... 108 Area of a Trapezoid,............................................. 112 Area of a Quadrilateral,.......................................... 112 Area of a Polygon,............................................... 113 Area of a Regular Polygon,..................................... 114 To find the Circumference of a Circle,............................ 116 To find the Diameter of a Circle,............................... 116 To find the length of an Arc,..................................... 117 Area of a Circle................................................. 117 Area of a Sector,................................................ 118 Area of a Segment,.............................................. 118 Area of a Circular Ring.......................................... 119 viii CONTENTS. PAGE. Area of the Surfaoe of a Prism,.................................. 120 Area of the Surface of a Pyramid,................................ 120 Area of the Frustum of a Cone,................................ 121 Area of the Surface of a Sphere,................................ 122 Area of a Zone,................................................ 122 Area of a Spherical Polygon,.................................... 123 Volume of a Prism,............................................... 124 Volume of a Pyramid,............................................ 124 Volume of the Frustum of a Pyramid,............................ 125 Volume of a Sphere,............................................. 126 Volume of a Wedge,........................................... 127 Volume of a Prismoid,........................................ 128 Volumes of Regular Polyedrons,................................. 132 ELEMENTS O F GEOMETRY. INTRODUCTION. 1. QUANTITY is anything that can be measured. To measure a thing is to find out how many times it contains some other thing of the same kind, taken as a standard. The assumed standard is called the unit of measure. 2. Since the unit of measure is of the same kind as the thing measured, there are as many species of quantity as there are kinds of units. In Geometry, there are four kinds of units, viz.: Units of Length, UTzits of Surface, Units of Volume, and Units of Angular lJfeasure; and consequently, four species of quantity, viz.: LINES, SURFACES, VOLUMES, and ANGLES. These are called GEOMETRICAL MAGNITUDES. 3. GEOMErRY is that branch of Mathematics which treats of the properties and relations of Geometrical Magnitudes. 4. In Geometry, the quantities considered are generally represented by pictorial symbols. The operations to be performed upon them, and the relations between them, are indicated by signs, as in Analysis. 10 GEOMETRY. The following are the principal signs employed' The Sign of Addition, +, called plus: Thus, A + B, indicates that B is to be added to A. The Sign of Subtraction, -, called minus: Thus, A - B, indicates that B is to be subtracted from A. The Sign of Midtiplication, x: Thus, A x B, indicates that A is to be multiplied by B. The Sign of Division,: Thus, A B, or, A, indicates that A is to be divided by -B. The Exponential Sign: Thus, A3, indicates that A is to be taken three times as a factor, or raised to the third power. The Radical Sign, /: Thus, A-, 3B, indicate that the square root of A, and the cube root of B, are to be taken. When a compound quantity is to be operated upon as a single quantity, its parts are connected by a vinculum or by a parenthesis: Thus, A + B x (7, indicates that the sum of A ajid B is to be multiplied by C; and (A 4- B) ~ C, indi-.:tites that the sum of A and -B is to be divided bv C. A number written before a quantity, shows how Inan:i t iies it is to be taken. rThus, 3(A + B), indicates that the sum of A and B is to be taken three times. The Sign of Equality, -: Thus, A = B + C, indicates that A is equal to the sum of B and C. INTRODUCTION. 11 The expression, A = B + C, is called an equation. The part on the left of the sign of equality, is called the first member; that on the right, the second member. The Sign of Inequality, <: i'Thus, /A <.B, indicates that the square root of A is less than the cube root of B. The opening of the sign is towards the greater quantity. The sign,.'. is used as an abbreviation of the word hence, or consequently. 5. The general truths of Geometry are deduced by a course of logical reasoning, the premises being definitions and principles previously established. The course of reasoning employed in establishing any truth or principle, is called a demonstration. 6. A THEOREM is a truth requiring demonstration. 7. An AXIOM is a self-evident truth. 8. A PRoBLEM is a question requiring a solution. 9. A POSTuLATE is a problem whose solution is selfevident. Theorems, Axioms, Problems, and Postulates, are all called Pro}'ositions. 10. A LEMMA is an auxiliary proposition. 11. A COROLLARY is an obvious consequence of one or more propositions. 12. A SCuoLIUM is a remark made upon one or more propositions, with reference to their connection, their use, their extent, or their limitation. 12 GEOMETRY. 13. An HYPOTHESIS is a supposition made, either in the statement of a proposition, or in the course of a demonstration. 14. Two magnitudes are equal, when they are equal in measure. When they may be so placed as to coincide through. out their whole extent, they are equal in all their parts. ELEMENTS OF GEOMETRY. BOOK I. ELE3MENTARY PRINCIPLES. DEFINITIONS. 1. GEOMETRY is that branch of Mathematics which treats of the properties and relations of Geometrical Magnitudes. 2. A POINT is that which has position, but not magnitude. 3. A LINE is that which has length, but neither breadth nor thickness. Lines are divided into two classes, straight and curved. 4. A STRAIGHT LINE is one which does not change its direction at any point. 5. A CURVED LINE is one which changes its direction at every point. The word line, alone, is used for straight line; and the word curve, alone, for curved line. 6. A line made up of straight lines, is called a broken line. 7. A SURFAcE is that which has length and breadth, without thickness. 14 GEOMETRY. Surfaces are divided into two classes, plane and curved surfaces. 8. A PLANE is a surface, such, that if any two of its points be joined by a straight line, that line will lie wholly in the surface. 9. A CURVED SURFACE is a surface which is neither a plane nor composed of planes. 10. A PLANE ANGLE is the amount of divergence of two lines lying in the same plane. Thus, the amount of divergence of the lines AB and A C, is an angle. The lines AB and A C are called sides, and A their common point A, is called the vertex. An angle is designated by naming its sides, or some. times by simply nlaming its vertex; thus, the above is called the angle BACG, or simply, the angle A. 11. When one straight line meets another the two angles which they form D are called adjacent angles. Thus, the A B angles AB-D and D1BC are adjacent. 12. A RIGHT ANGLE is fbrmed by one straight line meeting another so as to make the adjacent angles equal. The first line is then said to be perpendicular to the second. 13. An OBLIQUE ANGLE is formed by one straight line meeting another so as to make the adjacent angles zunequal. Oblique angles are subdivided into two classes, acute aniles, and obtuse angles. 14. An ACUTE ANGLE is less than a right angle BOOK I. 15 15. An OBTUSE ANGLE is greater than,/ a right angle. 16. Two straight lines are parallel, when they lie in the same plane and cannot meet, how far soever, either way, both may be produced. They then have the same direction. 17. A PLANE FIGURE is a portion of a plane bounded by lines, either straighlt or curved. 18. A POYGON is a plane figure bounded by straight lines. The bounding lines are called sides of the polygon. The broken line, made up of all the sides of the polygon,'is called the perimeter of the polygon. The angles formed by thte sides, are called angles of the polygon. 19. Polygons are classified according to the number of their sides or angles. A Polygon of thlree sides is called a trianzgle; one of four sides, a quadcrilateral; one of five sides, a pentagYon; one of six sides, a Lhexagaon; one of seven sides, a ]heptagon; one of eight sides, an octagon; one of ten sides, a decagon; one of twelve sides, a dodecagon, &c. 20. An EQUILATERAL POLYGON, is one whose sides arie all equal. An EQUIANGULAR POLYGON, is one whose angles are:il equal. A REGULAR POLYGON, is one which is both equilateral and equiangular. 21. Two polygons are euilctferal, or mutzually equilateral, when their sides, taken in the same order, are equal, each to each: that is, following their perimeters in the sanae 16 GEOMETRY. direction, the first'side of the one is equal to the first side of the other, the second side of the one, to the second side of the other, and so on. 22. Two polygons are equiangular, or mutually equiargulcar, when their angles, taken in the same order, are equal, each to each. 23. A DIAGONAL of a polygon is a line joining the vertices of two angles, not consecutive. 24. A BASE of a polygon is any one of its sides on which the polygon is supposed to stand. 25. Triangles may be classified with reference either to their sides, or their angles. When classified Fwith reference to their sides, there arp two classes: scalene and isosceles. 1st. A SCALENE TRIAANGLE is one which has no two of its sides equal. 2d. An ISOSCELES TRIANGLE is one which has two of its sides equal. When all of the sides are equal, the triangle is EQUILATERAL. When classified with reference to their angles, there are are two classes: right-angled and oblique-angled. 1st. A RIGHT-ANGLED TRIANGLE is one that has one right angle. The side opposite the right angle, is called the h]ypothlenuse. 2d. An OBLIQUE-ANGLED TRIANGLE is one whose angles are all oblique. BOOK I. 17 If one angle of an oblique-angled triangle is obtuse, the triangle is said to be OBTUSE-ANGLED. If all of the anglies are acute, the triangle is said to be ACUTE-AcNGLED. 26. Quadrilaterals are classified with reference to the relative directions of their sides. There are then two cl; s: the first class embraces those which have no two sides parallel; the second class embraces those which have two sides parallel. Quadrilaterals of the first class, are called trapezizms. Quadrilaterals of the seconlt class, are divided into two species: trapezoids and parallelograms. 27. A TRAP.PEZOID is a quadrilateral which has only tw-o of its sides parallel. 28. A PARALLELOGRAM is a quadrilateral which has its opposite sides parallel, two and two. There are two varieties of parallelograms': rectangles and rhomboids. Ist. A RECrANGLE is a1 parallelogram whose angles are all right angles. A SQUARnE is an equilateral rectangle. 2d. A RHOMBOID is a parallelogram whose angles are all oblique. A Riom-Bus is an equilateral rhomboid. 2 18 GEOMETRY. AXIOMS. 1. Things which are equal to the same thing, are equal to each other. 2. If equals be added to equals, the suTns will be equal. 3. If equals be subtracted from equals, the remainders will be equal. 4. If equals be added to unequals, the sums will be unequal. 5. If equals be subtracted from unequals, the remainders will be unequal. 6. If equals be multiplied by equals, the products will be equal. 7. If equals be divided by equals, the quotients will be equal. 8. The whole is greater than any of its parts. 9. The whole is equal to the sum of all its parts. 10. All right angles are equal. 11. Only one straight line can be drawn between two points. 12. The shortest distance between any two points is measured on the straight line which joins them. 13. Through the same point, only one line can be drawn parallel to a given line. POSTULATES. 1. A straight line can be drawn between any two points. 2. A straight line may be prolonged to any length. 3. If two lines are unequal, the length of the less may be laid off on the greater BOOK I. 19 4. A line may be bisected; that is, divided into two equal parts. 5. An angle may be bisected. 6. A perpendicular may be drawn to a given line, either from a point without, or from a point on the line. 7. A line may be drawn, making with a given line an angle equal to a given angle. 8. A line may be drawn through a given point, parallel to a given line. NOTE. In making references, the following abbreviations are employed, viz.: A. for Axiom; B. for Book; C. for Corollary; D. for Definition; I. for Introduction; P. for Proposition; Prob. for Problem; Post. for Postulate; and S. for Scholium. In referring to the same Book, the number of the Book is not given; in referring to any other Book, the number of the Book is given. av) GEOMETRY. PROPOSITION I. THEORE]M. f a straight line meet another straight line, the sum of the adjacent angles will be equal to two right angles. Let DC meet AB at C: then will the sum of the angles E D D CA and D CB be equal to two right angles. As C, let CE be drawn per- A B pendicular to AlB (Post. 6); then, C by definition (D. 12), the angles ECA and ECB will both be right angles, and consequently, their sum will be equal to two right angles. The angle D CA is equal to the sum of the angles ECA and ECD (A. 9); hence, DCA + DCB = ECA + ECD +D CB; But, ECD + D CB is equal to ECB (A. 9); hence, D CA + D CB = ECA + YECB. The sum of the angles ECA and ECB, is equal to two right angles; consequently, its equal, that is, the sum of the angles D CA and D CB, must also be equal to two right angles; which was to be proved. Cor. 1. If one of the angles DCCA, -DCB, is a right angle, the other must also be a right angle. Cor. 2. The sum of the angles BACA CGA, DC AE, EAE, C D formed about a given point on E the same side of a straight line BF, is equal to two right an- B A gles. For, their sum is equal to BOOK I. 21 the sum of the angles EAB and -EAF; which, from the proposition just demonstrated, is equal to two right angles. DEFINITIONS. If two straight lines intersect each other, they form four angles about the point of intersection, which have received different names, with respect to each other. 1~. ADJACENT ANGLES are those which lie on the same side A of one line, and on opposite sides of the other; thus, A CE and C ECI, or ACE and ACD, are 3 adjacent angles. A~. OPPOSITE, or VER1ICAL ANGLES, are those which lie on opposite sides of both lines; thus, A CE and D CB, or A CD and ECB, are opposite angles. From the proposition just demonstrated, the sum of any two adjacent angles is equal to two right angles. PROPOSITION II. THEOREM. If two straight lines intersect each other, the opposite or vertical angles will be equal. Let AB and -DE intersect at C: then will the opposite A or vertical angles be equal. The sum of the adjacent angles A CE and A CD, is equal to two right angles (P. I.): the sum of the adjacent angles A CE and ECB, is also equal to two right angles. But things which are equal to the same thing, are equal to each other (A. 1); hence, 22 GEOMETRY. ACE + A CD = A CE + ECB; Taking from both the common A angle ACE (A. 3), there remains, C A CD- = ECB. D B D B In like manner, we find, ACD + ACE = ACD + DCB; and, taking away the common angle A CD, we have, A C-E = D CB. Hence, the proposition is proved. Cor. 1. If one of' the angles about C is a right angle, all of the others will be eight angles also. For, (P. I., C. 1), each of its adjacent angles will be a right angle; and from the D proposition just demonstrated, its opposite angle will also be a right A C B angle. Cor. 2. If one line DE, is E perpendicular to another AB, then will the second line AB be perpendicular to the first DE. For, the angles D CA and D CB are right angles, by definition (D. 12); and from what has just been proved, the angles A CE and B CE are also right angles. Hence, the two lines are mutually perpendicular to each other. B Cor. 3. The sum of all the A angles ACB, BCD, DCE, ECE,, D FCA, that can be formed about a point, is equal to four right angles. BOOK I. 23 For, if two lines be drawn through the point, mutually perpendicular to each other, the sum of the angles which they form will be equal to four right angles, and it will also be equal to the sum of the given angles (A. 9). Hence, the sum of the given angles is equal to four right angles. PROPOSITION III. THEOREM. If two straight lines have two points in common, they will coincide throughout their whole extent, and fornm one and the same line. Let A and B be two points E common to two lines: then will A B C the lines coincide throughout. D Between A and B they must coincide (A. 11). Suppose, now, that they begin to separate at some point C, beyond AB, the one becoming A C]E, and the other A CD. If the lines do separate at C, one or the other must change direction at this point; but tlhis is contrary to the definition of a straight line (D. 4): hence, the supposition that they separate at any point is absurd. They must, therefore, coincide throughout; which was to be proved. Cor. Two straight lines can intersect in only' one point. NoTE.-The method of demonstration employed above, is called the reductio ad absurdum. It consists in assuming an hypothesis which is the contradictory of the proposition to be proved, and then continuing the reasoning until the assumed hypothesis is shown to be false. Its contradictory is tlhus proved to be true. This method of demonstration is often used in Geometry. 24 GEOMETRY. PROPOSITION IV. THEOREM. If a straight line meet two other straight lines at a comrnWon point, making the sum of the contiguous angles equal to two right angles, the two lines met will form one and the same straight line. Let D d meet AC and BC at C, mlaking the sum of the A A angles D CA and 1) CB equal C B to two right angles: then will COB be the prolongation of A C. For, if not, suppose CE to be the prolongation of A O;', then will the sum of the angles D CA and D CE be equal to two right angles (P. I.): We shall, consequently, have (A. 1), DC.A + DCB = DCA + DCE; Taking from both the common angle D CA, there remains, DC2B = D CE, which is impossible, since a part cannot be equal to the whole (A. 8 ). Hence, CB must be the prolongation of A C; which was to be proved. PROPOSITION V. THEOREM. If two triangles have two sides and the included angle of the one equal to two sides and the included angle of the other, each to each, the triangles will be equal in all their parts. In the triangles AB C and D1)EF, let AB be equal BOOK I. 25 to -DE, A C to D)F, and the angle A to the angle D: then will the triangles be equal in all their parts. For, let ABC be applied to D)EF, in A D such a manner that the angle A shall coincide with the anlge D, the side AB taking B C E F the direction DE, and the side A C the direction DE Then, because AB is equal to DJE, the vertex B will coincide with the vertex E; and because A C is equal to DF, the vertex C will coincide with the vertex F; consequently, the side B C will coincide with the side ET (A. 11). The two triangles, therefore, coincide throughout, and are consequently equal in all their parts (I., D. 14); which was to be proved. PROPOSITION VI. THEOREM. If two triangles have twoo angles and the included side of the one equal to two angles and the included side of the other, each to each, the triangles will be equial in all their parts. In the triangles ABC and DE:F, let A D the angle B be equal to the angle E, the angle C to the angle F, and the side BC B C E F to the side ET: then will the triangles be equal in all their parts. For, let ABC be applied to DEE/; in such a manner that the angle B shall coincide with the angle E, the side 26 GEOMETRY. BC taking the direction E, and the side BA the direce tion ED. Then, because B C is equal to EP, the vertex C will coincide with the vertex F; and because the angle C is equal to the angle F, the side CA will take the direction FD. Now, the vertex A being at the same time on the lines EED and PFD, it must be at their intersection D (P. III., C.): hence, the triangles coincide throughout, and are therefore equal in all their parts (I., D. 14); which was to be proved. PROPOSITION VII. THEOREM. The sum of any two sides of a triangle is greater than tha third side. Let AB C be a triangle: then will B the sum of any two sides, as AB, B C, be greater than the third side A C. For, the distance from A to C, A -C measured on any broken line AB, BC, is greater than the distance measured on the straight line AC (A. 12): hence, the sum of AB and BC is greater than A C; which was to be proved. Cor. If from both members of the inequality, AC < AB + BC, we take away either of the sides AB, B C(, as BC, for example, there will remain (A. 5), AC - BC < AB; that is, the di/ference between any two sides of a triangle is less than the third side. Scholium. In order that any three given lines may re BOOK I. 27 present the sides of a triangle, the sum of any two must be greater than the third, and the difference of any two must be less than the third. PROPOSITION VIII. THEOREM. If from any point within a triangle two straight lines be drawn to the extremities of any side, their sumn will be less than that of the two rema-ining sides of the triangle. Let 0 be any point within the triangle BA C, and let the lines OB, O C, be drawn to the extremities of any side, as B C: A then will the sum of B 0 and OC be less than the sum of the sides -0 BA and AC. Prolong one of the lines, as B 0, till it meets the side A C in D; then, from Prop. VII., we shall have, OC < OD + DC; adding BO to both members of this inequality, recollecting that the sum of BO and OD is equal to BD, we have (A. 4), BO +OC < B1 - + C. From the triangle -BAD, we have (P. VII.), BD < BA + AD); adding DC to both members of this inequality, recollectini that the sum of AD and DC is equal to A C, we have, B-D + DC < BA + A C. But it was shown that B 0 + O C is less than BD + DC; still more, then, is BO + 0 C less than BA + -AC; which was to be proved. 28 GEOMETRY. PROPOSITION IX. THEOREMI. If two triangles have two sides of the one equal to two sides of the other, each to each, and the included angles unequal, the third sides will be unequal; and the greater side will belong to the triangle which has the greater included angle. In the triangles BA C and DE1/; let AB be equal to DE, AC to _D:F and the angle A greater than the angle D: then will B C be greater than lE-. Let the line A G be drawn, making the angle CA G equal to the angle D (Post. 7); make A G equal to DE, and draw G C. Then will the triangles A G C and DEFl have two sides and the included angle of the one equal to two sides and the included angle of the other, each to each; consequently, GC is equal to EZ (P. V.). Now, the point G may be without the triangle ABUi, it may be on the side B C, or it may be within the triangle ABC. Each case will be considered separately. A D 10. - When G is without the triangle ABC. In the triangles GI7C and AIB, we have, (P. VII.), G E GI + IC > GC, and BI-IA > AB; whence, by addition, recollecting that the sum of BI and IC is equal to B C, and the sum of GI and IA, to GA, we have, AG + BC > A - 4- GO BOOK I. 29 Or, since AG -- AB, and GC = EFE we have, AB + BC > AB + EF. Taking away the common part AB, there remains (A. 5), BC > EF. 20. When G is on B C. A D In this case, it is obvious that GC is less than BC; or, since G C - EF, we have, BC > EL B G C E F 30. When G is within the triangle ABC. From Proposition VIII., we have, i D BA + BC > GCA + GC; or, since GA =BA, and G = EFI BG,/ we have, BA + B C > BA + FE\ Taking away the common part AB, there remains, C F BC > EF. Hence, in each case, B C is greater than EF; which teas to be proved. Conversely: If in. two triangles AB C and 1DEE tile side AB is equal to the side IDE, the side A C to DE, and B C greater than E1, then will the angle -BA C be greater than the angle EDE For, if not, BA C must either be equal to, or less than, EDF. In the former case, B C would be equal to EF (P. V.), and in the latter case, B C would be less than EF; either of which would be contrary to the hypothesis hence, BA C must be greater than EDE. 80 GEOMETRY. PROPOSITION X. THEOREM. If two triangles have the three sides of the one equal to the three sides of the other, each to each, the triangles will be equal in all their parts. In the triangles ABC and DEEi let AB be equal t6 DE, AC to DF, and BC to EF: then will the triangles be equal in all their parts. For, since the sides AB, AC, are equal to A D DE, tE, each to each, if the angle A were greater than D, it would follow, by the last Pro- B C E F position, that the side B C would be greater than BET; and if the angle A were less than D, the side B C would be less than EF. But BC is equal to EF, by hypothesis; therefbre, the angle A can neither be greater nor less than D: hence, it must be equal to it. The two triangles have, therefore, two sides and the included angle of the one equal to two sides and the inclu. ded angle of the other, each to each; and, consequently, they are equal in all their parts (P. V.); which was to be proved. Scholium. In triangles, equal in all their parts, the equal sides lie opposite the equal angles; and conversely. PROPOSITION XI. THEOR:EM. In an isosceles triangle the angles opposite the equal sides are equal. Let BA C be an isosceles triangle, having the side AB equal to the side A C: then will the angle C be equal to the angle B. BOOK I. 31 Join the vertex A and the middle point D of the base BC. Then, AB is equal to A C, by hypothesis, AD common, and BD equal to D C, by construction: hence, the triangles BAD, A atld DA C, have the three sides of the one equal to those of the other, each to each; therefore, by the last Proposition, B -- tle angle B is equal to the angle C; which was to be proved. Cor. 1. An equilateral triangle is equiangular. Cor. 2. The angle BAD is equal to DA C, and BDA to C-DA: hence, the last two are right angles. Consequently, a line drawn from the vertex of an isosceles triangle to the middle of the base, bisects the vertical angle, and is perpe#dicuiar to the base. PROPOSITION XII. THEOREM. If two angles of a triangle are equal, the sides opposite to them are also equal, and consequently, the triangle is isosceles. In the triangle ABC, let the angle A AB C be equal to the angle A CB: then will A C be equal to AB, and consequently, the triangle will be isosceles. For, if AB and AC are not equal, B suppose one of them, as AB, to be the greater. On this, take BD equal to A C (Post. 3), and draw D C. Then, in the triangles ABC, DB C, we have the side BD equal to A C, by construction, the side B C common, and the included angle A CB equal to the included angle DB C, by hypothesis: hence, the two triangles are equal 32 GEOMETRY. in all their parts (P. V.). But this is impossible, because a part cannot be equal to the whole (A. 8): hence, the hypothesis that AB and A C are unequal, is false. They must, therefore, be equal; which was to be proved. Cor. An equiangular triangle is equilateral. PROPOSITION XIII. THEOREM. ln any triangle, the greater side is opposite the greater angle; and, conversely, the greater angle is opposite the greater side. In the triangle ABC, let the angle A CB be greater than the angle ABC: A then will the side AB be greater than D the side A C. For, draw CD1, making the angle C H BC.D equal to the angle B (Post. 7): then, in the triangle D CB, we have the angles D CB and DB C equal: hence, the opposite sides 1)B and DC are equal (P. XII.). In the triangle A CD, we have (P. VII.), AD + DC > AC; or, since DC = D/B, and AD + DB = AB, we have, AB >AC; which was to be proved. Conversely: Let AB be greater than AC: then will the angle A CB be greater than the angle ABC. For, if AC'B were less than ABC, the side AB would be less than the side AC, from what has just been proved; if A CB were equal to AB C, the side AB would be equal to A C, by Prop. XII.; but both conditions are contrary BOO K I. 33 to the hypothesis: hence, A CB can neither be less tllh.l, nor equal to, ABC; it must, thereuore, be greater; whicth was to be proved. PROPOSITION XIV. THEOREM. lrom a given point only one perpendicular can be drawn to a given straight line. Let A be a given point, and AB A a perpendicular to DE: then can no other perpendicular to _DE be drawn through A. D c B E For, suppose a second perpendicular A C to be drawn. Prolong AB till BF is equal to AB, and draw CF. Then, the triangles ABC and PFB C will have AB equal to B1P, by construction, CGB common, and the includced angles ABC and PB C equal, because both are right angles: hence, the angles ACB and _FCB are equal (P. V.) But A CGB is, by a hypothesis, a right angle: hence, FCB must also be a right angle, and consequently, the line ACF must be a straight line (P. IV.). But this is impossible (A. 11). The hypothesis that two perpendiculars can be drawn is, therefore, absurd; consequently, only one such perpendicular can be drawn; which was to be proterd. If the given point is on the given line, the propositim, is equally true. For, if through A two perpendiculars -I B/ and A C could be drawn to D)E, we should have BAE and CAE each equal to a right angle; and consequently, equal to each other; which is absurd (A. 8). D A 3~~~ 34 GEOMETRY. PROPOSITION XV. THEOREM. If from a point without a straight li2e a perpendicular be let fail on the line, and oblique lines be drawn to different points of it' 1~. The perpendicular will be shorter than any oblique line: 2~. Any two oblique lines that meet the given line at points equally distant from the foot of the perpendicular, will be equal: 3~. Of two oblique lines that meet the given line at points unequally distant from the foot of the perpendicular, the one which meets it at the greater distance will be the longer. Let A be a given point, DE a A given straight line, AB a perpendicular to DE, and AD, A C, AE oblique lines, BC being equal to BE, and BD D C B E greater than B C. Then will AB be less than any of the oblique lines, AC~ will be equal to AF, and AD greater than A C. Prolong AB until BE is equal to AB, and draw FC, FD. 10. In the triangles ABC', FB C, we have the side AB equal to BF, by construction, the side BC common, and the included angles ABC and PB C equal, because both are right angles: hence, FC is equal to A C (P. V.). But, AF is shorter than A CE (A. 12): hence, AB, the half of AF, is shorter than A C. the half of A CF; which was to be proved. 2~. In the triangles ABC and ABE, we have the side B C equal to BE, by hypothesis, the side AB common, and the included angles ABC and ABE equal, BOOK I. 35 because both are right angles: hence, AC is equal to AE'; which was to be proved. 3~. It may be shown, as in the first case, that AD is equal to )F. Then, because the point C lies within the triangle ADTi, the sum of the lines AD and DF will be greater than the sum of the lines A C and CF (P. VIII.): hence, AD, the half of ADJi is greater than A C, the half of A CF; which was to be proved. Cor. 1. The perpendicular is the shortest distance from a point to a line. Cor. 2. From a given point to a given straight line, only two equal straight lines can be drawn; for, if there could be more, there would be at least two equal oblique lines on the same side of the perpendicular; which is impossible. PROPOSITION XVI. THEOREM. If a perpendicular be drawn to a given straight line at its middle point: 1~. Any point of the perpendicular will be equally distant from the extremities of the line: 20. Any point, without the perpendicular, will be unequally distant from the extremities. Let AB be a given straight line, C F its middle point, and EF the perpendicular. Then will any point of ET be equally distaut fiom A and /B; and any point without ERiF, will be unequally distant from A and B. A c B 1~. From any point of El, as D, draw the lines -DA and D)B. Then will DA E and DB be equal (P. XV.): hence, D is equally distant from A and B; which was to be proved. 36 GEOMETRY. 2~. From any point without E:, as I, draw IA and IB. One of these lines, as Iz, will cut EF in some point D; draw DB/'. Then, from what has just been shown, DA. and D)B will F be equal; but IB is less than the sum of ID and D)B (P. VII.); and because the sum of ID and DB is equal to the sum of ID and DA, or IA, we have C IB less than IA: hence, I is unequally distant from A and B; which was to be proved. Cor. If a straight line EF have two of its points E and 17 equally distant from A and B, it will be perpendicular to the line lAB at its middle point. PROPOSITION XVII. THEOREM. If two right-angled triangles have the hypothenuse and a side of the one equal to the hypotiheznse and a side of the other, each to each, the triangles will be equal in all their parts. Let the right-angled tri- A D angles ABC and DEEIS have the hypothenuse i C equal to DF, and the side AB G C E F equal to D/E: then will the triangles be equal in all their parts. If the side B C is equal to EF, the triangles will be equal, in accordance with Proposition X. Let us suppose then, that B C and EF are unequal, and that B C is the longer.- On BC lay off BG equal to EF, and draw *UG. T1 triangles ABG and D)EF have AB equal to DE, by hypothesis, BG equal to EF, by construction, and BOOK I. 37 the angles B and E equal, because both are right angles; consequently, A G is equal to SDF (P. V.) But, AC is equal to DPF, by hypothesis: hence, AG i ", I.,' —; i \.,11 // a z Let the angle X1k be a common unit; and suppose, for example, that this unit is contained 7 times in the angle A CB, and 4 times in the angle DOE. Then,- suppose A CB be divided into 7 angles, by the radii Cm, C(I, Cp, &c.; and I) OE into 4 angles, by the radii Ox, Oy, and Oz, each equal to the unit M. BOOK II. 75 From the last proposition, the arcs Am, inn, &c., Dx, Zy, &c., are equal to each other; and because there are 7 of these arcs in AB, and 4 in DE, we shall have, are AB~ are`DEE'7 4. But, by hypothesis, we have, angle A CB ~ angle DOR'7 4; hence, from (B. II., P. IV.), we have, angle A CB angle DOE: arc AB: arc DE If any other numbers than 7 and 4 had been used, the same proportion would have been found; which was to be proved. Cor. If the intercepted arcs are commensurable, they will be proportional to the corresponding angles at the centre, as may be shown by changing the order of the couplets in the above proportion. PROPOSITION- XVII. THEOREM. In equal circles, incommensurable angles are proportional to their intercepted arcs. In the equal circles, whose C O centres are C an'W' let ACB and FOH be:ineommensurable: then will they A \ be proportional to the arcsB F AB and FtIL For, let the less angle FiO0f, be placed upon the greater angle A CB, so that it shall take the position A CD. 76 GEOMETRY. Then, if the proposition is not O true, let us suppose that the angle A CB is to the angle _FO/l, or its equal A CD, A as the are AB is to an arc D A 0, greater than _Fi; or its equal AD1; whence, angle A CB angle A CiD arc AB arc A0. Conceive the are AB to be divided into equal parts, each less than DDO: there will be at least one point of division between _D and 0; let I be that point; alnd draw CI. Then the arcs AB, AT, will be commensuralble, and we shall have (P. XVI.), angle AC:lC angle A CI:: arc AB: are AT. Comparing the two proportions, we see that the antecedents are the same in both: hence, the consequents are proportional (B. II., P. IV., C.); hence, angle ACD: angle ACI:: are AO. are Al. But, 210 is greater than Al: hence, if this proportion is true, the angle A CD must be greater than the angle A CI. On the contrary, it is less: hence, the fourth term of the proportion cannot be greater than AD). In a similar manner, it may be shown that the fourth term cannot be less than AD: hence, it must be equal to AD; therefore, we have, angle A Cl: angle ACD:: arc AB: arc AD; which was to be proved. Cor. 1. The intercepted arcs are proportional to the cor BOOK III. 77 responding angles at thle centre, as may be shown by change ing the order of the couplets in the preceding proportion. Cor. 2. In equal circles, angles at the centre are proportional to their intercepted are cs; and the reverse, whether they are commensurable or incommensurable. Cor 3. In equal circles, sectors are proportional to their angles, and also to their arcs. Scholium. Since the intercepted arcs are proportional to the corresponding angles at the centre, the arcs may be taken as the measures of the angles. That is, if a circumference be described from the vertex of any angle, as a centre, and with a fixed radius, the are intercepted between the sides of the angle may be taken as the measure of the angle. In Geometry, the right angle which is measured by a quarter of a circumference, or a quadrant, is taken as a unit. If, therefore, any angle be measured by one-half or two-thirds of a quadrant, it will be equal to one-half or two-thirds of a right angle. PROPOSITION XVIII. THEOREM. An inscribed angle is measured by half of the ar.e included between its sides. There may be three cases: the centre of the circle may lie on one of the sides of the angle; it may lie within' the angle; or, it may A lie without the angle. 10. Let EAD be an inscribed angle, one of whose sides AE passes through the centre: then will it be measured by half of the are DE. E 78 GEOMETRY. For, draw the radius CD). The external angle D)CE; of the triangle D CA, is equal to the sum of the opposite interior angles CJAD and C-DA (B. I., P. XXV., C. 6). But, the triangle D)CA being isosceles, the angles D and A are equal; A therefore, the angle _DCE is double the angle DAE. Because D) CE is at the centre, it is measured by the arc _DE (P. XVII., S.): hence, the, angle DAE is measured by half of. the are DE; which was to be proved. E 2~. Let DAB be an inscribed angle, and let the centre lie within it: then will the angle be measured by half of the arc BE)D. For, draw the diameter AE. Then, from what has just been proved, the angle DAE is measured by half of DL), and the angle EAB by half of EB: hence, BAD, which is the sum of E4-AB and DA-E, is measured by half of the sum of DE and EB, or by half of BED; which was to be proved. 3~. Let BAD be an inscribed angle, and let the centre lie without it: then will it be measured by half of the are arc BD. For, draw the diameter AE. Then, A from what precedes, the angle DAE is measured by half of DE, and the angle BAE by half of BE: hence, BAI), which is the difference of BAE B and DAE, is measured by half of the difference of BE and D)E or by D half of the arc BD;' wohich was to be proved. BOOK III. 79 D Cor. 1. All the angles BA, C, BD C, BEC, inscribed in the same segment, are equal; because they are each measured by half of the same B arce BOC. Cor. 2. Any angle BAD, in- A scribed in a semi-circle, is a right angle; because it is measured by half B the semi-circumference B O-D, or by a quadrant (P. XVII., S.). Cor. 3. Any angle BA C, inscribed in a segment greater than a semi-circle, is acute; for it is measured by half the arc BOG, less than a semi-circumference. lB a Any angle B O C, inscribed in a segment less than a semi-circle, is obtuse; for it is measured by half the are BA C, greater than a semi-circumference. Cor. 4. The opposite angles A B and C, of all inscribed quadrilateral ABCGD, are together equal to two right angles; for the angle DAB is measured by half the arc DCGB, the angle D GB by half the arc DAB: hence, the two angles, taken together, are mea. sured by half the circumference: hence, their sum is equal to two right angles. 80 GE O M ETRY. PROPOSITION XIX. T1I EO IE. Any angle formzed by thwo chords, zwhich intersect, is nmea sured by half the sum of the included aTres. Let -DEB be an angle formed by the intersection of the chords AD and CD: then will it be measured by half the sum of the arcs A G and DB. For, draw AlF parallel to D C: CC theln, the are _DF will be equal to A C (P. X.), and the angle FAD} j\ equal to the angle DEB (B. I., P. XX., C. 3). But the angle FI]AD is mleasured by half the are PEDB (P. XVIII.); therefore, DEFB is measured by half of __DD; that is, by half the suim of FD and DB, or by half the sum of A C and DB; which'was to be proved. PROPOSITION XX. THEIOREM. The angle formed by two secanzts, is measured by haly the diference of the included arcs. Let AD3, A C, be two secants: then will the angle BAC be measured by half the. difference of the arcs B C and 2D2.A Draw /DE parallel to A C: the arc EC n-ill be equal to DDF (P. X.), andi the angle B)DE equal to the angle BAC (B. I., P. XX., C. 3.). But BDE is measured by half the are BE (P. XVIII.): hence, BA C is also measured by half the are BE;'CJ/ that is, by half the difference of B C and.EC, or by half the difference of BBC and 13T; which was to be proved. BOOK III. 81 PROPOSITION XXI. TH1EOREM. An. angle formed by a tan;gent and a chord mneeting it at the point of contact, is measured by half the incilde&darc. Let BE be tangent to the circle AAIC, and let A C' be a chord drawn from the point of contact A: then will the angle BAC be measured by half of the are AMIC. For, draw the diameter AD. D The angle BAD is a right angle (P. IX.), and is measured by half the semi-circumference AJID (P. XVII., S.); the angle DA C is measured by half of the are D C B A (P. XVIII.): hence, the angle BA C, which is equal to the sum of the angles BAD and DA C, is measured by half the sum of the arcs AlID and D)C, or by half of the arec AMC; which was to be proved. Scholium. The angle CAE, which is the difference of DAE and 1DAC, is measured by half the difference of the arcs D CA and D C, or by half the arc CA. 6 PRACTICAL APPLICATIONS. PROBLEM I. To bisect a given straight line. Let AB be a given straight line. From A and B, as centres, with a radius greater than one half of AB, describe arcs intersecting at E and F: join E' and X1, by the straight line EF. lien will -EF bisect the A A C t given line AB. For, E and F are each equally distant from A and B; and consequently, the line EF' bisects AB (B. I., P. XVI., C.). PROBLEM II. To erect a perpendicular to a given straight line, at a given point of that line. Let B C be a given line, and let A be a given point on that line. Lay off from A the equal distances AB and A C; from B and C, as centres, with a radius greater than one half of B C, describe arcs intersecting B A - BOOK III. 88 at D; draw the line AD: then will AD be the perpendicular required. For, 1D and A are each equally distant from B and C; consequently, -DA is perpendicular to BC at its middle point A (B. I., P. XVI., C.). PROBLEM II. To draw a perpendicular to a given straight line, from a given point without that line. Let BD be the given line, and A the given point. From A, as a centre, with a radius sufficiently great, describe an arc A cutting BD) in two points, B and D'; with B and Z) as centres, and C a radius greater than one-half of BD, D describe arcs intersecting at E; draw AlE: then will AE be the perpendicular required. For, A and E are each equally distant from B and D: hence, AE is perpendicular to BD (B. L, P. XVI., C.). PROBLEM IV. At a point on a given line, to construct an angle equal to a given angle. Let A be the given point, AB the given line, and IKL the given angle. From the vertex KI as a L centre, with any radius If1, describe the are IL, terminat- - K B ing in the sides of the angle. From A as a centre, with a radius AB, equal to liI, 84 GEOMETRY. describe the indefinite arc B 0; then, with a radius equal to the chord LI, from.B as a centre, describe an arc cutting the arc B0 in.D; draw AD: then will BAD L/ be equal to the angle K. For, the arcs BD, IL, K ItA B have equal radii and equal chords: hence, they are equal (P. IV.); therefore, the angles BAD, IKL, measured by them, are also equal (P. XV.). PROBLEM V. To bisect a given arc, or a given angle. 10. Let AEB be a given are, and C its centre. Draw the chord AB; through C, draw CD perpendicular to AB (Prob. C III.): then will C-D bisect the are AEB (P. VI.). 20. Let A CB be a given angle. A With C as a centre, and any radius C-B, describe the arc BA; bisect it by the line CD, as just explained: then will CD bisect the angle A CB. For, the arcs AE and fEB are equal, from what was just shown; consequently, the angles A CE andC E/CB are also equal (P. XV.). Scholium. If each half of an arc or angle be bisected, the original arc or angle will be divided into four equal parts; and if each of these be bisected, the original arc or angle will be divided into eight equal parts; and so on. BOOK III. 85 PROBLEM VI. Through a given point, to draw a line parallel to a givne line. Let A be a given point, and B C a given line. From the point A as a centre, with a radius AE, greater than the B F E C shortest distance firom A to _Ba, describe an indefinite are PO; from \ P as a centre, with the same ra- A o dius, describe the are AF; lay off ED equal to AF, and draw AD: then will AD be the parallel required. For, drawing AE, the angles APF, EAD, are equal (P. XV.); therefore, the lines AD, EP are parallel (B. I., P. XIX., C. 1.). PROBLEMI VII. Given, two angles of a triangle, to construct the third angle. Let A and B be given angles of a triangle. Draw a line D)F, and at somle point of it, as E, construct the an- C\ H gle.TEP1 equal to A, and IIEC equal to B. Then, will CED -be F e (ulal to the required angle. 1For, the sum of the three angles at E is equal to two rigllt angles (B. I., P. I., C. 3), as is also the sum of the three angles of a triangle (B. I., P. XXV.). Consequently, the third angle CED must be equal to the third angle of the triangle. 86 GEOMETRY. PROBLEaM VIII. Given, two sides and the included angle of a triangle, to construct the triangle. Let B and C denote the given sides, and A the given angle. Draw the indefilite line DE, and at D construct an angle Z FDE, equal to the angle A; on / D_1, lay off D)H equal to the D B side C, and on DE, lay off C DG equal to the side B; draw GH: then will ZDGff be the required triangle (B. I., P. V.). PROBLEM IX. Given, one side and two angles of a triangle, to construct the triangle. The two angles may be either both adjacent to the given side, or one may be adjacent and'the other opposite to it. In the latter case, construct the third angle by Problem VII. We' shall then have two angles and their included side. Draw a straight line, and on it lay off DE equal to the given G Z, side; at D construct an angle equal to one of the adjacent angles, and at X construct an angle D E equal to the other adjacent angle; produce the sides D)' and CEG till they intersect at I: then will -DEH be the triangle required (B. I., P. VI.). BOOK III. 8q PROBLEM X. Given, the three sides of a triangle, to construct the triangle. Let A, B, and C, be the given sides. Draw 1DE, and make it equal F to the side A; from.D as a centre, with a radius equal to the E side B, describe an are; from E Al as a centre, with a radius equal to the side C, describe an arc intersecting the former at F; draw DF and EF: then will DEE be the triangle required (B. I., P. X.). Scholium. In order that the construction may be possible, any one of the given sides must be less than the sum of the other two, and greater than their difference (B. I., P. VII., S.). PROBLEM XI. Given, two sides of a triangle, and the angle opposite one of them, to construct the triangle. Let A and -B be the given sides, and (C the given angle. Draw an indefinite line D G, Ai -.. and at some point of it, as D), Bi construct an angle GDEY equal E to the given angle; on one side of this angle lay off the distance D DE equal to the side B adjacent to the given angle; from E as a centre, with a radius equal to the side opposite the given angle, describe an arc cutting the side DG at G; draw ER. Then will DE)G be the required triangle. 88 G E OMETRY. For, the sides DE and EG are equal to the given sides, and the angle D, opposite one of them, is equal to the given angle. Scholiunzm. When the side opposite the given angle is greater than the other given side, there will be but one solution. When the given angle is acute, and the side opposite the given angle is less than the other given side, and Ad glreater than the shortest dis- B H - E tance from E to )GC, there will be two solutions, 2DEG D and /DEE. When the side F\ opposite the given angle is equal to the shortest distance from E to D G, the arce will be tangent to DE, the angle opposite /DE will be a right angle, and there will be but one solution. W}hen the side opposite the given angle is shorter than the distance firom E to DG, there will be no solution. PROBLEM XII. Given, two adcjacent sides of a parallelogram and their inclhded an/le, to construct the parallelogram. Let A and B be the given sides, and C the given angle. Draw the line D)II, and F at some point as D), construct the angle lII)F equal to the angle C. Lay off D)E equal D to the side A, and 2D)i equal Ai to the side B; draw CFG B - C parallel to DE, and EG parallel to DIF: then will DCGE be the parallelogram required. BOOK III. 89 For, the opposite sides are parallel by construction; and consequently, the figure is a parallelogram (D. 28); it is also formed with the given sides and given angle. PROBLEM XIII. To find the centre of a given circumference. Take any three points A, B, and C, on the circumference E or are, and join themn by the' F chords AB, BC; bisect these (V-;/ chords by the perpendiculars /DE \ and FG: then will their point of intersection 0, be the centre A required (P. VII.). Scholium. The same construction enables us to pass a circumference through any three points not in a straight line. If the points are vertices of a triangle, the circle will be circumscribed about it. PROBLEM XIV. Through a given point, to draw a tangent to a given circle. There may be two cases: the given point may. lie on the circumference of the given circle, or it may lie without the given circle. 10. Let C be the centre of the A D given circle, and A a point on the circumference, through which the tan- I gent is to be drawn. Draw the radius CA, and at A draw AD perpendicular to AC: then will AD) be the tangent required (P. IX.). 90 GEOMETRY. 2~. Let C be the centre of the given circle, and A a point without the circle, through which the tangent is to be drawn. Draw the line A C; bisect it at 0, and from 0 as a centre, with a radius OC, describe the circumference B;;- ABOCD; join the point A with the' " points of intersection D and B: then will both A1D and AB be 0 tangent to the given circle, and there/ will be two solutions. For, the angles A BC and ADC A are right angles (P. XVIII., C. 2): hence, each of the lines AB and AD is perpendicular to a radius at its extremity; and consequently, they are tangent to the given circle (P. IX.). Scholium. The right-angled triangles ABC and AD C, have a common hypothenuse A C, and the side B C equal to D C(7; and consequently, they are equal in all their parts (B. I., P. XVII.): hence, A2B is equal to AD, and the angle CUAB is equal to the angle C-4AD. The tan. gents are therefore equal, and the line A C bisects the angle between them. PROBLEM XV. To inscribe a circle in a given triangle. Let AB C be the given B triangle. ~ Bisect the angles A and B, by the lines A 0 and BO, meeting in the point 0.. (I~rob. V.); from the point 0 A F C BOOK III. 91 let fall the perpendiculars OD, OE, OF, on the sides of the triangle: these perpendiculars will all be equal. For, in the triangles B0OD and B 0X, the angles O0BE and OBD are equal, by construction; the angles 0DB and OEB are equal, because both are right angles; and consequently, the angles B OD and B OR are also equal (B. I., P. XXV., C. 2), and the side OB is common; an.d therefore, the triangles are equal in.fil their parts (B. I., P. V.): hence, OD is equal to OE. In like manner, it may be shown that OD is equal to OE. From 0 as a centre, with a radius OD, describe a circle, and it will be the circle required. For, each side is perpendicular to a radius at its extremity, and is therefore tangent to the circle. Scholium. The lines that bisect the three angles of a triangle all meet in one point. PROBLEM XVI. On a given line, to construct a segment that shall contain a given angle. Let AB be the given line. M 1E E G,,D A G F Produce 1AB towards D; at B construct the angle J))BR equal to the given angle; draw BO perpendicular 92 GEOMETRY. to BE, and at the middle point G, of AB, draw G O perpendicular to AB; from their point of intersection 0, as a centre, with a radius OB, describe the arc AJM3B: then will the segment AMB be the segment required. M E E AEG D K F K For, the angle ABE, equal to EBD, is measured by half of the arc AKB (P. XXI.); and the inscribed angle A31~IB is measured by half of the same arc: hence, the angle AJJIB is equal to the angle EB1D, and consequently, to the given angle. BOOK IV. MEASUREMENT AND RELATION OF POLYGONS. DEFINITIONS. 1. SIMILAR POLYGONS, are polygons which are mutually equiangular, and which have the sides about the equal angles, taken in the same order, proportional. 2. In similar polygons, the part's which are similarly placed in each, are called homologous. The corresponding angles are homologoues angles, the corresponding sides are homologous sides, the corresponding diagonals are homologous diagonals, and so on. 3. SIMILAR Arcs, SECTORS, or SEGMENTS are those which correspond to equal angles at the centre. Thus, if the angles A and 0 are A equal, the arcs.BFC and D)GE are similar, the sectors BA C and ] DOEi are similar, and the segments BFC B D E and D GE are similar. F 4. The ALTITUDE OF A TRIANGLE, is the perpendicular distance from the vertex of either angle to the opposite side, or the opposite side produced. The vertex of the angle from which the distance is measured, is called the vertex of the triangle, and the opposite side, is called the base of the triangle. 94 G EOMETRY.. 5. The ALTITUDE OF A PAnALLELOGRAM, is the perpen. dicular distance between two opposite sides. These sides are called bases; one the upper, and the other, the lower base. 6. The ALTITUDE OF A TRAPEZOID, is the perpendicular distance between its parallel sides. These sides are called bases; one the / upper, and the other, the lower base. 7. The AREA OF A SURFACE, is its numerical value expressed in terms of some other surface taken as a unit. The unit adopted is a square described on the linear unit, as a side. PROPOSITION I. THEOR:]EM. Parallelograms which have equal bases and equal altitudes, aere equal. Let the parallelograms AB CD and EFGff have equal bases and equal altitudes: then will the parallelograms be equal. For, let them be so placed that their lower bases shall D H C G H coincide; then, because they have the same altitude, their A f' upper bases will be in the same line DC, parallel to AB. The triangles _DATl and C2BG, have the sides AD and B C equal, because they are opposite sides of the parallelogramn A C (B. I., P. XXVIII.); the sides AI and BGO equal, because they are opposite sides of the parallelogram AG; the angles DAlI and C3BG equal, because their BOOK IV. 95 sides are parallel and lie in the same direction (B. I., P. XXIV.): hence, the triangles are equal (B. I., P. V.). If from the quadrilateral AtBGD, we take away the triangle -DA-I, there will remain the parallelogram A G; if from the same quadrilateral ABGGD, we take away the tritriangle CBBG, there will remain the parallelogram A C hence, the parallelogram 4 C is equal, to the parallelogram EG (A. 3); which was to be proved. PROPOSITION II. THEOREM. A triangle is equal to one-half of a parallelogram having an equal base and an, equal altitude. Let the triangle ABC, and the parallelogram ABTD, have equal bases and equal altitudes: then will the triangle be equal to one-half of the parallelogram. For, let them be so placed that the base of D E F C C the triangle shall coinaide with the lower base A B' A B of the parallelogram; then, because they have equal altitudes, the vertex of the triangle will lie in the upper base of the parallelogram, or in the prolongation of that base. From A, draw AE parallel to B C, forming the parallelogram AB CE,. This parallelogram will be equal to the parallelogram ABED), from Proposition I. But the triangle ABC is equal to half of the parallelogram ABCE (B. I., P. XXVIII., C. 1): hence, it is equal\ to half of the parallelogram AB2 ED (A. 7); which was to be proved. Cor. Triangles having equal bases and equal altitudes are equal, for they are halves of equal parallelograms. 96 GE O 3I ETR Y. PEROPOSITION III. THIEOREM3'. Rectangles having equal altitudes, are proportional to their bases. There may be two cases: the bases may be commensurable, or they may be incommensiraable. 10. Let AB2CD and IIEFFJi, be two rectangles whose altitudes 1AD and FHK are equal, and whose bases AB and HE are commensurable: then will the areas of the rectangles be proportional to their bases. D C K F i * i * A B HI E Suppose that AB is to HE, as 7 is to 4. Conceive AB to be divided into 7 equal parts, and HE into 4 equal parts, and at the points of division, let perpendiculars be drawn to AB and HE. Then will ABCD lbe divided into 7, and HEFITr into 4 rectangles, all of which will be equal, because they have equal bases and equal altitudes (P. I.): hence, we have, AB C: ~ ItIE:FK: 7: 4. But we have, by hypothesis, AB: HE:: ~ 7: 4. From these proportions, we have (B. II., P. IV.), AB CD: HEFK:: AB: HE. Had any other numbers than 7 and 4 been used, the same proportion would have been found; wihich was to be proved. BOOK IV. 97 20. Let the bases of the rectangles be incommensurable: thlen will the rectangles be proportional to their blises. For, place the rectangle /HEKIr upon the rectangle AB CD, so that D lF' t' C it shall take the positionl AEil?. Then, if the rectangles are not proportional to their bases, let us sup- 1A -) 13 pose that ABD C AD ~ D AiE):: AB AO in which A 0 is greater than l E. Divide A 0 into equal parts, each less than OE; at least one point of division, as I,, will fall between E and 0; at this point, draw IKC perpendicular to A2B. Then, because AB and Al are commensurable, we shall have, from what has just been shown, AB CD AIfID AB AI. The above proportions have their antecedents the same in each; hence (B. II., P. IV., C.), A EtIf) D AIKD:: A O AI. The rectangle A. EPF.)D is less than AIKD; and if the above pr9portion were true, the line A 0 would be less than Al; whereas, it is greater. The fourth term of the proportion, therefore, cannot be greater than A-. In like manner, it may be shown that it cannot be less than. A-; consequently, it must be equal to AE: hence, AB C.D AED A A AE; thich was to be prozed. Cor. If rectangles have equal bases, they are to eaeh other as their altitudes.'1 98 GEOMETRY. PROPOSITION IV. THEOREM. AJny two rectangles are to each other as the products of their bases and altitudes. Let AB CD and A1EGx be two rectangles: then will ABBCD be to AEGiF, as AB x AD is to AE x AF. For, place the rectangles so that the angles DAB and EAF -, _ __C shall be opposite or vertical; sthen, produce the sides CD E A and GE till they meet in -. The rectangles AB CD and G F ADHR-E have the same altitude A.D': hence (P. III.), AB CD: ADfE: AB AE. The rectangles ADHE and AEGF have the same altitude AE: hence, ADHE:~ AEGF::: AD AF Multiplying these proportions, term by term (B. II., P. XHII.), and omitting the common factor ADHE (B. H., P. VII.), we have, ABCD AEGF:: AB x AD: AE x AF; wle4ch was to be proved. Scholium 1. If we suppose AP, and AF, each to be equal to the linear unit, the rectangle A-EGF will be the superficial unit, and we shall have, AB CD: 1:: AB x AD: 1; BOOK IV. 90 ABCD = AB x AD): hence, the area of a rectangle is equal to the product of its base and altitude; that is, the number of superficial units in the rectangle, is equal to the product of the number of linear units in its base by the number of linear units in its altitude. Scholium 2. The product of two lines is sometimes called the rectangle of the lines, because the product is equal to the area of a rectangle constructed with the lines as sides. PROPOSITION V. THEOREM. The area of a parallelogram is equal to the product of its base and altitude. Let ABC)D be a parallelogram, AB its base, and BE its altitude: then will the area of AB CD be equal to AB x BE. For, construct the rectangle D E C ABEF,, having the same base and altitude: then will the rectangle be equal to the parallelogram (P. I.); but the area of the A 1 rectangle is equal to AB x BE/: hence, the area of the parallelogram is also equal to AB x BE; uhich was to be proved. Cor. Parallelograms are to each other as the products of their bases and altitudes. If their altitudes are equal, they are to each other as their bases. If their bases are equal, they are to each other as their altitudes. 100 GEOMETRY. PROPOSITION VI. THEOREM. The area of a triangle is equal to half the product of its base and altitude. Let AB be a triangle, B C its base, and AD its altitude: then will the area of the triangle be equal to i-3 C x AD, For, from C, draw CE E A parallel to BA, and from A, /'. —! draw AE parallel to CB. The area of the parallelogram BCEA/ is BC x AD (P. V.); but the C B1 D triangle AB C is half of the parallelogram B CEA: hence, its area is equal to -B C x AD; which wqas to be proved. Cor. 1. Triangles are to each other, as the products of their bases and altitudes (B. II., P. VII.). If their altitudes are equal, they are to each other as their bases. If their bases are equal, they are to each other as their altitudes. Cor. 2. The area of a triangle is equal to half the product of its perimeter and the radius of the inscribed circle. For, let DEFE be a circle inscribed in the triangle ABC. B Draw OD, OE, and Oi, to E the points of contact, and OA, D. / OB, and 0C, to the vertioes. I. The area of OBC will be A F equal to -Ow x BC; the area of OA C will be equal to 4OF x A C; and the area BOOK IV. 101 of OAB will be equal to OD x A; and since OD, 0E, and OF, are equal, the. area of the triangle AB C (A. 9), will be equal to OD (AB + BC + CA). PROPOSITION VII. THIEOREM. The area of a trapezoid is equal to the product of its dvi-. tucle and half the sum of its parallel sides. Let AB CD be a trapezoid, )DE its altitude, and A/B and D)C its parallel sides: then will its area be equal to lIE x -(AB + DC). For, draw the diagonal AC, forming the triangles ABC and A CD. D C The altitude of each of these triangles is equal to DE. The area of ABC is equal to -}AB x DE (P. A E B VI.); the area of A CD) is equal to -PDC x RD: hence, the area of the trapezoid, which is the sum of the triangles, is equal to the sum of JAB 2x DE and IDC x DE, or to DE x ~L(AB + DC); twhich was to be proved. PROPOSITION VIII. THEOREM. l'he squacre described on the sum of twoo lines is equal to th.e siou of the squares described on the lines, increased by tzwice the rectangle of the lines. Let A1B and B C be two lines, E I D allclnd A C their sum: then will F I 0 Ac2 = - AB2 + B2 + 2AB x BC]. On A C, construct the square A CDE; from B, draw BH- par 102 GEOMETRY. allel to AE; lay off AP equal to AB, and from F, draw FG parallel to A C: then will IG and IXl be each equal to B C; and IB and IF, to AB. The square A CDE is composed of four parts. The part ABIF is E H H a square described on AB; the part I IGCDH[ is equal to a square described F G on B C'; the part B CGI is equal to the rectangle of AB and B C; and the part PFITE1 is also equal to A B C the rectangle of AB and B C: and because the whole is equal to the sum of all its parts (A. 9), we have, AC2 = AB2B-C2+ 2AB x BC; which cwas to be proved. Cor. If the lines AB and B C are equal, the four parts of the square on A C will also be equal: hence, the square described on a line is equal to four times the square described on half the line. PROPOSITION IX. THEOREM. The square cdescribed on the diJffrence of two lines is equal to thle sum, of thle squares described on the lines, diminished by tzoice the rectangle of the lines. Let AB and B C be two lines, and A C their difference: then will ACZ = AB 2 +BC2 _ 2AB x BC. On AB construct the square ABIF; from C draw CC parallel to BI; lay off CD) equal to A C, and from D draw ZDI parallel and equal to BA; complete BOOK IV. 103 the square LFLK: then will EEK be equal to B C, and'EFLK will be equal to the square of B C. The whole figure ABILKE is equal to the sum of the squares described on AB and B C. The KLE part C-BIG is equal to the rect-. angle of AB and B C; the part _D UGLK~ is also equal to the rect- A CB angle of AB and B C. If from the whole figure ABIYIEJ the two parts C-BIG and D GCLK] be taken, there will remain the part A CDE, which is equal to the square of A C: hence, A-2 = AB 4- C2 - 2AB x BC; which was to be proved. PROPOSITION X. THEOREM. The rectangle contcainecl by the sum and dffcrence of two lines, is equal to the diCference of their squares. Let AB and B C be two lines, of which AB1 is the greater: then will (AB + B C) (AB - BC) = A2 -B C On AB, construct the square ABIF; prolong AB, and make BK equal to BC; then will A E -- | —be equal to AB +:BC; from K, draw ]KL parallel to BI, and l make it equal to A C; draw LLE parallel to KEA, and CG parallel A (3 K to BI: then DCG is equal to B C, and the figure DIIIG is equal to the square o(, B C, and EY)DGF is equal to BIfILt~ 104 GEOETTRY. If we add to the figure ABHE, the rectangle BIKLH, we shall have the rectangle A LiLE, which is equal to the the rectangle of AB + BC and AZB - BC. If to the same figure F A-BITE, we add the rectangle DG) E, equal to BKLII, we EH L shall have the figure ABffID GF, wt4ich is equal to the difference of tile squares of AB and BC. But the sums of equals are equal (A. 2), lhence, (AB + BC) (AB-BC) - AB2-BC2; which was to be proved. PROPOSITION XI. THEOREM. The square described on the hypothenucse of a'igght-angleb d triangle, is equal to the sum of thle squares described o0n the other two sides. Let A/BC be a triangle, right-angled at A: then will BCo - A-B2 + A C. Construct the square B G on the side B C, the square Ail on the side AB, and the square 1I on the side K AC; from. A draw AD L perpendicular to BC, and,- - prolong it to E: then will /,, iDE be parallel to BF; - draw AF and ITC. B D In the triangles HfB C and ABF, we have lB f equal to AB, because they are sides of the same square; F E G BOOK IV. 105 B C equal to BE, for the same reason, and the included angles IB C and ABE equal, because each is equal to the anlgle AB C plus a right angle: hence, the triangles are equal in all their parts' (B. I., P. V.). The triangle ABE, and the rectangle BE, have the smlle base Bl1, and because DE is the prolongation of DA, their altitudes are equal: hence, the triangle ABE' is equal to half the rectangle BE (P. II.). The triangle JIB C, and the square BL, have the same base BH,, and because AC is the prolongation of AL (B. I., P. IV.), their altitudes are equal: hence, the triangle JIBC is equal to half the square of All But, the triangles A1BF and JfB C are equal: hence, the rectangle BE is equal to the square All. In the same manner, it may be shown that the rectangle DG is equal to the square AI: hence, the sum of the rectangles BE and DG, or the square BG, is equal to the sum of the squares AXl and AI; or, BC2 AB- + A wC2; which was to be proved. Cor. 1. The square of either side about the right angle is equal to the square of the hypothenuse diminished by the square of the other side: thus, AB- = BC2 - AC2; or, AC2 - BC2 - AB2. Cor. 2. If from the vertex of the right angle, a perpendicular be drawn to the hypothenuse, dividing it into two segments, BBD and D C, the square of the hypothenuse will be to the square of either of the other sides, as the hypothenuse is to the segment adjacent to that side. For, the square B G, is to the rectangle BE, as B C to B-D (P. III.); but the rectangle BE is equal to the square AEZr: hence, B. C2: AB2 B: C: BD. i36 GEOMETRY. In like manner, we have, BC2: AC2 BC DC. Cor. 3. The squares of the sides about the right angle are to each othzer as the adjacent segments of the hypothenuse. A For, by combining the proportions of the preceding corollary (B. II., P. IV., C.), we have, 3B D A-2: A:C2 B}D: C. Cor. 4. The square described on the diagonal of a square is clouble the given square. For, the square of the diagonal is H D G equal to the sum of the squares of the two sides; but the square of each side A C is equal to the given square: hence, E B F AC2 2A-'2; or, AC2 = 2BC2. Cor. 5. Fromn the last corollary, we have, A 2::: 2: 1; hence, by extracting the square root of each term, we have, A C: AB:: 1; that is, thie diagonal of a square is to the side, as the square root of two to one; consequently, the dicgonal am~i the side of a square are incommensurable. BO OK IV. 107 PROPOSITION XII. THEOREIM. In any triangle, the square of a side opposite an acute angle, is equal to the sum of the squares of the base antd the other side, diminished by twice the rectangle of the base and the distance from, the vertex of the acute angle to the foot of the perpendicular cdrawn from the vertex of the opposite angle to the base, or to the base procduced. Let AB C be a triangle, C one A of its acute angles, B C its base, and AD the perpendicular drawn from A to B C, or B C produced; then will A-B2 = C + AC -2B C D. B D C For, whether the perpendicular meets the base, or tile base produced, we have 2BD equal to the difference of B C and CD: hence (P. IX.), _ _ _ A BD2 = BC2 + C1D2-2BC x CD. Adding AjD2 to both members, wre have, D ]).. C B-2 + A~D = BC2 + CD2 + Az - 2BC x CD. But, B-2 + A —2 = A-B2, and CD2 + A A)2 = 1ff2 hence, A-2 = B1C2 + AC2 - 2B C x CD; zwhich was to be proved. 108 GEOMETRY. PROPOSITION XIII. THEOREM. Inz any obtuse-angled triangle, the square of the side opposite the obtuse angle is equal to the sum of the squares of the base and the other side, increased by twice the rcctangle of the base and the distance fron the vertex of the obtuse angle to the foot of the perpendicular drawn fJorn the vertex of the opposite angle to the base produced. Let ABC be an obtuse-angled triangle, B its obtuse angle, B C its base, and AD the perpendicular drawn from A to BC produced; then will AC - BC 2 + AB2~2 D 2B C x BD For, C'D is the sum of B BC A and B-D: hence (P. VIII.), CU -B = C2 B-2 + 2 BC x IBD. Adding A-D2 to both members, )B C and reducing, we have, A2 _2 J= 2 + A-2 Jr 2BC x BD; which was to be proved. Scholium. The right-angled triangle is the only one in which the sum of the squares described on two sides is equal to the square described on the third side. PROPOSITION XIV. THEOREM. liz any triangle, the sumn of the squares described on 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 point of that side to the vertex of the opposite angle. Let ABC be any triangle, and EA a line drawn from BOOK IV. 109 the middle of the base B C to the vertex A: then will AB2 + A+2 - 2BE2 + 2EA2. Draw AD perpendicular to B C; then, from Proposition XII., w e have, AC2C2 _ ~E_ + i2 _ 2EC X ED. From Proposition XIII., we have, j — _BE2 + I-2: + 2BE x ED. B:. Adding these equations, member to member (A. 2), recollecting that BE is equal to EC, we have, AB2 - A2 = 2BE2+ 2'A2; which was to be proved. Cor. Let AB ICD be a parallelogram, and BD, A C, its diagonals. Then, since the diagonals mutually bisect each other (B. I., P. B C XXXI.), we shall have, ABp2 + f -f- _ 2A4, + ~2BE-2'; and, A- D G-2 + -DA = 2G 2 + 2DE2; whence, by addition, recollecting that AE is equal to CE, and BE to D)E, we have, A-B2 + -B2 + CD-2 ~D A-12 = 4CGE2 + 4DE2; but, 4C-E2 is equal to A C2, and 4D)E'2 to BD2 (P. VIII., C.): hence, jABj ~ + G D2 + A-t - + BDi = A - +. That is, th the sarumes of the suaides of the ss of a parcallelogram, is equal to the sum of the squares of its diagonals. 110 GEOMETRY. PROPOSITION XV. THEOREM. In any triangle, a line drawn parallel to the base divides the other sides proportionally. Let AB C be a triangle, and D)E a line parallel to the base BC. then AD: DB:: CE. Draw EB and D C. Then, because A the triangles AED and FDEB have their bases in the same line AB, and their vertices at the same point E, they will D have a common altitude: hence, (P. VI., C.) AED: DEB:: AD: DB. B C The triangles AED and EDC, have their bases in the same line A C, and their vertices at the same point D); they have, therefore, a common altitude; hence, AED: ED C:: AE: EC. But the triangles DEB and EDC have a common base DiE, and their vertices in the line ABC, parallel to DE; they are, therefore, equal: hence, the two preceding proportions have a couplet in each equal; and consequently, the remaining terms are proportional (B. II., P. IV.), hence, AD: 1B:: AE: EC; wohich was to be proved. Cor. 1. We have, by composition (B. II., P. VI.), AD + DB: AD:: AE +EC: AE; BOOK IV. 111 or, AB AD: AC: AE; and, in like manner, AB: DB AC: A EC. Cor. 2. If any number of parallels be drawn cuttingtwo lines, they will divide the lines proportionally. For, let 0 be the point where AS and CD meet. In the triangle OEF, o the line A C being parallel to the base EF, we shall have, A C OE: AE: OF: CF. E In the triangle OGH1 we shall have, G OE: EG:: OF:; /F\ hence (B. II., P. IV., C.), AE: EG:: CF: Fi. In like manner, EG GB:: HD E.i; and so on. PROPOSITION XVI. TH]EOR.EM. If a line divides two sides of a triangle proportionally, it will be parallel to the third side. Let AB C be a triangle, and let -DE A divide AB and AC, so that AD:DB:: AE EC; D then will DE be parallel to B C. Draw DC and EB. Then the tri- B 112 GEOMETRY. angles ADE and DEB will have a common altitude; and consequently, we shall have, ARDE: DEB:: AD: DB. A The triangles ADE and ED C have also Ia common altitude; and consequently, we D shall have, AD)E ED) C:: AE EC; B C but, by hypothesis, AD: DB A: E EC; hence (B. II., P. IV.), ADE DE /)B:ADE ED C. The antecedents of this proportion being equal, the consequents will be equal; that is, the triangles DEIB and ED C are equal. But these triangles have a common base DE: hence, their altitudes are equal (P. VI., C.); that is, the points B and C, of the line BCC, are equally distant firom D)E, or DE prolonged: hence, B C and DE are parallel (B. I., P. XXX., C.); which was to be proved. PROPOSITION XVII. THEOREMI. The line which bisects the vertical angle of a triangle, divides the base into segments proportional to the adjacent sides. Let AD bisect the vertical angle A of the triangle BA C: then will the segmepts _BD and DC be propor. tional to the adjacent sides BA and CA. From C, draw CE parallel to _DA, and produce it BOOK IV. 113 until it meets BA prolonged, at E. Then, because CE and DA are parallel, the angles BAD and A-EC are equal (B. I., P. XX., C. 3); the angles DAC and ACE are E also equal (B. I., P. XX., C. 2). But, BAD and DA C are Ap. equal, by hypothesis; consequently, AEC and A CE are equal:/ \ hence, the triangle A CE is isosceles, AE being equal to B D C AC. In the triangle BEC, the line AD is parallel to the base EC: hence (P. XV.), BA: AE:: BBD: DC; or, substituting A C for its equal AE, BA C:: BD: DCC; which was to be proved. PROPOSITION XVIII. THEOREM. Triangles which are mutually equiangular, are similar. Let the triangles AB C and DEF have the angle A equal to the angle D), the angle B to the angle E, and the angle C to the angle -F: then will they be similar. For, place the triangle DEF upon the triangle D AB C, so that the angle E shall coincide with the angle B; then will the point PX fall at some B H C E F point 1; of BC; the point D at some point G, of BA; 8 114 GEOMETRY. the side DPR will take the position GH, and.BGH will be equal to EDE. Since the angle BHG A is equal to B CA, G G D will be parallel' to A C (B. I., P. XIX., C. 2); and consequently, we shall B H C E F have (P. XV.), B BA BG: BC BH; or, since BG is equal to ED, and BH to EF, BA ED:: BC EF. In like manner, it may be shown that B C EF:: CA PFD; and also, CA: _FD: AB: DE; hence, the sides about the equal angles, taken in the same order, are proportional; and consequently, the triangles are similar (D. 1); which was to be proved. Cor. If two triangles have two angles in one, equal to two angles in the other, each to each, they will be similar (B. I., P. XXV., C. 2). PROPOSITION XIX. THEOREM. Triangles which have their corresponding sides proportional, are similar. In the triangles ABC and DEF, let the corresponding sides be proportional;. that is, let 1BOOK IV. 115 AB: DE:: BC: EF:: CA: D; then will the triangles be similar. For, on BA lay off BG equal to ED; on BC lay off IBH equal to EF, A and draw GI. Then, D because B G is equal to DE, and BH to EF, we have, B UH C BA B::: B.C BH; hence, GH is parallel to A C (P. XVI.); and consequently, the triangles BA C and BGCH are equiangular, and therefore similar: hence, from Prop. XVIII., we have, BC:C B:: CA: CG. But, by hypothesis, BC: EF CA ED; hence (B. II., P. IV., C.), we have, B:t EF:::IG: _F. But, BMiL is equal to EF; hence, HJG is equal to FD. The triangles BHG and EFD have, therefore, their sides equal, each to each, and consequently, they are equal in all their parts. Now, it has just been shown that BHG and BCGA are similar: hence, EFiD and BCA are also simi. Jar; which was to be proved. Scholium. In order that polygons may be similar, they must fulfill two conditions: they must be mutually equiangular, and the corresponding sides must be proportional. In the case of triangles, either of these conditions involves the other, which is not true of any other species of polygons. 116 GEOMETRY. PROPOSITION XX. THEOREM. Triangles which have an angle in each equal, and the including sides proportional, are similar. In the triangles ABC and -DEF, let the angle B be equal to the angle E; and suppose that BA: ED: B C EF; then will the triangles be similar. For, place the angle E A upon its equal B; F G will fall at some point of BC, as Hr; D will fall at some point of BA, as B H C E F G; DYF will take the position GC, and the triangle DEIF will coincide with GCBH, and consequently, will be equal to it. But, from the assumed proportion, and because B G is equal to ED, and B1H to EF we have, _BA BG:: BC: B; hence, Gff is parallel to AC; and consequently, BA C and BGIl are equiangular, and therefore similar. But, EDF is equal to B G2ff: hence, it is also similar to BA C; which was to be proved. PROPOSITION XXI. THEOREM. Triangles which have their sides parallel, each to each, or perpendicular, each to each, are similar. 1~. Let the- triangles ABC and DlEF have the side AB parallel to DE, B C to E,: and CA to FD: then will they be similar. BOOK IV. 117 For, since the side AB is parallel to DE/ and B C to EF, the angle B is equal to the angle E (B. L, P. XXIV.); in like manner, A the angle C is equal to D the angle E; and the angle A to the angle D); the triangles are, therefore, mutually equiangular, and B consequently, are similar (P. XVIII.); which was to,: proved. 2~. Let the triangles ABBC and _DEF have the side AB perpendicular to DE, B C to E, and CA to ED: then will they be similar. For, prolong the sides of the tri- A angle )DET till they meet the sides of the triangle ABC. The sum of the interior angles of the quadrilateral NE BIEGG is equal to four right angles / (B. I., P. XXVI.); but, the angles B G C EIB and EGB are each right angles, by hypothesis; hence, the sum of the angles IEG IBG is equal to two right angles; the sum of the angles IEG and.DEF is equal to two right angles, because they are adjacent; and since things which are equal to the same thing are equal to each other, the sum of the angles IEG and IBG is equal to the sum of the angles IEG and DEF; or, taking away the common part IEG, we have the angle IB G equal to the angle DEE. In like manner, the angle G CH may be proved equal to the angle EFDI, and the angle HAI to the angle EDEF'; the triangles ABC and DEF are, therefore, mutually equiangular, and consequently, similar; which was to be proved. Cor. 1. In the first case, the parallel sides are homolo 118 GEOMETRY. gous; in the second case, the perpendicular sides are homologous. Cor. 2. The homologous angles are those included by sides respectively parallel or perpendicular to each other. Scholium. When two triangles have their sides perpendicular, each to each, they may have a different relative position from that shown in the figure. But we can always construct a triangle within the triangle ABC, whose sides shall be parallel to those of the other triangle, and then the demonstration will be the same as above. PROPOSITION XXII. THEOREM. If a line be drawn parallel to the base of a triangle, and lines be drawn fronm the vertex of the triangle to points of the base, these lines will divide the base and the parallel proportionally. Let AB C be a triangle, B C its base, A its vertex, DE parallel to B C, and AI, AG, AH; lines drawn from A to points of the base: then will DI: BF: IK: JFG' l: GI::'LE: HC. For, the triangles AID and A AEB, being similar (P. XXI.), we have, AI:AF::DI BF; and, the triangles AIfK and AG, F being similar, we have, AI AF IK: FG; hence, (B. II., P. IY.), we have, BOOK IV. 119 DI: BF:: IK F G. In like manner, IK FG:': KL G/f, and, KL: G:H:: LE: HC; hence (B. II., P. IV.), 1DI: BF::IK E: FG:: KL GH:: LE H; which was to be proved. Cor. If B C is divided into equal parts at F, G, and H, then will DE be divided into equal parts, at I, K, and L. PROPOSITION XXIII. THEOREM. If, in a right-angled triangle, a perpendicular be drawn from the vertex of the right angle to the hypothenuse: 1~. The triangles on each side of the perpendicular will be similar to the given triangle, and to each other: 20. Each side about the right angle will be a mean proportional between the hypothenuse and the adjacent segment: 3~. The perpendicular will be a mean proportional between the two segments of the hypothenuse. 1~. Let ABC be a right-angled triangle, A the vertex of the right angle, B C the hypothenuse, and AD perpendicular to BC: then will ADB and ADC be similar to AB C, and consequently, similar to each other. C The triangles A/DB and ABC have the angle B common, and the angles A:DB and 120 GEOMETRY. BAlC equal, because both are right angles; they are, therefbre, similar (P. XVIII., C). In like manner, it may be shown that the triangles AD C and ABC are similar; and since ADB and ADC are both similar to AB C, they are similar to each other; which was to be proved. 2~. AB will be a mean pro- A portional between B C and BD); and A C will be a mean proportional between CB and C.D. For, the triangles ADB and 33 ) C Bl C being similar, their homologous sides are proportional: hence, BC: AB:: AB: B~D. In like manner, BC: AC:: AC: DC; which was to be proved. 30~. AD will be a mean proportional between BD and DC. For, the triangles ADB and ADDC being similar, their homologous sides are proportional; hence, BD: AD:: AD DC; which was to be proved. Cor. 1. From the proportions, BC: AB:: AsB: BD, and, BC: AC:: AC: SC, we have (B. II., P. I.), B-2 = BC x BD, and, AC2 BC x DC; BOOK IV. 121 whence, by addition, B2.- + X C2 = BC (BD + DC); or, AB2+ - C2 = BC2; as was shown in Proposition XI. Cor. 2. If from any point A, in a semi-circumference BA C, chords be drawn to the extremities B and C of the diameter BC, and a perpendicular AD be drawn to the diameter: then B C will ABC be a right-angled triangle, right-angled at A; and from what was proved above, each chord will be a mean proportional between the diameter and the adjacent segment; and, the perpendicular will be a mean proportional between the segments of the diameter. PROPOSITION XXIV. THEOREM. Triangles which have an angle in each equal, are to each other as the rectangles of the including sides. Let the triangles GHK and AB C have the angles G and A equal: then will they be to each other as tlhe rectangles of the sides about these angles. For, lay off AD equal to GH, AE to GlK, and G draw DE; then will the D triangles ADE and GHf H -i. E/ be equal in all their parts. Draw EB. C 122 GEOMETRY. The triangles ADE and ABE have their bases in the same line AB, and a common vertex E; therefore, they have the same altitude, and consequently, are to each other as their bases; that is, ADE ABE AD ~ AB. The triangles ABE and G A AB C, have their bases in the same line AC, and a H E common vertex B; hence, K ABE: AB C: AE AB C multiplying these proportions, term by term, and omitting the common factor ABE (B. II., P. VII.), we have, ADE: ABC:: AD x AE AB x AC; substituting for APE, its equal, GHK, and for AD x AE, its equal, Gff x GK, we have, GHffK~ ABC ~:: GTx GK: AB x AC; which was to be proved. Cor. If ADE and ABC are similar, the angles D and B being homologous, DE will be parallel to B C, and we shall have, AD: AB::AE AC; A hence (B. II., P. IV.), wj have, ADE ABE ABE ABC; that is, ABE is a mean proportional between ADE and ABC. BO0OK IV. 123 PROPOSITION XXV. THEOREM. Similar triangles are to each other as the squares of their homologous sides. Let the triangles ABC and IDEF be similar, the angle A being equal to the angle D), B to E, and C to F: then will the triangles be to each other as the squares of any two homologous sides. Because the angles A and D are equal, we have (P. XXIV.), AB C D: J)E:: AB x A C: -DE x EE; and, because the triangles A are similar, we have, AB: DE:: AC: DF; multiplying the terms of B E F Z:)C BE F this proportion by the corresponding terms of the proportion, A C: A C DE, we have (B. II., P. XII.), AB x AC: DE x DF:: AC2: F2; combining this, with the first proportion (B. II., P. IV.), we have, ABC 1)EF A C2 ~. In like manner, it may be shown that the triangles are to each other as the squares of AB and DE, or of B C and EF; which was to be proved. 124 GEOMETRY. PROPOSITION XXVI. THEOREM. Similar polygons may be divided into the same number of triangles, similar, each to each, and similarly placed. Let ABCD.E and F-'IIIK be two similar polygons, the angle A being equal to the angle F, B to G, C to H, and so on: then can they be divided into the same number of similar triangles, similarly placed. For, from A draw the diagonals AC, C AD, and from F, B 7 G homologous with A,,'" draw the diagonals A P-,11 FI, to the ver- K tices ff and I, hon- E ologous with C and -D. Because the polygons are similar, the triangles ABC and FGfH have the angles B and G equal, and the sides about these angles proportional; they are, therefore, similar (P. XX.). Since these triangles are similar, we have the angle A CB equal to FJG, and the sides ACt and FIPH, proportional to B C and G1, or to CD) and Hl. The angle B CD being equal to the angle GHI/, if we take from the first the angle A CB, and from the second the equal angle PFCG, we shall have the angle A C)D equal to the angle EP1I: hence, the triangles A CD and EIH have an angle in each equal, and the including sides proportional; they are therefore similar. In like manner, it may be shown that ADE and FIE are similar; which was to be proved. Cor. 1. The corresponding triangles in the two polygons are homologous triangles, and the corresponding diagonals are homologous diagonals. BOO0K IV. 125 Cor. 2. Any two homologovs triangles are like parts of the polygons to which they belong. For, ABC and FGH being similar, we have, AB C: FGH:: AC2.; and, for a like reason, Ac~ FHI AC2 ~ -; whence, AB C FGI:: A CD: FPHI; and, in like manner, A C)D PHfI:: AD E: IKE Cor. 3. If two polygons are made up of similar triangles, similarly placed, the polygons themselves will be similar. PROPOSITION XXVII. THEOREM. Tfhe perimeters of similar polygons are to each other as any two homologous sides; and the polygons are to each other as the squares of any two homologous sides. 10. Let ABCDE and FGMHIK be similar polygons: then will their perimeters be to each other as any two homologous sides. For, any two homologous sides, as AB B and FG, are like parts of the perimeters to A which they belong: K hence (B. II., P. IX.), E the perimeters of the polygons are to each other as AB to FG, or as any other two homologous sides; which was to be proved. 126 GEOMETRY. 20. The polygons will be to each other as the squares of any two homologous sides. For, let the poly- C gons be divided into B G H homologous triangles (P. XXVI., C. 1); then, because the K homologous triangles E ABC and FGIT are like parts of the polygons to which they belong, the polygons will be to each other as these triangles; but these triangles, being similar, are to each other as the squares of AB and PG: hence, the polygons are to each other as the squares of AB and FG, or as the squares of any other two homologous sides; which was to be proved. Cor. 1. Perimeters of similar polygons are to each other as their homologous diagonals, or as any other homologous lines; and the polygons are to each other as the squares of their homologous diagonals, or as the squares of any other homologous lines. Cor. 2. If the three sides of a right-angled triangle be made homologous sides of three similar polygons, these polygons will be to each other as the squares of the sides of the triangle. But the square of the hypothenuse is equal to the sum of the squares of the other sides, and consequently, the polygon on the hypothenuse will be equal to the sum of the polygons on the other sides. PROPOSITION XXVIII. THEOREM. If two chords intersect in a circle, their segments will be reciprocally proportional. Let the chords AB and CD intersect at 0: then BOOK IV. 127 will their segments be reciprocally proportional; that is, one segment of the first will be to one segment of the second, as the remaining segment of the second is to the remaining segment of the first. For, draw CA and BD. Then C B will the angles OD)B and OA C be equal, because each is measured by half / I of the arc CB (B. III., P. XVIII.). A The angles OBD and O CA, will also be equal, because each is measured by half of the arc AD: hence, the triangles OBD and OCA are similar (P. XIX., C.), and consequently, their homologous sides are proportional: hence, DO: AO:: OB: OC; which was to be proved. Cor. From the above proportion, we have, DO x OC = AO x OB; that is, the rectangle of the segments of one chord is equal to the rectangle of the segments of the other. PROPOSITION XXIX. THEOREM. If from a point without a circle, two secants be drawn terminating in the concave are, they will be reciprocally proportional to their external segments. Let OB and O C be two secants terminating in the concave arc of the circle B CD: then will OB: OC:: OD: OA. 128 GEOMETRY. For, draw A C and D-B. The triangles ODB and OA C have the angle O common, and the angles OBD and 0 CA equal, because each is measured by half of the arc AD: hence, they are similar, and consequently, their homologous sides are proportional; whence, OB 0C:: OD OA; which was to be proved. Cor. From the above proportion, we have, OB x OA = OC x OD; that is, the rectangles of each secant and its external segment are equal. PROPOSITION XXX. THEOREM. If from a point without a circle, a tangent and a secant be drawn, the secant terminating in the concave are, the tangent will be a mean proportional between the secant and its external segment. Let AD C be a circle, O C a secant, and OA a tan. gent: then will OC OA:: OD. For, draw AD and A C. The triangles OAD and OA C will have the angle 0 common, and the angles OAD and A CD equal, because each is mea- A sured by half of the arc AD (B. Ill., P. XVIII., P. XXI.); the triangles are therefore similar, and consequently, their BOOK IV. 129 homologous sides ore proportional: hence, OC: OA:: OA: OD; which was to be proved. Cor. From the above proportion, we have, AO2 = OC x OD; that is, the square of the tangent is equal to the rectangle of the secant and its external segment. PRACTICAL APPLICATIONS. PROBLEM I. To aivide a given line into parts proportional tb given lines, also into equal parts. 1~. Let AB be a given line, and let it be required to divide it into parts proportional to the lines P, Q and 1R. From one extremity A, I F B draw the indefinite line AG, A making any angle with AB;, lay off AC equal to P, CD equal to Q, and DE equal to B; draw EB, and D, from the points C and D, G draw CI and DF parallel to EB: then will Al, -IF: and FB, be proportional to P, Q, and R (P XV., C. 2). 9 130 GEOMETRY. 20. Let AH be a given line, and let it be required to divide it into any number of equal parts, say five. From one extremity I- draw the indefinite A C D E F H line AG; take Al equal' to any convenient line, k and lay off 1K, KL, LT, Lf/, and JMB, each equal to AI. Draw G BH, and from I, K, L, and M, draw the lines IC, KD, LE, and 1E-i, parallel to BH: then will AH be divided into equal parts at C, D, E, and F (P. XV., C. 2). PROBLEg II. To construct a fourth proportional to three given lines. Let A, B, and C, be D the given lines. Draw.DE and DF, making Al any convenient angle with each other. Lay off DA equal to A, -DB equal E/\F to B, and D C equal to C; draw AC, and from B draw BX parallel to A C: then will DX be the fourth proportional required. For (P. XV., C.), we have, DA: DB:: DC: f)X; or, A: B:: C: DX. Cor. If DC is made equal to DB, DX will be a third proportional to )DA and DB, or to A and B. BOOK IV. 131 PROBLEM III. To construct a mean proportional between two given linee. Let A and B be the given G lines. On an indefinite line, lay off DE equal to A, and EFl equal D to B; on DDF as a diameter de- E scribe the semi-circle D)GF, and Al-H draw EG perpendicular to DlF: then will EG be the mean proportional required. For (P. XXIII., C. 2), we have, DE: EG::: EF; or, A: G:: EG: B. PROBLEM IV. To divide a given line into two such parts, that the greater part shall be a mean proportional between the whole line and the other part. Let AB be the given line. At the extremity B, draw BC perpendicular to AB, and make it equal to half of AB. With C as a centre, and C7B / [ as a radius, describe the are A F B DBE; draw AC, and produce it till it terminates in the concave are at E; with A as centre and AD as radius, describe the arc DF: then will Ali be the greater part required. 132 GEOMETRY. For, AB being perpendicular to CB at B, is tangent to the arc DBE: hence (P. XXX.), AE: AB:: AB: AD; D and, by division (B. II., P. VI.), A AE- AB: AB AB-AD: AD. But, DE is equal to twice CB, or to AB: hence, AE- AB is equal to AD1, or to AF; and AB - AD is equal to AB - A-, or to FB: hence, by substitution, AF: AB: B: AF; and, by inversion (B. II., P. V.), AB AF:: A: FB. AScholium. When a line is divided so that the greater segment is a mean proportional between the whole line and the less segment, it is said to be divided in extreme and moean ratio. Since AB and DE are equal, the line AE is divided mi extreme and mean ratio at D; for we have, from the first of the above proportions, by substitution, AE: DE:: DE::'A-D. BOOK IV. 133 PROBLEM V. Through a given point, in a given angle, to draw a line so that the segments between the point and the sides of the angle shall be equal. Let B CD be the given angle, and A the given point. Through A, draw AE parallel to D C; lay off EF equal to CE, and draw FAD: then will AF and AD be the segments required. F For (P. XV.), we have, FA: A:: FE: EC; AD but, FE is equal to EC; hence, FA is equal to AD. PROBLEM VI. To construct a triangle equal to a, given polygon. Let AB CDE be the given polygon. Draw CA; produce PA, and c draw B G parallel to CA; draw the line CG. Then the triangles / BAC and GAC have the com- / mon base A C, and because their G A E F vertices B and G lie in the same line BG parallel to the base, their altitudes are equal, and consequently, the triangles are equal: hence, the polygon G CDE is equal to the polygon AB CDE. Again, draw CE; produce AP and draw DF parallel to C7E; draw also CF; then will the triangles FUE and D CE be equal: hence, the triangle G CF is equal to the polygon G CDE, and consequently, to the given polygon. In like manner, a triangle may be constructed equal to any other given polygon. 134 GEOMETRY. PROBLEM VII. To costruct a square equal to a given triangle. Let ABC be the given triangle, AD its altitude, ancl BC its base. Construct a mean pro- A portional between AD and half of BC (Prob. III.). Let XY be that / mean proportional, and on B D C it, as a side, construct a square: then will this be the square required. For, from the construction, XY2 = ~BC x AD = area ABC. &cholium. By means of Problems VI. and VII., a square may be constructed equal to any given polygon. PROBLEM VIII. On a given line, to construct a polygon similar to a given polygon. Let FG be the given line, and AB CUDE the given polygon. Draw A C and AD. At _, construct the angle GCEH equal B H to BAC, and at C A GG the angle FGf equal A>D to ABC; then will CFGI be similar to E ABC (P. XVIII., C.) BOOK IV. 135 In like manner, construct the triangle FPII similar to A CID, and FI similar to ADE; then will the polygon.FGHIK be similar to the polygon AB CDE (P. XXVI., C.). PROBLEM IX. To construct a square equal to the.sum of two given squares, also a square equal to the difference of two given squares. 10. Let A and B be the sides of the given squares, and let A be the greater. Construct a right angle C A B CDE; make DE equal B to A, and D C equal to B; draw CE, and on it D E construct a square: this square will be equal to the sum of the given squares (P. XI.). 20. Construct a right angle CIDE. Lay off D C equal to B; with C C as a centre, and CE, equal to A, as a radius, describe an arc cutting DE at E; draw CE, and on DE construct D /E a square: this square will be equal to the difference of the given squares (P. XI., C. 1). Scholium. By means of Probs. VI., VII., VIII., and IN.. a polygon may be constructed similar to two given polygolns, and equal to their sum, or to their difference (P. XXVII., C.). BOOK V. REGULAR POLYGONS.-AREA OF THE CIRCLE, DEFINITION. 1. A REGULAR POLYGON is a polygon which is both equilateral and equiangular. PROPOSITION I. THEOREM. Regular polygons of the same number of sides are similar. Let ABCD.EF and abedef be regular polygons of the same number of sides: then will they be similar. For, the corresponding angles in each are equal, B D because any angle in either polygon is equal F C to twice as many right angles as the polygon has sides, less four, divided by the number of angles (B. I., P. XXVI., C. 4); and further, the corresponding sides are proportional, because all the sides of either polygon are equal (D. 1): hence, the po!ygons are similar (B. IV., D. 1); which was to be proved BOOK V. 137 PROPOSITION II. THEOREM. The circumference of a circle may be circumscribed about any regular polygon; a circle may also be inscribed within it. 1~. Let ABCF be a regular polygon: then can the circumference of a circle be circumscribed about it. For, through three consecutive vertices A, B, C, describe the circum- B ference of a circle (B. II., Problem A c XIII., S.). Its centre 0 will lie on PO, drawn perpendicular to B C, H at its middle point P; draw OA and OD. G Let the quadrilateral OPCD be F turned about the line OP, until PC falls on PB; then, because the angle C is equal to B, the side C-D will take the direction BA; and because C)D is equal to BA, the vertex D, will fall upon the vertex A; and consequently, the line OD will coincide with OA, and is, therefore, equal to it: hence, the circumference which passes through A, B, and C, will pass through D. In like manner, it may be shown that it will pass through all of the other vertices: hence, it is circumscribed about the polygon; which was to be proved. 20. 4 circle may be inscribed within the polygon. For, the sides AB, B C, &c., being equal chords of the circumscribed circle, are equidistant from the centre 0: hence, if a circle be described from 0 as a centre, with OP as a radius, it will be tangent to all of the sides of the polygon, and consequently, will be inscribed within it; which was to be proved. 138 GEOMETRY. Scholium. If the circumference of a circle be divided into equal arcs, the chords of these arcs will be sides of a regular inscribed polygon. For, the sides are equal, because they are chords of equal arcs, and the angles are equal, because they are measured by halves of equal arcs. If the vertices A, B, C, &c., e D of a regular inscribed polygon be joined with the centre 0, the tri- \ angles thus formed will be equal, H C because their sides are equal, each to each: hence, all of the angles about the point O are equal to each other. DEFINITIONS. 1. The CENTRE OF A REGULAR POLYGON, is the common centre of the circumscribed and inscribed circles. 2. The ANGLE AT THE CENTRE, is the angle formed by drawing lines from the centre to the extremities of either side. The angle at the centre is equal to four right angles divided by the number of sides of the polygon. 3. The APOTHEr, is the distance from the centre to either side. The apothem is equal to the radius of the inscribed circle. BOOK V. 139 PROPOSITION III. PROBLEM. To inscribe a square in a given circle. Let ABCD be the given cir- B cle. Draw any two diameters A C and BD) perpendicular to each other; they will divide the circum- A C ference into four equal arcs (B. III., P. XVII., S.). Draw the chords AB, BC, CD, and DA,4: then D will the figure AB CD be the square required (P. II., S.). Scholium. The radius is to the side of the inscribed square as 1 is to /.T PROPOSITION IV. THEOREM. If a regular hexagon be inscribed in a circle, any side will be equal to the radius of the circle. Let ABD be a circle, and AB C(DEH a regular inscribed hexagon: then will any side, as AB, be equal to the radius of the circle. Draw the radii OA and OB. Then will the angle A OB be A equal to one-sixth of four right angles, or to two-thirds of one H C —-— C right angle, because it is an an- gle at the centre (P. II., D. 2). The sum of the two angles OAB A B and OBA is, consequently, equal to four-thirds of a right angle (B. I., P. XXV., C. 1); but, the angles OAB and OBA are equal, because the opposite sides OB and OA are equal: hence, each is equal 140 GEOMETRY. two-thirds of a right angle. The three angles of the triangle A OB are therefore, equal, and consequently, the triangle is equilateral: hence, AB is equal to OA; which was to be proved. PROPOSITION V. PROBLEM. To inscribe a regular hexagon in a given circle. Let ABE be a circle, and 0 its centre. Beginning at any point of B the circumference, as A, apply the radius OA six times as a chord; then will AB CDET be the hexagon required (P. IV.). \ Cor. 1. If the alternate vertices of the regular hexagon be joined by the lines A C, B CfE, and -EA, the inscribed triangle A CE will be equilateral (P. II., S.). Cor. 2. If we draw the radii OA and 0OC, the figure A 0 CB will be a rhombus, because its sides are equal: hence (B. IV., P. IV., C.), we have, PB2 + ]V2 + 0A2 -C2 =A2 +;oB or, taking away from the first member the quantity O-Aa, and from the second its equal OB2, and reducing, we have, 30A2 = A-2; whence (B. II., P II.), A-C2: OA 3: 1; BOOK V. 141 or (B. II., P. XII., C. 2), AC: OA::: 1; that is, the side of an inscriber equilateral triangle is to thle radius, as the square root of 3 is to 1. PROPOSITION VI. THEOREM. If the radius of a circle be divided in extreme and mean ratio, the greater segment will be equal to one side of a regular inscribed decagon. Let A CG be a circle, OA its radius, and AB, equal to OMf the greater segment of OA when divided in extreme and mean ratio: then will AB be equal to the side of a regular inscribed decagon. Draw OB and BSK. We have, by hypothesis, AO 01:: OM: AM; / or, since A-B is equal to OM, we have, AO: AB:: AB: AM; M hence, the triangles OAB and BAM have the sides B about their common angle BA3I1, proportional;/ they are, therefore, similar (B. IV., P. XX.). But, the triangle OAB is isosceles; hence, BAII is also isosceles, and consequently, the side B2 is equal to AB. But, AB is equal to Oath, by hypothesis: hence, BM is equal to 02, and consequently, the angles KlOB 142 GEOMETRY. and MB O are equal. The angle AXIB being an exterior angle of the triangle 0OB, is equal to the sum of the angles JI00B and XB 0, or to twice the angle MOB; G and because AMB is equal to OAB, and also to OBA, the sum of the angles OAB and I OBA is equal to four times the angle A OB: hence, A OB is equal to one-fifth of two M right angles, or to one-tenth of four right angles; and conse- A C quently, the arc AB is equal B to one-tenth of the circumference: hence, the chord AB is equal to the side of a regular inscribed decagon; which was to be proved. Cor. 1. If AB be applied ten times as a chord, the resulting polygon will be a regular inscribed decagon. Cor. 2. If the vertices A, C, E, G, and I, of the alternate angles of the decagon be joined by straight lines, the resulting figure will be a regular inscribed pentagon. Scholium 1. If the arcs subtended by the sides of any regular inscribed polygon be bisected, and chords of the semiarcs be drawn, the resulting figure will be a regular inscribed polygon of double the number of sides. Scholium 2. The area of any regular inscribed polygon is less than that of a regular inscribed pclygon of double the number of sides, because a part is less than the whole. BOOK V. 143 PROPOSITION VII. PROBLEM. To circumscribe a polygon about a circle which shall be similar to a given regular inscribed polygon. Let TNQ be a circle, 0 its centre, and AB CDEF, a regular inscribed polygon. At the middle points u T G T, Nr; P, &c., of the arcs subtended by the sides of N the inscribed polygon, draw tangents to the circle, and prolongr them till they in- C tersect; then will the re- // sulting fignre be the poly- / gon r equil ed. 10. The side HG be- K Q L ing parallel to _BA, and JIL to B C, the angle HT is equal to the angle B.' In like manner, it may be shown that any other angle of the circumscribed polygon is equal to the corresponding angle of the inscribed polygon: hence, the circumscribed polygon is equiangzlar. 2~. Draw the lines OG, OT, OH, ON, and 01. Then, because the lines IIT and HNV are tangent to the circle, OH will bisect the angle NHT, and also the angle NOT (B. III., Prob. XIV., S.); consequently, it will pass through the middle point B of the arc NB T. In like manner, it may be shown that the line drawn from the centre to the vertex of any other angle of the circumscribed polygon, will pass through the corresponding vertex of the inscribed polygon. The triangles OIIG and O0I have the angles OHG 144 GEOMETRY. and O.HI equal, from what has just been shown; the angles G OH and. O0I equal, because they are measured by the equal arcs AB and BC, and the side OHI H T G common; they are, therefore, equal in all their parts: hence, GHr is equal to HI. In like nmanner, it may be shown that HI is equal to IK, IX to KZ, and so on: hence, the circumscribed polygon is equilateral. K L The circumscribed polygon being both equiangular and equilateral, is regular; and since it has the same number of sides as the inscribed polygon, it is similar to it. Cor. 1. If lines be drawn from the centre of a regular circumscribed polygon to its vertices, and the consecutive points in which they intersect the circumference be joined by chords, the resulting figure' will be a regular inscribed polygon similar to the given polygon. Cor.' 2. The sum of the lines _HT and HrN is equal to the sum of -TT and TG, or to HTG; that is, to o - of the sides of the circumscribed polygon. Cor. 3. If at the vertices A, B, C, &c., of the inscribed polygon, tangents be drawn to the circle and prolonged till they meet the sides of the circumscribed polygon, the resulting figure will be a circumscribed polygon of double the number of sides. Cor. 4. The area of any regular circumscribed polygon BOOK V. 145 is greater than that of a regular circumscribed polygon of double the number of sides, because the whole is greater than any of its parts. Scholium. By means of a circumscribed and inscribed square, we may construct, in succession, regular circumsciibcd and inscribed polygons of 8, 16, 32, &c., sides. By means of the regular hexagon, we may, in like manner, construct regular polygons of 12, 24, 48, &c., sides. By means of the decagon, we may construct regular polygons of 20, 40, 80, &c., sides. PROPOSITION VIII. THEOREM. The area of a regular polygon is equal to half the product of its perimeter and apothem. Let GHIK be a regular polygon, O its centre, and OT its apothem, or the radius of the inscribed circle: then will the area of the polygon be equal to half the product of the perimeter and the apothem. For, draw lines from the centre to the vertices of the polygon. H T G These lines will divide the polygon into triangles whose bases will be N the sides of the polygon, and I whose altitudes will be equal to O the apothem. Now, the area of any triangle, as OHG, is equal to half the product of the side HG K and the apothem: hence, the area of the polygon is equal to half the product of the perimeter and the apothem; which was to be proved. 10 146 GEOMETRY. PROPOSITION IX. THEOREM. The perimeters of similar regular polygons are to each other as the radii of their circumscribed or inscribcd(T circles; and their areas are to each other as the squares of those radii. 10. Let ABC and XKIM be similar regular polygons. Let OA and QK be the radii of their circumscribed, 01D and QR be the radii of their inscribed circles: then Awill the perimeters of the polygons be to each other as OA is to QK, or as OD is to QR. For, the lines OA and QK are A B homologous lines of the polygons to which they belong, as are also the lines 0) and QR hence, the M perimeter of AB C is to the perimeter of KILMV, as OA is to QI, or as OD is to QR (B. IV., P. XXVII., C. 2); which was to be proved. 2~. The areas of the polygons will be to each other as OA2 is to Q_2, or as 02 is to Q-R. For, OA being homologous with QK, and OD with QR, we have, the area of ABC is to the area of ILXnf, as jT2 is to QiW2 or as 052 is to &n2 (B. IV., P. XXVII., C. 2); which was to be proved. BOOK V. 147 PROPOSITION X. THEOREM. Two regular polygons of the same number of sides can be constructed, the one circumscribed about a circle and the other inscribed in it, which shall differ from each other by less than any given surface. Let AB CE be a circle, 0 its centre, and Q the side of a square which is less than the given surface; then can two similar regular polygons be constructed, the one circum. scribed about, and the other inscribed within the given circle, which shall differ from each other by less than the square of Q, and consequently, by less than the given surface. Inscribe a square in the given circle (P. III.), and by means of it, inscribe, in succes- b d sion, regular polygons of 8, 16, K 32, &c., sides (P. VII., S.), un- A til one is found whose side is \ less than Q; let AB be the side of such a polygon. Conlstruct a similar circum. spribed polygon abode: then will these polygons differ from each other by less than tel square of Q. For, from a and b, draw the lines a O and b O; they willS pass through the points A and B. Draw also OK to the point of contact K; it will bisect AB at I and be perpendicular to it. Prolong A 0 to E. Let P denote the circumscribed, and p the inscribed polygon; then, because they are regular and similar, we shall have (P. IX.), 1A~8 GEOMETRY. P p ~ 6OA OK or 02 0-12; hence, by division (B. II., P. VI.), we have, P P —p:: OA. 02_ O I-2; P P-p' ii1 A0A DI2. Multiplying the terms of the DB second couplet by 4 (B. II., P. VII), we have, ae: P-p: 40- 42: 4A-2; wilence (B. IV., P. VIII., C.), IP: P-p:: A —2C: AB2. But P is less than the square of AE (P. VII., 0. 4); hence, P- p is less than the square of AB, and conse quently, less than the square of Q, or than the given surfkce; which was to be proved. Cor. 1. If the number of sides of the polygons be made greater than any assignable number; that is, infinite, the difference between their areas will be less than any assignable surface; that is, it will be zero*. Cor. 2. When the number of sides of the polygons is infinite, either polygon differs from the circle by less than any assignable quantity; for, the circumference of the circle lies between the perimeters of the polygons: hence, the circle differs from either polygon by less than they differ from each other. * Univ. Algebra, Arts. 72,'13. Bourdon, Art.'71. BOOK V. 1 Scholium 1. The circle may be regarded as the limit of the inscribed polygons; that is, it is a figure towards which a polygon may be made to approach as near as desirable, but beyond which it cannot be made to pass. Scholium 2. The circle may be regarded as a regular polygon of an infinite number of sides, and because of the principle, that whzatever is true of a whole class, is true of every individual of that class, we may affirm that whatever is true of regular polygons, is also true of circles. Scholium 3. When the circle is regarded as a regul:a polygon, the circumference is to be regarded as its perinmeter, and the radius as its apothem. PROPOSITION XI. PROBLEM. The area of a regular inscribed polygon, and that of X similar circumscribed polygon being given, to find the, areas of the regular inscribed and circumscribed polygons having double the number of sides. Iyet AB be the side of the given inscribed, and EFl that of the given circumscribed polygon. Let C be their common centre, AXB a portion of the circumference of the circle, and J1 the middle point of the are AMB. Draw the chord AM, and at A and B draw the tangents E P M F AP and BQ; then will Aa D be the side of the inscribed polygon, and PQ the side of the circumscribed polygon of double the number of sides (P. VII.).', Draw CE, CP, C," and CR. C 150 GEOMETRY. Denote the area of the given inscribed polygon by p, the area of the given circumscribed polygon by P, and the areas of the inscribed and circumscribed polygons having double the number of sides, respectively by p' and P'. 1~. The triangles CAD, CAJM, E _P M F and CE1i_, are like parts of the polygons to which they belong: hence, they are proportional to the polygons themselves. But CAMf is a mean proportional between C'AD and CEMr (B. IV., P. XXIV., C. A); consequently p' -' is a mean proportional between p and P: hence, p'= Vp X P....... (1.) 2~. Because the triangles CPl and CP'E have the common altitude CM], they are to each other as their bases: hence, CPM: CPE:: PM: PE; and because CP bisects the angle A CM, we have (B. IV., P. XVII.), PM E: PE: CM: CE:: CD: CA; hence (B. II., P. II.), CPM: CPE:: CD: CA or CM. But, the triangles CAD and CAM have the common altitude AD; they are, therefore, to each other as their bases: hence, CAD: CA:: CD CM; BOOK V. 151 or, because CAD and CA2MI are to each other as the polygons to which they belong, p: p':: CD CM; hence (B. II., P. IV.), we have, CPM: CPE: p:P', and, by composition, CPR1: CPMf + CPE or CME ~ p p+ p'; hence (B. II., P. VII.), 2 CPM or CMPA: CME:: 2 + p'. But, CMIPA and CME are like p:arts of P' and P, hence, P P:: 2p: p+p' or, p2p... (2.) P +P, Scholium. By means of Equation ( 1), we can find p', and then, by means of Equation (2), we can find P'. PROPOSITION XII. PROBLEM. To find the approximate area of a circle whose radius is 1. The area of an inscribed square is equal to twice the square of the radius, or 2 (P. III., S.), and the area of:a circumscribed square is 4. Making p equal to 2, antl P equal to 4, we have, from Equations (1) and (2) of Proposition XI., p' = a = 2.8284271... inscribed octagon; PI - 2 - -= 3,3137085... circumscribed octagon, 1 S2 GEOMETRY. Mlaking p equal to 2.8284271, and P equal to 3.3137085, we have, from the same equations, p' - 3.0614674... inscribed polygon of 16 sides. P' = 3.1825979... circumscribed polygon of 16 sides. By a continued application of these equations, we find th}e areas indicated in the following TABLE. NUMBER OF SIDES. INSEIMBED POLYGONS. CIRCUMSCRIBED POLYGONS. 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.. 8.1403311.. 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 16384.. 3.1415925.. 3.1415927 Now, the areas of the last two polygons differ from each other by less than the millionth part of a unit, but the area of the circle differs from either by less than they dif'er from each other; hence, the value of the area of either will differ friom that of the circle by less than a millionth part of a Ilnit, Taking the figures as far as they agree, and denoting the number of units in the required area by ~r, we have, approximately, -- = 3.141592; that is, the area of a circle whose radius is 1, is 3.141592. 6Scholium. For practical computation, the value of 7r is taken equal to 3.1416. BOOK V. 153 PROPOSITION XIII. THIIEOREMI. The circumferences of circles are to each other as their radii, and the ereas are to each other as the sqpuares of their radii. Let C and 0 be the centres of twvo circles whose radii are CA and 01 then will the circumferences be to each other as their radii, and the areas will be to each other as the squares of their radii. For, let similar regular polygons NVPST and EFCGKL be inscribed in the circles: then will the perimeters of these polygons be to each other as their apothems, and the areas will be,to each other as the squares of their apothems, whatever may be the number of their sides (P. IX.). If the number of sides be made infinite (P. VII., S.), the polygons will coincide with the circles, the perimeters with the circumferences, and the apothems with the radii: hence, the circumferences of the circles are to each other as their radii, and the areas are to each other as the squares of the radii; which was to be proved. Cor. 1. Diameters of circles are proportional to their radii: hence, the circumferences of circles are proportional to their diameters, and the areas are proportional to the sqpares of the diameters. 154 GEOMETRY. Cor. 2. Similar arcs, as AB and DE, are like parts of the circumferences to which they belong, and similar sectors, A B as ACB and DOE, are like parts of the circles to wlfich they belong: hence, similar arcs are to each other as their C radii, and similar sectors are to each other as the squares of their radii. Scholium. The term infinite, employed above, is to be understood in its limited technical sense. When it is proposed to make the number of sides of the polygons infinite, by the method indicated in the Scholium of Proposition VII., it is simply meant to make that number so great that the difference between the areas of the circle and polygon shall be less than any appreciable quantity. We have seen (P. XII.), that when the number of sides 16384, the areas differ by less than the millionth part of a unit. By increasing the number of sides, a still closer approximation may be had. PROPOSITION XIV. THEOREM. The area of a circle is equal to half the product of its circumference and radius. Let 0 be the centre of a circle, O C its radius, and A CDE its circumference: then will the area of the circle be eqtuil to half c the product of the circumferejace and radius. For, inscribe in it a regular polygon A CUDE. Then will the area of this polygon be equal to half the pro BOOK V. 155 duct of its perimeter and apothem, whatever may be the number of its sides (P. VIII.). If the number of sides be maCd! infinite, the polygon will coincide with the circle, the perimeter with the circumference, and the apothem with the radius: hence, the area of the circle is equal to half the product of its circumference and radius; which was to be proved. Cor. 1. The area of a sector is equal to half the product of its arc and radius. Cor. 2. The area of a sector is to the area of the circle, as the arc of the sector to the circumference. PROPOSITION XV. PROBLEM. To find an expression for the area of any circle in terms of its radius. Let C be the centre of a circle, and CA its radius. Denote its area by area CA, its radius by R, and the area of a circle whose radius is 1, by w (P. XII., S.). Then, because the areas of circles are to each other as the squares of A their radii (P. XIII.), we have, area CA ~: R2: 1; whence, area CA =- rR2. That is, the area of any circle is 3.1416 times the sjtuptre of the radius. PROPOSITION XVI. PROBLEM. To find an expression for the circumference of a circle, in terms of its radiucs, or diameter. Let C be the centre of a circle, and CA its radius. 156 GEOMETRY. Denote its circumference by circ. CA, its radius by 1, and its diameter by 1). From the last Proposition, we have, area CA = r~B2; and, from Proposition XIVT., we have, area CA = J-circ. CA x R; A hence, Icirc. CA x R = ~.R2; whence, by reduction, circ. CA = 2,.rI, or, circ.'CA = -D. That is, the circumference of any circle is equal to 3.1416 times its diameter. Scholium 1. The abstract number e, equal to 3.1416, denotes the number of times that the diameter of a circle is contained in the circumference, and also the number of times that the square constructed on the radius is contained in the area of the circle (P. XV.). Now, it has been.proved by the methods of Higher Mathematics, that the value of: is incommensurable with 1; hence, it is impossible to express, by means of numbers, the exact length of a circumference in terms of the radius, or the exact area in terms of the square described on the radius. We may also infer that it is impossible to square the circle; that is, to construct a square whose area shall be exactly equal to that of the circle. Scholiurm 2. Besides the approximate value of mr, 3.1416, usually employed, the fractions _2a and -55 are also used, when great accuracy is not required. BOOK VI. PLANES AND POLYEDRAL ANGLES. DEFINITIONS. 1. A straight line is PERPE-NDICULAR TO A PLANE, when it is perpendicular to every line of the plane which passes through its FOOT; that is, through the point in which it meets the plane. In this case, the plane is also perpendicular to the line. 2. A straight line is PARALLEL TO A PLANE, when it cannot meet the plane, how far soever both may be produced. In this case, the plane is also parallel to the line. 3. Two PLANES ARE PARALLEL, when they cannot meet, how far soever both may be produced. 4. A DIEDRAL ANGLE is the amount of divergence of two planes. The line in which the planes meet, is called the edge of the angle, and the planes themselves are called faces qf the angle. The measure of a diedral angle is the same as that of a plane angle formed by two lines, one drawn in each face, and both perpendicular to the edge at the same point. A diedral angle may be acute, obtuse, or a right angle. In the latter case, the faces are perpendicular to each other. 158 GEOMETRY. 5. A POLYEDRAL ANGLE is the amount of div -gence of several planes meeting at a common point. This point is called the vertex of the angle; Jae lines in which the planes meet are called edges of the angle, and the portions of the planes lying between the edges are called faces of the angle. Thus, S is the vertex of the polyedral angle, whose edges are SA, SB, S C, SD, and whose faces are ASB, / BSC, CSD, DSA.-. C A polyedral angle which has but / three faces, is called a triedral A angle. POSTULATE. A line may be drawn perpendicular to a p ne from any point of the plane, or from any point without the plane. PROPOSITION I. THEOREM. If a straight line has two of its points in d plane, it will lie wholly in that plane. For, by definition, a plane is a surface suvm, that if any two of its points be joined by a straight line, that line will lie wholly in the surface (B. I., D. 8). Cor. Through any point of a plane, an infinite number of straight lines may be drawn which will lie in the plane. For, if a line be drawn from the given point to any other point of the plane, that line will lie wholly in the plane. Scholium. If any two points of a plane be joined by a straight line, the plane may be turned about that line as an BOOK VI. 159 axis, so as to take an infinite number of positions. Hence, we infer that an infinite number of planes may be passed through a given line. PROPOSITION II. THEOREM. Through three points, not in the same straight line, one plane can be passed, and only one. Let A, B, and C be the three points: then can one plane be passed through them, and only one. Join two of the points, as A and B, by the line AB. Through AB let a plane be passed, and let this plane be turned around AB until it contains the point C; in this position it will pass through the three points A, B, and C. If now, the plane be turned about AB, in either direction, it will no longer contain the point C: hence, one plane can always be passed through three points, and only one; which was to be proved. Cor. 1. Three points, not in a straight line, determine the position of a plane, because only one plane can be passed through them. Cor. 2. A straight line and a point without that line, determine the position of a plane, because only one plane can be passed through them. Cor. 3. Two straight lines which intersect, determine the position of a plane. For, let AB and A C intersect at A: then will either line, as AB, and one point of the other, as C, determine the position of a plane. Cor. 4. Two parallel lines determine the position of a 160 GEOMETRY. plane. For, let AB and CD be parallel. By definition (B. I., D. 16) two parallel lines always lie in the same plane. But either line, as AB, and any point of the other, as F, determine the posi- A B tion of a plane: hence, two parallels -D determine the position of a plane. F PROPOSITION III. THEOREM. The intersection of two planes is a straight line. Let AB and CD be two planes: then will their intersection be a straight line. For, let E and F be any two D B points common to the planes; draw the straight line EF. This line having two points in the plane AB, will lie wholly in that plane; and having two points in the plane CD, will lie wholly in that plane: hence, every point of ]FF is common to both planes. Furthermore, the planes can have no common point lying without El, otherwise there would be two planes passing through a straight line and a point lying without it, which is impossible (P. II., C. 2); hence, the intersection of the two planes is a straight line; which was to be proved. PROPOSITION IV. THEOREM. If a straight line is perpendicular to two straight lines at their point of intersection, it is perpendicular to the plane of those lines. Let MN be the plane of the two lines BB, CC, and let AP be perpendicular to these lines at P: then will BOOK VI. 161 AP be perlpendicular to every line of the plane whidh passes through P, and consequenlty, to the plane itself. For, through P, draw inl thle plane llNT, any line PQ; through any point of this line, as Q, draw the line B(C, so / that BQ shall be equal to QC / Q / (B. IV., Prob. V.); draw AB, B AQ, and AC. The base B (J, of the triangle -BP C, being bisected at Q, we have (B. IV., P. XIV.), P' + PB p2 - 2PQ+~ 2QC2. In like manner, we have, from the triangle ABC, A C2 + AB2 = 2A Q2 + 2 QC2. Subtracting the first of there equations from the second, member from member, we have, A2 _p2 4- AB2 B_ = 2AQ2 2PQ2. A-C - P C + A-B~ - F2Y= 2 - 2,PQ~. But, from Proposition XI., C. 1, Book IV., we have, 2_A- - = lp2, and -A 2 -_ p _ A-p2 hence, by substitution, 2AP2 2AQ2 _ 2PQ2; whence, A-2 -; or, AP2 + P2 Q2. The triangle APQ is, therefore, right-angled at P (B. IV., P. XIII., S.), and consequently, AP is perpendicular to _PQ: hence, AP is perpendicular to every line of the p.lane 1VIN passing through P, and consequently, to the plane itself; which woas to be proved. 11 162 GEOM ETR Y, Cor. 1. Only one perpendicular can be drawn to a plane firom a point without the plane. For, suppose two perpendiculars,:as AP and A Q, could be N drawn from the point A to the [)lane fIN]. Draw PQ; then tihe triangle APQ would have two right angles, APQ and A QP; which is impossible (B. I., P. XXV., C. 3). Cor. 2. Only one perpendicular can be drawn to a plane from a point of that plane. For, suppose that two perpendliculars could be drawn to the plane I7;Y, from the point P. Pass a plane through the perpendiculars, and let P Q be its intersection with /~N; then we should have two perpendiculars drawn to the same straight line from a point of that line; which is impossible (B. I., P. XIV., C.). PROPOSITION V. THEOREM. If from a point without a plane, a perpendicular be drawcn to t/he plane, and oblique lines be drawn to different points of the plane: 10. The perpendicular will be shorter than any oblique line: 2~. Oblique lines which meet the plaqne at equal cdistances from the foot of the perpendicular, will be equal: 3.~ Of two oblique lines which meet the plane at unequal distances from the foot of the perpendicular, the ohne which meets it at the greater distance will be the longer. Let A be a point without the plane JIVN; let AP be perpendicular to the plane; let A C, AD, be any two oblique lines meeting the plane at equal distances from the foot of the perpendicular; and let A C and AE be any BOOK VI. 163 two oblique lines meeting the plane at unequal distances from the foot of the perpendicular: 1~. AP will be shorter A than any oblique line A C. For, draw PC; then will N A1P be less than A C (B. I., P. XV.); which was to / - B be proved. M 2~0. A C and AD will be equal. For, draw PD; then the right-angled triangles AP C, APD, will lhave the side AP common, and the sides PC, PD, equal: hence, the triangles are equal in all their parts, and consequently, A C and AD will be equal; which was to be proved. 30~. AP will be greater than A C. For, draw PE, and take PB equal to PC; draw AB: then will AX be greater than AB (B. I., P. XV.); but ARB and A C are equal: hence, AE is greater than A C; which cwas to be proved. Cor. The equal oblique lines AB, A C, AD, meet the plane -rl1 in the circumference of a circle, whose centre is P, and whose radius is PB: hence, to draw a perpendicular to a given plane irVY, from a point A, without that plane, find three points B, C, D, of the plane equally distant from A, and then find the centre P, of the circle whose circumference passes through these points: then will AP be the perpendicular required. Scholiuzm. The angle ABP is called the inclination qof the oblique line AB to the plane ~liaV. The equal oblique lines AB, A C, AD, are all equally inclined to the plane MEYIrY The inclination of AE is less than the inclination of any shorter line AB. 164 GEO METRY. PROPOSITION VI. THEOREM. If from the foot of a perpendicular to a plane, a line be drawn at right angles to any line of that plane, and the point of intersection be joined with any point of the per. pendicular, the last line will be perpendicultar to the line of the plane. Let AP be perpendicular to the plane 3JIK, P its foot, B C the given line, and A any point of the perpendicular; draw P1) at right angles to B C, and join the point D) with A: then will AD be perpendicular to B C. For, lay off D1B equal to D)C, and draw PB, P C, AP, A and A C. Because PD is perpendicular to BC, and )B N equal to D)C, we have, PB equal to PC (B. I., P. XV.); B and because AP is perpendicu- M tar to the plane JIIN, and PB equal to PC, we have AB equal to A C (P. V.). The lhie AD has, therefore, two of its points A and D), each equally distant from B and C': hence, it is perpendicular to B C (B. I., P. XVI., S.); which was to be proved. Cor. 1. The line -B C is perpendicular to the plane of the triangle APD); because it is perpendicular to AD and PD, at D (P. IV.). Cor. 2. The shortest distance between AP and BC, is measured on PD, perpendicular to both. For, draw BkE] between any other points of the lines: then will BE be greater than P.B, and PB will be greater than PD: hence, PD is less than BE. BOOK VI. 165 Scholium. The lines AP and BBC, though not in the same plane, are considered perpendicular to each other. In general, any two straight lines not in the same plane, are considered as making an angle with each other, which angle is equal to that formed by drawing through a given point, two lines respectively parallel to the given lines. PROPOSITION VII. THEOREM. If one of two parallels is perpenclicular to a planze, the other one is also perpenclicular to the same planze. Let AP and ED be two parallels, and let AIP be perpendicular to the plane r1N: then will ED b6 also perpendicular to the plane JJN. For, pass a plane through the E parallels; its intersection with uIfiN will be PD); draw AD, C and in the plane X/NV draw D B C perpendicular to _PD at D. Now, B-D is perpendicular M to the plane APD9E (P. VI., C.); the angle BDE is consequently a right angle; but the am gcle EDP is a right angle, because ED is parallel to 1AP (B. I., P. XX., C. 1): hence,,ED is perpendicular to _B) and PD, at their point of intersection, and consequently, to their plane XjNV (P. IV.); which was to be proved. Cor. 1. If the lines AP and ED are perpendicular to the plane AllY, they are parallel to each other. For, if not, draw through D a line parallel to.PA; it will be perpendicular to the plane MJN, from what has just been proved; we shall, therefore, have two perpendiculars to the the plane A1KN, at the same point; which is impossible (P. IV., C. 2). 166 GE O MIETRY. Cor. 2. If two lines, A and B, are parallel to a tUird line C, they are parallel to each other. For, pass a plane perpendicular to C; it will be perpendicular to both A and B: hence, A and B are parallel. PROPOSITION VIII. THIEOREfM. Y.f a line is parallel to a line of a plane, it is parallel to that plane. Let the line AB be parallel to the line CD of the plane MYIN; then will AB be parallel to the plane N.<* For, through AB and CD pass a plane (P. II., C. 4); CD A B will be its intersection with N the plane lN/Y. Now, since AB lies in this plane, if it can meet /c i/ the plane JNIv, it will be at 5I some point of CG); but this is impossible, because AB and CD are parallel: hence, AB cannot meet the plane M/IN, and consequently, it is parallel to it; which was to be proved. PROPOSITION IX. THEOREM. If two planes are perpendicular to the same straight line, they are parallel to each other. Let the planes MU-YN and PQ Q be perpendicular to the line AB, B,-. / at the points A and B: then will they be parallel to each M other. At — For, if they are not parallel, / NI BOOK VI. 167 they will meet; and let 0 be a point common to both. From O draw the lines OA and OB: then, since OA lies in the plane 2IMN, it will be perpendicular to BA at A (D. 1). For a like reason, OB will be perpendicular to AB at B hence, the triangle OAB will have two right angles, which is impossible; consequently, the planes cannot meet, and are therefore parallel; which was to be proved. PROPOSITION X. THEOREM. If a plane intersect two parallel placnes, the lines of intersection will be parallel. Let the plane ElI intersect the parallel planes JliV and PQ, in the lines ET; and GIt: then will EF and GU/ be parallel. For, if they are not parallel, H Q they will meet if sufficiently prolonged, because they lie in the P / same plane; but if the lines meet, the planes MN and P Q, in F N which they lie, will also meet; but this is impossible, because M E these planes are parallel: hence, the lines EF and GH7 cannot meet; they are, therefore, parallel; which was to be proved. PROPOSITION XI. THEOREM. If a straight line is perpendiculcar to one of two paral.lel planes, it is also perpendicular to the other. Let M1~N and PQ be two parallel planes, and let thle line AB be perpendicular to PQ: then will it also be perpendicular to MkN~1 168 G E O M1 E T R Y. For, through AB pass any plane; its intersections with J1K7~ and PQ will be parallel (P. X.); but, its intersection iwiith P9Q is perpendicular to AB at B (D. 1); hence, its intersection with lMNV is also perpendicular to AB at A (1B. I., P. XX., C. 1):hence, 1 B is perpendicular to every line of the plane /YK through N 4, and is, therefore, perpendicu- A lar to that plane; which was to M be proved. PROPOSITION XII. THEOREM. Parallel lines inzcluded between parallel planes, are equal. Let'EG and Eill be any two parallel lines included between the parallel planes J1KT and P Q: then will they be equal. Through the parallels conceive a plane to be passed; it will intersect the plane 1/NV in the line E1, and PQ in the line Gil; and these lines will be E N )arallel (Prop. X.). The iigure iZIFJ'IG is, therefore, a parallelo- F gram: hence, GE and ITF are equal (B. I., P. XXVIII.); which was to be proved. Cor. 1. The distance between two parallel planes is measirled on a perpependicular to both; but any two perpendiculars between the planes are equal: hence, parallel planes are everywhere equally distant. Cor. 2. If a line G~T is parallel to any plane h1i7, then can a plane be passed through Gil parallel to Ji1K: hence, if a line is parallel to a plane, all of its points are eqlqally distant from thlat plane. BOOK VI. 169 PROPOSITION XIII. THEOREMI If two angles, not situated in the same pl2ane, have their sides parallel and lying in the same direction, the angles will be equal acnd their planes parallel. Let CAE' and DBF be two angles lying in the planes 31V and P Q, and let the sides A C alld AE' be respectively parallel to -BID and Bi_, and lying in the same direction: then will the angles CAE and DBF be equal, and the planes 3lN and PQ will be parallel. Take any two points of AC and AE, as C and ], and make BD) equal to A C, and BE1' to AE; draw CE, DEF, c H AB, C-D, and iE/. 1i. The angles CA_1 and A DBF will be equal. For, A:E and BE being parallel and equal, the figure / AIlBEE is a parallelogram (B. / 1., P. XXX.); hence, EF is parallel and equal to AB. For a like reason, C6D is parallel and equal to AB: hence, ('D and EF are parallel and equal to each other, and consequently, CE and D)F are also parallel and equal to each other. The triangles CAE and )BE2' have, therefore, their corresponding sides equal, and consequently, the corresponding angles CAE and DBF are equal; which qcas to be proved. 2~. The planes of the angles ~i1V and P Q are paralleL For, if not, pass a plane through A parallel to _PQ, and suppose it to cut the lines C'D and ElF in G and H. Then will the lines GD and lcl be equal respect. 170 GEOMETR Y. ively to AB (P. XII.), and consequently, GD will be equal to CD, and HiP to El; which is impossible: hence, the planes 31NX and PQ must be parallel; which was to be proved. Cor. If two parallel planes 1kYN and PQ, are met by two other planes AD and AF, the angles CAE and DB1 formed by their intersections, will be equal. PROPOSITION XIV. THEOREM. If three straight lines, not situated in the same plane, are equal and parallel, the triangles formed by joining the extremities of these lines will be equal, and thzeir ilcanes parallel. Let AB, CD, and PF be equal parallel lines not in the same plane: then will the triangles ACE and BDF be equal, and their planes parallel. For, AB being equal and parallel to FiF, the figure ABFEP is a parallelogram, and consequently, A1E is equal and par- /DF allel to B7i. For a like reason, A C is equal and parallel to BD': hence, the included angles CGAPE and DB-P are equal and their planes parallel (P. XIII.). E Now, the triangles CAE and JDBF have two sides and their included angles eqnal, each to each: hence, they are equal in all their parts. The triancles are, therefore, equal and their planes parallel; which was to be proved-. BOOK VI. 171 PROPOSITION XV. TIEORPEM. rf two straight lines are cut by three parallel planes, they will be divided proportionally. Let the lines AB and CGD be cut by the parallel planes 1N, P Q, and -2S, in the points A, E, B, and C, _F D; then A EB:: CF: ED. For, draw the line AD, and suppose it to pierce the plane PQ in G; draw AC, B, D, E G, and GC1 R / The plane AB1D intersects Q thle parallel planes BfS and p Q E / in the lines BiD and CG; p 1 consequently, these lines are par- N allel (P. X.): hence (B. IV., / - \ / P. XV.), / A A: EB:: A G: GD. The plane A CD intersects the parallel planes 3JN and P Q, in the parallel lines A C and GE: hence, AGC: GD:: CF: ED. Combining these proportions (B. II., P. IV.), we have, A E: EB:: C: FD; which was to be proved. Cor. 1. If two lines are cut by any number of parallel planes th.ey will be divided proportionally. Cor. 2. If any number of lines are cut by three parallel planes, they will be cdividecd 1),!),olrtionally. 172 (GEOMETRY. PROPOSITION XVI. THEOREM. If a line is perpendicular to a plane, every plane passed tArough the line will also be perpenzdicular to that plane.. Let AP be perpendicular to the plane IN, and let BF be a plane passed through AP: then will BEF be perpendicular to JIV. F In the plane JWN, draw PD A perpendicular to B C, the intersection of BE and JN. Since AP - is perpendicular to JNIY, it is perpendicular to BC and DP (D. 1); B D/ anld since AP and DP, in the planes BEF and AIT5, are perpendicular to the intersection of these planes at the sanme point, the angle which they form is equal to the angle formed by the planes (D. 4); but this angle is a right angle: lhence, BF is perpendicular to JliN; which was to be proved. Cor. If three lines AP, BP, and DP, are perpendicular to each other at a common point P, each line will be perpendicular to the plane of the other two, and the three planes will be perpendicular to each other. PROPOSITION XVII. THEOREM. If two planes are perpendicular to eaclh other, a line drawn int one of them, perpendiceular to their intersection, wilg be perendicular to the other. Let the planes BF and _l/N be perpendicular to each other, and let the line AP, drawn in the plane B17~ be perpendicular to the intersection B C; then will AP be perpendicular to the plane JuiN. BOOK VI. 173 For, in the plane INV, draw PD perpendicular to B C at P. Then because the planes BF and MIN are perpendicular to each other, the angle AP2D will be a right angle: hence, AP is A perpendicular to the two lines PD 1 | and B C, at their intersection, and - tN consequently, is perpendicular to their I7 plane MN; which was to be proved. / B Cor. If the plane BZE: is perpendicular to the plane MA1N, and if at a point P of their intersection, we erect a perpendicular to the plane 21~N, that perpendicular will be in the plane 3BE. For, if not, draw in the plane BE1; P11 perpendicular to PC, the common intersection; AP will be perpendicular to the plane M2JN~, by the theorem; therefore, at the same point P, there are two perpendiculars to the plane 2MN; which is impossible (P. IV., C. 2). PROPOSITION XVIII. THEOREM. If two planes cut each other, and are perpendicular to a third plane, their intersection is also perpendicular to that plane. Let the planes BF, -DfI, be perpendicular to INXV: then will their intersection.AP be perpendicular to XIlY. For, at the point P, erect a per- F pendicular to the plane IlEN; that A perpendicular must be in the plane N BF,1 and also in the plane D "- --- (P. XVII., C.); therefore, it is their common intersection AiP: which was Z5 _ to be proved. M 174 GEOMETRY. PROPOSITION XIX. THEOREM. The sum of any twzo of the plane angles formeed by the edges of a triedral angle, is greater than the third. Let SA, SB, and S C, be the edges of a triedral angle: then will the sum of any two of the plane angles formed by them, as ASC and CSB, be greater than the third ASB. If the plane angle ASB is equal to, or less than, either of the other two, the truth of the proposition is evident. Let us suppose, then, that AS-B is greater than either. In the plane AS-B, construct the angle.BSD equal to BSC; S draw AB-L in that plane, at pleasure; lay off SC equal to S/D, and draw A C and C-B. The triangles.BS-D and BSC have..\ — the side SC equal to SD, by construction, the side SB common, and the included angles BSED and BSC equal, by construction; the triangles are therefore equal in all their parts: hence, BD) is equal to B C. But, from Proposition VII., Book I., we have, BC + CA > _BD + IDA. Taking away the equal parts B C and.BD, we have, CA > DA; hence (B. I., P. IX., C.), we have, angle ASC > angle ASD; and, adding the equal angles BSC and?BSD, BOOK VI. 175 angle ASC + angle CSB > angle ASD + angle D)SB; or, angle ASC + angle CSB > angle ASB; which was to be proved PROPOSITION XX. THEOREM. The sum of the plane angles formed by the edges of any polyedrcal angle, is less than foyur right angles. Let S be the vertex of any polyedral angle whose edges are SA, SB, S,, SD, and SE; then will the sum of the angles about S be less than four right angles. For, pass a plane cutting the edges in the points A, B', C, D, and L, wiand the faces in the lines AB, BC0, CDN, JE, and EA. From any point within the polygoin thus formed, as 0, /. draw the straight lines OA, OB, 0 C, OD, and OF. We then have two sets of triangles, S ( one set having a common vertex S, the other having a common vertex 0, and both having conmmon bases ABl, B (, C(D, DE, BEA. Now, in the set which has the common vertex S, the sum of all the angles is equal to the sum of all the plane angles formed by the edges of the polyedral angle whose vertex is S, together with the sum of all the angles at the bases viz., SAAB, SBA, SB C, &c.; and the entire sum is equal to twice as many right angles as there are triangles. In the set whose common vertex is 0, the sum of all the angles is equal to the four right angles about 0, together with the interior angles of the polygon, and this sum is equal to twice as many right angles as there are triangles. Sinco 176 G E OM ETR Y. the number of triaanoles, in each set, is the same, it follows tlhLat these sums are equal. But inll the triedral angle wlnhose vertex is B, we have (P. XIX.), ABS + SBC > A-BC; ancd the like may be shown at each of the other vertices, C, XD, -:, A: hence, the sum of the angles at the bases, in the triangles whose common A vertex is 8, is gfeater than the sum >1 of the angles at the bases, in the set B C whose common vertex is 0: therefore, the sum of the vertical angles about S, is less than the sum of the angles about 0: that is, less than four right. angles; which was to be proved. Scholium. The above demonstration is made on the sup. position that the polyedral angle is convex, that is, that the diedral angles between the consecutive faces are each less than two right angles. PROPOSITION XXI., THEOREM. If the plane angles formed by the edges of two triedral angles are equal, each to each, the planes of the equal angles are equally inclined to each other. Let S and T be the vertices of two triedral angles, and let the angle ASC be equal to DTi ASB to iDTE,:aned 1BSC to ET.: then will the planes of the equal angles be equally inclined to each other. For, take any point of S-B, as B, and from it draw in the two faces ASB and CSB, the lines BA and B(C, respectively perpendicular to SB: then will the angle AJBC measure the inclination of these faces. Lay off TE equal BOOK VI. 177 to SB, and from E draw in the faces DTE and METL, the lines ED and E1 respectiveiy perpepdicular to TPE; then will the angle DEE measure the inclination of these \' faces. Draw A C and DE F The right-angled triangles A- C / SBA and TED, have the B D side- SB equal to TL, and the angle ASB equal to _DTE; hence, AB is equal to DiE, and AS to TD. In like manner, it may be shown that B C' is equal to EF, and CS to iFT. The triangles ASC and )DTIF have the angle ASC equal to DTF, by hypothesis, the side AS equal to DT, and the side CS to FT, from what has just been shown; hence, the triangles are equal iln all their parts, and consequently, A C is equal to AD)E. Now, the triangles ABC and DIEF have their sides equal, each to each, and consequently, the corresponding angles are also equal; that is, the angle ABC is equal to DE': hence, the inclination of the planes ASB and CSB, is equal to the inclination of the planes D)TE and FTE. In like manner, it may be shown that the planes of the other equal angles are equally inclined; which was to be proved. Scholizmz. If the planes of the equal plane angles are like placed, the triedral angles are equal in all respects, for they may be placed so as to coincide. If the planes of the equal angles are not similarly placed, the triedral angles are equal by symmetry. In this case, they may be placed so that two of the homologous faces shall coincide, the triedial angles lying on opposite sides of the plane, which is then called a plane of symmetry. In this position, for every point on one side of the plane of symmetry, there is a corresponding point on the other side. 12 BOOK VII. P 0 L Y E D R 0 N S. DEFINITIONS. 1. A POLYEDRON is a volume bounded by polygons. The bounding polygons are called faces of the polyedron; the lines in which the polyedrons meet, are called edges of the polyedron; the points in which the edges meet, are called' vertices of the polyedron. 2. A PRISM is a polyedron; two of whose faces are equal polygons having their homo- i.! logous sides parallel, the other faces being / parallelogramss. The equal polygons ale callecl bases of the plrism; on~ the upper, and the other the.lower base; the parallelograms taken together make up the lateral or convex surface of the pirism; the'lines in which the lateral faces meet, are called lateral edge8 of the prism. 3. The ALTITUD:E of a prism is the perpendicular distance between the planes of its bases. 4. A RIGHT P.TsSr is one whose lateral edges are perpendicular to the planes of the bases In this case,,any lateral edge is equal to the altitude. BOOK VII. 179 5. An OBLIQUE PRISM is one whose lateral edges are oblique to the planes of the bases. In this case, any lateral edge is greater than the altitude. 6. Prisms are named from- the number of sides of their bases; a triangular prism is one whose bases are triangles; a quadrangular prism is one whose bases are quadrilaterals; a pentangular prism is one whose bases are pentagons, and so on. 7. A PAnIALLELOPIPEDON is a prism whose bases are parallelograms. A -Rectangular Parallelopipedon is a right parallelopipedon, all of whose faces are rectangles; a cube is a rectangular parallelopipedon, all of whose faces are squares. 8. A PYRAMID is a polyedron bounded by a polygon called the base, and by triangles meeting at a common point, called the vertex of the pyramid. The triangles taken together make up the /. lateral or convex surface of the pyramid; the lines in which the lateral faces meet, are called the lateral edges of the pyramid. 9. Pyramids are named from the number of sides of their bases; a triangular pyramid is one whose base is a triangle; a quadrangular pyramid is one whose base is a quadrilateral, and so on. 10. The ALTITUDE Of a pyramid is the perpendicular distance from the vertex of the pyramid to the plane of its base. 180 GEOMETRY. 11. A RIGHT PYRAMID is one whose base is a regular polygon, and in which the perpendicular drawn from the vertex to the plane of the base, passes through the centre of the base. This perpendicular is called the axis of the pyramid. 12. The SLANT HEIGHT of a right pyramid, is the perpendicular distance from the vertex to any side of the base. 13. A TRUNCATED PYRAMID is that portion of a pyramid included between the base and any plane which cuts the / 4 i pyramid.,\ WVhen the cutting plane is parallel to the base, the truncated pyramid is called a FRUSTUM OF A PYRAMID, and the intersection of the cutting plane with the pyramid, is called the tpper base of the frustum; the base of the pyramid is called the lower base of the frustum. 14. The ALTITUDE of a frustum of a pyramid, is the peryendicular distance between the planes of its bases. 15. The SLANr HEIGHT of a frustum of a right pyramid, is that portion of the slant height of the pyramid which lies between the planes of its upper and lower bases. 16, SIMILAR POLYEDRONS are those which are bounded by Aimilar polygons, similarly placed. Parts which are similarly placed, whether faces, edges, or angles, are called homologous. 17. A DIAGONAL of a polyedron, is a straight line joining the vertices of two polyedral angles not in the same &-e. BOOK VII. 181 18. The VOLUME OF A POLYEDRON is its numerical value expressed in terms of some other polyedron as a unit. The unit generally employed is a cube constructed on the linear unit as an edge. PROPOSITION I. THEOREM. The convex sulface of a right prism is equal to the perimeeter of either base multiplied by the altitude. Let AlBC.D-E-KI be a right prismn: then is its convex surface equal to, (AB + B C + C) + RDE + EA) x AF. For, the convex surface is equal to the sum of all the rectangles AG, Bf, C, D1ffi, E,, which compose it. Now, l\F i the altitude of each of the rectangles AF, BG, OCH, &c., is equal to the altitude of the prism, and the area of each rectangle is equal to its base multiplied by its altitude (B. IV., P. V.): A B hlence, the sum of these rectangles, or the convex surface of the prism, is equal to, (AB + B C + CD + DE+.EA) x AF; that is, to the perimeter of the base multiplied by the altitude; which was to be proved. Cor. If two right prisms have the same altitude, their convex surfaces are to each other as the perimeters of their bases. 182 GEOMETRY. PROPOSITION II. THEOREM. In any prism, the sections made by parallel planes are equal polygons. Let the prism Af be intersected by the parallel planes 2NP, SV: then are the sections NOPQR, STVXY, equal polygons. For, the sides NO, ST, are parallel, K being the intersections of parallel planes F / with a third plane ABGEF; these sides, F / NO, ST, are included between the par- S - ix allels NS, OT: hence, NO is equal to ST (B. I., P. XXVIII., C. 2). For like / reasons, the sides OP, PQ, Q1i, &c., / / P of NOPQQR, are equal to the sides TV, VX, &c., of STVXY, each to A. C each; and since the equal sides are par- B allel, each to each, it follows that the angles NOP, OP Q, &c., of the first section, are equal to the angles STV, T:VX, &c., of the second section, each to each (B. VI., P. XIII.): hence, the two sections NOPQR, STVXY, are equal polygons; which was to be proved. Cor. Every section of a prism, parallel to the bases, is equal to either base. PROPOSITION III. THEOREiM. If a pyramid be cut by a plane parallel to the base: 1~. The edges and the altitude will be divided proportionally: by. The section will be a polygon similar to the base. Let the pyramid S-AlBCDE, whose altitude is SO, be cut by the plane abcde, parallel to the base ABC (IDE. BOOK VII. 183 10. The edges and altitude will be divided proportionally. For, conceive a plane to be passed through the vertey S, parallel to the plane of the base; then will the edges and the altitude 1e cut S by three parallel planes, and consequently they will be divided proportionally (B. VI., \\ P. XV., C. 2); which was to be proved. a e' 2~. The section abcde, will be similar -, \ to the base AldCDE. For, ab is par- A.',' allel to D14, and be to B C (B. VI., P. X.): hence, the angle abc is equal to B the angle AB C. In like manner, it may be shown that each angle of the polygon abcde is equal to the corresponding angle of the base: hence, tile two polygons are mutually equiangular. Again, because ab is parallel to ADB, we have, ab A sb SB; End, because bce is parallel to.B C, we have, be DB C:: sb SB; hence (B. II., t. IV.), we have, ab: AB:: be B C. In like manner, it may be shown that all the sides of abcde are proportional to the corresponding sides of the polygon A BC1DE: hence, the section abede is similar to the base AD CDE (B. IV., D. 1); which was to be proved. Cor. 1. If two pyramids S-AB CDLE, and S-X~Y-., having a common vertex S, and their bases in the sale plane, be cut by a plane abe, parallel to the plane of their bases, the sections will be to each other as the base (. 184 GE OMETRY. For, the polygons abed and AB CD, being similar, are to each other as the squares of their homologous sides ab and AB (B. IV., P. XXVII); but, ab2 A1X2:: Sa2 SIA2: o2 2 S hence (B. II., P. IV.), we have, abccle: AB3C D): so: ~ -02. In like manner, we have, / ryz: XYZ S02 o SO2; A hence, B y abede AB CDE:: xyz: XYZ. Cor. 2. If the bases are equal, any sections at equal distances from the bases will be equal. Cor. 3. The area of any section parallel to the base, is proportional to the square of its distance from the vertex. PROPOSITION IV. THEOREM. The convex surface of a right pyramid is equal to the perimeter of its base multiplied by half the slant height. Let S be the vertex, A13 CDE the S base, and S1, perpendicular to EA, the slIint height of a right pyramid: then will / the convex surface be equal to, (A BBC + ~CD + DEz + EA) x SF / Draw SO perpendicular to the plane of the base. A BOOK VII. 185 From the definition of a right pyramid, the point 0 is the centre of the base (D. 11): hence, the lateral edges, SA, SB, &c., are all equal (B. VI., P. V.); but the sides of the base are all equal, being sides of a regular polygon: hence, the lateral faces are all equal, and consequently their altitudes are all equal, each being equal to the slant height of the pyramid. Now, the area of any lateral face, as SEA, is equal to its base EA, multiplied by half its altitude SF: hence, the sum of the areas of the lateral faces, or the convex surface of the pyramid, is equal to, (AB + B C + CD + RDE + EA) x iSF; which was to be proved. Scholium. The convex suiwface of a frustum of a right pyramid is equal to half the sumn of the perimeters of its upper and lower bases, multiplied by the slant height. Let AB CDE-e be a frustum of a right S pyramid, whose vertex is S: then will the section abode be similar to the base AB CDL eE'l - and their homologous sides will be parallel, (P. III.). Any lateral face of the firustum, as A.Eea, is a. trapezoid, whose altitude is C equal to Ff, the slant height of the frustum; hence, its area is equal to ~(EA lea) x 2vf (B. IV., P. VII.). But the area of the con- A vex surface of the frustum is equal to the sum of the areas of its lateral faces; it is, therefore, equal to the half sum of the perimeters of its upper and lower bases, multiplied by half the slant height. 186 GEOMETRY. PROPOSITION V. THEOREM. If the three faces which include a triedral angle qf a prism are equal to the three faces which include a triedral angle of a second prism, each to each, and. are like placed, the two prisma are equal in all their parts. Let B and b be the vertices of two triedral angles, included by faces respectively equal to each other, and similarly placed: then will the prism AB CD-E-K be equal to the prism abeade-k, in all of its parts. For, place the base abcde upon the equal K k base AB CDE, so that they shall coincide; then I because the triedral angles whose vertices are / b and.B, are equal, the parallelogram bh will coincide with BET, and the parallelogram bf with B C b c BF: hence, the two sides fg and gh, of one upper base, will coincide with the homologous sides of the other upper base; and because the upper bases are equal, they must coincide throughout; consequently, each of the lateral faces of one prism will coincide with the corresponding lateral face of the other prism: the prisms, therefore, coincide throughout, and are therefore equal in all their parts; which was to be proved. Cor. If two right prisms have their bases equal in all their parts, and have also equal altitudes, the prisms themselves will be equal in all their parts. For, the faces which include any triedral angle of the one, will be equal to the faces which include the corresponding triedral angle of the other, each to each, and they will be similarly placed. BOOK VII. 187 PROPOSITION VI. THEOREM. In any parallelopipedon, the opposite faces are equal, each to each, and their planes are parallel. Let AB C.D-Ht be a parallelopipedon: then will its opposite faces be equal and their planes will be parallel. For, the bases, AB C/D and EFGHff are equal, and their planes parallel by E I{ definition (D. 7). The opposite faces AEJID and.BFGC, have the sides AE and BF parallel, because they are oppo- A site sides of the parallelogram;BE B C and the sides ifH/ and P7G parallel, because they are opposite sides of the parallelogram EG; and consequently, the angles AEI~i and _B3G are equal (B. VI., P. XIII.). But the side AL is equal to BT, and the side -EH to CFG; hence, the faces A4EJID and BFG C are equal; and because AE is parallel to BIE, and -EIt to FG, the planes of the faces are parallel (B. VI., P. XIII.). In like manner, it may be shown that the parallelograms ABFEL and )DCG,II equal and their planes parallel: hence, the opposite faces are equal, each to each, and their planes are parallel; which was to be proved. Cor. 1. Any two opposite facees of a parallelopipedon may be taken as bases. Cor. 2. In a rectangular parallelopipedon, the square of either of the diagonals is equal to the sumn of the, - squares of the three edges which meet / at the same vertex. For, let FID be either of the diagonals, and draw PFlL 188 GEO METRY. Then, in the right-angled triangle PHD, we have, b2 = DH2 + EI2. But D~II is equal to PFB, and Fg2 A / A is equal to iA2 plus A-2 or C2: IC hence, 2s=e 7- -2 + S -2 PD2P2-~ + FA2 + C B Cor. 3. A parallelopipedon may be constructed on three lines AB, AD, and JAE, intersecting in a common point and not lying in the same plane. For, pass through the extremity of each line, a plane parallel to the plane of the other two lines; then will these planes, together with the planes of the given lines, determine a patallelopipedon. PROPOSITION VII. THIEORE3I. If a plane be passed through the diagonally opposite edges of a parallelopipedon, it will divide the parallelopipeclon into two equal triangular prisms. Let ABCD-A4 be a parallelopipedon, and let a plane be passed through the edges BF and DH:: then will the prisms ABD-II and B CD-fl be equal in volume. II For, through the vertices F and B let planes be passed perpendicular to e FiB, the fbrmer cutting the other lateral edges in the points e, h, g, and the F latter cutting those edges produced, in A the points a, d, and c. The sections Fehg and Badc will be parallelograms, BOOK VII. 189 because their opposite sides are parallel, each to each (B. VI., P. X.); they will also be equal (P. II.): hence, the polyedrlon cBadc-g is a right prism (D. 2, 4), as are also the polyedrons -Bad-h and Bcd-h. Place the triangle Feh upon -Bad, so that F shall coincide with B, e with a, and Ih with d; then, because eF, h/l/, are perpendicular to the plane _Z1R, and aA, dD, to the plane -Bad, the line eE will take thle direction aA, and the line hft the direction diD. The lines AE and ae are equal, because each is equal to _BE (B. I., P. XXVIII.). If we take away from the line aE the part ae, there will remain the part eE; and if from the same line, we take away the part AE, there will remain the part Ala: hence, eE and aA are equal (A. 3); for a like reason hHf is equal to dD': hence, the point E will coincide with A, and the point ff with D, and consequently, the polyedrons ]eh-l~ and Bad —D will coincide throughout, and are therefore equal. If from the polyedron _Bad-11, we take away the part Bad-D, there will remain the prism BA-D-H; and if from the same polyedron we take away the part ]Eh-1, there will remain the prism Bad-h: hence, these prisms are equal in volume. In like manner, it may be shown that the prisms B CD-It and Bcd-h are equal in volume. The prisms.Bad-h, and Bcd-hl, have equal bases, because these bases are halves of equal parallelograms (B. I., P. XXVIII., C. 1); they have also equal altitudes; they are therefore equal (P. V., C.): hence, the prisms BA-D-H- and B CD-tI are equal (A. 1); which was to be proved. Cor. Any triangular prism ABD-H-, is equal to half of the parallelopipedon A G, which has the same triedral angle A, and the same edges AB, AD, and AE. 190 GEOMETRY. PROPOSITION VIII. THEOREM. If two parallelopipedons have a common lower base, and their upper bases between the same parallels, they are equal in volume. Let the parallelopipedons A G and AL have the comrn non lower base AB CD, and their upper bases EFGOH and ITJL, between the same parallels BEK and IIL: then will they be equal in volume. For, the lines E and 1H eTM L IK are equal, because each is equal to AB; hence, the sum of EF and lY,: | or El, is equal to the sum of FI and ICK, or D FK. In the triangular A B prisms AEI-3I and BFK~-L, we have the line AE equal and parallel to BP', and Er equal to F1K; hence, the faoe AEI is equal to BTRIK. The line EBH is equal and parallel to FG, and EI is equal and parallel to _IK; hence, the face E131II is equal to KEKL G: the faces A E/rD and BFG C are also equal (P. VI.): hence, the prisms are equal (P. V.). If from the polyedron ABIBE-I, we take away the prism BFIK-L, there will remain the parallelopipedon A G; and if from the same polyedron we take away the prism AEj~l-3l, there will remain the parallelopipedon AL: hence, these parallelopipedons are equal in volume (A. 3); which was to be proved. BOOK VII. 191 PROPOSITION IX. THEOREM. If two parallelopipedons have a common lower base and the same altitude, they will be equal in volume. Let the parallelopipedoIls AG and AL have the common lower base AB GCD) and the same altitude: then will they be equal in volume. Because they have the same altitude, their upper bases will lie in the same plane. Let the sides Il_~ and KTL _ P _____r be prolonged, and also the N / sides FE and GUH; these -' pi'oloncgations will form a' i)p rallelogram OQ, which will be equal to the com- / mon base of the given par- / allelopipedons, because its / sides are respectively parallel c:and equal to the correspond- A B ing sides of that base. Now, if a third parallelopipedon be constructed, having for its lower base the parallelogram AB C), and for its upper base NOP Q, this third parallelopipedon will be equal in volume to the parallelopipedon A G, since they have the same lower base, and their upper bases between the same parallels, QG, NF (P. VIII.). For a like reason, this tlhird parallelopipedon will also be equal in volume to the parallelopipedon AL: hence, the two' parallelopipedons A G, AL, are equal in volume; which was to be proved. Cor. Any oblique parallelopipedon is equal in volume to a right parallelopipedon, having the same base and an equal altitude. 192 GEOMETRY. PROPOSITION X. PROBLEM. To construct a rectangular parallelopipedon which shall be equal in volume to a right parallelopipedon whose base is any parallelogram. Let AB CID-I1 be a right parallelopipedon, having for its base the parallelogram AB CD. Through the edges Al and iBK pass M Q L P the planes A Q and BP, respectively perpendicular to the plane AEft, the for- mer meeting the plane DL in OQ, and the latter meeting that plane produced in NP: then will the polyedron AP be a, rectangular parallelopipedon equal to the D C i. given parallelopipedon. It will be a rectangular parallelopipedon, because all of its faces are rectangles, and it will be equal to the given parallelopipedon, because the two mlay be regarded as havilng the common base AK (P. VI., C. 1), and an equal altitude A O (P. IX.). The rectangle AN is equal to the parallelogram AC (B. IV., P. I.). Cor. 1 A right parallelopipedon, whose base is any parallelogram, is equal in volume to a rectangular parallelopipedon having an equal base and the same altitude. For, the base ANV is equal to the base A C (B. IV., P. I.); and the altitude AI is common. Cor. 2. An oblique parallelopipedon is equal in volume to a rectangular parallelopipedon, having an equal base and an equal altitude. Cor. 3. Any two parallelopipedons are equal in volume, when they have equal bases and equal altitudes. BOOK VII. 193 PROPOSITION XI. THEOREM. Two rectangular parallelopipedons having a common lower base, are to each other as their altittcdes. Let the parallelopipedons A G and AL have the common lower base AB CD: then will they be to each other as their altitudes AE and AI. 10. Let the altitudes be commensurable, and suppose, for example, that AE is to AlI, as 15 is to 8. Conceive AE to be divided into 15 equal parts, of which AI will contain 8; through the points of division let planes be passed parallel to AB CD. These planes will divide the parallelopipedon A G into 15 parallelopipedons, which have equal bases (P. II. C.) and equal altitudes; hence, they are equal (P. X., Cor. 3). Now, A G contains 15, and AL 8 E H of these equal parallelopipedons; hence, ACG is to AL, as 15 is to 8, or as AE is to AL. In like manner, it may 1m be shown that A G is to AL, as AE; is to Al, when the altitudes are to each z E. other as any other whole numbers.: 2~. Let the altitudes be incommensur- B C able. Now, if AG is not to AL, as AE is to AI, let us suppose that, A G: AL:: AE: A O, in which AO is greater than AI. Divide AlE into equal parts, such that each shall be ess than OI; there will be at least one point of division 13 [ 494. G E O M ETRY. m, between 0 and I. Let P denote the parallelopipcdon, -hose base is AB CD, and altitude Am; since the altitudes AE, Am, are to each other as two whole nninbet's, we have, E H A G P A Am. But, by hypothesis, we have, O AG AL: Al A O; therefore (B. II., P. IV., C.), y AL: P:: AO Am. B C But A 0 is greater than Am; hence, if the proportion is true, AL must be greater than P. On the contrary, it is less; consequently, the fourth term of the proportion cannot be greater than Al. In like mannelr, it may be shown that the fourth term cannot be less than Al; it is, therefore, equal to AI. Inr this case, therefore, AG is to AL, as AE is to AI. Hence, in all cases, the given parallelopipedons are to each other as their altitudes; which was to be proved. Scholium. Any two rectangular parallelopipedons whose bases are equal, are to each other as their altitudes. PROPOSITION XII. THEOREM. Two rectangular parallelopipedons having equal altitudes, are to each other as their bases. Let the rectangular parallelopipedons A G and AKX have the same altitude AlE: then will they be to each other as their bases. BOOK VII 195 For, place them as shown in the figure, and produce the plane of the face 1V/L, until it intersects the plane of the I E H face IGC, in PQ; we shall K thus form a third rectangular parallelopipedon A Q. F The parallelopipedons A G and A Q have a common base AHir; they are therefore to each other as their altitudes AB and AO M - ----- (P. XI.): hence, we have the proportion, B C vol. A G: vol. A Q:: AB A O. The parallelopipedons A Q and AK have the common base AL; they are therefore to each other as their altitudes AD and AJf: hence, vol. AQ: vol. AK: A AD: AM. Multiplying these proportions, term by term (B. II., P. XII.), and omitting the common factor, vol. AQ, we have, ol. vol. vol,. AK:: AB x AD AO x AM. But AB x AD is equal to the area of the base ABC2D; and A 0 x AMJ is equal to the area of the base AJINO: hence, two rectangular parallelopipedons having equal altitudes, are to each other as their bases; which was to be proved. 196 GEOMETRY. PROPOSITION XIII. THEOREM. Any two rectangular pacrallelopipedons are to each other as the products of their bases and altitudes; that is, as the products of their three dimensions. Let AZ and A1G be I E H any two - rectangular parallelopipedons, placed as shown in the figure: then will they be to each other as Y thb products of their three Z dimensions. For, produce the faces necessary to complete the M rectangular parallelopipedon N AK. The parallelopipedons AZ and AK have a comB (l mon base AN; hence (P. XI.), vol. AZ: vol. AK:: AX: AE. The parallelopipecdons AT and A G have a common;litude AF; hence (P. XII.), vol. AK: vol. AG:: AMNO: ABC.D. Multiplying these proportions, term by term, and omitting the common factor, vol. AK, we have, vol. AZ: ol. AG:: AMNO x AX: AB CD x AE; or, since AMNO is equal to AM X AO, and AB CD to AB x AD, v. AZ: vol. A G:: A x AO x AX: AB x AD x AE; wthich was to be proved. BOOK VII. 19'7 Cor. 1. If we make the three edges AM, A 0, and A4X, each equal to the linear unit, the parallelopipedon AZ will be a cube constructed on that unit, as an edge; and consequently, it will be the unit of volume. Under this supposition, the last proposition becomes, 1 vol. AG:: 1: AB x AD x AE; whence, vol. AG = AB x AD x AE. Hence, the volume of anzy rectangular parallelopipedon is equal to the product of its three dimensions; that is, the number of times which it contains the unit of volume, is equal to the number of linear units i5i its length, by the number of linear units in its breadth, by the number of linear units in its height. Cor. 2. Thle volume' of a rectangular parallelopipedon is equal to the product of its base and alZtitude; that is, the number of times which it contains the unit of volume, is equal to the number of superficial units in its base, multiplied by the number of linear units in its altitude. Cor. 3. The volume of any parallelopipedon is equal to the product of its base and altitude (P. X., C. 2). PROPOSITION XIV. THEOREM.'The volume of any prism is equal to the product of its base and altitucle. Let AB CDE-iT be any prism: then is its volume equal to the produzcl of its base and altitude. For, through any lateral edge, as AF, pass the planes AH, AI, dividing it into triangular prisms. These prisms will all have a common altitude equal to that of the given prism. 198 GEOMIETRY. Now, the volume of any one of the triangular prisms, as ABC-t, is equal to half that of a parallelopipedon constructed on the edges BA, B C, B G (P. VII., C.); but the volume of this plarallelopipedon is equal to the product of its - base and altitude (P. XIII., C. 3); and F because the base of the prism is half that of the parallelopipedon, the volule.. of the prism is also equal to the pro- C duct of its base and altitude: hence, A B the suml of the triangular prisms, -which make up the given prism, is equal to the sum of their bases, which make up the base of the given prism, into their common altitude; which was to be proved. Cor. Any two prisms are to eaoh other as the products of their bases and altitudes. Prisms having equal bases are to each other as their altitudes. Prisms having equal altitudes are to each other as their bases. PROPOSITION XV. THIEORE3f. Two triangqular pyramids having equal base:~ and equal altitudes, are equal isn volume. Let S-AB C, and S-abc, be two pyramids having their equal bases ABC and abc in the same plane, and let AT be their common altitude: then will they be equal in volume. For, if they are not equal in volumle, suppose one of them, as S-AB C, to be the greater, and let their difference be equal to a prism whose base is AB C, and whose altitude is Aa. BOOK VII. 199 Divide the altitude AT into equal parts Ax, xy, &c., each of which is less than Act, and let k denote one of these parts; through the points of division pass planes parallel to the plane of the bases; the sections of the two pyramids, by each of these planes, will be equal, namely, DE'F to def, GHI to git, &c. (P. II., C. 2). T S n / Z__ K yf G f On the triangles ABC,, DEF, &c., taken as lower bases, construct exterior prisms whose edges shall be parallel to AS, and whose altitudes shall be equal to D: and on the triangles def, ghi, &c., taken as upper bases, construct interior prisms, whose edges shall be parallel to Sa, and whose altitudes shall be equal to lo. It is evident that the sumn of the exterior prisms is greater than the pyramid S-AB C, and also that the sum of the interior prisms is less than the pyramid S-abe: hence, the difference between the sum of the exterior and the sum of the interior prisms, is greater than the difference between the two pyramids. Now, beginninng at the bases, the second exterior prism iEFID-G, is equal to the first, interior prism efd-a, 200 GEO0METRY. because they have the same altitude c, and their bases RFE7D, efd, are equal: for a like reason, the third exterior prism IIIG-K, and the second interior prism hig-c, are eqtual, and so on to the last in each set: hence, each of the exterior prisms, excepting the first B CA-1D, has an equal corresponding interior prism; the prism B CAl-D, is, therefore, the difference between the sum of all the exterior prisms, and the sum of all the interior prisms. But the difference between these two sets of prisms is greater than that between the two pyramids, which latter difference was supposed to be equal to a prism whose base is _B CA, and whose altitude is equal to Act, greater than k; consequently, the prism B (IA-D is greater than a prism havilln the same base and a greater altitude, which is impossible: hence, the -supposed inequality between the two pyramidis cannot exist; they arle, therefore, equal in volume; which was to be proved. PROPOSITION XVI. THEOREM. Any triangular prism may be divided into three triangular pyramids, equal to each other in volume. Let ABUC-D be a triangular E D prism: then can it be divided into -three equal triangular pyramids. For, through the edge A C, pass the plane ACFi, and through the edge ]liF lass the plane EFC. The pyramids A CE-F and ECD-F,, have their bases A CE A and ECD equal, because they are halves of the same parallelogram A CDE; and they have a common BOOK VII. 201 altitude, because their bases are in the same plane AD, and their vertices at the same point F; hence, they are equal in volume (P. XV.). The pyramids AlBC-F- and -DEF_-C, have their bases AB C and DEE, equal, because they are the bases of the given prism, and their altitudes are equal because each is equal to the altitude of the prism; they are, therefore, equal in volume: hence, the three pyramids into which the prism is divided, are all equal in volume; which was to be proved. Cor. 1. A triangular pyramid is one-third of a prism, having an equal base and an equal altitude. Cor. 2. The volume of a triangular pyramid is equal to one-third of the product of its base and altitude. PROPOSITIONl XVII. THEOREM. The volume of cany pyramid is equal to one-third of the product of its base and altitude. Let S-AB CDE, be any pyramid: then is its volume equal to one-third of the product of its base and altitude. For, through any lateral edge, as SE, pass the planes SEB, SEC, dividing the pyramid into triangular pyramids. The altitudes of these pyramids will be equal to / each other, because each is equal to that of the given pyramid. Now, the volume of each triangular pyramid is equal to one- /Et third of the product of its base and alti- A'C tude (P. XVI., C. 2); hence, the sum of the volumes of the triangular pyramids, is equal to one-third of the product of the sum of their bases 202 GEOMETRY. by their common altitude. But the sum of the triangular pyramids is equal to the given pyramid, and the sum of their bases is equal to the base of the given pyramid: hence, the volume of the given pyramid is equal to onethird of the product of its base and altitude; which, was to be proved. Cor. 1. The volume of a pyramid is equal to one-third of' the volume of a prism having an equal base and an equal altitude. Cor. 2. Any two pyramids are to each other as the products of their bases and altitudes. Pyramids having equal bases are to each other as their altitudes. Pyramids having equal altitudes are to each other as their bases. Scholium. The volume of a polyedron may be found by dividing it into triangular pyramids, and computing their volumes separately. The sum of these volumes will be equal to the volume of the polyedron. PROPOSITION XVIII. THEOREMI. The volume of a frustum of any triangular pyramid is equal to the sumn of the volzumes of three pyramzids whose common altitude is that of the frustum, and twhose bases are the lower base of the frustmn, the uppe]r blasc of the frustum, and a mnean proportional between the twt; bases. Let FGPH-h be a frustum of any triangular pyramnid: then will its volutle be equal to that of three pyramids whose common altitude is that of the frustum, and iwhose bases are the lower base G&ir, the upper base fgh, and a mean proportional between their bases. BOOK VII. 203 For, through the edge FHJ; pass the plane FJTg, and through the edge fg, pass the plane fgti[, dividing the frustum into three pyramids. The pyramid g-_FGII, has for its base the lower f h base EG// of the frustum, and its al- g titude is equal to that of the frustumn, because its vertex g, is in the plane of the upper base. The pyramid Ht-fg\h, has for its base the upper base fgh of F L —--- H the frustum, tInd its altitude is equal to that of the frustum, because its vertex G lies in the plane of the lower base. The remaining pyramlid may be regarded as hlaving the triangle /FfH for its base, and the point g for its vertex. From g, draw gHi- parallel to f_, and draw also JKut and HKf. Then will the pyramids H/-FfHf and g-yf-l, be equal; for they have a common base, and their altitudes are equal, because their vertices l/~ and g are in a line parallel to the base (B. VI., P. XII., C. 2). Now, the pyramid H~-FflI may be regarded as having _IHT for its base and f for its vertex. From _iK, draw KIL parallel to GII; it will be parallel to ghA: then will the triangle._IEHL be equal to Pfh, for the side PFHI is equal to fg, the angle F to the angle f, and the angle Its to the angle g. But, _FIII is a. mean proportional between FHLL and FiG7I (B. IV., P. XXIV., C.), or between fJgh and FG/I. The pyramid f-P7JCl;, has, therefore, for its base a mean proportional between the upper and lower bases of the frustum, and its altitude is equal to that of the filrs tum; but the pyramid f-PFJtII is equal in volume to the pyramid g-FfII: hence, the volume of the given frustum is equal to that of three pyramids whose common altitude is equal to that of the frustum, and whose bases are the upper base, the lower base, and a mean proportional betweenl them; which was to be proved. 204 GEOMETRY. Cor. The volume of the frustum of any pyramid is equal to the sum of the volumnes of three pyramids whose common altitude is that of the frustuzm, and whose bases aree the lower base of the frustum, the uopper base of the ftrutstltm, and a mean proportional between them. For, let AB CDE-e be a frustum of S any pyramid. Through any lateral edge, as eE, pass the planes eyEBb, eEUCc, dividinog it into triangular frustums. Now, the a C sum of the volumes of the triangular frus- \ tums is equal to the sum of three sets of pyramids, whose common altitude is that of A the given frustum. The bases of the first set make up the lower base of the given firustum, the bases of the second set make up the upper base of the given frustum, and the bases of the third set make up a mean proportional between the upper and lower base of the given frustum: hence, the sum of the volumes of the first set is equal to that of a pyramid whose altitude is that of the frustum, and whose base is' the lower base of' of the frustum; the sum of the volumes of the second set is equal to that of a pyramid whose altitude is that of the frustum, and whose base is the upper base of the 2?."stum; and, the sumn of the third set is equal to that of a pyres mid whose altitude is that of the frustum, and whose base is a mean proportional between the two bases. PROPOSITION XIX. THEOREAM. Similar triangular prisms are to each other as the cubes of their homologous edges. Let CBBD-P, cbd-p, be two similar triangular prisms, and let B C, be, be any two homologous edges: then will the prism CBD-P be to the prism cbd-p, as B C3 to'be. BOOK VII. 205 For, the homologous angles B and b are equal, and the faces which bound them are similar (D. 16): hence, these triedral angles may be applied, one to the other, so that the angle cbd will coinP p cide with CBD, the edge ba a — aa with BA. In this case, the D} d. prism cbd-p will take the position Bcd-p. From A C C B draw AH perpendicular to the common base of the prisms: then will the plane BAH be perpendicular to the plane of the common base (B. VI., P. XVI.). From a, in the plane BAH, draw ah perpendicular to BH: then will ah also be perpendicular to the base B DC (B. VI., P. XVII.); and AHf, ah, will be the altitudes of the two prisms. Since the bases CBD, cbd, are similar, we have (B. IV,, P. XXV.), base CBD: base cbd:: C'' 2 ~ c:d. Now, because of the similar triangles ABJr, aBh, and of the similar parallelograms A C, ac, we have, AT: ah:: CB~ cb; hence, multiplying these proportions term by term, we have, base CBD x All: base cbd x ah:: CB3: cb3. But, base CBD x All is equal to the volume of the prism CDB-A, and base cbd x ah is equal to the volume of the prism cbd-p; hence, prism C(DB-P: prism cbd-p:::-b3; which was to be proved. 206 GEO METRY. Cor. 1. Any two similar prisms are to each other as the cubes of their homologotus edges. For, since the prisms are similar, their bases are similar polygons (D. 16); and these similar polygons may each be divided into the same number of similar triangles, similarly placed (B. IV., P. XXVI.); therefore, each prism may be divided into the same number of triangular prisms, having their faces similar and like placed; consequently, the triangular prisms are similar (D. 16). But these triangular prisms are to each other as the cubes of their homologous edges, and being like parts of the polygonal prisms, the polygonal prisms themselves are to each other as the cubes of their homologous edges. Cor. 2. Similar prisms are to each other as the cubes of their altitudes, or as the cubes of any other homologous lines. PROPOSITION XX. THEOREM. Similar pyramids are to each other as the cubes of their homologous edges. Let S-A/B CDLE, and S-abcde, be two similar pyramids, so placed that their homologous angles at the vertex shall coincide, and let AB and ab be any two homologous edges: then will the pyramids be to each other as the cubes of AB and ab. For, the face SAB, being similar to aI Sab, the edge AiB is parallel to theb \ edge ab, and the face SB C being simi-, lar to Sbc, the edge -BC is parallel to A C be; hence, the planes of the bases are B parallel (B. VI., P. XIII.). BOOK VII. 207 Draw SO perpendicular to the base AB CODE; it will also be perpendicular to the base abccde. Let it pierce that plane at the point o: then will SO be to So, as SA is to Sa (P. III.), S or as ABi is to ab; hence, ISO: -So:: AB: ab. a But the bases being similar polygons, we have (B. IV., P. XXVII.), A C base AB CDE: base abcde B: AB: a2. B Multiplying these proportions, term by term, we have, base ABCDE x SO: base abede x ISo:: AB3: abs. But, base ABCIDE x ISO is equal to the volume of the pyramid S-ABC DE, and base abcde x 0So is equal to the volume of the pyramid S-abcde; hence, pyramid S-ABCIDE: pyramid S-abcde:: A3: bs; which was to be proved. Cor. Similar pyramids are to each other as the cubes of their altitudes, or as the cubes of any other homologous lines. 208 GEOMETRY. GENERAL FOItMULAS. If we denote the volume of any prismn by V, its base by _B, and its altitude by If; we shall have (P. XIV.), V =B x I.. (1.) If we denote the volume of any pyramid by V, its base by B, and its altitude by 11, we have (P. XVII.), V = t,,B x ff..... (2.) If we denote the volume of the frustum of any pyramid by V, its lower base by 2B, its upper base by b, alld its altitude by 1; we shall have (P. XVIII., C.), V = (B + b + OBx b) x. ~ ~ (3.) RIEGULAR POLYEDRONS; A REGULAR POLYEDnRON is one whose faces are all equal regular polygons. There are five regular polyedrons, namely: 1. The TETrAEDRON, or regular pyramid-a polyedron bounded by four equal equilateral triangles. 2. The HEXAEDRON, or cube-a polyedron bounded by six equal squares. 3. The OcTAEDROIN-a polyedron bounded by eight equal equilateral triangles. 4. The DODECAEDRDON-a polyedron bounded by twelve equal and regular pentagons. BOOK VII. 205 5. The IcosEnDRooN —a polyedron bounded by twenty equal equilateral triangles. In the Tetraedron, the triangles are grouped about the polyedral angles in sets of three, in the Octaedron they are grouped in sets of four, and in the Icosaedron they'are grouped in sets of five. Now, a greater number of equilateral triangles cannot be grouped so as to form a salient polyedral angle; for, if they could, the sum of the plane angles formed by the edges would be equal to, or greater than, four right angles, which is impossible (B. VI., P. XX.). In the Ilexaedron, the squares are grouped about the polyedral angles in sets of three. Now, a greater number of squares cannot be grouped so as to form a salient polyedlral angle; for the samie reason as before. In the Dodeeaedron, the regular pentagons are grouped about the polyedral angles in sets of three, and for the same reason as before, they cannot be grouped in any greater number, so as to form a' salient polyedral angle. Furthermore, no other regular polygons can be grouped so as to form a salient polyedral angle; therefore, Only fire regular polyedrons can be formed. 14 BOOK VIII. THE CYLINDER, THE CONE, AND THE SPHER]2:. DEFINITIONS. 1. A CYLINDER is a volume which may be generated by a rectangle revolving about one of its sides as an axis. Thus, if the rectangle ABCD be turned about the side AB, as an axis, it will generate the cylinder FG CQ-P. The fixed line AB is called the axis of the cylinder; the curved surface generated P by the side CD, opposite the axis, is called E the convex surface of the cylinder; the equal!,. circles PCGCQ, and EIIDP, generated by M K the remaining sides BC and AD, are called bases of the cylinder; and the perpendicular F distance between the planes of the bases, is G called the altitude of the cylinder. The line D C, which generates the convex surface, is, in any position, called an element of the surface; the elements are all perpendicular to the planes of the bases, and anyv one of them is equal to the altitude of the cylinder. Any line of the generating rectangle A BCD, as Iid, which is perpendicular to the axis, will generate a circle whose plane is perpendicular to the axis, and which is equal to either base: hence, any section of a cylinder by a plane perpendicular to the axis, is a circle equal to either base. Any section, FCaDE, made by a plane through the axis, is a rectangle double the generating rectangle, BOOK VIII. 211 2. SIMILAR CYLINDERS are those which may be generated by similar rectangles revolving about homologous sides. The axes of similar cylinders are proportional to the radlii of their bases (B. IV., D. 1); they are also proportional to any other homologous lines of the cylinders. 3. A prism is said to be inscribed in a cylinder, when its bases are inscribed in the bases of the cylinder. In this case, the cylinder is said to ] be circumscribed about the prism. The lateral edges of the inscribed prism are elements of the surface of the circumscribing cylinder. 4. A prism is said to be circumscribed about a cylinder, when its bases are circumscribed about the bases of the cylinder. In this case, the cylinder is said to be inscribed in the prism. The lines which join the corresponding points of contact in the upper and lower bases, are common to the surface of the cylinder and to the lateral faces of the prism, and they are the only lines which are common. a I Tile lateral faces of the prism are said to be tangent to the cylinder along these lines, which are then called eleiments of contact. 5. A CONE is a volume which may be generated by a right-angled triangle revolving about one of the sides adja. cent to the right angle, as an axis. 212 GEO METRY. Thus, if the triangle SAB, right-angled at A, be turned about the side SA, as an axis, it will generate the cone S- CDBE. The fixed line SAl, is called the S axis of the conee; the curved surface generated by the hypothenuse S-B, is \ called the conzvex szurfcace of the cone; F 3 the circle generated by the side AB, is called the base of the cone; and the point S, is called the vertex of C A B t4ie cone; the distance from the vertex D to any point in the circumference of the base, is called the slant height of the cone; and the perpendicular distance from the vertex to the plane of the base, is called the altituede of the cone. The line SB, which generates the convex surface, is, in any position, called an element of the surface; the elements are all equal, and any one is equal to the slant height; the axis is equal to the altitude. Any line of the generating triangle SAB, as GC, wvhich is perpendicular to the axis, generates a circle whose plane is perpendicular to the axis: hence, any section of a cone by a plane perpendicular to the axis, is a circle. Any section SB C, made by a plane through the axis, is an isosceles triangle, double the generating triangle. 6. A TRUNCATED CONF, is that portion of a cone included between the base and any plane which cuts the cone. When the cutting plane is parallel to the plane of the base, the truncated cone is called a FRUSTUM OF A CONE, and the intersection of the cutting plane with the cone is called the upper base of the frustuml; the base of the cone is called the lower base of the frustum. BOOK VIII. 213 If the trapezoid tHGAB, right-an- F gled A and G, be revolved about -I C, as an axis, it will generate a frus- E turn of a cone, whose bases are ECDB C ---— i-A and FiAT"ll whose altitude is A G, and D whose slant height is Btl.. 7. SIMILAR CONES are those which may be generated by similar right-angled triangles revolving about homologous sides. The axes of similar cones are proportional to the radii of their bases (B. IV., D. 1); they are also proportional to any other homologous lines of the cones. 8. A pyramid is said to be in- S scribed in a cone, when its base is inscribed in the base of the cone, and when its vertex coincides with that of. the cone. / The lateral edges of the inscribed A pyramid are elements of the sur face of the circumscribing cone. B 9. A pyramid is said to be cir'cumscribed about ca cone, when its base is circumlscribed about the base of the cone, and when its vertex coincides with that of the cone. In this case, the cone is said to be inscribed in the )yramid. The lateral faces of the circumscribing pyramid are tangent to the surface of the inscribed cone, along lines which are called elements of contact. 10. A frustum of a pyramid is inscribed in a frustum 214 GEOMETRY. of a cone, when its bases are inscribed in the bases of the frustum of the cone. The lateral edges of the inscribed frustum of a pyramid are elements of the surface of the circumscribing frustum of a cone. 11. A frustum of a pyramid is circumscribed about a frustum of a cone, when its bases are circumscribed about those of the frustum of the cone. Its lateral faces are tangent to the surface of the frustum of the cone, along lines which are called elements of contact. 12. A SPHERE is a volume bounded by a surface, every point of which is equally distant from a point within called the centre. A sphere may be generated by a semicircle revolving about its diameter as an axis. 13. A RADIUS of a sphere is a straight line drawn from the centre to any point of the surface. A DIArETERiSis any straight line drlawn through the centre and limited at both extremities by the surface. All the radii of a sphere are equal: the diameters are also equal, and each is double the radius. 14. A SPHERICAL SEcTOR is a volume generated by a sector of a circle revolving about a diameter of the circle lying without it. The surface generated by the arc is called the base of the sector. 15. A plane is TANGENT TO A SPHERE when it touches it in a single point. 16. A ZONE is a portion of the surface of a sphere included between two parallel planes. The bounding lines BOOK VIII. 215 of the sections are called bases of the zone, and the distance between the planes is called the altitude of the zone. If one of the planes is tangent to the sphere, the zone las but one base. 17. A SPHErICAL SEGMENT is a portion of a sphere included between two parallel planes. The sections made by the planes are called bases of the segment, and the distance between them is called the altitude of the sefgment. If one of the planes is tangent to the sphere, the segnent has but one base. The CYLINDER, the CONE, and the SPHERE, are sometimes called THE THREE ROUND BODIES PROPOSITION I. THEOREM. The convex surface of a cylinder is equal to the circumference of its base multil29ied by the altitude. Let ABD be the base of a cylinder whose altitude is H: then will its convex surface be equal to the circumference of its base multiplied by the altitude. For, inscribe within the cylinder a prism whose base is a regular polygon. The convex surface of this prism will be equal to the perimeter of its base multiplied by its altitude (B. VII., P. I.), whatever may be the number of sides of its base. But, when the number of sides is infinite (B. V., P. X., C. 1), the T convex surface of the prism coineides with A D that of the cylinder, the p)-elrilneter of 216 G EGEOMETRY. the base of the prism coincides with the circumnference of the base of the cylinder, and the altitude of the prlism is the same as that of the cylinder: hence, the convex surface of the cylinder is equal to the circumference of its base multiplied by the altitude; which twas to be proveCd. Cor. The convex surfaces of cylinders having equal altitudes are to each other as the circumferences of their bases. PROPOSITION II. THIEOREM. The volume of a cylinder is equal to the product of its base and altitude. Let ABD be the base of a cylinder whose altitude is H; then will its volume be equal to the product of its base and altitude. For, inscribe within it a pr ism whose base is a regular polygon. The volume of this prisl is equal to the product of its base and altitude (B. VII., P. XIV.), whatever may be the number of sides of its base. But, wlhen the nthum- ber of sides is infinite, the prism coin-: cides with the cylinder,'the base of the: C D) prism with the base of the cylinder, and the altitude of the prislll is the same as that of the cylinder: hence, the volume of the cylinder is equal to the product of its base and altitude; cwhich wcas to be proved. Cor. 1. Cylinders are to each other as the products ot their bases and altitudes; cylinders having equal bases are to each other as their altitudes; cylinders having equal altitudes are to each other as their bases. BOOK VIII. 217 Cor. 2. Similar cylinders are to each other as the cubes of their altitudes, or as the cubes of the radii of their bases. For, the bases are as the squares of their radii (B. V., P. XIII.), and the cylinders being similar, these radii are to each other as their altitudes (D. 2): hence, the bases are as the squares of the altitudes; therefore, the bases multiplied by the altitudes, or the cylinders themselves, are as the cubes of the altitudes. PROPOSITION III. THEOREM. The convex surface of a cone is equal to the circznfe-rence of its base multiplied by hal'f the slant height. Let S-A CD be a cone whose base is A CD, and whose slant height is SA: then will its convex surface be equal to the circumference of its base multiplied by half the slant height. For, inscribe within it a right pyramid. The convex surface of this pyramid is equal to the perimeter of its base multiplied by half the slant height (B. VII., P. IV.), whatever may be the number f of sides of its base. But when the num- A ber of sides of the base is infinite, the C convex surface coincides with that of the cone, the perimeter of the base of the pyramid coincides with the ciceumlni'eence of the base of the cone, and the slant height of the pyramid is equal to the slant height of the cone: hence, the convex surface of the cone is equal to the circumference of its base multiplied by half the slant height; twhi'ch was to be proved. 218 GEOMETRY. PROPOSITION IV. THEOREM. The convex surface of a frustum of a cone is equal to half the sum of the circumferences of its two bases multipliedZ by the slant height. Let BIA-D be a frustum of a cone, _BIA and EGD its two bases, and EB its slant height: then is its convex surface equal to half the sum of the circumferences of its two bases multiplied by its slant height. For, inscribe within it the frustum E~ GD of a right pyramid. The convex surface cf this frustum is equal to half the sum of the perimeters of its bases, multiplied by the slant height (B. VII., B- - -i A P. IV., C.), whatever may be the 0 I number of its lateral faces. But when [ the number of these faces is infinite, the convex surface of the frustum of the pyramid coincides with that of the cone, the perimeters of its bases coincide with the circumferences of the bases of the frustum of the cone, and its slant height is equal to that of the cone: hence, the convex surface of the frustum of a cone is equal to half the sum of the circumferences of its bases multiplied by the slant height; which was to be proved. Scholium. From the extremities A and ), and fi'(ern the middle point 1, of a line AD, let the lines AO, P(C, and lK, be drawn perpendicular to a line O C: then will 1Ki be equal to half the sum of A 0 and ) C. For, draw PDd and li, perpendicular to A0: then, because Al is equal to lD, we shall have Ai equal to id (B. IV., P. XV.), and consequently to Is; that is, AO exceeds 1Ik BOOK VIII. 219 as much as It exceeds P C: hence, 1K is equal to the half sum of AO and DC. Now, if the line AD be revolved about O C, as an axis, it will generate the surf:ace of a frustum of a cone whose slant height is AD; the point I will generate a circumference which is equal to half the sum of thle circumferences generated by A and D: hence, if a straight line be revolved about another straight line, it will generate a surface whose measure is equrcal to the product of the generating line and thle circumference generated by its middle point. This proposition holds true when the line AD meets OC, and also when AD is parallel to OC. PROPOSITION V. THEOREM. The volume of a cone is equal to its base multiplied by one-third of its altitude. Let ABI)E be the base of a cone whose vertex is S, and whose altitude is So: then will its volume be equal to the base multiplied by one-third of the altitude. For, inscribe in the cone a right pyramid. The volume of this pyramid is equal to its base multiplied by onethird of its altitude (B. VII., P. XVII.), whatever may be the number of its lateral faces. But, when the number A'.l. 0 of lateral faces is infinite, the pyramid coincides with the cone, the base of B the pyramid coincides with that of the cone, and their altitudes are equal: hence, the volume of a cone is equal to the base multiplied by one-third of the altitude; which was to be proved. 220 GEOMETRY. Cor. 1. A cone is equal to one-third of a cylinder having an equal base and an equal altitude. Cor. 2. Cones are to each other as the products of their bases and altitudes. Cones having equal bases are to each other as their altitudes. Cones having equal altitudes are to each other as their bases. PROPOSITION VI. THEOREM. The volumre of a frustum of a cone is equal to the sum of the volumes of three cones, having for a common altitude the altitudle of the frustum, and for bases the lower base of the frustum, the upper base of the frustum, and a mneaon proportional between the bases. Let BIA be the lower base of a frustum of a cone, EG-D its upper base, and O C its altitude: then will its volume be equal to the sum of three cones whose common altitude is O C, and whose bases are the lower base, the upper base, and a mean proportional between them. For, inscribe a frustum of a right pyramid in the given frustum. The volume of this frustum is equal to the sum of the volumes of three pyramids whose common altitude is -- that of the frustum, and whose bases B - A are the lower base, the upper base, and a mean proportional between the two (B. VII., P. XVIII.), whatever may be the number of lateral faces. But when the number of faces is infinite, the frustum of the pyramid coincides with the frustum of the cone, its bases with the bases of the cone, the three pyramids become cones, and their altitudes BOOK VIII. 221 are equal to that of the frustum; hence, the volume of the frustum of a cone is equal to the sum of the volumes of three cones whose common altitude is that of the frustum, and whose bases are the lower base of the frustum, the upper base of the frustum, andi a mean proportional between them; which was to be proved. PROPOSITION VII. THEOREM. Any section of a sphere inade by a plane, is a circle. Let C be the centre of a sphere, CA one of its radii, and AJAB any section made by a plane: then will this section be a circle. For, draw a radius C O perpen- D dicular to the cutting plane, and let I it pierce the plane of the section at -B —----- -- 0. Draw radii of the sphere to any M two points 2/,, 11', of the curve which \ bounds the section, and join these points with 0: then, because the radii CGUi, CJ~'' are equal, the points XM, 3', will be equally distant from 0 (B. VI.,'P. V., C.); hence, the section is a circle; which was to be proved. Cor. 1. When the cutting plane passes through the centre of the sphere, the radius of the section is equal to that of the sphere; when the cutting plane does not pass through the centre of the sphere, the radius of the section will be less than that of the sphere. A section whose plane passes through the centre of the sphere, is called a great circle of the sphere. A section whose plane does not pass through the centre of the sphere, 222 GEOMETRY. is called a small circle of the sphere. All great circles of the same, or of equal spheres, are equal. Cor. 2. Any great circle divides the sphere, and also the surface of the sphere, into equal parts. For, the parts may be so placed as to coincide, otherwise there would b)e some points of the surface unequally distant from the centre, which is impossible. Cor. 3. The centre of a sphere, and the centre of any small circle of that sphere, are in a straight line perpendicular to the plane of the circle. Cor. 4. The square of the radius of any small circle is equal to the square of the radius of the sphere diminished by the square of the distance from the centre of the sphere to the plane of the circle (B. IV., P. XI., C. 1): hence, circles which are equally distant from the centre, are equal; and of two circles which are unequally distant from the centre, that one is the less whose plane is at the greater distance from the centre. Cor. 5. The circumference of a great circle may always be made to pass through any two points on the surface of a sphere. For, a plane can always be passed through these points and the centre of the sphere (B. VI., P. II.), and its section will be a great circle. If the two points are the extremities of a diameter, an infinite number of planes can be passed through them and the centre of the sphere (B. VI., P. I., S.); in this case, an infinite number of great circles can be made to pass through the two points. Cor. 6. The bases of a zone are the circumferences of circles (D. 16), and the bases of a segment of a sphere are circles. BOOK XVIII. 223 PROPOSITION VIII. THEOREM. Any plane perpendicular to a radius of a sphere at its extremity, is tangent to the sphere at that point. Let C be the centre of a sphere, CA any radius, and FAG a plane perpendicular to CA at A: then will the plane FA G be tangent to'the sphere at A. For, from any other point of the F plane, as M, draw the line 3IC: A M then because C.A is a perpendicular to the plane, and C(M an oblique line, C-3I will be greater than CA (B. VI., P. V.): hence, the point M lies without the sphere. The plane fal G, therefore, touches the sphere at A, and consequently is tangent to it at that point; which was to be proved. Scholium. It may be shown, by a course of reasoning analogous to that employed in Book III., Propositions XI., XII., XIII., and XIV., that two spheres may have any one of six positions with respect to each other, viz.: i. When the distance between their centres is greater than the sum of their radii, they are external, one to the other: 2~. When the distance is equal to the sum of their radii, they are tangent, externally: 3~. When this distance is less than the sum, and greater than the difference of their radii, they intersect each other: 40. When this distance is equal to the difference of their radii, they are tangent internzally: 50. When this distance is less than the difference of their radii, one is wholly within the other: 6~. When this distance is equal to zero, they have a oommon centre, or, are concentric. 224 GEO MET RY. DEFINITIONS. 1~. If a semi-circumference be diiided into equal arcs, the chords of these arcs form half of the perimeter of a regular illscribed polygon; this half perimeter is called a regular semi-perimneter. The figure bounded by the regular semiperimeter and the diameter of the semi-circumference is called a reguilarc semi-polygon. The diameter itself is called the axis of the semi-polygon. 2~. If lines be drawn from the extremi- A ties of any side, and perpendicular to the B.. E axis, the intercepted portion of the axis is called the projection of that side. C K The broken line ABCDGP is a regcular semi-perimeter; the figure bounded by it and the diameter AP, is a regular semi-polygon, AP is its axis, PTK is the G projection of the side.B C, and the axis, AP, is the projection of the entire semi-perimeter. PROPOSITION IX. LEMMA. If a regular semi-polygon, be revolved about its axis, the surface generated by the semi-perimeter will be equal to the axis multiplied by the circumnference of the inscribed circle. Let AB CDEF be a regular semi-polygon, AF its axis, and ON its apothem: then will the surface generated by the regular semi-perimeter be equal to AT x circ. ON. From the extremities of any side, as iDE, draw DI and EHI perpendicular to AF; draw also NiM[ perpendicular to AtF and EEK perpendicular to DI. Now, the surface generated by ED is equal to DE x cire. N-. BOOK VIII. 225 (P. IV., S.). But, because the triangles EDK_ and O0niI~ are similar (B. IV., P. XXI.), we have, DE: EK or I:: ON: NI:: circ. ON: circ.Ne; whence, DE x circ. XNAI= IXH x circ.ON; __ that is, the surface generated by any side D....... I is equal to the projection of that side o multiplied by the circumference of the inscribed circle: hence, the surface genlerated by the entire semi-perimeter is equal to the sum of the projections of its sides, A or the axis, multiplied by the circumference of the inscribed circle; which was to be proved. Cor. The surface generated by any portion of the perimeter, as C-DE, is equal to its projection PH; multiplied by the circumference of the inscribed circle. PROPOSITION X. THEOREM. The sulface of a sphere is equal to its diameter multiplied by the circumfierence of a great circle. Let AB CZ)D be a semi-circumference, E O its centre, and AE its diameter: then D will the surface of the sphere generated by revolving the semi-circumiference about AE, be equal to AE x circ. OE. For, the semi-circumference may be re- B garded as a regular semi-perimeter with an infinite number of sides, whose axis is AE, A and the radius of whose inscribed circle is OE: hence (P. IX.), the surface generated by it is equal to ABE x circ. OE; which was to be proved. 15 226 GEOMETRY. Cor. 1. The circumference of a great circle is equal to 2 0E (B. V., P. XVI.): hence, the area of the surface of the sphere is equal to 20 E X 2 OE, or to 40k. o2; that is, the area of the surface of a sphere is equal to foa' great circles. Cor. 2. The surface generated by any arc of the semicircle, as B C, will be a B. zone, whose altitude is equal to the projection of that arc on the diameter. But, the are N C is a portion of a semiperimeter having an infinite number of sides, and the radius of whose inscribed circle is equal to that of the sphere: hence (P. IX., C.), the surface of a zone is equal to its altitude multiplied by the circumference of a great circle of the sphere. Cor. 3. Zones, on the same sphere, or on equal sphere8, are to each other as their altitudes. PROPOSITION XI. LEMMA. If a triangle and a rectangle having the same base and equal altitudes, be revolved about the common base, the volume generated by t/he triangle will be one-third of' that generated by the rectangle. Let ABC be a triangle, and EWBC a rectangle, having fhe same base B C, and an equal altitude AD, and let tilem both be revolved about B C: then will the volume generated by ABC be one-third of that generated by FFB C. For, the cone generated by the right-angled triangle ADB, is equal to one-third of the cylinder generated by BOOK VIII. 227 the rectangle ADBF (P. V., C. 1); and the cone generated by the triangle ADC, is equal to one-third of the cylinder generated by the rectangle ADC-E. But, when AD falls within the F A E trianle, the sum of the cones generated by ADB and ADC, is equal to the volume generated by the triangle ABC; and the CP sum of the cylinders generated by A1DBF and ADCE, is equal to the volume generated by the rectangle EFB C. WVhen AD falls without the triangle, the difference of the cones generated by ADB and ADC, is equal to the volume generated by ABC; and the difference of the cylinders generated by AD1BF and AD CE, is equai to the volume generated by _EFB C: hence, in either case, the volume generated by the triangle AB C, is equal to one-third of the volume generated by the rectangle EFB C; which was to be proved. Cor. The volume of the cylinder generated by E7FB C, is equal to the product of its base and altitude, or to 4-AD2 x BC: hence, the volume generated by the triangle AB C, is equal to 3 AD2 x B C. PROPOSITION XII. LEIMMA. If an isosceles triangle be revolved about a straight line passing through its vertex, the volume generated will be equal to the surface generated by the base multiplied by one-thibrd of the altitude. Let CAB be an isosceles triangle, C its vertex, AB its base, CI its altitude, and let it be revolved about the line CD, as an axis: then will the volume generated be equal to surf. AB x 3 CI. 228 GEOMETRY. There may be two cases: the base, or base produced, may meet the axis; or, the base may be parallel to the axis. 10. Suppose the base, when A produced, to meet the axis at D; draw AM2l, 1K, and BE_, /o' B perpendicular to CD, and BO' parallel to D C. Now, the' volume generated by CAB is CKN'D equal to the difference of the volumes generated by CAD and CBD; hence (P. XI., C.), vol CAB =rA- li2x CD —r3BN2 x CD=-l — (A2 -B2-) x C(7D. But, AM' - BN2 is equal to (All+ ~BNe) (AX - BA), (B. IV., P. X.); and because AX + BN is equal to 2IK (P. IV., S.), and AX - BAT to A0, we have, vol. CAB - 23 IIR x AO x CD. But, the right-angled triangles A OB and C2DI are simslar (B. IV., P. XXI.); hence, AO: AB:: CI: CG9; or, A0 x CD = AB x CZI. Substituting, and changing the order of the factors, we have, vol. CAB = AB X 2 rIK x ~CI. But, AB x 2 rIK is equal to the surface generated by AB; hence, vol. CAB = surf. AB x CI. This demonstration holds good when the axis CD coincides with one side of the triangle CAB. 2~. Suppose the base of the triangle to be parallel to the axis. BOOK VIII. 229 Draw A2[ and BN2 perpendicular to the axis. The volume generated by CGAB, is equal A I to the cylinder generated by the rectangle ABNAJI, diminished by the sum of the cones generated by the triangles CA1~T and B CN; hence, M N vol. CAB= _~ ~IU x AB - -I' x Al — ~rc I x lB. But the sum of Al and IB is equal to AB: hence, we have, by reducing, and changing the order of the factors, vol. CAB AB= AB x 2r CI x CI. But AB x 2 CI is equal to the surface generated by AB; consequently, vol. CAB - surf. AB x 3 CI; hence, in all cases, the volume generated by CAB is equal to szuf. AB x ~C I; which was to be proved. PROPOSITION XIIT. LEMI3IA. If a regular semi-polygonz be revolved about its axis, the volume generated zoill be equal to the suwface generated by the semi-plerimeter multiplied by one-third of the apothem. Let FB DG be a regular semi-poly- F gon, FG its axis, 01 its apothem, and let the sefli-polycgon be revolved about.F'': then will the volume generated lbe equal to suzrf. FDB G x 01. _ For, draw lines froll the vertices to the centre 0. These lines will divide tle semi-pqlygon into isosceles triangles whose bases are sides of the semi-polygon, and whose altitudes are equal to 01 230 G EO M3 E T R Y. Now, the sum of the volumes generated by these triangles is equal to the volume generated by the semi-polygon. But, the volume generated by any triangle, as OAB, is equal to surf. AB x OI (P. XII;) hence, the volume generated by the semi-polygon is equal to surf. FBnD G x ~ OI; which was to be proved. Cor. The volume generated by a portion of the semipolygon, OAB C, limited by radii O C, OA, is equal to surf. ABC x ~O. PROPOSITION XIV. THEOREM. The volume of a sphere is equal to its surface multiplied by one-third of its radius. Let A CE be a semicircle, AE its diameter, 0 its centre, and let the semi- B circle be revolved about AE: then will the volume generated be equal to the surface generated by the semi-circumfer- - ence multiplied by one-third of the radius OA. For, the semicircle may be regarded as a regular semi-polygon having an infinite number of sides, whose semi-perimeter coincides with the semi-circumierence, and whose apothem is equal to the radius: hence (P. XIII.), the volume generated by the semicircle is equal to the surface generated by the semi-circuinference multiplied by one-third of the radius; which was to be proved. Cor. 1. Any portion of the semicircle, as OB6C, bounded by two radii, will generate a volume equal to the surface BOOK VIII. 231 generated by the are B C multiplied by one-third of the radius (P. XIII., C.). But this portion of the semicircle is a circular sector, the volume which it generates is a spherical sector, and the surface generated by the are is a zone: hence, the volume of a spherical sector is equal to thee zone which forms its base mnultiplied by one-third of the racli'us Cor. 2. If we denote the volume of a sphere by V, and its radius by 1?, the area of the surface will be equal to 4 ~i P2 (P. X., C. 1), and the volume of the sphere -ill be equal to 4~R 2 x - R; consequently, we have, V-= 4R3. Again, if we denote the diameter of the spllere by -D, we shall have 1B equal to I D, and BR3 equal to -1D3, and consequently, V-=' D3; hence, the volumes of spheres are to each other as the cube* of'their racldii, or as the cubes of their diameters. Scholihnzm. If the figure E BD, A formed by drawing lines from the ex- B tremities of the are 1BD perpendicular M to CA, be revolved about CA, as an D axis, it will generate a segment of a sphere whose volume may be found by C adding to the spherical sector generated by CiDB, the cone generated by C-BE, and subtracting firom their sum the cone generated by CDE. If the are BD is so taklen that the points BE and F fall on opposite sides of the centre C, the latter cone must be added, instead of subtracted: hence, segmelnt EBDF= zone 1BD x I CD +~,r x 3I CE —-FrD2 x 3 C-F. 232 GEOMETRY. PROPOSITION XV. THEOREM. The surface of a sphere is to the entire suiface of the circumscribed cylinder, including its bases, as 2 is to 3: and the volumes are to each other in the same ratio. Let PJIQ be a semicircle, and PAD Q a rectangle, whose sides PA and QD are tangent to the semicircle at P and Q, and whose side AD, is tangent to the semicircle at M. If the semicircle and the rectangle be revolved about P Q, as an axis, the former will generate a sphere, and the latter a circumscribed cylinder. 1i. The surface of the sphere is to the entire surface of the cylinder, as 2 is to 3. For, the surface of the sphere is equal to four great circles (P. X., C. 1), D C the conxex surface of the cylinder is equal to the circumference of its base multiplied by its altitude (P. I.); that is, it is equal to the circumference of a great circle multiplied by its diameter, or to four great circles (B. V., P. XV.); adding to this the two bases, each of which is equal to a great circle, we have the entire surface of the cylinder equal to six great circles: hence, the surface of the sphere is to the entire surface of he circumscribed cylinder, as 4 is to 6, or as 2 is to 3; which was to be proved. 20. The, volume of the sphere is to the volume of the cylinder as 2 is to 3. For, the volume of the sphere is equal to 4 e 123 (P. XIV., C. 2); the volume of the cylinder is equal to its base multiplied by its altitude (P. II.); that is, it is equal to BOOK VIII. 233 -rPj X 2R, or to 1.R3: hence, the volume of the sphere is to that of the cylinder as 4 is to 6, or as 2 is to 3; (which was to be proved. Cor. The surface of a sphere is to the entire surface of a circumscribed cylinder, as the volume of the sphere is to volume of the cylinder. Scholium. Any polyedron which is circumscribed about a sphere, that is, whose faces are all tangent to the sphere, may be regarded as made up of pyramids, whose bases are the faces of the polyedron, whose common vertex is at the centre of the sphere, and each of whose altitudes is equal to the radius of the sphere. But, the volume of any one of these pyramids is equal to its base multiplied by onethird of its altitude: hence, the volume of a circumscribed polyedcllon is equal to its surface multiplied by one-third of the radius of the inscribed sphere. Now, because the volume of the sphere is also equal to its surface multiplied by one-third of its radius, it follows that the volume of a sphere is to the volume of any circumscribed polyedron, as the surface of the sphere is to the surface of the polyedron. Polyedrons circumscribed about the same, or about equal spheres, are proportional to their surfaces. GENERAL FORMULAS. If we denote the convex surface of a cylinder by S, its volume by V, the radius of its base by 1, and its alti. tude by 11, we have (P. I., II.), S = 2R X ~II........ (1.) V = R2 x Hff....... ( 2.) 234 GEOMETRY. If we denote the convex surface of a cone by S, its volume by V, the radius of its base by _B, its altitude by IH; and its slant height by HA', we have (P. III., V.), S = lx II'........ (3.) V- ='-t~UxH[ ~......... (4.) If we denote the convex surface of a frustum of a cone by S, its Avolulne by V, the radius of its lower base by A, the radius of its upper base by 1', its altitude by H,; and its slant height by GI', we have (P. IV., VI.), -S = ~(R +') x II' * * * * * (5.) V = 3~~(Xj2'+ Ale x _LU') x I ~ (6.) If we denote the surface of a sphere by S, its volume by V, its radius by _R, and its diameter by D, we have (P. X., C. 1, XIV., C. 2, XIV., C. 1), S = 4 2.......... (7.) v-F _' 4 =..... ~ ~ (.) If we denote the radius of a sphere by B2, the area of any zone of the sphere by S, its altitude by II, and the volume of the corresponding spherical sector by T, we shall have (P. X., C. 2), S = 2/R~ x H.... ~ ~ ~ (0.) V- 2= Zr2 X H......... (10.) If we denote the volume of the corresponding spheric:al segment by V, the radius of its lower base by!', the radius of its upper base by B", the distance of its lower base from the centre by H/', and the distance of its upper base from the centre by Ii", we have (P. XIV., S.), v = 3 R(2X2 X I1+ B' 2 X' J:, r,"2 x II") ~ ( 1.) BOOK IX. SPH E I CAL GEOMET R Y. DEFINITIONS. 1 A SPrERICAL ANGLE is an angle included between the arcs of two great circles of a sphere meeting at a point. The arcs'are called sides of the angle, and the point at which they meet is called the vertex of the angle. The measure of a spherical angle is the same as that of the diedral angle included between the planes of its sides. Spherical angles may be acute, right, or obtusv. 2. A SPHERICAL POLYGON is a portion of the surface of a sphere bounded by arcs of three or more great circles. The bounding arcs are called sides of the polygon, and the points in which the sides meet are called vc;'tices of the polygon. Each side is supposed to be less than a semi-circumference. Spherical polygons are classifed in the same manner as plane polygons. 3. A SPHERICAi TRIANGLE is a spherical polygon of three sides. Sphericli triangles are classified in the same manner as plane triangles. 4. A LuNE is a portion of the surface of a slphere bounded by semi-circumferences of two great circles. 5. A SPHERmICL WVEDGE is a portion of a sphere bounded by a lune and t-wo semnicirc!es meeting in a diaimet:r of the sphere. 236 GEOMETRY. 6. A SPHERICAL PYRAMID is a portion of a sphere bounded by a spherical polygon and sectors of circles whose common centre is the centre of the sphere. The spherical polygon is called the base of the pyramid, and the centre of the sphere is called the vertex of the pyramid. 7. A POLE OF A CIRCLE is a point on the surface of the sphere, equally distant from all the points of the circumference of the circle. 8. A DIAGONAL of a spherical polygon is an are of a great circle joining the vertices of any two angles which are not consecutive. PROPOSITION I. THEOREM. Any side of a spherical triangle is less than the sumz of the other two. Let AB C be a spherical triangle situated on a sphere whose centre is 0: then will any side, as AB, be less than the sum of the sides AC and B C. For, draw the radii OA, OB, and O C: these radii *form the edges of a tlriedral angle whose vertex is 0, and the plane angles included between them are measured by the arcs AB, A C, and BC (B. III., P. XVII., Sch.). But any plane angle, as A O-B, is less than the sum of tile plane angles AOC 0 and B O(7 (B. VI., P. XIX.): hence, the arc AB is less tlhan the sunm of the arcs A C and BC; which was to be proved. BOO 0 IX. 237 Cor. 1. Any side AB, of a spherical polygon AB 2iDE, is less than the sum of all the other sides. For, draw the diagonals A C and AD, dividing the polygon into triangles. The are AB is less than the sumn of A C and B C, the are A C is less than the sum of AD and D C, D and the are AD is less than the sum of DE and PA; hence, AB E is less than the sum of B C, CD, DE, and EA. A Cor. 2. The are of a great circle joining any two points on the surface of a sphere, is less than the arc of a small circle joining the same points. For, divide the arc of the small circle into equal parts, and through the extremities of each part pass the are of a great circle. The are of the great circle joining the given points will be less than the sum of these arcs (C. 1), whatever may be their number. But when this number is infinite, the arcs of the great circle coincide with the corresponding arcs of the small circle, and their sum is equal to the entire are of the small circle. Cor. 3. The shortest distance between two points on the surface of a sphere, is measured on the are of a great circle joining them. PROPOSITION II. THEOREM. The sum of the sides of a spherical polygon is less than the circumference of a great circle. Let AB CDE be a spherical polygon situated on a sphere whose centre is 0: then will the sum of its sides be less than the circumference of a great circle. 238 GEO METRY. For, draw the radii OA, OB, O C, OD, and OE': these radii form the edges of a polyedral angle whose vertex is at 0, and the angles included between them are measured by the arcs AB, B C, E CD, LE~, and EaL. But the sum of A B these angles is less than four right angles / / (B. VI., P. XX.): hence, the sum of the arcs which measure them is less than the a circumference of a great circle; which was 0 to be proved. PROPOSITION III. THEOREM. If a diameter of a sphere be cldrawn perpendicular to the plane of any circle of the sphere, its extremities will be poles of that circle. Let C be the centre of a sphere, ENWG any circle of the sphere, and -DE a diameter of the sphere perpendicular to the plane of FiNG: then will the extremities D and E, be poles of the circle ENG. The diameter DE, being D perpendicular to the plane of iFNG, nmust pass through F _ the centre 0 (B. VIII., " -—. P. VII., C. 3). If arcs of great circles DN, D DC, )BG, &c., be drawn from -D to \. different points of the circumference ]iNG, and chords of these arcs be drawn, these E chords will be equal (B. VI., P. V.), consequently, the arcs themselves will be equal. But these arcs are the shortest lines that can be drawn from the BOOK IX. 239 point D, to the different points of the circumference (P. I., C. 2): hence, the point 1D, is equally distant from all the points of the circumference, and consequently is a pole of the circle (D. 7). In like manner, it may be shown that the point EV is also a pole of the circle: hence, both D, and E, are poles of the circle _FNG; which was to be proved. Cor. 1. Let A2IIB be a great circle perpendicular to D)E: then will the angles D CU1, ECAU, &c., be right angles; and consequently, the arcs Dint, WEIA, &e., will each be equal to a quadrant (B. III., P. XVII., S.): hence, the two poles of a great circle are at equal distances from the circumference. Cor. 2. The two poles of a small circle are at unequal distances from the circumference, the sum of the distances being equal to a semi-circumference. Cor. 3. The line D C being perpendicular to the plane _AMB, any plane, as D11AJ~, passed through it, will also be perpendicular to the plane A3I~B: hence, the spherical angle D)llA, is a right-angle; that is, if any point, in the circumference of a great circle, be joined with either pole by the are of a great circle, such are will be perpendicular to the circumference of the given circle. Cor. 4. If the distance of a point D, firom each of the points A and 21, in the circumference of a great circle, is equal to a quadrant, the point D, is the pole of the are A21. For, let C be the centre of the sphere, and drawv the radii CD, CA, CM. Since the angles A CiD,.~CD, are right angles, the line CD is perpendicular to the two straight lines CA, CMr: it is, therefore, perpendicular to their 240 GEO MfETR Y. plane (B. VI., P. IV.): hence, the point D, is the pole of the arc AJ.f. Scholium. The properties of these poles enable us to describe arcs of a circle on the surface of a sphere, with the same facility as on a plane surface. For, by turning the arc DE about the point D), the extremity Ty will tdescribe the small circle TNG; and by turning the quad. iant DPFA round the point D, its extremity A will describe an arc of a great circle. PROPOSITION IV. THEOREM. The angle formed by two arcs of great circles, is equal to that formed by the tangents to these arcs at their point of intersection, and is mnze-astred by the arc of a great circle described from the vertex as a pole, and limited by the sides, produced if necessary. Let the angle BA C be formed by the two arcs AB, AC: then is it equal to the angle FAt formed by the tangents A1F A G, and is measured by the arc DIE of a great circle, described about A as a pole. For, the tangent AI, drawn in the plane of the are AB, is perpendicular to the radius A0; and the tangent F -4 G, drawn in the plane of the are B A C, is perpendicular to the same radius A O: hence, the angle FA G is equal to the angle contained by the planes D A BDIt, ACEl/ (B. VI., D. 4); which is that of the arcs AB?, A C. Now, if the arcs AD and AE are both quadrants, the lines OD, OE, are perpendicular to OA, and BOOK IX. 241 the angle DOE is equal to the angle of the planes ABDI, A CEH: hence, the arc DE is the measure of the angle contained by these planes, or of the angle CAB; -which was to be proved. Cor. 1. The angles of spherical triangles may be compared by means of the arcs of great circles described from their vertices as poles, and included between their sides. A spherical angle can always be constructed equal to a given spherical angle. Cor. 2. Vertical angles, such as o A CO and B N are equal; for either of them is the angle formed by the two planes A CB, O CN. When two arcs A CB, OCN, intersect, the sum of two adjacent angles, as AGO, OCB, is equal to two right angles. PROPOSITION V. THEOREM. If from the vertices of the angles of a spherical triangle, as poles, arcs be described forming a spherical triangle, the vertices of the angles of this second triangle will be respectively poles of the sides of the first. From the vertices A, B, C, as poles, let the arcs EF, F'D, ED, be described, forming the A triangle DFE: then will the vertices, E, and F, be respectively poles of the sides B B C, AC, AB. For, the point A being 16 242 GEOMETRY. the pole of the arc EX, the distance P4E, is a quadrant; the point C being the pole of the arc DE, the distance (CE is likewise a quadrant: hence, the point E is at a quadl'ant's distance from the points A and C: hence, it is the pole of the arc AC (P. III., C. 4). It may be shown, ill like manner, that -D is the pole of the arc B C, and F that of the arc AB; which was to be proved. Scholium. The triangle AB C, may be described by means of DEF, as D)EF is described by means of ABC. Triangles thus related are called polar triangles, or supplemental triangles. PROPOSITION VI. THEOREM. Any angle, in one of two polar triangles, is measured by a semi-circumtference, minus the side lying opposite to it in the other triangle. Let ABC, and EFXD, be any two polar triangles: then will any angle in either triangle be measured by a semi-circumference, minus the side lying opposite to it in the other triangle. For, produce the sides AB, D A (C, if necessary, till they meet EP, in G and ff. The point A being the pole of M the arc GC; the angle A is measured by that arc (P. IV.). E But, since E is the pole of a -A 11, the arc EH is a quad- li rant; and since F is the pole of A G, FG is a quadrant: hence, the sum of the arcs fEH and GC, is equal to a semi-circumference. But, BOOK IX. 243 the sum of the arcs EH and GF, is equal to the sum of the arcs EiE and GT: hence, the are GC, which measures the angle A, is equal to a semi-circumference, minus the arc Ei. In like manner, it may be shown, that any other angle, in either triangle, is measured by a semicircumference, minus the side lying opposite to it ii the other triangle; which was to be proved. Scholium. Besides the triangle DEF,,three others may be formed by the' intersection of the arcs DlE, E', DF. But the proposition is applicable only Y ts to the central triangle, which is distinguished from the other three by the' —..d circumstance, that the two vertices, A and D, lie on the same side of _B C; the two vertices, B and E, on the same side of A C; and the two vertices, C and 1, on the same side of AB. PROPOSITION VII. THEOREM. If from the vertices of any two angles of a spherical triangle, as poles, arcs of circles be described passing through the vertex of the third angle; and if from the second point in which these.arcs intersect, arcs of great circles be drawn to the vertices, used as poles, the parts of the triangle thus formed will be equal to those of the given triangle, each to each. Let ABC be a spherical triangle situated on a sphere whose centre is 0, C.ED and CFD arcs of circles described about B and A as poles, and let DA and DB be arcs of great circles: then will the parts of the 244 GEOMETRY. triangle ABD be equal to those of the given triangle ABC, each to each. ID For, by construction, the side AD is equal to A C, the side ZDB is equal to B C, and the side AB is B common: hence, the sides are equal, " each to each. Draw the radii OA, / OB, 0 C, and OD. The radii OA, OB, and 0 C, will form the edges of a triedral angle whose vertex is 0 0; and the radii OA, OB, and OD, will form the edges of a second triedral angle whose vertex is also at O; and the plane angles formed by these edges will be equal, each to each: hence, the planes of the equal angles are equally inclined to each other (B. VI., P. XXI.). But, the angles made by these planes are equal to the corresponding spherical angles; consequently, the angle BAD is equal to BAG, the angle ABD to ABCG, and the angle ADB to A CB: hence, the parts of the triangle ABD are equal to the parts of the triangle A CB, each to each; which was to be proved. Scholiurm 1. The triangles ABC and ABD, are not, in general, capable of superposition, but their parts are symmetrically disposed with respect to AB. Triangles which can be so placed are called symmetrical triangles. Scholium 2. If symmetrical triangles are isosceles, they can be so placed as to coincide throughout: hence, they are equal in area. BOOK IX. 245 PROPOSITION VIII. THEOREIM. If two spherical triangles, on the same, or on equal spheres, have two sides and the iicluded angle of the one equal to two sides and the included angle of the other, each to each, the remaining parts are equal, each to each. Let the spherical triangles AB C and EFG, have the side EF equal to AB, the side EG equal to A C, and the angle FEGC equal to BA C: then will the side FG be equal to B C, the angle FG to ABC, and the angle EGCF to A CB. For, the triangle EFG may A E be placed upon AB C, or upon its symmetrical triangle ADB, so as to coincide with it throughout, as may be shown by the same D o G course of reasoning as that employed in Book I., Proposition V.: B F hence, the side FG is equal to B C, the angle EEG to ABC, and the angle ECG to A CB; which was to be proved. PROPOSITION IX. THEOREM. If two spherical triangles on the same, or on equal spheres, have two angles and the included side of the one equal to two angles and the included side of the other; each to each, t]he remaining parts will be equal, each to each. Let the spherical triangles ABC and EFG, have the angle FEYEG equal to BA C, the angle EFG equal to AB C, and the side EF equal to AB: then will the 246 GEOMETRY. side EG be equal to A C, the side FG to BC, and the angle FGE to B CA. For, the triangle EEG may A E be- placed upon ABC, or upon its symmetrical triangle ADB, so as to coincide with it throughout, as may be shown by the same D C G course of reasoning as that employed in Book I., Proposition B F VI.: hence, the side EG is equal to A C, the side FG to B C, and the angle FGE to B CA; which was to be proved. PROPOSITION X. THEOREM. If two spherical triangles on the same, or on equal spheres, have their sides equal, each to each, their angles will be equal, each to each, the equal angles lying opposite the equal sides. Let the spherical triangles EFG and ABC have the side FEE equal to AB, the side EG equal to A C, and the side FG equal to BC: then will the angle FEG be equal to BAC, the angle EFG to ABC, and the angle BEGF to A CB, and the equal angles will lie opposite the equal sides. For, it may be shown by the A E same course of reasoning as that employed in B. I., P. X., that the triangle EFG is equal in all respects, either to the triangle D 0 G G ABC, or to its symmetrical tri- B F angle ABD: hence, the angle FEG, opposite to the side FG, is equal to the angle BA C, BOOK IX. 247 opposite to BC; the angle EFG, opposite to EG, is equal to the angle ABC, opposite to AC; and the angle EGE, opposite to PF, is equal to the angle A CB, opposite to AB; which was to be proved. PROPOSITION XI. THEOREM. In any isosceles spherical triangle, the angles opposite the equal sides are equal; and conversely, if two angles of a spherical triangle are equal, the triangle is isosceles. 1~. Let AB C be a spherical triangle, having the side AB equal to A C: then will the angle C be equal to the angle B. For, draw the arc of a great circle A from the vertex A, to the middle point D, of the base B C: then in the two triangles ADB and AD C, we shall have the side AB equal to A C, by hypothe- B C sis, the side BD) equal to D)C, by con- D ktruction, and the side A.D) common; consequently, the triangles have their angles equal, each to each (P. X.): hence, the angle C is equal to the angle B; which was to be proved. 2~. Let AB C be a spherical triangle having the angle C equal to the angle B: then will the side AB be equal to the side A C, and consequently the triangle wil be isosceles. For, suppose that AB and A C are not equal, but that one of them, as AB, is the greater. On AB lay off the arc B 0 equal to A C, and draw the are of a great circle from 0 to C: then in the triangles A CB and OB C, owe shall have the side A C equal to OB, by construction, 248 GE O METRY. the side.BC common, and the included angle A CB equal to the included angle OB C, by hypothesis: hence, the remaining parts of the triangles are equal, each to each, and consequently, the ancle A O CB is equal to the angle AB C. But, the angle A CB is equal to ABC, by hypothesis, and therefore, the angle O CB is equal to A CB, or a part is equal to the whole, which is' impossible: hence, the 1D supposition that AB and A C are unequal, is absurd; they are therefore equal, and consequently, the triangle ABC is isosceles; which was to be pro'ved. Cor. The triangles ADB and ADC, having all of their parts equal, each to each, the angle D2.DB is equal to AD C, and the angle DiAB is equal to 1A C; that is, if a~n arc of a great circle be drawn from the vertex of an isosceles spherical triangle to the middle of its base, it will be perpendicular to the base, and will bisect the vertical angle of the triangle. PROPOSITION XII. THEOREM. In any spherical triangle, the greater side is opposite the greater angle; and conversely, the greater angle is'optusite the greater side. 1~. Let ABC be a spherical triangle, in which the angle A is greater than the angle B: then will the side B C be greater than the side AC. For, draw the arc AD, A making the angle BAD) equal to AB/D: then will AD be equal to BD (P. XI.). But, B the sum of AD and DC is D BOOK IX. 249 greater than AC (P. I.); or, putting for AD its equal BD), we have the sum of BD and D C, or B 2C, greater than AC; which was to be proved. 2~. In the triangle AB C, let the side 7BC be greater than A C: then will the angle A be greater than the angle B. For, if the angles A and B were equal, the sides B C and A C would be equal; or if the angle A was less than the angle B, the side B C would be less than A C, either of which conclusions is absurd: hence, the angle A is greater than the angle B; which was to be provecd. PROPOSITION XIII. THEOREM. If two triangles on the same, or on egual spheres, are mutzally equiangular, they are also mutually equilateral. Let the spherical triangles A and B, be mutually equiangular: then will they also be mutually equilateral. For, let P be the polar triangle of A, and Q the polar triangle of B: then, because the triangles A and B are mutually A equiangular, their polar triangles P and Q, must be mutually equilateral (P. VI.), and consequently mutually equiangular (P. X.). But, Q the triangles P and Q being mutually equiangular, their polar triangles A and B, are mutually equilateral (P. VI.); which was to be provcc. Scholium. This proposition does not hold good for plane triangles, for all similar plane triangles are mutually equiangular, but not necessarily mutually equilateral. Two spherical triangles on the same or on equal spheres, cannot be similar without being equal. 250 GEOMETRY. PROPOSITION XIV. THEOREM. The stun of the angles of a spherical triangle is less than six right angles, and greater than two right angles. Let ABC be a spherical triangle, and DE-P its polar triangle: then will the sum of the angles A, B, and C, be less than six right angles and greater than two. For, any angle, as A, beincg measured by a semi-circumference, minus the side EFl (P. VI.), is less than two right A angles: hence, the sum of the three right angles is less than six right angles; and because the measure of each angle is C equal to a semi-circumference, F minus the side lying opposite to it, in the polar triangle, the measure of the sum of the three angles is equal to three semi-circumferences, minus the sum of the sides of the polar triangle aDEE But the latter sum is less than a circumference; consequently, the neasure of the sum of the angles A, B, and C, is greater than a semi-circumference, and therefore the sum of the angles is greater than two right angles: hence, the sum of the angles A, B, and C, is less than six right angles, and greater than two; which was to be proved. Cor. 1. The sum of the three angles of a spherical triangle is not constant, like that of the angles of a rectilineal triangle, but varies between two right angles and six, witfhout ever reaching either of these limits. Two angles, therefore, do not serve to determine the third. BOOK IX. 251 Cor. 2. A spherical triangle may have two, or even three of its angles right angles; also two, or even three of its angles obtuse. Cor. 3. If the triangle ABC is bi-rectan- A gular, that is, has two right angles _ and C, the opposite sides of the polar triangle will be quadrants, and their point of intersection will be the pole of the other side (P. B III., C. 4). The angles opposite the equal sides are right angles (P. III., C. 3): hence, the sides AB and A C are quadrants. If the angle A is also a right angle, the triangle ABC is tri-rectangular r; each of its angles is a right angle, and its sides are quadrants. Fo'ur tri-rectangular triangles make up the surface of a hemisphere, and eight the entire surface of a sphere. Scholium. The right angle is taken as the unit of measure of spherical angles, and is denoted by 1. The excess of the sumn of the angles of a spherical triangle over two right angles, is called the spherical excess. If we denote the spherical excess by E, and the three angles expressed in terms of the right angle, as a unit, by A, B, and C, we shall have, E =A B + C - 2. The spherical excess of any spherical polygon is equal to the excess of the sum of its ancles over two right angles taken as many times as the polygon has sides, less two. If we denote the spherical excess by E, the sum of the angles by S, and the number of sides by n, we shall have, E= S- 2(n - 2) = S- 2n+4. 252 GEOMETRY. PROPOSITION XV. TIHEOREM..The area of a lune, is to the szurfjce of the sphere, as the angle of the late is to four right angles, or as tlhe are iwhich measures that angle is to the circumference of a great circle. Let AIB BN be a lune, and IMCY1 the angle of the lune: then will the area of the lune be to the surface of the sphere, as the arc B1 Now, if b < B, there will evidently be two solutions, c the sides including the given angle, in the one case, being supplements of those which include the given angle, in the other case. If b = B, the triangle will be bi-rectangular, and there will be but a single solution. If b > B, the triangle cannot be constructed, that is, there will be no solution. SOLUTION OF RIGHT-ANGLED SPHERICAL TRIANGLES. 76: In a right-angled spherical triangle, the right angle is always known. If any two of the other parts are given, the remaining parts may be found by Napier's rules for circular parts. Six cases may arise. There may be given, I. The hypothenuse and one side. II. The hypothenuse and one oblique angle. III. The two sides about the right angle. IV. One side and its adjacent angle. V. One side and its opposite angle. VI. The two oblique angles. In any one of these cases, we select that part which is either adjacent to, or separated from, each of the other given parts, and calling it the middle part, we employ that one of Napier's rules which is applicable. Having determined a third part, the other two may then be found in a similar manner. TRIGONOMETRY. 81 It is to be observed, that the formulas employed are to be rendered homogeneous, in terms of R, as explained in Art. 30. The method of proceeding will be readily understood from a few examples. EXAMPLES. 1. Given a - 1050 17' 29", and C b - 380 47' 11", to find c, B, and C. 1 Since a > 90~, b and c must be of different species, that is, c > 90g; A C for the same reason, C > 90~. OPERATION. From Formula (10), Art. 74, we have, log cos C - log cot a + log tan 6b - 10; log cot a (1050 17' 29") 9.436811 log tan b ( 38~ 47' 11") 9.905055 log cos C... 9.341866.'. C = 1020411 33". Fr'om Formula (2), Art. 74, we have, log sin c = log sin ca + log sin C- 10; log sin a (105~ 17' 29") 9.984346 log sin C (102~ 41' 33") 9.989256 log sin c..... 9.973602.'. c = 1090 46' 32". From Formula (4), we have, log cos B = log sin C + log cos b - 10; log sin C (1020 41' 33") 9.989256 log cos b (380 47' 11") 9.891808 log cos B.... 9.881064.'. B = 400 29' 50" Ans. c = 1090 46' 32", B = 40~ 29' 50", C = 1020 41' 33". 82 SPHERICAL 2. Given 6b = 510 30', and B = 58~ 35', to find a, c, and C. Because b < B, there are two solutions. OPERATION, From Formula ( 7 ), we have, log sin c = log tan b + log cot B - 10; log tan b (51~ 30') ~ 10.099395 log cot B (580 35') ~ 9.785900 log sin c.... 9.885295.'. c = 50~ 09' 51", and c = 1290 50' 09". From Formula (1), we have, log cos a = log cos b + log cos c - 10; log cos b (510 30')~ ~ 9.794150 log cos c (500 09' 51") 9.806580 log cos a.... 9.600730.'. a = 660 29' 54", and a = 1130 30' 06". From Formula ( 10 ), we have, log cos C = log tan b + log cot a- 10; log tan b (510 30') ~ 10.099395 log cot a (660 29' 54") 9.638336 log cos C. 9.737731.. C = 56~ 51' 38", and C = 1230 08' 22". In a similar manner, all other cases may be solved. 3. Given a = 860 51', and -B = 18~ 03' 32", to find b, c, and C. Ans. b = 180 01' 50", c = 860 41' 14", C = 880 58' 25". TRIGONOMETRY. $3 4. Given b = 1550 27' 54", and c = 290 46'08", to find a, B, and C. Ans. a = 1420~ 09' 13", B = 137~ 24' 21", - 540~ 01' 16". 5. Given c = 730~ 41' 35", and B = 990~ 17' 33", to find a, b, and C. Ans. a = 920 42' 17", b = -990 40' 30", C = 730 54' 47". 6. Given b = 1150 20', and B = 910 01' 477", to find a, c, and C. { 640 41' 11", f 1770 49' 27", 1 1770 35' 36". 115~ 18' 49", 20 10' 33", 20 24' 24". 7. Given B = 47~0 13' 48", and C = 126~ 40' 24", to and a, b, and c. Ans. a = 1330 32' 26', b = 320 08' 56", c = 1440 27' 03". In certain cases, it may be necessary to find but a single part. This may be effected, either by one of the formulas given in Art. 74, or by a slight transformation of one of them. Thus, let a and B be given, to find C. Regarding 900 - a, as a middle part, we have, cos a = cot B cot C; whence, COS a cot C = cot B; and, by the application of logarithms, log cotC -- log cos a + (a. c.)log cot B; from which C may be foulrd. In like manner, other cases may be treated. 84 SPHERICAL QUADRANTAL SPHERICAL TRIANGLES. 77. A QUADRANTAL SPHERICAL TRIANGLE is one in which one side is equal to 90~. To solve such a triangle, we pass to its polar triangle, by subtracting each side and each a, lg'le from 1800 (B. IX., P. VI.). The resulting polar tri-:tngle will be right-angled, and may be solved by the rules already given. The polar triangle of any quadrantal triangle lbeing solved, the parts of the given triangle may be found by subtracting each part of the polar triangle from 180~. EXAMPLE. Let A'B' C' be a quadrantal C triangle, in which B'C' = 900, B' ='75~ 42', and c' = 180 37'. Passing to the polar triangle, we have, A A - 90, b 6 104~ 18', and C = 161~ 2.3'. Solving this triangle by previous rules, we find, a = 760 25' 11", c = 1610 55' 20", B - 940 31' 21"; hence, the required parts of the given quadrantal triangle are, A' = 103~ 34' 49", C1' 180 04' 40", b' = 85~ 28' 39". In a similar manner, other quadrantal triangles may be solved. TRIGONOMETRY. 85 FORMULAS USED IN SOLVING OBLIQUE-ANGLED SPHERICAL TRIANGLES. 78. Let ABC represent an oblique-angled spherical tri. angle. From either vertex, C, draw the arc of a great circle C CB', perpendicular to the opposite side. The two triangles A CB' and BBCB' will be right- B angled at B'. A o B_ B From the triangle A CB', we have Formula ('2), Art. 74, sin CB' = sin A sin b. From the triangle B CB', we have, sin CB' = sin B sin a. Equating these values of sin CB', we have, sin A sin b = sin B sin a; from which results the proportion, sin a: sin b:: sin A: sinB... (1.) In like manner, we may deduce, sin a: sin c:: sinA: sin C.. (2.) sin b: sin c:: sinB: sin C... (3.) That is, in any spherical triangle, the sines of the sicdes are proportional to the sines of their opposite angles. Had the perpendicular fallen on the prolongation of AB, the same relation would have been found. 86 SPHERICAL 79. Let ABC represent any spherical triangle, and O the centre of the sphere on which it is situated. Draw the C radii OA, OB, and 0 C; from C draw CP perpendicular to b the plane AOB; from P, the 0 A B foot of this perpendicular, draw B PD and PPE respectively per- D A pendicular to OA and OB; join CD and CE, these lines will be respectively perpendicular to OA and OB (B. VI., P. VI.), and the angles CODP and C.EP will be equal to the angles A.and B respectively. Draw D)L and PQ, the one perpendicular, and the other parallel to OB. We then have, OE = cos a, 2 DC = sin b, OD = cos b. WNe have from the figure, OE = OL + QP. (1.) In the right-angled triangle OLD, OL = OD1 cos -POL = cos b cos c. The right-angled triangle PQD has its sides respectively perpendicular to those of O-LD; it is, therefore, similar to it, and the angle QDP is equal to c, and we have, QP = PD sin QDP = PD sine ~ ~ ~ (2.) The right-angled triangle CPD gives, PD = CD cos CDP = sin b cos A; substituting this value in ( 2 ), we have, QP = sin b sin c cos A; TRIGONOMETRY. 87 and now substituting these values of OE, OL, and QPo, in (1), we have, cos a -_ cos b cos c + sin b sin c cos A ~ ~ (3.) In the same way, we may deduce, cos b = cosa os c + sin a sinl c cos B (4.) cos c = cos a cos b + sinl a sin b cos C ~ ~ (5.) That is, the cosine of either side of a spherical triangle is equal to the rectangle of the cosines of the other two sides plus the rectangle of the sines of these sides into the cosine of their included angle. 80. If we represent the angles of the polar triangle of AB C, by A', B', and C', and the sides by a', b', and c', we have (B. IX., P. VI.), a = 1800 -A', b = 1800-B', 1800, - C', A = 180~- a', B = 180~0 - b', C = 1800~-c'. Substituting these values in Equation (3), of the preceding article, and recollecting that, cos (1800-A') = - cos A', sin (180~-B') = sin B', &c., we have, - cos A' = cos B' cos C' - sin B' sin C' cos a'; or, changing the signs and omitting the primes (since the preceding result is true for any triangle), cos A = sin B sin C cos a —cos B cos C ~ (1.) 88 SPHERICAL In the same way, we may deduce, cos B = sin A sin C cos b - cos A cos C ~ (2.) cos C = sin A sin B cos c~-cos A cos B ~ (3.) That is, the cosine of either angle of a spherical triangle is equal to the rectangle of the sines of the other two angles into the cosine of their included side, minus the rectangle of the cosines of these angles. 81. From Equation (3), Art. 79, we deduce, cos a - cos b cos c cosA =.. (1.) sin b sin c If we add this equation, member by member, to the number 1, and recollect that 1 + cos A, in the first member, is equal to 2 cos2 ~A (Art. 66), and reduce, we have, sin b sin c + cos a- cos b cos c 2 cos2~A -- sin b sine c or, Formula ((), Art. 66, 2 cos2~A cos a- cos (b + c)...) sin b sin c And since, Formula ( 1), Art. 67, cos a - cos (b + c) = 2 sin'(a + b + e) sin i(b + c- a), Equation (2) becomes, after dividing both members by 2, sin ~(a - b + c) sin (b + c - a) sin b sin c TRIGONOMET RY. 89 If, in this we make, (a + b + c) = is; whence, -(b + c —a) = As-a, and extract the square root of both members, we have,:sin ~s sin (Is —a) cos'A =AV I sin... (3.) VOS ~ sin b sin c That is, the cosine of one-half of either angle of a spherical triangle, is equal to the square root of the sine of one-half of the sum of the three sides, into the sine of one-half this sum minnus the side opposite the angle, divided by the rectangle of the sines of the adjacent sides. If we subtract Equation (1), of the preceding article, member by member, from the number 1, and recollect that, 1 -cos A = 2 sin2 WA, we find, after reduction, sin _A -= v/ sin -In sin(s-) ~ (4.) smb sin c Dividing the preceding value of sin 4 A, by cos ~4A, we obtain, tan -A = /sin (~ s -6) sin (s -s c) sill S sin (is - a) 82. If the angles and sides of the polar triangle of AB C be represented as in Art. 80, we have, A = 1800~-a', b = 1800~-B', c = 180~-C', is = 2700 - ~(A'+B'+ C'), is - a = 900~-(B'+ C'-A'). 90 SPHERICAL Substituting these values in (3), Art. 81, and reducing by the aid of the formulas in Table III., Art. 63, we find, sin) a=' /- cos ~(A'+.B'+ C') cos (B'+ C' —A') sin B2' sin C' Placing ~(A'+_B'+ C') = IS; whence, ~(B'+ C'-A') = S - A'. Substituting and omitting the primes, we have, sin 2' -- — ~s ~ os (t~ — A) 1.) V sin sin C In a similar way, we may deduce from (4), Art. 81. COS -qa CjyJ \ahl-ul uvu\Nvl(2.) cos a -= /cs (~S-B) cos (S- ) * (2.) sin B sin C and thence, co- cos (S -A) (3)- A) tan a 2= \/cos (-S-B) cos (~S- C) 83. From Equation (1), Art. 80, we have, sin A cos A + cos B cos C = sin B sin C cos a = sin C sin a b cos a; sin a (1.) since, fiom Proportion (1), Art. 78, we have, sin A sin B a sin b. sin a Also, from Equatiol (2), Art. 80, we have, cos B + cos.4A cosC - sin sinA sin C cos b _ sinC. sin a cos b. sin a (2.) TRIGONOMETRY. 91 Adding (1) and (2), and dividing by sin C, we obtain, 1 + cos C sin A (cos A + cos B) Wio C sin a in (a + b). (3.) The proportion, sin A: sin B: sin a: sin b, taken first by composition, and then by division, gives, sin A + sin B = *sina (sin a + sinb)' ~ (4.) sin A sin A - sin B = a (sin a-sin b) (5.) sin a' Dividing (4) and (5), in succession, by (3), we obtain, sin A +-sin B sin C sin a + sin b x (6.) cosA +- cosB 1 + cos C sin (a + b) sin A - sin B sin C sin a -sin b x (7.) cosA+ cosB 1+ cos C sin (a + b) But, by Formulas (2) and (4), Art. 67, and Formula (1"), Art. 66, Equation (6) becomes, tan ~(A + B) = cot IcosT - 6); * (8.) cos~(a+6b) and, by the similar Formulas (3) and (5), of Art. 67, Equation ('7) becomes, tan I(A - B) = cot,0 sinC (a - b) sin~I(a) (9') These last two formulas give the proportions known'as the first set of Napier's Analogies. cos(a + b): cos (a —b):: cot C: tan (A+B). (10.) sin (a+b): sin (a —b):: cotIC: tan~(A-B). (11.) ('2 SPHERICAL If in these we substitute the values of a, b, C, A, and B, in terms of the corresponding parts of the polar triangle, as expressed in Art. 80, we obtain, cos -(A+B): cos (A —B):: tan~c: tan (a+b). (12.) sin (A +B): sin (A —B):: tan ic: tan (a-b). (13.) the second set of Napier's Analogies. In applying logarithms to any of the preceding formulas, they must be made homogeneous, in terms of R, as explained in Art. 30. SOLUTION OF OBLIQUE-ANGLED SPHERICAL TRIANGLES. 84. In the solution of oblique-angled triangles six different cases may arise: viz., there may be given, I. Two sides and an angle opposite one of them. II. Two angles and a side opposite one of them. III. Two sides and their included angle. IV. Two angles andXtheir included side. V. The three sides. VI. The three angles. CASE I. Given two sides and an angle opposite one of them. 85. The solution, in this case, is commenced by finding the angle opposite the second given side, for which purpose Formula'( 1), Art. 78, is employed. As' this angle is found by means of its sine, and because the same sine corresponds to two different arcs, there would seem to be two different solutions. To ascertain when there are two soluttions, when one solution, and when no solution at all, it becomes necessary to examine the relations which TRIGO O NOMETRY. 93 may exist between the given parts. Two cases may arise, viz., the given angle may be acute, or it may be obtuse. We shall consider each case separately (B. IX., P. XIX., Gen. Scholium). first CGase. Let A be C the given angle, and let a b and b be the given sides. A Prolong the arcs A C and AB till they meet at A', B B' forming the lune AA'; andB from C, draw the are CB' perpendicular to ABA4'. From C, as a pole, and with the are a, describe the arc of a small circle BB. If this circle cuts ABA', in two points between A and A', there will be two solutions; for if C be joined with each point of intersection by the are of a great circle, we shall have two triangles ABC, both of which will conform to the conditions of the problem. If only one point of intersection lies between A and A', or if the small circle is tangent to ABA', X there will be but one solu- l A IB tion. If there is no point of intersection, or if there are points of intersection which do not lie between A and A', there will be no solution. From Formula (2), Art. 72, we have, sin CB' - sin b sin A, from which the perpendicular, which will be less than 900, will be found. Denote its value by p. By inspection of the figure, we find the following relations: 23 94 SPHERICAL 1. WThen a is greater than p, and at the same time less than both b and 1800 - b, there?oill be two solutions. 2. When a is greater than p, and intermediate in nalue between b and 1800 - b; or, when a is equal to p, there will be but one solution. If a = b, one of the points of intersection will be at A, and in this case, there will, ill like manner, be but one solution. 3. When a is greater than p, and at the same time greater than both b and 180~ - b; or, when a is less than p, there will be no solution. Second Case. Adopt the same construction as before. In this case, the perpendicu- A / lar will be greater than 900, X and greater also than any other arc CA, CB, CA', that can be drawn from C to ABA'. By a course of reasoning entirely analogous to that in the preceding case, we have the following principles: 4. When a is less than p, and at the same time greater than both b and 1800 - b, there will be two.solutions. 5. When a is less than p, and intermediate in.alute between b and 180~ - b; or, when a is equal to p, there will be but one solution. 6. When a is less than p, and at the same time less than both b and 180~ - b; or, when a is greater than p, there will be no solution. Having found the angle or angles opposite the second side, the solution may be completed by means of Napier's Analogies. TRIGONOMETRY. 9 EXAMPLES. 1. Given a = 430 27' 36", b = 820 58' 17", and A = 290 32' 29", to find B, C, and c. We see at a glance, that a > p, since p cannot exceed A; we see further, that a is less than both b and 1800 - b; hence, from the first condition there will be two solutions. Applying logarithms to Formula (1), Art. 78, we have, log sinB = log sin b + log sin A + (a. c.) log sin a — 10; log sin b ~ ~ (82~ 58' 17") ~ ~ ~ 9.996724 log sinA * ~ (29~32'29") ~ ~ ~ 9.692893 (a. c.) log sin a ~ ~ (43~ 27'36") ~ ~ ~ 0.162508 log sin B......... 9.852125 B = 450 21' 01", and B = 134~ 38' 59". From the first of Napier's Analogies (10), Art. 83, we find, log cot jC = log cos (a +J b) + log tan ~(A+B) + (a. c.) log cos (a - b)- 10. Taking the first value of B, we have, ~(A + B) = 37~ 26' 45"; also, ~(a + b) = 630 12' 56"; and, j(a- b) = 19~ 45' 20". log cos'(a + b) (630~ 12' 56") ~ 9.653825 log tan -(A+ -B) ~ (37~ 26'45") ~ 9.884130 (a. c.) log cos J(a - b) ~ (190 45' 20") ~ 0.026344 log cot jC........ 9.564299.. C = 690~ 51'45", and C = 139043'30". 96 SPHERICAL The side c may be found by means of Formula (12), Art. 83, or by means of Formula (2), Art. 78. Applying logarithms to the proportion, sin A sin C:: sin a sin c, we have, log sin c = log sin a + log sin C + (a. c.) log sin A - 10; logsin a ( 43~27'36") 9.837492 log sinC (139~ 43' 30") 9.810539 (a. c.) log sinA ( 29~ 32' 29") 0.307107 log sin c. * * * 9.955138.. c = 1150~ 35'48". We take the greater value of c, because the angle C, being greater than the angle B, requires that the side c should be greater than the side b. By using the second value of B, we may find, in a similar manner, C = 320 20' 28", and c = 450 16' 18". 2. Given a = 97~ 35', b = 27~0 08' 22", and A = 40~ 51' 18", to find B, C, and c. Ans. B = 17~0 31' 09", C = 1440 48' 10", c = 1190 08' 25". 3. Given a = 1150 20' 10", b = 57~0 30' 06", and A = 1260 37' 30", to find B, C, and c. Ans. B = 480 29' 48", C = 610 40' 16", c - 820 34' 04". CASE II. Given two angles and a side opposite one of thenm. 86. The solution, in this case, is commenced by finding the side opposite the second given angle, by means of Formula (1), Art. 78. The solution is completed as in Case I. TRIGONOMETRY. 97 Since the second side is found by means of its sine, there may be two solutions. To investigate this case, we pass to the polar triangle, by substituting for each part its supplement. In this triangle, there will be given two sides and an angle opposite one; it may therefore be discussed as in the preceding case. When the polar triangle has two solutions, one solution, or no solution, the given triangle will, in like manner, have two solutions, one solution, or no solution. The conditions may be written out from those of the preceding case, by simply changing angles into sides, and the reverse; and greater into less, and the reverse. Let the given parts be A, B, C and a, and let p be an are computed from the equation, A B sin p = sin a sin B. There will be two cases a may be greater than 900~; or, a may be less than 900~. In the first case, 1. When A is less than p, and at the same time greater than both B and 1800 - B, there will be two solutions. 2. TWhen A is less than p, and intermediate in value between B and 180~ - B; or, when A is equal to p, there will be but one solution. 3. ~When A is less than p, and at the same time less than both -B and 1800 - B; or, when A i greater than p, there will be no solution. 98 SPHERICAL In the second case, 4. When A is greater than p, and at the same less than both B and 1800 - B, there will be two solutions. 5. When A is greater than p, and intermediate in value between B and 1800 - B; or, when A is equal to p, there will be but one solution. 6. When A is greater than p, and at the same time greater than both B and 180~- B; or, when A is less than p, there will be no solution. E XAM PLES. 1. Given A = 950 16', B - 800 42' 10", and a = 57~ 38', to find a, b, and C. Computing p, from the formula, log sinp = log sin B + log sin a- 10; we have, p = 56~ 27' 52". The smaller value of p is taken, because a is less than 90~. Because A > p, and intermediate between 800 42' 10" and 990 17' 50", there will, from the fifth condition, be but a single solution. Applying logarithms to Proportion (1), Art. 78, we have, log sin b = log sinB + log sin a + (a. c.) log sin A - 10; log sin B (80042' 10") 9.994257'log sin a (57~ 38') 9-.926671 (a.c.) log sinA (950 16') 0.001837 log sin b * * 9.922765.'. b = 560~ 49' 57". TRIGONOMETRY. 99 We take the smaller value of b, for the reason that A, being greater than B, requires that a should be greater than b. Applying logarithms to Proportion (12), Art. 83, we have, log tan jc = log cos J(A + B) + log tan 4(a + b) + (a. c.) log cos ~(A - B) - 10; we have, J(A + B) = 870 59' 05", ~(a + b) 57~0 13' 58", and,, (A - B) = 7~ 16' 55". log cos (A + B) ~ (87~ 59' 05") ~ 8.546124 log tan'(a + b) ~ (57~0 13' 58") ~ 10.191352 (a. c.) log cos I(A - B) ~ ( 7 16' 55") ~ 0.003517 log tan jc......... 8.740993 c = 30 09' 09", and c - 60 18' 18". Applying logarithms to the proportion, sin a: sin c:: sin A sin C, we have, log sin C = log sin c + log sin A +- (a. c.) log sin a - 10; log sin c (60 18' 18"). 9.040685 log sin A (950 16') ~ ~ 9.998163 (a. c.) log sin a (57~0 38') ~ ~ 0.073329 log sin...... 9.112177.'. -= 7~ 26' 21". The smaller value of C is taken, for the same reason as before. 2. Given A - 500 12', B = 580 08', and a = 62~42', to find b, c, and C. { 790 12' 10", {1190 03' 26", f 1300 54' 28", 100 47' 50", C- 15 4' 8", 156 15' 1000 471 50", 1520 14' 18", 1560 15' 06". 100 SPHERICAL CASE III. Given two sides and their included angle. 87. The remaining angles are found by means of Napier's Analogies, and the remaining side, as in the preceding cases. E X A M P L ES. 1. Given a = 620 38', b = 100 13' 19", and (7 = 1500 24' 12", to find c, A, and B. Applying logarithms to Proportions (10) and (11), Art. 83, we have, log tan {(A+B) = log cos (a -b) + log cot I C + (a. c.) log cos (a + b) - 10o; log tan ~(A-B) = log sin (a - b) + log cot ~C + (a. c.) log sin'(a + 6b) - 10; we have, -(a- b) = 26~ 12' 20", ~C ='750 12' 06", and, J(a + b) ='360 25' 39". 19g cos (a - b) ~ (26~ 12' 20") ~ 9.952897 log cot 2-C ~ ~ ~ (75~ 12' 06") 9.421901 (a. c.) log cos -(a + b) ~ (36~ 25' 39") ~ 0.094415 log tan J(A + B). * 9.469213 (A + B) = 160 24' 51". log sin J(a - b) * (26~ 12' 20") 0 9.645022 log cot EC * * * (750 12' 06") * 9.421901 (a. c.) log sin (a +6b) * (36~ 25' 39") 0.226356 log tan (A- B)..... 9.293279 (A - B) = 110 06' 53". TRIGONOMETRY. 101 The' greater angle is equal to the half sum plus the half difference, and the less is equal to the half sum minus the half difference. Hence, we have, A 27~0 31' 44", and 2B = 50 17' 58". Applying logarithms to the Proportion (13), Art. 83, we have, log tan ~ec = log sin -(A + 2B) + log tan ~(a - b) + (a. c.) log sin ~(A - B) - 10; log sin I(A + B) ~ (160 24' 51") ~ 9.451139 log tan,(a - b) ~ (260 12' 20") ~ 9.692125 (a. c.) log sin I (A- B) ~ (11 06' 53") ~ 0.714952 log tan -ic......... 9.858216.. = 350 48' 33", and c = 710 37' 06". 2. Given a = 680 46' 02", b = 37~ 10', and a = 390 23' 23", to find c, A, and B. Ans. A = 1200 59' 47", B = 330 45' 03", c - 430 37' 38". 3. Given a = 84~ 14' 29", b = 440 13' 45", and C = 360 45' 28", to find A and -B. Ans. A = 130~ 05' 22", B = 320 26' 06". CASE IV. Given two angles and their included side. 88. The solution of this case is entirely analogous to that of Case III. Applying logarithms to Proportions ( 12 ) and ( 13 ), Art. 83, and to Proportion (11), Art. 83, we have, 102 SPHERICAL log tan 4(a + b) = log cos (A -B ) + log tan tc + (a. c.) log cos (A + B) - 10o; log tan'(a- b) = log sin ~(A-B) + log tan lc + (a. c.) log sin ~(A ~ B) - 10; log cot ~C = log sin I(a + b) + log tan'(A- B) + (a. c.) log sin (ca - 6) - 10; The application of these formulas a.xe sufficient for the solution of all cases. EXAMPLES. 1. Given A = 810 38' 20",.B -- 700 09' 38", and e = 590 16' 22", to find C, a, and b. Ans. C = 640 46' 24", a = 70~ 04' 17"', b = 630 21' 27". 2. Given A =- 340 15' 03", B = 42~ 15' 13", and c = 760 35' 36", to find C, a, and b. Ans. C = 121 36' 12", a = 40~ 0' 10", b = 500 10' 30". CASE V. Given the three sides, to find the remaining parts. 89. The angles may be found by means of Formula (3), Art. 81; or, one angle being found by that formula, the other two may be found by means of Napier's Analogies. EXAMP LES. 1. Given ac = 74 23', b = 350 46' 14", and c = 100039', to find A, B, and C. TRIGONOMETRY. 103 Applying logarithms to Formula (3), Art. 81, we have, log cos JA = 10 + j[log sin js + log sin (is - a) + (a. c.) log sin b + (a. c.) log sin c - 20]; or, log cos ~A = ~[log sin is + log sin (-s - a) + (a. c.) log sin b + (a. c.) log sin c], we have, =s = 105~ 24' 07", and i-s - a = 310 01' 07". log sin is ~. (105 24' 07"). 9.984116 log sin (-s - a) ~ ( 31~ 01' 07") ~ 9.712074 (a. c.) log sin b. ( 350~ 46' 14") ~ 0.233185 (a. c.) log sin c..... (1000 39') 0.007546 2)19.936921 log cos A.....* 9.968460'.. A= 210 34' 23", and A = 430 08' 46". Using the same formula as before, and substituting _B ior A, b for a, and a for b, and recollecting that is-b = 690 37' 53", we have, log sin Is ~ ~ ~ (105~ 24' 07"). 9.984116 log sin (~s - b) ~ ( 69 37' 53") ~ 9.971958 (a. c.) log sin a. * * *. (740 23') ~ ~ 0.016336 (a. c.) log sin c.. * (100~ 39') * * 0.007546 2)19.979956 log cos jB. * 9.989978.'. B =- 12~ 15' 43", and.B = 240 31' 26'. Using the same formuLla, substituting C for A, c for a, and a for c, recollecting that -s - c = 40 45' 07", we have, 104 SPHERICAL TRIGONOMETRY. log sin is ~ ~ (105~ 24' 07") 9.984116 log sin (is - c) ~ (40 45' 07") ~ ~ 8.918250 (a. c.) log sin a.... (74" 23') * * 0.016336 (a. c.) log sin b ~ ~ ~ (350 46' 14") ~ ~ 9.23318.5 2)19.151887 log coS iC......... 9.575943.. = 67~ 52' 25", and C = 1350 44' 50". 2. Given a = 56~ 40', b = 830 13', and c = 114" 30'. Ans. A = 48~ 31' 18", B = 620 55' 44", C = 125" 18' 56". CASE VI. The three angles being given, to find the sides. 90. The solution in this case is entirely analogous to the preceding one. Applying logarithms to Formula (2), Art. 82, we have, log cos 2a -= [log cos (1S - B) + log cos (IS - C) + (a. c.) log sin B + (a. c.) log sin C]. In the same manner as before, we change the letters, to suit each case. EXAMPLES. 1. Given A = 48~ 30', B = 125~ 20', and C = 620 54'. Ans. a = 56B 39' 30", b = 1140 29' 58", c - 83012' 06". 2. Given A = 109" 55' 42", B = 116~ 38' 33", and C = 120~ 43' 37", to find a, b, and c. Ans. a = 98~ 21' 40", b - 109" 50' 22", c = 115~13' 28". MENSURATION. 91. MENSURATION is that branch of Mathematics which treats of the measurement of Geometrical Macrnitudes. 92. The measurement of a quantity is the operation of finding how many times it contains another quantity of the same kind, taken as a standard. This standard is called the unit of mneasure. 93. The unit of measure for surfaces is a square, one of whose sides is, the linear unit. The unit of measure for volumes is a cube, one of whose edges is the linear unit.' If the linear unit is one foot, the superficial unit is one square foot, and the unit of volume is one czbic foot. If the linear unit is one yard, the superficial unit is osne sguare yard, and the unit of volume is one cubic yard. 94. In Mensuration, the term product of two lines, is used to denote the product obtained by multiplying the number of linear units in one line by the number of linear units in the other. The term product of three lines, is used to denote the continued product of the number of linear units in each of the three lines. Thus, when we say that the area of a parallelogram is equal to the product of its base and altitude, we mean that the number of superficial units in the parallelogram is equal to the number of linear units in the base, multiplied by the number of linear units in the altitude. In like manner, the 106 MENSURATION number of units of volume, in a rectangular parallelopipedon, is equal to the number of superficial units in its base multiplied by the number of linear units in its altitude, and so on. NiENSURATION OF PLANE FIGURES. To find the area of a parallelogram. 95. From the principle demonstrated in Book IV., Prop. V., we have the following PUL E. iMultiply the base by the altitude; the product will be the area?required. E X A M P LE S. 1. Find the area of a parallelogram, whose base is 12.25, and whose altitude is 8.5. Ans. 104.125. 2. What is the area of a square, whose side is 204.3 feet? Ans. 41738.49 sq. ft. 3, How many square yards are there in a rectangle, whose base is 66.3 feet, and altitude 33.3 feet? Ans. 245.31 sq. yd. 4. What is the area of a rectangular board, whose length is 12} feet, and breadth 9 inches? 98 sq. ft. 5. What is the number of square yards in a parallelogram, whose base is 37 feet, and altitude 5 feet 3 inches? Ans. 21T-7' To find the area of a plane triangle. 96. ]F'rst Case. When the base and altitude are given. OF SURFACES. 107 From the principle demonstrated in Book X., Prop. VI., we may write the following RULE. Multiply the base by half the altitude; the product oillU be the area required. EXAMPLES. 1. Find the area of a triangle, whose base is 625, and altitude 520 feet. Ans. 162500 sq. ft. 2. Find the area of a triangle, in square yards, whose base is 40, and altitude 30 feet. Ans. 662. 3. Find the area of a triangle, in square yards, whose base is 49, and altitude 25} feet. Ans. 68.7361. Second Case. When two sides and their included angle are given. Let ABC represent a plane tri- A angle, in which the side AB =c, BC = a, and the angle B, are given. From A draw IAD perpen- / B D dicular to B C; this will be the altitude of the triangle. From Formula ( 1 ), Art. 37, Plane. Trigonometry, we have, AD = c sin B. Denoting the area of the triangle by Q, and applying the rule last given, we have, ac sinB 2 _ as n or, 2Q = ac sin B. sin B Substituting for sin B, R (Trig., Art. 30), and applying logarithms, we have, log (2Q) = log a 4- logc + log'sin B - 10; 108 MENSURATION hence, we may write the following Pt U L E. Add together the logarithms of the two sides and the logarithmic sine of their inzcluled angle; fr'om this Sumt subtract 10; the remainder will be the logarithm of closdble the area of the triangle. _Find, from the table, the number answering to this logarithmz, and divide it by 2; the quotient will be the required area. EXAMIPLES. 1. What is the area of a triangle, in which two sides a and b, are respectively equal to 125.81, and 57.65, and whose included angle C, is 57~ 25'? Ans. 2Q - 6111.4, and Q = 3055.7 Alns. 2. What is the area of a triangle, whose sides are 30 and 40, and their included angle 28~ 57'? Ans. 290.427. 3. What is the number of square yards in a triangle, of which the sides are 25 feet and 21.25 feet, and their included angle 450? Ans. 20.8694. L E M M A. To find half an angle, when the three sides of a plane triangle are given. 97. Let AB C be a plane tri- C C angle, the angles and sides being denoted as in the figure. We have (B. IV., P. XII., XIII.),' Bc/ B a2 = b2 + c2 ~- 2C. Ai.... (1.) When the angle A is acute, we have (Art. 37), AD = b cos A; when obtuse, AD' = b cos CAD'. OF SURFACES. 109 But as CAD' is the supplement of the obtuse angle A, cos CAD' = - cos A, and AD' - b cos A. Either of these values, being substituted for AD, in (1), gives, a2 = b2 + C2 2be cos A; whence, b2 - c2 -a a2 cosA = C2- ( 2.) 2be If we add 1 to both members, and recollect that I + cosA = 2 cos2 -A (Art. 66), Equation (4), we have, 2 cos2 A 2bc + 62 + C2 - a2 2be (b + C)2- ca2 (b t c + a) (b + c-a) 2bc 2be or, cos2'A = (b - c +a) (b + c -a 3 Tbe'' If we put b -k c + a -- s, we have, b +ca bn -c-a 2 s, and, is - a; 2 2 t Substituting in (3), and extracting the square root, cos,. (4.) the plus sign, only, being used, since JA < 900; hence, The cosine of half' of either angle of a plane triang/le, is equal to the square root of half the suem of the three sides, into half that sum minus the side opposite the agf/le, divided by the rectangle of the adjacezt sides. By applying logarithms, we have, log cos'-A =' [llog s og (s - ) + (a. c.) log, + (a. c.) log c]. ~ (a.) 24 110 MENSURATION If we subtract both members of Equation (2), from 1, and recollect that 1 - cos A = 2 sin2'~A (Art. 37), we have, 2bc - 6b2 - c2 + a2 2 sin2 A - 2c 2be a2-(b-c)2 _ (a +b-c) (a - b (5 c) - 2bc 2bc Placing, as before, a + b + c = s, we have, a6 b - c b + c;2 -s-c, and, 2bc is - b. Substituting in ( 5 ), and reducing, we have, hsin ~A -= - sb) (., -.. (6.) hence, 6 The sine of either angle of a plane triangle, is eqzutl to the square root of half the sum of the three sides, imnus one of the adjacent sides, into the half sum minus the other adjacent side, divided by the rectangle of the adjacent sides. Applying logarithms, we have, log sin A = tlog (s-) + log (s - C) + (a..) log b + (a. c.)log c]. (lW.) Third Case. To find the area of a triangle, when the hree sides are given. Let ABC represent a triangle B whose sides a, b, and c are given. From the,principle demonstrated in the last case, we have, A -C Q -- -be sinlA. OF SURFACES. 111 But, from Formula (2N'), Trig., Art. 66, we have, sinA = 2 sin'A cos ~A; whence, Q = be sin 2A cos A. Substituting for sin -2A and cos ~A, their values, taken from Lemma, and reducing, we have, Q = /Is (s -a) (s - b) (s — c); hence, we may write the following R U L E. _Find half the sum of the three sides, and from it subtract each side separately. Find the continued product of the half sum and the three remainders, and extract its square root' the result will be the area required. It is generally more convenient to employ logarithms; for this purpose, applying logarithms to the last equation, we have, log Q - I [log Is + log (s - a) + log (s - b) + log (5s-c)] hence, we have the following RULE. Find the half sum and the three remainders as before, then find the half sum of their logarithms; the number corresponding to the resulting logarithm will be the area required. EXAMPLES. 1. Find the area of a triangle, whose sides are 20, 30, and 40. WVe have, s = 45, is-a = 25, is-b = 15, ~s-c = 5, By the first rule, Q = v45 25 155 = 290.4737 Ans. 112 MENSURATION By the second rule, log Is.... (45).. 1.653213 log (~s - a) ~ ~ (25)... 1.397940 log (-s - b) ~ ~ (15). 1.176091 log (s - c) ( 5). 0.698970 2 )4.926214 log Q.......... 2.463107.'. Q = 290'4737 Ans. 2. How many square yards are there in a triangle, whose sides are 30, 40, and 50 feet? Ans. 663. To find the area of a trapezoid. 98. From the principle demonstrated in Book IV., Prop. VII., we may write the following RULL E. Find half the smn, of the parallel sides, and multiply it by the altitude; the product will be the area required.'E X AM P L E S. 1. In a trapezoid the parallel sides are 750 and 1225, and the perpendicular distance between them is 1540; what is the area? Ans. 1520750. 2. How many square feet are contained in a plank, whose length is 12 feet 6 inches, the breadth at the greater end 15 inches, and at the less end 11 inches? Ans. 13I-j. 3. How many square yards are there in a trapezoid, whose parallel sides are 240 feet, 320 feet, and altitude 66 feet? Ans. 2053- sq. yd. To find the area of any quadrilateral. 99. From what precedes, we deduce the following OF SURFACES, 113 RULE. Join the vertices of two opposite angles by a diagonal; fronm? each of the other vertices let fall per2endiculars upon this diagonal; multiply thle diagonal by half of the sum of the peirpendiculars, and the product will be the area required. E X AM PLE S. 1. What is the area of the quadrilateral ABC-D, the diagonal A C being 42, and the perpendiculars Dg, C Bb, equal to 18 and 16 feet? Ans. 714 sq. ft. B 2. How many square yards of paving are there in the quadrilateral, whose diagonal is 65 feet, and the two perpendiculars let fall on it 28 ancd 331 feet? Ans. 222-. To find the area of any polygon. 100. From what precedes, -we have the following P U L E. Draw diagonals dividing the proposed polygon into trapezoicds and triangles: then tfind thle areas of these figures separately, and add them together for the area of t/he whole polygon, E X A P L E. 1. Let it be required to determine the area of the polygon AB CDDE, having five sides. E > Let us suppose that we have nicasured the diagonals and perpendiculars, and found AC - 36.21, EC = 39.11, Bb = 4, )d -= 7.26, Aa = 4.18: required the area. Ans. 296.1292. 114 MENSURATION To find the area of a regular polygon. 101. Let AB, denoted by s, represent one side of a regular polygon, whose centre is C. Draw CA and C CB, and from C draw CD) perpendicular to AB. Then will C-D be the apothem, and we shall have A-D = BD. Denote the number of sides of the polygon by n; then 360~ will the angle A CB, at the centre, be equal to 6 (B. V., Page 138, D. 2), and the angle A CD, which is half 180~ of A CB, will be equal to In the right-angled triangle AD C, we shall have, Formula (3), Art. 37, Trig., C-D = s tan CYAD. But CA), being the complement of A CD, we have, tan CAD = cot AC-D; 1800 hence, CD = -ls cot 180, a formula by means of which the apothem may be computed. But the area is equal to the perimeter multiplied by half the apothem (Book V., Prop. VIII.): hence the following RULE. _Find the apothem, by the preceding formula; nzultiply the perimeter by half the ajpothenm; the product' will be the area required. E X A M P L E S. 1. What is the area of a regular hexagon, each of whose sides is 20? We have, CD = 10 x cot 300; or, log CI) = log 10- +log cot 30~-10. log ~s... (10 ). 1.000000 180~ log cot — (300) * 10.238561 log CD. 1.238561.'. CD) - 17.3205. OF SURFACES. 115 The perimeter is equal to 120: hence, denoting the area by Q, 120 x 17.3205 Q = = 1039.23 Ans. 2. What is the area of an octagon, one of whose sides is 20? Ans. 1931.36886. The areas of some of the most important of the regular polygons have been computed by the preceding method, on the supposition that each side is equal to 1, and the results are given in the following TABLE. I NAMIES. SIDES. AREAS. NAMES. SIDES. AREAS. Triangle,. 3.. 0.4330127 Octagon,.. 8.. 4.8284271 Square,. 4 1.0000000 Nonagon,. 9 6.181842 Pentagon,.. 5.. 1.7204774 Decagon,..10.. 7.6942088 Hexagon.. 6.. 2.5980762 Undecagon,.11.. 9.',8G5639! Heptagon.. 7.. 3.6339124 Dodecagon,.12.. 1 1.19; 15 The areas of similar polygons are to each other as tile squares of their homologous sides (Book IV., Prop. XXVII.). Denoting the area of a regular polygon whose side is s, by Q, and that of a similar polygon whose side is 1, by T, the tabular area, we have, Q: T' s2 12;.. Q T2; hence, the. following R u L E. Multiply the corresponding tabzlar area by the square of the given side; the product will be the area required. EXAMPLES. 1. What is the area of a regular hexagon, each of whose sides is 20? We have, T_ 2.5980762, and s2 = 400: hence, Q = 2.5980762 x 400 = 1039.23048 Ans. 116 MENSURATION 2. Find the area of a pentagon, whose side is 25. Anzs. 1075.298375. 3. Find the area of a decagon, whose side is 20. Ans. 3077.68352.:To fied the circumference of a circle, when the diameter is given. 102. From the principle demonstrated in Book V., Prop. XVI., we may write the following RULE. ][ucltiply the givenz diameter by 3.1416; the product will be the circumnference requirel. E X AM P L E S. 1. Wlhat is the circumference of a circle, whose diameter is 25? Alns. 78.54. 2. If the diameter of the earth is 7921 miles, what is the circumference? Ans. 24884.6136. Io find the diameter of a circle, when the circtiumference 2s given. 103. From the preceding case, we may write the following U LE. lDivide the given circzumference by 3.1416; the quotient will be the diameter required. E XA M P L ES.' 1. VqWhat is the diameter of a orcle, whose circumference is 11652.1944? Ans. 3709. 2. VWhat is the diameter of a circle, whose circumference is 6850? Ans. 2180.41. OF SURFACES. 117 l2b find the length of an arc containing any number of degrees. 104. The length of an arc of 10, in a circle whose diameter is 1, is equal to the ciroumference, or 3.3416 divided by 360; that is, it is equal to 0.0087266: hence, the length of an arce of n degrees, will be, n x 0.0087266. To find the length of an are containing n degrees, when the diameter is d, we employ the principle demonstrated in Book V., Prop. XII., C. 1: hence, we may write the following P U L E. Alultiply t/he number of degrees in t/he arc by.0087266, and the product by the diameter of the circle; the result will -be the length required. E X A P L E S. 1. What is the length of an arc of 30 degrees, the diameter being 18 feet? Ans. 4.712364 ft. 2. What is the length of an arc of 12~ 10', or 12-~, the diameter being 20 feet? Ans. 2.123472 ft. To find the area of a circle. 105. From the principle demonstrated in Book V., Prop. XV., we may write the following RU LE. Multiply the square of the radius by 3.1416; the pro. duct will be the area required. EXA MPLES. 1. Find the area of a circle, whose diameter is 10, and circumference 31.416. Ans. 78.54. 2. How many square yards in a circle whose diameter is 3~ feet? Ans. 1.069016. 3. What is the area of a circle whose circumference is 1.2 feet? Ans. 11.4595. 118 MENSURATION To find the, area of a circular sector. 106. From the principle demonstrated in Book V.,. Prop. XJV., C. 1 and 2, we may write the following R U L E. I..Alultiply half the arce by the raclius; or, II. Findc the area of thle whole circle, by the last rule; then write the proportion, as 360 is to the number of degrees in the sector, so is the area of the circle to the area of thze sector. E X A I P L E S. 1. Find the area of a circular sector, whose arc contains 180, the diameter of the circle being 3 feet. 0.35343 sq. ft. 2. Find the area of a sector, whose arc is 20 feet, the radius being 10. Ans. 100. 3. Required the area of a sector, whose arc is 147~ 29', arnd radius 25 feet. Ans. 804.3986 sq. ft. To find the area of a circular segment. 107. Let AB represent the chord C corresponding to the two segments A CB and AFB. Draw APE and BE. The segment A CB is equal to the sector EA CB, mninus the triangle AE-B. The segment AFB is equal to the sector EAFB, plus the triangle AEB. Hence, we have the following RULE. Find the area of the corresponding sector, and also of thee triangle formed by the chord of the segment and the two extreme radii of the sector; subtract the latter from the former when the segment is less than a semicircle, and take their sunm when. the segment is greater than a semicircle; the result will be the area required. OF SURFiACES. 119 EXAMPLE S. 1. Find the area of a segment, whose chord is 12 and the radius 10. Solving the triangle AEB, we find the angle AEB]3U is equal to'730 44', the area of the sector EA -CB equal to 64.35, and the area of the triangle A-E'B equal to 48; hence, the segment A CB is equal to 16.35 Ans. 2. Find the area' of a segment, whose height is 18, the diameter of the circle being 50. Ans. 636.4834. 3. Required the area of a segment, whose chord is 16, the diameter being 20. -Ans. 44.764. To find the area of a circular ring contained between the circumferences of two concentric circles. 108. Let X- and r denote the area of the two circles, B being greater than r. The area of the outer circle is B2 x 3.1416, and that of the inner circle is r2 x 3.1416; hence, the area of the ring is equal to (22 - r2) x 3.1416. Hence, the following RULE. FPinsd the dcference of the squares of the radii of the two circles, and multiply it by 3.1416; the product will be the area required. EXAM1PLES. 1. The diameters of two concentric circles being 10 and 6, required the area of the ring contained between their circumferences. Ans. 50.2656. 2. What is the area of the ring, when the diameters of the circles are 10 and 20? Anzs. 235.62. 120 MENSURATION MENSURATION OF BROKEN AND CURVED SURFACES. To find the area of the entire smurface of a riight prism. 109. From the principle demonstrated in Book VII., Prop, I., we may write the following RULE. 3lultiply the perimeter of the base by the altitude, the reodect will be the area of the convex suz2face; to this add the areas of the two bases; the result will be the area required. E XA P L E S. 1. Find the surface of a cube, the length of each side being 20 feet. Ans. 2400 sq. ft. 2. Find the whole surface of a triangular prism, whose base is an equilateral triangle, having each of its sides equal to 18 inches, and altitude 20 feet. Ans. 91.949 sq. ft. To find the area of the entire su/fcace of a right pyramitd. 110. From the principle demonstrated in Book VII., Prop. IV., we may write the following RULE. iaulltip2ly the perimeter of the base by the slant height; the product will be the area of the convex surface; to this acdd the area of the base; the result will be the area required. E X A 3 P L E S. 1. Find the convex surface of a right triangular pyramid, the slant height being 20 feet, and eachl side of the base 3 feet. Ans. 90 sq. ft. 2. What is the entire surface of a right pyramid, whose slant height is 15 feet, and the base a pentagon, of which each side is 25 feet? Ans. 2012.798 sq. ft. OF SURFACES. 121 To,find the area of the convex szpuface of a frustum of a right pyramid. 111. From the principle demonstrated in Book XII., Prop. IV., C., we may write the following R U LE..M.fiiply the ha;f sum of the perimeters of the tco bases by the slaht height; the product will be the area required. E:X A P L E S. 1. How many square feet are there in the convex surface of the frustum of a square pyramid, whose slant height is 10 feet, each side of the lower base 3 feet 4 inches, and each side of the upper base 2 feet 2 inches? Ans. 110 sq. ft. 2. What is the convex surface of the frustum of a heptagonal pyramid, whose slant height is 55. feet, each side of the lower base 8 feet, and each side of the upper base 4 feet? Ans. 2310 sq. ft. 112. Since a cylinder may be regarded as a prism whose base has an infinite number of sides, and a cone as a pyramid whose base has an infinite number of sides, the rules just given, may be applied to find the areas of the surfaces of right cylinders, cones, and frustums of cones, by simply changing the terrm perimeter, to circumference. E X A I P L E S. 1. What is the convex surface of a' cylinder, the diameter of whose base is 20, and whose altitude 50? Ans. 3141.6. 2. What is the entire surface of a cylinder, the altitude being 20, and diameter of the base 2 feet? 131.9472 sq. ft. 3. Required the convex surface of a cone, whose slant height is 50 feet, and the diameter of its base 8~ feet. Ans. 667.59 sq. ft. 122 MENSURATION 4. Required the entire,surface of a cone, whose slant height is 36, and the diameter of its base 18 feet. Ans. 1272.348 sq. ft. 5. Find the convex surface of the fiustum of a cone, the slant height of the frustum being 12~ feet, and the circum. ferences of the bases 8.4 feet and 6 feet. Ans. 90 sq. ft. K Find the entire surface of the frustum of a cone, the slant height being 16 feet, and the radii of the bases 3 feet, and 2 feet. Ans. 292.1688 sq. ft. To find the area of the surface of a sphere. 113. From the principle demonstrated in Book VIII., Priop. X., C. 1, we, may write the following R ULE. Find the area of one of its great circles, and multiply it by 4; the prodluct will be the area require. EXA MPLES. 1. What is the area of the surface of a sphere, whose radius is 16? Ans. 3216.9984. 2. What is the area of the surface of a sphere, whose radius is 27.25 Ans. 9331.3374.'ob find the area of a zone. 114. From the principle demonstrated in Book VIII., PIrojp. X., C. 2, we maly lwrite the following RULE. F'ind the circ,'o~e /rence of a great circle of the sphere, and multiply it bT/ the altitude of the zone; the product will be the area r'equired. OF SURFACES. 123 EXAM P LES. 1. The diameter of a sphere being 42 inches, what is the area of the surface of a zone whose altitude is 9 inches. SAns. 1187.5248 sq. in. 2. If the diameter of a sphere is 12- feet, what will be the surface of a zone whose altitude is 2 feet?' 78.54 sq. ft. To find the area of a spherical polygon. 115. From the principle demonstrated in Book IX., Prop. XIX., we may write the following RU LE. From the sum of the angles of the polygon, subtract 1800 taken as many times as the polygon has sides, less two, and divide the remainder by 90~; the quotient will be the spherical excess. Find the area of a great circle of the sphere, and divide it by 2; the quotient will be the area of a tri-rectangular triangle. lzultiply the area of the trirectangular triangle by the spherical excess, and the product will be the area required. This rule applies to the spherical triangle, as well as to any other spherical polygon. E XAM]I PLE S. 1. Required the area of a triangle described on a sphere, whose diameter is 30 feet, the angles being 1400, 92~, and - 68~. Ans. 471.24 sq. ft. 2. What is the area of a polygon of seven sides, described on a sphere whose diameter is 17 feet, the sum of, the angles being 1080~? Ans. 226.98. 3. What is the area of a regular polygon of eight sides, described on a sphere whose diameter is 30 yards, each angle of the polygon being 140~? Ans. 157.08 sq. yds. 124 Ml ENSURATION MENSURATION OF VOLUMES. To find the volume of a prism. 116. From the principle demonstrated in Book VII., Prop. XIV., we may write the following RULE. JIultiply the area of the base by the altitude; the,product will be the volume required. EXAMPLE S. 1. What is the volume of a cube, whose side is 24 inches? Ans. 13824 cu. in. 2. How many cubic feet in a block of marble, of which the length is 3 feet 2 inches, breadth 2 feet 8 inches, and height or thickness 2 feet 6 inches? Ans. 215 cu. ft. 3. Required the volume of a triangular prism, whose height is 10 feet, and the three sides of its triangrular base 3, 4, and 5 feet. Ans. 60. To find the volume of a pyramid. 117. From the principle demonstrated in Book VII., Prop. XVII., we may write the following RULE. lMultiply the area of the base by one-third of the altitude; the product will be the volume required. E X A P L E S. 1. Required the volume of a square pyramid, each side of its base being 30, and the altitude 25. Ans. 7500. 2. Find the volumle of a triangular pyramid, whose altitude is 30, and each side of tile base 3 feet. 38.9711 cu. ft. OF VOLUMES. 125 3. What is the volume of a pentagonal pyramid, its altitude being 12 feet, and eac;h side of' its base 2 feet. An's. 27.5276 cu. ft. 4. What is the volume of an hexacgonal pyramid, w liose altitude is 6.4 feet, and each side of its base 6 inches? Ans. 1.38564 (c. it. To find the volume of a frustum of a pyramifd. 118. From the principle demonstrated in Book VII., Prop., XVIII., C., we may write the following R ULE. Find the sum of the upper base, the lower base, and a mean proportional between theyn; multiply the result by onethird of the altitude; the product vwill be the volume required. EX AMPLE S. 1. Find the number of cubic feet in a piece of timber, whose bases are sqaares, each side of the lower base being 15 inches, and each side of the upper base 6 inches, the altitude being 24 feet. Ans. 19.5. 2. Required the volume of a pentagonal firustum, whose altitude is 5 feet, each side of the lower base 18 inches, and each side of the upper base 6 inches. Ans. 9.31925 cu. ft. 119. Since cylinders and cones are limiting cases of prisms and pyramids, the three preceding rules are equally applicable to them. EX A i P LES. 1. Required the volume of a cylinder whose altitude is 12 feet, and the diameter of its base 15 feet. Ans. 2120.58 cu. ft. 2. Required the volume of a aylinder whose altitude is 20 feet, and the circumference of whose base is 5 feet 6 inches. Ans. 48.144 cu. ft. 25 126 MENSURATION 3. Required the volume of a cone whose altitude Is 27 feet, and the diameter of the base 10 feet. Ans. 706.86 cu. ft. 4. Required the volume of a cone whose altitude is 1 0 feet, and the circumference of its base 9 feet. Ans. 22.56 cu. ft. 5. Find the volume of the frustum of a cone, the altitudle tbeing 18, the diameter of the lower base 8, and that of the upper base 4. Ans. 527.7888. 6. What is the volume of the frustum of a cone, the altitude being 25, the circumference of the lower base 20, and that of the upper base 10? An's. 464.216. 7. If a cask, which is composed of two equal conic firlus tums joined together at their larger bases, have its bung diameter 28 inches, the head diameter 20 inches, and the lenith 40 inches, how many gallons of wine will it contain, there'being 231 cubic inches in a gallon? Ans. 79.0613. To find the volume of a sphere. 120. From the principle demonstrated in Book VIII., Prop. XIV., we may write the following RULE. Cube the diameter of the sphere, and multiply the result by 6r,, that is, by 0.5236; the product will be the volumee required. E XA I P L E S. 1. What is the volume of a sphere, whose diameter is 12? A2ns. 904.7808. 2. What is the volume of the earth, if the mean dialneter be taken equal to 7918.7 miles. Ants. 259992792083 cu. miles. OF VOLUMES. 127 To-find the volume of a wedge.. 121. A WEDGE is a volume bound- G I ed by a rectangle ABC)D, called the back, two trapezoids ABHG, D CI~G, called faces, and two triangles ADG, D CBtI, called ends. The, line GiE, in which the faces meet, is called the A B ecdge. It may happen that the edge is either greater or less than the length of the back. In deducing an expression for the volume, we shall' take'the latter case, premising that the result, so far as the volume is concerned, is the same in either case. Let AB CD-GIT represent R a wedge, and denote the length of the base by I, the length of the edge by 1, the breadth| of the base by i, and the alti- D | tude of the wedge by Ah. Through G, pass a plane A A B parallel to HfCB, dividing the wedge into the triangular prism GNI31-B, and the pyramid A~fIND- G. The altitude of the prism G1,1 is equal to i, and the area of its base CG/ff, is equal to ~bh; hence, its volume is equal to lbhl. The altitude GP, of the pyramnid, is equal to h, and its base AMJND is equal to b(L - ); hence, the volume of the pyramid is equal to ~bh(L - 1). The volume of the wedge, denoted by V, is equal to the sum of the volumes of the prism and pyramid; hence, V = ~bhl + I~bh(L- ) MI b + bJ - ~-bMl. Factoring and reducing, we have, v = bh(Wl + IL) =- bh(l + 2L); 128 MENSURATION hence, the following R ULE. Add twice the length of the back to the length of the. edge; multi2ply the sum by the breadth of the back, and t/haot result by one-sixth of the altitude; the final product will be the area required. E XA3 PLE S. 1. If the back of a wedge is 40 by 20 feet, the edge 35 feet, and the altitude 10 feet, what is the volume? Ans. 3833.33 cu. ft. 2. What is the volume of a wedge, whose back is 18 feet I y 9, edge 20 feet, and altitude 6 feet? 504 cu. ft. To find the volume of a prismoid. 1!2. A PRIsMroID is a frustum of a wedge. Let L and -B denote the length and breadth of the lower l' bash, 1 and b the length and breadth of the upper base, AM and m the length and breadth of the section equidistant from the bases, and h the altitude of the priginoid. B Through the edges L and 1', let a plane be passed, and it will divide the prismoid into two wedges, having for bases, the bases of the prismoid, and for edges the lines L and t1 The volume of the prismoid, denoted by V, will be equal to the sum of the volumes of the two wedges; hence, V- = Bh(l + 2L) +,bh(L + 21); V' = h(2BrL + 2bl +.Bl + bL); OF VOLUMES. 129 which may be written under the form, TT = h, [(BL + bl + L + b) + BL + bl]. ~ ~ (a.) Because the auxiliary section is midway between the bases, we have, 2M= L+l, and 2m = + b; hence, 4_?}m = (L + l) (B + b) = BL + bl -+ BL + bil. Substituting in (A), we have, V -- lh(B-L + bl + 4Km). But BL is the area of the lower base, or lower section, bl is the area of the upper base, or upper section, and Mm is the area of the middle section; hence, the following RULE. To find the volume of a prismoid, find the sum of the areas of the extreme sections and fouri times the middle section; multiply the result by one-sixth of the distance between the extreme sections; the result will be the volume required. This rule is used in computing volumes of earth-work in railroad cutting and embankment, and is of very extensive application. It may be shown that the same rule holds for every one of the volumes heretofore discussed in this work. Thus, in a pyramid, we may rSgard the base as one extreme section, and the vertex (whose area is 0), as the other extreme' their sum is equal to the area of the base. The area of a section midway between between them is equal to ole-fourth of the base: hence, four times the middle section is equal to the base. Multiplying the sum of these by onesixth of the altitude, gives the same result as that already found. The application of the rule to the case of cylinders, frustums of cones, spheres, &c., is left as an exercise for the student. 130 MENSURATION EXAMPLES. 1. One of the bases of a rectangular prismoid is 25 feet by 20, the other 15 feet by 10, and the altitude 12 feet; required the volume. Ans. 3700 cu. ft. 2. What is the volume of a stick of hewn timber, whose ends are 30 inches by 27, and 24 inches by 18, its length being 24 feet? Ans. 102 cu. ft. MENSURATION OF REGULAR POLYEDRONS. 123. A REGULAR POLYEDRON is a polyedron bounded by equal regular polygons. The polyedral angles of any regular polyedron are all equal. 124. There are five regular polyedrons (Book VII., Page 208). To find the diedral angle between the faces of a regular polyedron. 125. Let the vertex of any polyedral angle be taken as the centre of a sphere whose radius is 1: then will this sphere, by its intersections with the faces of the polyedral angle, determine a regular spherical polygon whose sides will be equal to the plane angles that bound the polyedral angle, and whose angles are equal to the diedral angles between the faces. It only remains to deduce a formula for finding one angle of a regular spherical polygon, when the sides are given. OF POLYEDRONS. 131 Let AB CDE represent a regular spherical polygon, and let P be the pole of a small circle passing through its vertices. Suppose P to be connected with each of the vertices by arcs of D great circles; there will thus be formed as many equal isosceles tri- E A angles as the polygon has sides, the P vertical angle in each being equal to 3600 divided by the number of sides. Through P draw PQ per- Q lpendicular to AB: then will A Q be equal to _BQ. If we denote the number of sides by ns, 360~ 1800 tile angle AP Q will be equal to 2, or In the right-angled spherical triangle APQ, we know the base AQ, and the vertical angle APQ; hence, by Napier's rules for circular parts, we have, sin (90 - APQ) = cos (900 - PAQ) cos AQ; or, by reduction, denoting the side AB by s, and the angle PAB, by- A, 1800 cos - sin ~A cos Is; 180~ cos whence, sin A =- - cos -s E XAMPLES. In the Tetraedron, 180~ - 60, and is = 30.. A - 00 31' 42", In the Hexaedron, 1800 600, and ~s 450.'. A 90. 132 MEN SURATI O N In the Octaedron, 1800 450, and Is = 300.. A = 1090 28' 18". f1 In the Dodecaedron, 180 - 60~, and ~s = 108~.'. A = 116~ 33' 54". In the Icosaedron, 180~ = 36~, and s - 30~.'. A = 138j 11' 23". To find the volume of a regular polyedron. 126. If planes be passed through the centre of the polyedron and each of the edges, they will divide the polyedron into as many equal right pyramids as the polyedron has faces. The common vertex of these pyramids will be at the centre of the polyedron, their bases will be the faces of the polyedrpn, and their lateral faces will bisect the diedral angles of the polyedron. The volume of each pyramid will be equal to its base into one-third of its altitude, and this multiplied by the number of faces, will be the volume of the polyedron. It only remains to deduce a formula for finding the distance from the centre to one face of the polyedron. Conceive a perpendicular to be drawn from the centre of the polyedron to one face; the foot of this perpendicular will be the centre of the face. From the foot of this perpendicular, draw a perpendicular to either side of the face in which it lies, and connect the point thus determined with the centre of the polyedron. \There will thus be formed a right-angled triangle, whose base is the apothem of the face, whose angle at the base is half the diedral angle of the polyedron, and whose altitude is the required altitude of the pyramid, or in other words, the radius of the inscribed sphere. OF POLYEDRONS. 133 Denoting the perpendicular by P, the base by. b, and the diedral angle by A, te have Formula (3), Art. 37, Trig., P = b tan A; liat b is the apothem of one face; if, therefore, we denote the number of sides in that face by n, and the length of each side by s, we shall have (Art. 101, Mens.), 180~ b'= Is cot whence, by substitution, - 1800 P -= s cot tan ~A; hence, the volume mnay be computed. The volumes of all the regular polyedrons have been computed on the supposition that their edges are each equal to 1, and the results are given in the following TABLE. NAMES. NO. OF FACES. VOLUMES. Tetraedron,. 4..... 0.1178513 Hexaedron, 6.1.0000000 Octaedron, 8.0.4714045 Dodecaedron,.12..... 7.6631189 Icosaedron,..... 20.....2.1816950 Fromr the principles demonstrated in Book VII., we may write the following RULE. To find the volume of any regular polyecron, mnultiply the cube of its edge by the corresponding tabular voluzme;. the product will be the volume required. 134 MENSURATION. EXAMPLES. 1. What is the volume of a tetraedron, whose edge is 15? Ans. 397.75. 2. What is the volume of a hexaedron, whose edge is 12? Ans. 1728. 3. What is the volume of a octaedron, whose edge is 20? Ans. 3771.236. 4. What is the volume of a dodecaedron, whose edge is 25? Ans. 119736.2328. 5. What is the volume of an icosaedron, whose edge is 20? Ans. 17453.56. A TABIE OF LOGARITHMS OF N UMBERS FROM 1 TO 10,000. W. Log. N. Log. N. Log. N. Log. I 0-000000 26 1 414973 51 I 707570 76 i 88o814 2 3o io3o 27 1I431364 52 I.7I6003 77 I188649I 3 0-477I2I 28 I 447158 53 1.724276 78 I. 892095 4 o- 60260 29 1.462398 54 I 732394 79 I*897627 5 0.698970 30 1.477I 2 55 1 740363 80 1.903090 6 0-778151 3I 1-49I362 56 I~748188 i8 1-908485 7 0.845098 32. 5o515o 57 1-755875 82 I91I3814 -O90309 33 I 5i8514 58 1 763428 83 I 9I9078 9 0.954243 34 x 531479 59 1 770852 84 924279 IO 000000oooooo 35. 1i544o06 60 I-778I5I 85 I-9294I9 lI I o041393 36 I.556303 6i 1.785330 86 1.934498 Il 1 07918I 37 I 568202 62 I 792392 87 I 939519 13 I I3943 38 1.579784 63 1 79934I 88 1-944483 14 I I46128 39 I 591o65 64 I 8o6i8i 89 1.949390 I5 1 17609I 40 1 602060 65 I.8129I3 90 1.954243 I6 I 204120 41 I 612784 66 I. 89544 91 1.95904I I7 1.230449 42 1.623249 67 I826075 92a I963788 I8 1-255273 43 I-633468 6 I-832509 93 i.968483 19 1.278754 44 I-643453 6o 1838849 94 1-973128 20 I.30Io30 45 I.653213 70 I 845098 95 1 977724 2 I1.322219 46 1, 662758 71 1.851258 96 1.98227I 22 I.342423 47 I.672098 72 I 857333 97 I'986772 23 I 1361728 48 1.681241 73 I 863323 98 1-991226 24 1.380211 49 1.690196 74 X 869232 99 I 995635 25 1.397940 50 I 698970 75 I.87506 100 2 -000000 REMARK. In the following table, in the nine right hand columns of each page, where the first or leading figures change from 9's to O's, points or dots are introduced instead of the O's, to catch the eye, and to indicate that from thence the two figures of the Logarithm to be taken from the secoind column, stand in the next line below. 2 A TABLE OF LOGARITHMS FROM 1 TO 10,000. N. o I 2 3 1 4 6 7 8 9 i 7! 10 00ooooo000 0434 o868 i3oi 1734 2166 2598 3029 346I 3891 432 I01 4321 4751 5i8I 5609 6038 6466 6894 7321 7748 8174 428 102 1 8600[ 9026 945I 9876 3oo'724 I I47 1570 1993 2415 424 103 1oI2837, 3259 3680 4100 4521 4940 5360 5779 6197 66i6 419 104 7033; 7451 7868 8284 8700 9116 9532 )947 036I *775 416 I05.021189 I603 2016 2428 284I 3252 3664 4075 4486 4896 412 io6 53061 5715 6125 6533 6942 7350 7757 8164 857 8978 408 107 93841 9789 e195 00oo 1004 I408 I812 2216 26I9 3021 404 I08 o33424 3826 4227 4628 5029 5430 583o 6230 6629 7028 400 109 74262 7825 8223 8620 9017 9414 9811 207 *602 0 *998 396 I10 o41393 I787 2182 2576 2969 3362 3755 4148 4540 4932 393 III 53231 57I4 6Io 6495 685 7275 7664 8o53 8442 883o 389 112 92181 6o6 9606 9993 *38 *766 II53 I538 I924 2309 2694 386 113 o53078 3463 3846 4 4230 4613 4996 5378 576o 6I42 5524 382 114 6905 7286 7666 8046 8426 8805 9185 9563 9942 *32o 379 115 o60698 1075 1452 I829 2206 2582 2958 333 3769 4o83 376 II6 4458 4832 5206 558o 5953 6326 6699 7071 7443 7815 372 117 8186 8557 8928 9298 9668 0*38 *407 0776 145 i514 369 1I8 071882 2250 26I7 2985 3352 37 8 4085 445I 48I6 5182 366 119 5547 5912 6276 6640 7004 7368 7731 8094 8457 8819 363 o20 7918I 9543 9904'266 *626 *987 I347 I707 2067 2426 36o 121 o82785 3 44 3503 3861 42I9 4576 4934 5291 5647 6004 357 122 636o 67I6 7071I 7426 778I 8136 8490 8845 9198 9552 355 123 9905| 258 *6iI *963 I3i5 I667 2018 2370 2721 3071 351 124 o93422 3772 4122 447I 4820 5I69 55I8 5866 62I5 6562 34 125 69o10 7257 7604 795I 8298 8644 8990 9335 9681 "'26 346 126 Ioo371 0715 I159 I4o3'747 2091 2434 2777 31I9 3462 343 127| 38o04 414 6 4487 4828 5I69 55io 585I 619i 653i 6871 340 128 7210o 7549 7888 8227 8565 8903 924I 9579.9916 J 253 338 I29 1i0590 0926 I263 1 599 I934| 2270 2605 2940 3275 3609 335 i3o II3943 4277 461I, 4944 5278 56Ii 5943 6276 6608 6940 333 13I 7271 7603 7934 8265 8595 8926 9256 9586 9915 *245 330 I32 120574 0903 1231 I56o i888 22I6 2544 287I 3198 3525 328 133 3852 4178 4504 4830 5156 548, 58o6& 6i3i 6456 678I 325 J34 7105 7429 7753 8076 8399 8722 9045 9368 9690 j0I2 323 5 i3o334 o655 0977 1 I298 16i9 1939 2260 2580 2900 32I9 321 i36 3539 3858 4177 4496 48I4 5I33 5451 5769 6o86 64o3 J,3i8 137 672I 7037 7354 7671 7987 83o3 86i8 8934 9249 9564 315 i38 9879J 1i94 *5o8 | 822 ii36 I450 I763 2076 2389 2702 314 I39I 143 oi 3327 3639 3951 4263 4574 4885 596 5507 58I8 3I 140 146128 6438 6748 7058 7367 7676 7985 8294 86o3 89 II 309 14' 9219 9527 9835 | *142`449 0756 io63 1370 1 676 1982 307 142 152288 2594 2900 3205 350o 3815 4120 4424 4728 5032 305 143 5336 5640 5943 6246 6549| 6852 7154 7457 7759 8o6I 303 144 8362 8664 8965 9266 9567 9868'i68 469 769 i068 301 145 |I6I368 1I667 I967 2266 2564 2863 316i 346o 3758 4055 299 146 43531 4650 4947 5244.5541 5838 6134 6430 6726 7022 297 147 7317 7613' 79o8 8203 8497 8792 9086 938o 9674 9968 295 I48 173262 0555 0848 1141 1434 1726 2019 23I 2603 2895 293'49 3i86 3478 3769 4o6o 4351 4641 4932 5222 5512 58o2 291 I50 I76o9I 638I 6670 6959 7248 7536 7825 8 93 | 8401 8689 289 I5I 89771 9264 9552 9839 |126 |413 0699 985 1272 I558 28 1 52 |I8I844|2123 2415 2 2700 2985 3270 3555 3839 4I23 4407 285 53 469, 497 5259 5542 5825 6o08 639I 6674 6956 7239 283 I54 7521 7 03 8o84 8366 8647 8928 9209 9490 9771 0 e5I 281 i55 1I9o332! 0612 08 2 1171 145I 1730 2010 2289 2567 2846 279 I56| 31251 34o3 3681 3959 4237 45I4 4792 569 5346 5623 278 157 5899 6I76 6453 6729 7oo| 7528I 7556 7832 8I7 8382 276 1581 8657 8932 9206 9481 9755 |29 303 577 |850 1 1I24 274 159 201I397 1670 1943 2216 2488 2761 3033 3305 3577 3848 272. O I I 2 3 | 4 5 6 7 8 9 D. A TABLE OF LOGARITHMS FROM 1 TO 10,000. 8 N. o 1 2 3 4 5 6 7 8 9 i6o 204I20 4391 4663 4934 5204 5475 5746 60o6 6286 6556 271 i61 6826 7096 7365 7634 7904 8173 844i 8710 8979 9247 269 162 95i 5 9783 *05I ~3 i9 "586 *853 1121 i 388 z65 1921 I 267 i63 2L2188 2454 2720 2986 3252 35I8 3783 4049 4314 4579 266i 164 48448 5109 5373 5638 5902 6i66 643o 6694 6957 7221 264 165 7484 7747 8oio 8273 8536 8798 9060 9323 9385 9846 262 i66 220108 o370 o631 I0892 1153 14194 1675 1936 2196 2456 26i 1 167 2716 2976 3236 3496 3755 4oI5 4274 4533 4792 505 I 259 168 5309 5568 5826 6084 6342 66oo 6858 7I I5 7372' 7630 258i 169 7887 8144 84o0 8657 8913 9170 9426 9682 9938 *193 256 170 230449 0704 og606o 1215 I470 I 724 I 979 2234 2488 2742 254 171 2996 3250 3504 3757 4011 I 4264 45I7 4770 5023 5276 2531 [72 5528 578I 6033 6285 6537 6789 7041 7292 7544 7795 252 173 80o46 8297 8548 8799 9049 9299 9550 98o00 **o *3oo 250 174 240549 0799 1048 I297 i546 I793 2044 2293 2541 2790 249 175 3038 3286 3534 3782 403o 4277 4525 4772 5o09 5266?48 176 5513 5759 6006 6252 6499 6745 6991 7237 7482 7728 246 177 7973 82I9 8464 8709 8954 gI98 9443 9687 9932 *176 245 178 250420 o664 ogo8 151 I395 I638 I 88I 2I25 2368 26Io 243 179 2853 3096 3338 3580 3822 4o64 43o6 4548 4790 5o3I 242 180 255273 55I4 5755 5996 6237 6477 67I8 6958 7198 7439 24I1 I81 7679 7918 8i58 8338 8637 8877 91I6 9355 9594 983 239 I82 260071 o3io o548 o787 1025 1263 i5oi 1739 I976 2214 238 I83 2451 2688 2925 3I62 3399 3636 3873 j409 4346 4582 2371 184 4818 5054 5290 5525 576 I 5996 6232 6467 6702 6937 235 185 7172 7406 764I 7875 8iio 8344 8578 88I2 9046 9279 234 i86 9513 9746 9980 *213'446'679 ~ 912 1144 1377 1609 233 187 271842 2074 2306 2538 2770 3ooi 3233 3464 3696 3927 232 188 4158 4389 4620 485o 508i 531t1 5542 | 5772.6002 6232 230 I89 6462 6692 6921 7151 7380 7609 7838 | 8067 8296 8525 229 190 278754 8982 9211 9439 9667 9895 *123 35i *578 *8o6 228 91 281033 126i I488 1715 1942 269 2396 2622 2849 3075 227 192 33oi 3527 3753 3979 4205 4431 4656 4882 5107 1 5332 226 193 5557 5782 6007 6232 6456 668I 6905 730 7354 7578 225 194 7802 8026 8249 8473 8696 8920 9143 9366 9589 9812 223 195 290035 0257 o480 0702 09og25 II47 J369 1591 8I3 2034 222 196 2256 2478 2699 2920 3I41 3363 3584 3804 4025 4246 221 I97 44661 4687 4907 5127 5347 5567 5787 6007 6226 6446 220 I98 6665| 6884 7I04 7323 7542 776I 7979 8198 8416 8635 219 199 8853 9071 9289 9507 9725 9943 *16I'378 1595 *83 2I8 200 3oio3o 1247 1464 i68I 1898 214. 233I 2547 2764 2980 217 201 3I96 3412 3628 3844 4059 /4275 4491 4706 4921 5I36 26 202 535i 5566 5781I 5996 6211 6425 J 6639 6854 7068 728,2 25 203 7496 7710 7924 8137 8351 8564 I 8778 8991 9204 9417 2I31 204 9630 9843 Ie56 268 e48I *693 9o6 I 118 I33o I542 2121 2o5 |311754 I966 2I77 2389 2600 281 3o23 3234 3445 3656 211 206 3867 4078 4289 4499 4710 4920 5i3o 5340 555i 5760 210o 207 5970 6I8o 6390 6599 68o9 7018 7227 7436 7646 7854 209 o 208 8o63 8272 848I 8689 8898 06 934 9522 9730 9938 20o81 2o9 0320146 o354 o562 0769 o977 184 1391 1598 i8o5 2012 207 2I0 3222I9 2426 2633 2839 3046 3252 3458 3665 3871 4077 2061 211 4282 4488 4694 4899 55 53io 5516 5721 5926 6i3 2051 212 6336 6541 6745 6950o 7155 7359 7563 7767 7972 8176 2041 213 838o 8583 8787 8991 9194 9398 9601 9805 j@@8 0211 203 214 330414 o067 0819 I1022 1225 I427 i630 832 2034 2236 202 215 2438 2640 2842 3044 3246 3447 3649 385o 4o5I 4253 202 216 44541 4655 14856 5057 5257 5458 5658 5859 6059 6260 201 217 6460o 6660 6860 7060 7260 7459 7659 7858 8o58 8257 200 218 8456 8656 8855 i 9054 9253 9451 9650 9849 @*47 *246 i 991 2I9 a340444 0642 o84I I039 1237 1435 1632 830 2028 2225 I98 N. 0 I 2 3 4 5 6 7 8. 4 A TABLE OF LOGARITHMS FROM 1 TO 10,000 N. |I 2 3 4 5 6 7 8 9 220 342423 2620 2817 3oi4 3212 3409 36o6 3802 3999 4I96 197 221 4392 4589 4785 498 5178 5374 5570 5766 5962 6157 196 222 6353, 6549 6744 6939 7135 7330 7525 7720 7915 8iio 195 223 830o5 85oo00 8694 8889 9083 9278 9472 9666 9860 e'54 t94 224 350248[ 0442 o636 0829 1023 12I6 I410 1603 1796 I989 t93, 225 2I831 2375 2568 276I 2954 3147 3339 3532 3724 3 i6 193 226 4Io08 430I 4493 4685 4816 5o68 526o 5452 5643 5834 I92 227 6026{ 6217 6408 6599 6790 6981 7I72 7363 7554 7744 191 228 7935i 8125 83i6 8506 8696 8886 9076 9266 9456 9646 Igo90 229 98351 "-25 2 I 5 40o4 593'783 *972 I i6 I350o 1539 189 230 361728 1917 2105 2294 2482 2671 2859 3o48 3236 3424 188 23i 3612 3800 3988 4176 4363 455, 4739 4926 5ii3 530I i88 232 5488 5675 5862 6049 6236 6423 66io 6796 6983 7169 I87 233 7356i 7542 7729 7915 8ioi 8287 8473 8659 8845 9030 186 234 92161 9401 9587 9772 9958'143'328 *5I3 J698 *883 I85 235 3710o681 1253 1437 I622 I8o6 1991 2175 2360 2544 2728 184 236 29I2 3096 3280 3464 3647! 383 40o5 4198 4382 4565 184 237 4748 4932 5115 5298 548 5664 5846 6029 6212 6394 i83 238 6577 6759 6942 7124 7306 7488 7670 7852 8034 8216 182 239 8398 858o 8761 8943 9124 9306 9487 9668 9849 *30 18I 240 380211 o392, 0573 0754 0934 1115 1296 1476 1656 1837 181 24I 20I7 2197 2377 2557 2737 297 3097 3277 3456 3636 i80 242 38i5 3995 4174 4353 4533 47I2 489I 5070 5249 5428 179 243 56o6 5785 5964 6142 6321 6499 6677 6856 7034 7212 178 244 7390 7568 7746 7923 810i 8279 8456 8634 8811 8989 I78 245 91661 9343 9520 9698 9875 |SI *228 4055 *582 759 177 246 |39o935 1112 1288 I464 1641 I8I7 1993 2169 2345 2521 176 247 2697 2873 3048 3224 3400 3575 37 51 3926 4IoI 4277 176 I248 |44322 4627 |4802 4977 5I52 5326 55o0 5676 5850 6025 175 249 6199 6374 6548 6722 6896 7071 7245 7419 7592 7766 174 250 397940o 814 8287 8461 8634 88o8 8981 9154 9328 9501 173 251 96741 9847 "20 192' 365 538'71 I'883 o056 I228 173 252 4oI40: 1573 1745 1917 2089 2261 2433 2605 2777 2949 172 253 3121 3292 3464 3635 3807 3978 4149 4320 4492 4663 171 254 4834 500oo 5176 5346 5517 5688 5858 6029 6199 6370 171 255 65401 67I 688i 7o51 722I 739I 7561 7731 79oi 8070 170 256 8240 8410 8579 8749 8918 9087 9257 9426 9595 9764 169 257 9933 *102'271 0440 06o9 1 777'946 4 i283 145, i69 258 4I620 I788 956 2124 229 2461 262 2796 2964 3132 I68 259 33001 3467 3635 3803 3970 4I37 430o 4472 4639 48o6 167 260 414973 5140 5307 5474 564, 58o8 5974 614i 63o8 6474 167 261 6641 68o7 6973 713 7306 7472 7638 78044 7970 8135 i66 262 8301o 8467 8633 879 8964 9129 9295 9460 9625 9 I91 i65 263 9956 *121'286'42D1 616'78I'945 1110 1275 1439 i65 264 |42I604 I788 I933 2097 226I 1 2426 2390 2754 2I8 3082 I64 265 3246 34o10 374 i 3737 3901 4065 4228 4392 4255 4718 164 266 4882] 5045 5208 5371 5534 5697 5860 6023 6I86 6349 I63 26t 65Ii 6674 6836 6999 716I 7324 7486 7648 1 I1 7973 162 268 8i35J 8297 8459 I 862I 8783 8944 9IO6 9268 9429 9'9 I6i 269 9752 9914 75 236 398 *559'720 188i 1042 1203 i6 270 43i364 i525 I1685 1846 2007 2I67 2328 2488 2649 2809 x6II 27I1 2969 3i3o 329o 3450 36io 3770 3930 4090 4249 4409 160o 272 45691 4729 4 883 5o48 5207 5367 5526 5685 5844 6004 1X59j 213 16I63 6322 648,I 6640 6798 6957 7116 7275 7433 7592 5; 274 775II 7909! 8067! 8226 8384 8542 870I 8859 9017 1 917 5 I5 1 275 9333 949 9648 906 9964 122 0279 *437'594 0752 1 58 276 440gog io66 1224 i38i I38 I 1695 1 I852 2009 2I66 2323 1 57 277 124801 2637 2793 2950o 3io6 3263 3419 3576 3732 3889 157 278 4o45 4201 4357 4513 4669 4825 498 5I37 5293 5449 i56 279 5604 576o 59i5 6071 6226 6382 6537 6692 |6848 703 |55 Ni N 2 3 4 5 | I6 j.1..3..Zi.~~i A TABLE OF LOGARITHIMS'FROM I TO 10,000. 5 N. 0 I 2 3 4 5 6 7 8 9 ID. 280 447158 7313 7468 7623 7778 7933 8088 8242 8397 8552 I55 28I 8706 886I goi5 9170 9324 9478 9633 9787 994I *"95 154 282 450249 0403 o557 0711 o865 I018 II72 I326 1479 I633 154 283 1I786 1940 2093 2247 2400 2553 2706 2859 30612 3r65 153 284 3318 347' 3624 3777 3930 4082 4235 4387 4540 4692 153 285 4845 4997 5150! 5302 5454 5606 5758 5910o 6062 6214 152 286 6366 65I8 6670 1 682I 6973 7125 7276 7428 7579 773I 152 287 7882 8033 8I84 8336 8487 8638 8789 8940 9091 9242 I51 288 9392 9543 9694 9845 9995 *146 *296 @447'597 *748 151 289 460898 10o48 1198 i 1348 I499 I 649 1799 1948 2098 2248 I50 290 462398 2548 2697 2847 2997 3146 3296 3445 3594 3744 I50 291 389 3 4042 419I 4340 4490 4639 4788 4936 5085 5234 I49 292 5383 5532 5680 5829 5977 6126 6274 6423 6571 6719 I49 293 6868 7016 7164 7312 7460 7608 7756 7904 8052 8200 148 294 8347 8495 8643 8790 8938 9085 9233 9380 9527 9675 148 295 9822 9969'I 6 *263'4Io @557 *704 *85I @998 1145 147 296 471292 I438 i585 1732 I878 2025 2171 2318 2464 2610 I46 297 2756 2903 3049 3I95 334i 3487 3633 3779 3925 4071 146 298 4216 4362 4508 4653 4799 4944 5090 5235 538I 5526 146 299 5671 58i6 5962 6107 6252 6397 6542 6687 6832 6976 145 300 477121 7266 74I 7555 7700 7844 7989 8133 8278 8422 145 3oi 8566 871 iI 8855 8999 9143 9287 943 I 9575 9719 9863 144 302 480007 oi051 0294 0438 0582 725 o869 1012 _1156 1299 144 303 1443 i586 1729 I872 2016 2159 2302 2445 e2588 273I 143 304 2874 3o16 3159 3302 3445 3587 3730 3872 4015 4157 143 305 4300 4442 4585 4727 4869 5oii 5153 5295 5437 5579 I42 3o6 5721 5863 6005 6147 6289 6430 6572 6714 6855 6997 142 307 7r38 7280 742r 7563 7704 7845 7986 8127 8269 84iO 141 308 855I 8692 8833 8974 9114 9255 9396 9537 9677 9818 141 309 9958 ogg99 239 *38o *520 *66i @8o0I 94I io8i 1222 140 3io 1491362 1502 I642 1782 1922 2062 2201 2341 2481 2621 140! 3 2760 29o00 3040o 3179 33 9 8 458 3597 3737 3876 4015 139 312 4155 4294 4433 4572 4711 4850 4989 5128 5267 5406 139 33 5544 5683 5822 5960 6099 6238 6376 65I5 6653 679I 39 314 6930 7068 7206 7344 7483 7621 7759 7897 8035 8173 138 315 831I 8448 8586 8724 8862 8999 9137 9275 9412 9550 138! 3i6 9687 9824 9962 @*99 @236 o374'51I 0648 0785 *922 137 317 501059 1196 i333 1470 I607 1744 i88o 2017 2154 229I 137 318 2427 2564 2700 2837 2973 3IO9 3246 3382 3518 3655 136 3I9 379I 3927 4063 4199 4335 4471 4607 4743 4878 50o4 136 320 505I50 5286 5421 5557 5693 5828 5964 6099 6234 6370 136 321 6505 6640 6776 691I 7046 718I 7316 7451 7586 7721 I35i 322 7856 799I 8I26 8260 8395 8530 8664 8799 8934 9o68 I35 323 9203 9337 9471 9606 9740 9874 "9 143 @277 *4I 134 324 5io545 0679 o8i3 0947 Io8i 1215 I349 1482 i616 1756 1341 325 I883 20I7 2151 2284 2418 2551 2684 2818 2951 30o84 133 326 3218 335i 3484 3617 3750 3883 4016 4149 4282 4414 133t 327 4548 4681 4813 4946 5079 5211 5344 5476 5609 5741 1331 328 5874 6006 6139 627I 6403 6535 6668 6800 6932 7064 1321 329 V; 7196 7328 7460 7592 7724 7855 7987 8119 825I 8382 132 33o,518514 8646 8777 8909 9040 917I 9303 9434 9566 9697 131 33I 9828 9959 e"9o 221 *353 *484 ~6I5 *745 *876 1007 131 332 52II38 1269 1400 I53o I66I 1792 1922 2053 2183 2314 131 333 2444 2575 2705 2835 2966 3096 3226 3356 3486 3616 i3o 334 37461 3876 4006 4136 4266 4396 4526 4656 4785 4915 |1301 335 50451 5I74 5304 5434 5563 5693 5822 595I 6o8i 62IO 129' 336 6339 6469 6598 6727 6856 6985 7114 7243 7372 750x 129 337 7630 7759 7888 8oi6 8145 8274 8402 853i 8660 8788 129 338 8917 9045 9174 9302 9430 9559 9687 98I5 9943 @72 128 339 530200oo 0328 0456 0584 0712 o840 0968 o096 I223 I35I 128 N. 0 1 I 2 3 4 5 6 7 7 8 9.........8. L j A TABLE OF LO)GARITHMS FRC M 1 TO 10,000. N. o I 2 3 5 6 7 8 9 D. 340 531479 1607 1734 i1862 1 990 2117 2245 2372 2500 2627 128 341 2754 2882 3009 3136 3264 339I 35i8 3645 3772 3899 127 342 4026: 4153 4280 4407 4534 4661 4787 4914 504I 5167 I27 343 5294 5421 5547 5674 5800 5927 60o53 618o 6306 6432 126 344 0558 6625 681I 6937 7063 7189 73I5 7441 I 7567 7693 126 345 7819F 7945 807I 8I97 8322 8448 8574 8699 8825 89 I 126 346 90o76 9202 9327 9452 9578 9703 9829 9954 079'204 1 25 347 540329, 0455 0580 o0705 o8 0955 io8o 1205 i33o I454 125 348 i579 1704 1829 1953 2078 2203 2327 2452 2576 270I 125 349 2825i 2950 3074 3I99 3323 3447 357I 3696 3820 3944 124 350!5440681 4192 4316 4440 4564 4688 4812 4936 5o6o 5i83 124 351 5307 543I 5555 5678 5802 5925 6049 6172 6296 6419 124 352 6543 6666 6789 6913 7036 7159 7282 7405 529 7652 123 353 7775 7898 8021 8144 8267 8389 8512 8635 8758 88I 123 354 9003 9126 9249 9371 9494 96I6 9739 986I 9984 Qio6 123 355 550228 o35I 0473 0595 0717 0840 0962 1084 1206 I328 I22 356 I45oq 1572 1694 I816 I938 2060 2181 2303 2425 2547 I 22 357 2668 2790 2911 3033 3155 3276 3398 35I9 3640 3762 I'21 358 3883 4004 4126 4247 4368 4489 4610 4731 4852 4973 121 359 5094 5215 5336 5457 5578 5699 5820 5940 6o6i 6I82 12I 360 556303 6423 6544 6664 67'85 6905 7026 7146 7267 7387 120 36I 75o7 7627 7748 7868 7988 80o8 8228 8349 8469 8589 120 362 8709 88 8948 9068 9188 9308 9428 9548 9667 9787 120 363 9907 *~2 *146 *265 *385 *5o4 *624'743.863 *982 I I9 364 56iioi I22I I340 I459 I578 I698 1817 I936 2055 2I74 119 365 2293 2412 2531 2650 2769 2887 3006 3125 3244 3362 II9 366 3481 3600oo 37I8 3837 3955 4074 4192 4311 4429 4548 9 167 4666 4784 4903 502i 5139 5257 5376 5494 5612' 5730 II8 168 5848 5966 6084 6202 6320 6437 6555 6673 679I 6909 0 IS9 j69 7026 7144 7262 7379 7497 76I4 7732 7849 7967 8084 ii8 170 568202 8319 8436 8554 8671 8788 8905 9023 9140 9257 117 171 9374 9491 9608 9725 9842 9959 "'76 *193 *309 *426 II7 172 570543 o66o 0776 o893 1O10 1126 1243 I359 I476 1592 117 173 1709 1825 1942 2058 2174 229I1 2407 2523 2639 2755 iI6 374 2872 2988 3I04 3220 3336 3452 3568 3684 38oo 39I5 II6 375 40o3 4147 4263 4379 4494 4610 4726 4841 4957 5072 iI6 376 5i88 5303 5419 5534 5650 5765 5880 5996 61I 6226 115 377 634I 6457 6572 6687 6802 69I7 7032 7I47 7262 17377 II5 378 7492 7607 7722 7836 795I 8o66 8i8I 8295 8410o 8525 II 379 8639 8754 8868 8983 9097 9212 9326 9441 9555 9669 114 380o 579784 9898 1 I2'126'24I'355'469'583 *697 *81I 1 I4 38,~ 580925 I39j II53 1267 i38I 1495 I608 I22 I836 I950 114 382 2063 2I77 2291 2404 2518 2631 2745 2858 2972 3085 II4 383 3199 3312 3426 3539 3652 3765 3879 3992 4105 42I8 113 384 4331 4444 4557 4670 4783 4896 5009 5122 5235 5348 1i3 385 546I 5574 5686 5799 5912 6024 6I37 6250 6362 6475 113 386 6587 6700 68I2 6925 7037 7i49 7262 7374 7486 7599 II2 387 77I 7823 7935 8047 8160 8272 8384 8496 8608 8720 I2 388 8832 8944 9056 9167 9279 9391 9503 9615 9726 9838 112 389 9950 006I'173'284'396 0507 96I9'730'842 9543 I12 390 59o1065 1176 1 287 I399 I5Io I I 1732 |I843 I955 2066 II 391 2177 2288 2399 2510 2621 2732 2843 2954 3o64 375 I I 392 3286 3397 35o8 36i8 3729 3840 3950 4o6I 4171 4282 III 393 43931 45o3 4614 4724 4834 4945 5055 5i65 5276 5386 IIo 394 54961 5606 57I7 5827 5937 6047 6I57 6267 6377 6487 IIO 3935 6597 6707 6817 6927 7037 7146 7256 7366 7476 7586 iio 396 7695/ 7805 7914 8024 8134 8243 8353 8462 8572 868i IIo 397 8791 8900 gdog 9119009 9228 9337 9446 9556 9665 9774 09 398 9883! 9992 *IOI'*210'319 428'537 *646 *755 0864 1o9 399 600973 1082 1191 1299 |14o8 I517 I625 1734 1843 1951 io9 NL o 2 3 4 5 6 7 8 9 D. A TABLE OF LOdARITHMS FROM 1 TO 10,000. 7 LN. 0' 12 3 4<56 6 7 1 99J9D 40 602o6o 6 2169 2277 2386 2494 2603 2711 2819 2928 3036 io8 40i 3144/ 3253 336i 3469 3577 3686 3704 3902 4010 4Iz8 io8 402 42261 4I334 4442 455o t6538 4766 4 874 4982 5089 597 o8 403 J53o5j 54i3 5521 5628 5'36 5844 29)1 6029 6166 6274 1 o8 404 633i 6489 6596 6704 611 6919 o7026 7133 7241 7348 I107 405 7455 7562 7669 7777 788t4 7991 8098 8205 8312 j 8419 107 406 8526 8633 8740 8847 8954 9061 9167 9274 9381 9488 71 40o7 98'94' 9701 98o8 9914 21 128 *234'341'447 4 17 4o8 6io66o 0767 o873 0979 [086 1192 1298 i4o5 151 i6i 7. 409 i I7231,829 9 24 L4 19 0 5I 6 1 936 2042 2148 2254 2360 2466 2572 267 6 1723o i29 2936 30/207233742 40!6127841 290 296 3102 3207 3313 34I9 3525 3630 3736 io610 41, 3842 3947 4033 4159 4204 4370 4475 4581 4686 4792 100I 41 8r!14 5003 412 4897 5oo3 5io8 5213 53I 9 5424 5529 5634 5740 5845 1 z 59 60o55 6i60 6265 6370 6476 658I 6686 6790 6895 0 4i4 7000 7105 72Io 7315 7420 7525 7629 7734 7839 79443 ioil 4 5 8048j 8i53 8257 78362 66 857 8676 878 8884 8989 icS, 4i6 9093 9198 9302 9406 911 18^5 e,1 9824 9928 "~32 9 o417 62oi0136 0240 0344 0448 052 o006 jio o0L4 u968'072 10j 41 11ii76 1280 1384 1488 I592 1695 1799 1903 2007 2O110 104 419 22[Z 23iS 2421 2525 2628 2732 2835 2939 3042 3146 104 L420 623249 3353 3456 3559 3663 3766 3869 3973 4076 4179 103 421 4282 4385 4488 4591 4695 4798 4901 5004 5107 5210 io3 422 5312 5415 5518 5621 5724 5827 5929 6032 6135 6238 io3 423 634o 6443 6546 6648 675I 6853 6956 7058 716I 7263 io3 424 7366 7468' 757I 7673 7775 7878 7980 8082 885 8287 102' 425 8389 8491 8593 8695 8797 8900 9002 9104 9206 9308 102 426 941o0 9512 9613 9715 9817 9919 002I'I23 0224 0326 102 427 630428 0530 063 0733 o835 o936 io38 1139 1241 1342 102 0733 34 10235 428 I444 1545 1647 1748 1849 1951 2052 2153 2255 2356 i01 429 2457 2559 2660 2761 2862 2963 3064 3i65 3266 3367 101 430 633468 3569 3670 3771 3872 3973 4074 4175 4276 4376 0oo 43I 4477 4578 469 4779 4880 498 50o8i 5182 5283 5383 ioo 432 5484 5584 5682 578 5886 5986 6087 6187 6287 6388 ioo 433 6488 6588 6688 6789 6889 6989 70o89 189 7290 7390 100 434 7490 7590 7690 7790 7890 7990 89 I9o 82o 8389 99 435 8489 8589 8689 8789 8888 8988 9088 9188 9287 9387 99 436 9486 9586 9686 9785 9885 9984 "84'i83'283'382 99 437 64o48, o58 o68o 0779 0879 0978 1o77 1177 1276 375 99 438 1474 o573 1672 1771 1871 1970 2069 2168 2267 2366 991 439 2465 2593 2662 2761 2860 2959 3058 3i56 3255 3354 99 440 643453 3551 365o 3749 3847 3946 4044 4143 4242 4340 981 1441 4439 4537 4636 4734 4832 4931 5029 5127 5226 5324 98, 442 54221 552I 56i9 57I2 5815 5913 6oii 6iio 6208 6306 98! 443 6404 6502 66oo 6698 6796 6894 6992 7089 7187 7285 981 444JA 73831 7481 7579 7676 7774 i872 7969 8067 86 8262 i445 836o, 8458 8555 8653 8750 8848 8945 9043 9140 9237 97: 446 9335 9432 9530 9627 9724 9821 9919, i6'ii3'210 97j 447 65o3o08 0405 3502 0599 0696 o0793 8o0oo 0987 io84 ii8i I 97 4448 1 1278 1375 1472 1569 1660 1762 1839 1956 2053 2150 97 449 2246 2343 2440 2536 2633 2730 282 2923 3019 11 97 45P j65,320~ 33o9 3405 3502 3598 3695 371 I 3888 3984 4080 96; 451 4177 4273 4369 4465 4562 4I 47 4 48~ ~ 4946 5042 96 452 5138. 5235 533i 5427 5523 5619 5715 58io 5906 6002 96 453 6o081 6194 6290 6386 6482 1 6577 6673 16769 6864 6960 9 8oi54 8x70 o6[ 7247 7343 7438 753 4 7629 7725 1820 796 96 455 8oi,6 8I7 7247 77 96 9 455 801I 8107 820, 8298 8393 8 488 8584 8679 o7z4 8870 91 456 8965 go9060 9155 920 9346 9441 9536 9631 9726 9821 95 457 9916 0'11 0106 ~20o 4296 3ozI480'58i'676'771 95 458 66o8651 0960 io55 115o 1245 139 I434 ]:529 1623 1718 95 459 18131 1907 2002 2096 2191 2286 2380 2475 2569 2663 95 0. o I 2 3 4 5 6 7 8 1 D. A TABLE OF LOGARITHMS FROM 1 TO 10,000.! t i ~ ~~ ~~~~~~~~~~~i % N. o 1 2 3 4 5 6 7 8 -9 46-0 662758 2852 294- 3o4 4i 3135 3230 3324 34,8 3512 3607 94i 461 37011 3795 I3889 3983 4078 4172 4266 4360 4454 4548 941 462 4642 4736 4830 4924 5o18 5112 5206 5299 5393 5487 94 463 5581i 5675 5769 5862 56956 6o50 643 6237 633 i 6424 94 464 6518 f612 67o5 6799 6892 6986 7079 7173 7266 7160 94 465 74531 7546 7640 7733 7826 7920 8oi3 8io6 8199 829? 93 466 8386 8479 8572 8665 8759 8852 8942 9038 9131 8 922.4 3 467 9317 9410 9503 9596 9689 9752 7 9967 006o 0153,3 468 670246 0339 0431 0524 0~617 0710 0502 0895 o988 i0o8o 93 469 1173 1265 135 1451 1543 i636 1728 1821 1913 2005 93 470 672098 2190 2283 2375 2467 2560 2652 2744.2836 2929 92 471 3021 3ii3 3205 3297 3390 3482 3574 3666 3758 38 90 472 3942 4034 4126 4218 43io 4402 4494 4586 4677 476 92 473 52~~~2~ 28 5320 40 4473 461 4953 5o45 5137 5228 5320 5412 55o03 5595 5,687 92 474 5778 5870 8962 6053 6145 6236 6328 6419 6511 6602 12 1475 [6694 6785 6676 6968 7059 7151 7242 7333 7424 7516 91 476 J 607 7698 7789 7881 7972 8063 8,54 8245 8336 8427 91 477 8518 8609 8700 8791 8882 89'3 9064 9155 9246 9337 I 473 9428 9519 9610o 9700 9791 9882 9973 0063 *x54 242 91 1 479 68o3361 0426 o517 0607 0698 0789 0879 0970 io6o 1151 9 480 681241 1332 1422 i5i,,603 1693 1784 1874 1964 2055 q90 i.48, 2145 2235 2326 2416 2506 2596 2686 2777 2867 2957 90 482 3047 3137 3227 3317 3407 3497 3587 3677 3767 3857 90 483 3047 4037 4127 4217 4307 4396 4486 4576 4666 4756 go 484 4845 4932 5025 51.14 5204 5294 5383 5473 5563 5652 90 485 5742 53 i 5921 6oio 6ioo 6189 6279 6368 6458 6547 89 486 6636 6726 68i5 6904 6994 7083 7172 7261 7351 7440 89 487 529 7618 7707 7796 7886 7975 8064 8,53 8242 833i 80i I488 4290 85o9 8598 8687 8776 8865 8953 9042 9131 9220 89 489 9309 9398 9486 9575 9664 9753 9841 9930 -19 o107 489 I 97~93 9394 948611.7 8 490 69o196 0285 0373 0462 o550 o639 0728 o8i6 o9o5 0993 8' 491 ioox 1170 1258 1347 1435 1524 1612 1700 1789 1877 88. 492 1965 2053 2142 2230 2318 2406 2494 2583 2671 2759 1 88I 493 2847 2935 3023 311i 3199 3287 3375 3463 3551 3639 88 494 3727 38i5 3903 3991 4078 4i66 4254 4342 4430 4517 88 495 46o5 4693 4781 4868 4956 5044 53i 5219 5307 5394 88 496 5482 5569 5657 5744 5832 5919 6007 6094 6182 6269 8~ 497 6356 6444 653i 66i8 6706 6793 6880 6968 7055 142 87 498 7229 7317 7404 749 7578 7665 7752 7839 7926 8o04 87 499 8ioi 8x88 8275 8362 8449 535 622 8709 876 8883 499[ 96883oj8~8 82 500 698970 9057 9144 9231 9317 9404 9491 9578 9664 9751 87 501 9838 9924 jeel.'98 *i84 *271 0358 0444 *53i 0617 2502 700704 0790 0877 0963 io5o ii36 1222 1309 1395 1482 I} 503 i568 1654 1741 1827 1913 1999 20o86 2172 2258 2344 86 504 2431 2517 2603 2689 2775 2861 2947 3033 3119 3205 86 505 3291 3377 3463 3549 3635 3721 3807 3893 3979 4065 %6 506 4i51 4236 4322 4408 4494 4579 4665 475 4837 4922 86 507 5oo8 5o094 5179 5265 535o 5436 5522 56ov 563 5778 6 5o8 5864 5949 6035 62o 62d6 629 6376 646 654 6632 509 6718 68o3 6888 6974 7059 7144 7229 7315 7400 7485 85 50io 707570 7655 7740 7826 7911 7996 808i 8166 8251 8336 85 511 8421 8506 8591 8676 8761 8846 8931 9015 9100 9185 95 i12 9270 9355 9440 9524 9609 9694 9779 9863 9948'*33 85 513 710117 0202 0287 0371 0456 05.40 0625 0710 0794 0879 85 I514 I4 0963 Io48 1132 1217 i3oi i385 1470 i554 1639 1723 84 5:5 1807 i89, 1976 12060 2144 2229 2313 2397 2481 2566 84 5r 2650o 2734 288 i 2902 2986 3070 3i54 3238 3323 3407 84 517 3491 3575 3559 3742 3826 39o10 3994 4078 4162 4246 8 I~ ~ ~ ~ ~ ~ ~ ~~~~~244 84~ 51 433o 4414 4497 458i 4665 4749 4833 4916 Sooo 5084 84' 519 5167 5251 5335 548 5502 5586 5669 15753 5836 5920 84 0 I 2 3 4 5 6 7 a n s'd 1 ya, ii-r A TABLE OF LOGARITHMS FROM 1 TO 10,000. 9 N. Li1 213 4 5 6 7 8 9 D. 520 716003 6087 6170 o 6254 6337 6421 65o4 6588 6671 6754 83 521 6838 6921 7004 7088 717I 7254 7338 7421 7504 7 587 83 522 7671J 7754 7837 7920 8003 8o86 8169 8253 8336 8419 83 523 8502 8585 8668 875I 8834 8917 9ooo 9083 9165 9248 83 524 9331 9414 9497 9580 9663 9745 9828 9911 9994 **77 83 525 720159 0242 o0325 0407 0490 0573 o655 0738 6821 3903 83 526 0986 io68 iI5I 1233 i3i6 1398 i481 i563 1646 1728 82 527 i8ii i893 1975 2058 2140 2222 2305 2387 2469 2552 84 528 2634/ 2716 2798 2881 2963 3045 3127 3209 3291 3374 82 529 3456 3538 3620 3702 3784 3866 3948 4030 4112 4194 82 i 530 724276 4358 4440 4522 4604 4685 4767 4849 4931 5oi3 82 i 53 50o95 576' 5258 5340 5422 5503 5585 5667 5748 5830 82 532 5912 5993 6075 6i56 6238 632~ 64O1 6483 6564 6646 82 533 6727 68o9 6890 6972 7053 7134 7216 7297 7379 746o 81 534 754I 7623 7704 7785 7866 7948 80298110 8191 8273 8i 535 8354 8435 8516 859 7 8678 8759 8841 8922 90oo3 9084 81 536 9165 9246 9327 9408 9489 9570 965 9732 09813 9893 8i 537 9974 *'55 *136'217 298 9378 *459 *54o 0621 702 8i 538 730782 o863 0944 1024 110o5 ii86 1266 1347 1428 15o8 81 539 1589 1669 1750 i83o 19ii 1991 2072 2152 2233 2313 81 540 732394 2474 2555 2635 2715 2796 2876 2956 3037 3117 80 541 3197 3278 3358 3438 35iS 3598 3679 3759 3839 39I9 80 542 3999 4079 4i6o 4240 4320 4400 4480 4560 4640 4720 80 543 48oo00 4880 4960 5o4o 5120 52oo00 5279 5359 5439 5519 80 544 5599 5679 5759 5838 5918 5998 6078 6157 6237 6317 80 I 545 6397 6476 6556 6635 6715 6795 6874 6954 7034 7113 80 546 7193 7272 7352 7431 751I 7590 7670 7749 7829 7908 79 547 79 7 8067 8146 8225 83o5 8384 8463 8543 8622 8701 79 548 8781 886o 8939 9d18 9097 9177 9256 9335 9414 943 79 549 9572 9651 9731 9810 9889 9968 @047 *126 0205!284 79 550 740363 0442 0521 o6oo 0678 0757 o836 0915 0994 1073 79 551 1152 1230 1309 0388 1467 1546 1624 17o0 1782 i86o 79 552 1939 2018 2096 2175 2254 2332 2411 248i 2568 2647 79 553 2725 2804 2882 2961 303 3ii8 3i96 3275 3353 3431 78 554 35io 3588 3667 3745 3823 3902 3980 4058 4136 4215 78 555 4293g 4371 4449 4528 46o6 4684 4762 4840 4919 4997 78 556 5075 553 5231 5309 5387 5465 5543.5621 5699 5777 78 557 5855 5933 6onI 6089 6167 6245 6323 64oi 6479 6556 78 558 6634 6712 6790 6868 6945 7023 7101 7179 256 7334 78 559 7412 7489 7567 7645 7722 7800 7878 7955 8033 81o10 78 560 74818S 8266 8343 8421 8498 8576 8653 8731 8808 8885 77 56i 8963 9040 9118 9195 9272 9350 9427 9504 9582 9659 77 562 9736 9814 9891 9968 ee 45 eI23 20oo *277 *354 *431 77 563 750508 0586 0663 0740 0817 0894 0971 io48 1125 122 77 564 1279 356 1433 151o 1587 1664 1741 18i8 1895 1972 77 565 2048 2125 2202 2279 2356 2433 2509 2586 2663 2740 77 566 286 2893 2970 3047 3123 3200 3277 3353 343o 3506 77 567 3583 3660 3736 3813 3889 3966 4042 4119 4195 4272 77 568 4348 4425 45o.4578 4654 4730 4807 4883 4960 5o36 76 569 5112 5189 5265 5341 5417 5494 5570 5646 5722 5799 76 570 755875 5951 6027 6io3 6i8o 6256 6.332 6408 6484 656o 76 571 6636 6712 6788 6864 6940 7o16 7092 7168 7244 7320 76 572 1396 7472 748 7624 7700 7775 851 7927 8oo003 8079 76 573 8i55 8230 8306 8382 458 8533 6o9 8685 8761 8836 76 574 8912 8988 9o63 9139 9214 9290 9366 944 i 9517 9592 76 575 9668, 9743 9819 9894 9970 *'45 2I 2196 *272 *347 75 576 760422 0498 0573 0649 0724 0799 0875 0900 1025 1101 75 577 1176 1251 1326 1402 1477 1522 1627 1702 1778 i853 75 578 1928 2003 2078 2153 2228 2303 2378 2453 2529 2604 75 579 2679 2754 2829 2904 2978 3053 3,28 3203 3278 3353 75 I 2 ~ 3 4 5 6 7 8 9 D. 10 A TABLE OF LO'GARITIIMS FROM 1 TO 10,000. N.o I 2 T34 5 678 96D. 58o 63428 350 3 3578 3653 3727 38o2 1 3877 3952 4027 I 4101 75 58i 4176 4251 4326 4400 4475 4550 4624 4699 4774 4848 75 582 4923 4998 5072 5147 5221 5296 5370 5445 5520 5594 75 583 5669 5743 58i8 5892 5966 604 I 61ii5 6190 6264 6338 74 584 6413 6487 6562 6636 670io 6785 6859 6933 7007 7082 74! 585 7156 7230 7304 379 7453 7527 76o0 7675 749 7823 74 586 7898 7972 8046 8120 8194 268 8342 8416 49o 8564 74 587 8638 8712 8786 8860 8934 9008 9082 9156 9230 9303 74 588 9377 9451 9525 9599 9673 9746 9820 9894 9968 *42 74j 589 770115 0189 0263 o336 04I0 0484 0557 o631 0705 0778 74 590 770852 0926 0999 1073 ii46 1I220 1293 1367 1440 ID14 74 591 1587 166i 1734 I8o8 i88i 1955 2028 2102 2175 2248 73 592 2322 2395 2468 2542 2615 2688 2762 2835 2908 2981 73 593 3055 3128 3201 3274 3348 3421 3494 3567 3640 3713 73 594 3786 386o 3933 4006 4079 4152 4225 4298 4371 4444 73 595 4517 4590 4663 4736 4809 4882 4955 5028 S5oo 51i3 73 596 5246 5319 5392 5465 5538 56io 5683 5756 5829 5902 73 597 5974 6047 6120 6193 6265 6338 64,1I 6483 6556 6629 731 598 67n1 6774 6846 6919 6992 7064 7137 7209 7282 7354 73 599 7427 7499 7572 7644 7717 7789 7862 7934 8oo6 8079 72 60oo 778151 8224 8296 8368 8441 8513 8585 8658 8730 8802 72 6oi 8874 8947 9019 9091 9163 9236 9308 9380 9452 9524 72 602 9596 9669 9741 9813 9885 9957 ee29 1O1 173 *245 72 603 780317 0389 o461 0533 o6o 0677 0749 0821 893 0965 72 604 1037 1109 ii8i 1253 1324 396 1468 1I 5 1612 684 72 605 1755 1827 1899 1971 2042 2114 2186 2258 2329 2401 72 606 2473 2544 2616 2688 275 283I 2902 974 346 3I7 72 607 3i89 3260 3332 3403 3475 3546 36i8 3689 3176 3832 71 608 3904 3975 4046 4118 4189 4261 4332 44o3 4475 4546 71 609 4617 4689 4760 483I 4902 4974 5045 5ii6 5187 5259 71 6io 785330 540oi 5472 5543 56i5 5686 5757 5828 5899 5970 71 6ii 6o41 6112 6i83 6254 6325 6396 6467 6538 6609 6680 71 612 6751 6822 6 93 6964 7035 7106 7177 7248 7319 7390 7 6,3 7460 7531 7602 7673 7744 7815 7882 7956 8027 8098 71 614 8i68 8239 83io 838i 8451 8522 8593 8663 8734 88o4 71 615 8875 8946 901oi6 9087 9157 9228 9299 9369 9440 9510 71 616 9581 9651 9722 9792 9863 9933 ee4 74 144 0215 70 617 790285 o356 0426 0496 0567 0637 0707 0778 o848 0918 70 6x8 o988 1029 I129 1199 I269 i340 410 1480 i55o 1620 70 619 1691 1761 1831 1901 1971 2o41 2111 2181 2252 2322 70 620 792392 2462 2532 2602 2672 2742 2812 2882 2952 3022 70 621 3092 3i62 3231 33oi 3371 344i 351I 358i 365i 372I 70 622 3790 3860 3930 4000 4070 4139 4209 4279 4349 44i8 70 623 4488 4558 4627 4697 4767 4836 4906 4976 5045 5115 70 624 5i85 5254 5324 5393 5463 5532 5602 5672 5741 5Sii 70 625 5880o 5949 6o019 6088 6i58 6227 6297 6366 6436 6So5 69 626 6574 6644 6713 6782 6852 6921 6990 7060 7129 7198 69 627 7268 7337 7406 7475 7545 7614 7683 7752 7821 780 69 628 7960! 8029 98 8167 8236 1 8305 8374 8443 85i3 8582 69 629 865I 872 8789 8858 8927 8996 9065 9134 9203 9272 69 63o 799341 940 9478 9547 9616 9685 9754 9823 9892 9961 69 631 800029 0098 0167 0236 0305 0373 0442 o5ii o580 o648 69 632 0717 0786 0854 0923 0992 io6i 1129 118 266 i335 69 633 1404 1472 1541 16o9 1678 1747 85 i884 952 20211 69 634 2089 2158 2226 2295 2363 2432 2500 2568 2637 92705 6 635 2774 2842 2910 2979 3047 3ii6 384 3252 332 3389 6 636 3457 3 3294 3662 3730o 37?8 3867 3935 4003 4071I 68 637 439 4208 4276 4344 4412 44 0 4548 466 4685 4753 68] 1638~~~~41 44821 4588 -7 53! 568 63 482 9 497 5025 5093 56i 5229 5297 5365 5433 68' 639 550i 5569 5637 5705 5773 5841 5908 5976 6044 6112 68 N. o 2 3 5 6 7 8 9 D A TABLE OF LOGARlrHMS FROM 1 TO 10,000. 11 N.I 2 3 4 5 6 9 I I 64o &-6i8o' 6248 63i6 6384 645i 6519 6587 6655 6723 6790 68 b4i 6858 6926 6994 7061 7129 7197 7264 7332 7400 7467 68 542 7535 7603 7670 7738 7806 7873 7941 8oo8 8076 8143 68 643 8211 8279 8346 84i4 8481 8549 86i6 8684 875I 8818 67 644 8886! 8953 9021 9088 9156 9223 9290 9358 9425 9492 6'7 645 9560! 9627 9694 9762 9829 9896 9964 e3el 0098 0 665 646 81o233 o3oo 0367 o434 o9oi 0569 o636 0703 0770 0837 67 6417 0go4 o97i 1039 iio6 1173 1240 1307 1374 1441 i508 67 648 I275 1642 1709 1776 I843 2' 1977 2044 211I 2178 67 649 2245 2312 2379 2445 2512 79 246 2713 2780 2847 67 49 ~ ~ ~ ~ ~ ~ 2: 12579 264 278o 687 65o0 812913 2980 3047 3,114 3181 3247 33{ 4 338i 3448 3514 67 651 358i 3648 3714 3781 3848 3914 3981 4048 4114 418i 67 652 4248 4314 4381 4447 4514 458i 4647 4714 4780 4847 67 653 4913 4980 5046 5113 5179 5246 5312 5378 5445 5511 66i 55 - 5578 5644 571 1 5777 5843 5910 5976 6042 60og 6175 661 655 62 6308 6374 6440 6506 6373 6639 6705 6771 6838 66; 656 6904 6970 7036 7102 7169 7235 7301 i 7367 7433 7499 661 657 7565 7631 7698 7764 7830 7896 7962 8028 8094 8,60 66 658 8226 8292 8358 8424 8490 8556 8622 8688 8754 8820 661 659 8885 8921 9017 9083 9149 9215 9281 9346 9412 9478 66 f6o 819544 9610 9676 974, 9807 9873 9939 0*4'70 r,36 66 b66i 820201 0267 0333 0o399 0464 o53o 0595 o66i 0727 0792 66 562 o858 0924 0959 lo 1120 i86 12 317 1382 144 66 563 iI514 1579 i645 1710 1775 1841 1906 1972 2037 2103 65 664 2168! 2233 2299 2364 2430 2495 2560 1 2626 2691 2756 65' 665 2822 2887 2932 3oi8 3o083 3148 3213 3279 3344 3409 65 666 3474 3539 3605 3670 3735 38oo 3865 3930 3996 406 65 567 4126 419, 4256 4321 4386 445i 45{ 6 i 481 4646 47I' 65 568 4776 484z 4906 4971 5o36 Sioi 5i66 523i 5296 536i 65 669 5426 549 5556 5621 5686 5751 58i5 588o 2942 6oio 650 70o 826075 6140 6204 6269 6334 6399 6464t 6528 6593 6658 65 671 6,23 6787 6852 6917 6981 7046 711 7175 7240 7305 65 672 [73691 7434 7499 7563 7628 7692 7757 7821 7886 7951 65 673 8 8o8o 8144 8209 8273 8338 8402 8467 853I 8595 64 b74' 866~18724 8789 8853 89'8 8982 9046 j 9111i 9175 9239 641 675 93041 9368 9432 9497 9561 9025 9690 9754 9818 9882 64 676 9947'01 *75 139 p204 268 l332 0396 460 1525 64 6 7783o589 0653 I 02 i i0 1 6ci7 83089 o653 0717 0781 0845 0909 0973 ~1037 i2 I66 64 678 1230 1294 i358 1,422 1486 I 35o I',4 1678 1742 i8o6 64 679 187o 1934 1998 2062 2126 2189 2253 2317 2381 2442 64 68o 832209 2573 2637 2700 2764 2828 2892 295o 3020 3083 64 (68, 3'47 3211 3275 3338 3402 3456 330 33 3657 3721 64 682 3-8' 3848 3912 3975 4039 }4io3 4166 4230 4294 1 4357 64 683 A42I 4484 4548 4611 4673 4739 4802 4866 4929 4993 64 684 50o56 5120 5i83 5247 53io 5373 5437 55oo 5564 5627 63 685 5691 5754 5817 588i 59,4/ 6007 6o7 i 6134 6197 6261 63 686 6324 6387 6451 65i4 6577 6641 6704 6767 6830 6894 63 67 69571 7020 7083 7399 746 687 6927 7020 7083 7146 7210 7273 7336 7399 7462 7525 63 688 7288 2 7715 7778 741 7904 7967 8030 8093 8i56 63 6898 8219 8282 8345 840 847 83 4 97 866o 8723 8786 63 9 gi 9227 9289 9352 945 63 6go 838849 1 8912 8975 9038 91019227 4 69i 9478 954 9604 9667 9729 I 9792 9855 9918 9981 I *043 63 692 84 6 o169 0232 094 o357 420 0482 O545 8 63 693 0733 0796 0859 0921 035 42 04 o 8 0 I7 63 0984 1,o46I 1109 ii172 1234 1 297 6 694 1359 1422 1485 1547 i6io 1672 1735 1797 86o 1922 63 695 298 2047 2 2172 2235 2297 236o0 24,22 2484'2547 62 696 2009 2672 2734 2795 2859 2921 2983 3o46 3o8 3170 62 697 3233 32:5 3357 3420 3482 3544i 3o6 3669 373' 3793 62 6,8 3855 3i918 3980 4042 41o4 4i66 4229 4291 4353 44i5 6 44771 4239 46o1 46k4 4726 4788 14850 4912 4974 5o36 61.,, oI, 2, 3 i- 6 87 iD IL~~~~~~~~~~~~~~45 12 A TABLE OF LOGARITHMS FROM 1 TO i0,000 N. o 1 2 3 4 5 6 7 8 9 D. 700 845098 5i60 5222 5284 5346 5408 5470 5532 5594 5656 62 701 5718 5780 5842 5904 5966 6028 6090 6i5i 6213 6275 62 702 6337 6399 646I 6523 6585 6646 6708 6770 6832 j 6894 621 703 6955 7017 7079 7141 7202 7264 7326 7388 7449 7511 62i 704 7573 7634 7696 7758 7819 7881 7943 8oo4 8o66 8128 62 705 8189 8251 8312 8374 8435 8497 8559 8620 8682 8743 62 706 88o0 8866 8928 8989 90o5I 911 2 9I74 9235 9297 9358 6i. 70, 19 9481 9542 9604 9665 9726 9788 9849 9911 9972 6i 708 8Joo33 0095 0156 0217 0279 o340 0401 0462 0524 o585 6I 709 o646 0707 0769 o830 0891 0952 1014 75 I I36 1197 6i 710 851258 1320 I38i 1442 i5o3 1564 1625 i686 I747 1809 61 711 I870 1931 1992 2053 21 I4 2175 2236 2297 2358 2419 61 712 2480 2341 2602 2663 2724 2785 2846 2907 2968 3029 61 713 3090 3150 321I 3272 3333 3394 3455 33i6. 3077 3637 6i 714 3698 3759 3820 3881 3941 4002 4063 4124 4185 4245 61i 715 43o6 4367 4428 4488 4549 46io 4670 473I 4792 4852 6i 716 4913 4974 5034 5o95 5156 52t6 5277 5337 5398 5459 61 7 7 5519 558o 5640 5701 576I 5822 5882 5943 6003 6064 6 i 7I8 6124 6185 6245 6306 6366 6427 6487 6548 66o8 6668 60 719 6729 6789 6850 6910 6970 7031 7091! 7152 7212 7272 60 720 857332 7393 7453 75I3 7574 7634 7694 7755 7815 7875 60 721 7935 7995 8056 8116 8176 8236 8297 8357 417- 8477 6o0 7?2 8537 8597 8657 8718 8778 8838 8898 8958 9018 9078 60 723 9138 9198 9258 9318 9379 9439 949 955 619 96 9679 60 724 9739 9799 98539 9918 9978 **38 i98'158 218 278 60 725,86o338 o39 0458 o518 0578 0637 0697 0757 0817 0877 60 726 o0937 0996 io56 ii16 1176 1236 295 1355 14,5 1475 6o 727 1534 I594 1654 I714 1773 I833 1893 1932 2012 2072 6o 728 2131 2191 2251 2310 2370 2430 248 2049 2608 2668 60 729 27281 2787 2S47 2906 2966 3025 3083 3144 3204 3263 60 730 8633231 3382 3442 3501o 356i 3620 3680 3739 3799 3858 59 731 3917 3977 4036 4096 4155 4214 4274 4333 4392 4452 59 732 401 4570 4630 4689 4748 48o8 14867 4926 4985 5045 59 733 5 io4 5i63 5222 5282 5341 540oo 5459 5319 5578 5637 59 734 56961 5755 5814 5874 5933 5992 6o51 6IIo 6169 6228 59 735 6287 6346 6405 6465 6524 6583 6642 6701 6760 6819 59 736 6878 6937 6996 7055 7114 7173 7232 7291 7350 7409 59 737 7467 7526 7085 7644 7703 7762 9821 939 59 738 8o56 8115 8174 8233 8292 8350 409 8468 8527 8586 59 739 8644j 8703 8762 8821 8879 8938 8997 9056 9114 9173 59 740 869232 9290 9349 9408 9466 9525 9584 9642 9701 9760 59 74I 9818 9877 9935 9994 0*53 1i1 |170 9 228 9287 34 5 9 742 870404 0462 o52 1 0579 o638 0696 0755 o813 o872 J0930 5 743 09891 I47 |io6 ii64 1223 1281 1339 1308 1456 Ij55 58 744 I573 I631 I69O I748 I806 I865 1923 1I 2040 1 2098 58 745 2156 2215 2273 2331 2389 2448 25306 264 2622 12681I 58 746 2739 2797 2855 2913 2972 3030 3o083 3146 3204 3262 58 747 3321 3379 3437 3495 3553 361i 3669 i 2727 3785 3844 58 74 3902 3960 4oi8 4076 434 4192 4250 1 4303 4366 4424 58 749 44821 4540 4598 1 4656 4714 4772 4830 4888 4945 500oo 58 750 875061I 5ii 5177 5235 5293 5351 5409 5466 5524 5582 58 75I 564o 569 5756 5813 587I 5929 5987 6045 6102 616o 58 752 6218i 6276 6333 6391 6449 6507 6364 6622 6680' 6737 58 753 6795 6853 69i o 6968 7026 7083 7141 7199 7256 7314 58 754 737 7429 7487 7544 7602 7659 7717 7774 7832 7889 58 755 79471 8004 8062 8119 8177 8234 8292 8349 8407 8464 57 756 85322 8579 8637 8694 8752 8809 8866 8924 1 8981 9039 57 757 9961 9153 9211 9268 9325' 9383 9414o 9497 955 9612 37 |58 9669 9726 9784 9841 9898 9956 @0I3 |*70 127 ~I8 5 759 8802421 0299 o0356 o413 o471 0528 oS0585 0642 0699 0756 57.. 1 2 3 4I V D. A TABLE OF LOGARITHMS FROM 1 TO 10,000. 18 N. i~~~~~~~~~~~~~~. N. o I 2 3 4 5 7 3 D. 760 88o0or4 0871 0928 0985 1042 99 I1 xi6 1213 1271 i328 57, 761 1385 1442 499 i 556 i6i3 1670 I727 1784 i84t 1898 571 762 1955 2012 2069 2126 2183 2240 2297 2354 2411 2468 57 763 2325 2581 2638 2695 2752 2809 2866 2923 2980 3037,7 764 3093 3i5o 320 3264 3321 3377 3434 349 3548 36o5 5 765 366i 3718 3775 3832 3888 I 3945 4002 4059 4115 4172 571 766 422 485 4342 4399 4455 4512 4569 4625 4682 4739 57 767 479 48 4909 4965 5022 5078 i350 5192 6248 53 57j 768 536i 5418 6474 5331 5587 5644 5700 5757 583 5870 571 769 5926 5983 6039 6096 6152 6209 6265 6321 6378 6434 56 77 88649! 6547 6604 666o 6716 6773 6829 6885.6942 6998 56 77 7034 7111 I 7167 7223 7280o 7336 73[2 7449 7505 756, 56 772 617 7674 7730 7786 7842 7898 -5 8oI 8067 8I23 561 773 8179 8236 8292 8348 84041 8460 851 i6 8573 8629 8685 56 774 8741 8797 8853 8909 8065 9021 9077 9134 919o 9246 56, 773 9302 9358 9414 9470 9526 9582 9638 9694 97o30 9806 561 770 9862 99'8 9974 I'3o *086 I141 1I 7 253'3oq j 365 56t 777 89o421 o477 o533 0589 0645 0700 0756 0812 o868 o0924 56J 778 0980 0io35 o1091 1147 1203 1259 i314 1370 1426 1482 56! 779 i537 1593 649 1705 1760 1816 1872 I928 983 2039 561 780 892095 2150 2206 2262 2317 2373 2429 2484 2540 2595 561 781 2651 2707 2762 2818 2873 2929 298 I 3o4o 3096 3i5i 561 782 3207 3262 33i8 3373 3429 3484 3 4o 355 365i 3706 56 783 3762 3817 3873 3928 3984 4039 4094 40 4205 426 551 784 4361 4371 4427 4482 4638 4593 4648 4704 4759 4814 55i 785 4870 4925 4o8o 5o36 509 5146 5201 5257 5312 5367 55 786. 5423 5478 533 5588 5644 5699 5754 5809 5864 5920 55 787 5975 6o3o 6o85 6140 61956 23 63o6 636i 6416 6471 55 7~ ~ ~ ~ ~ ~~~~~~~~~~~~~889 6857 6912 6972 788 6526 658r 6636 6692 6747 6802 857 692 667 7022 55i 789 7077 7132 7187 7242 7297 7352 7407 7462 717 7572 55 790 897627 7682 7737 7792 7847 7902 757 8012 8067 8122 55 7957 55; 791 8176 8231 8286 834' 8396 843' 85o6 856i 86i5 8670 55 792 8725 8780 8835 8890 8944 8999 9054 9109 9164 9218 55 793 9273 9328 9383 9437 9492 9547 9602 9656 97 766 55 194 9821 9875 993o 9986 *.39 *-94!49.23 *258 0312 561 19 9oo367 0422 o476 o53 o586 0695 0749 0804 0859 55 193 9003671 0~22 e~;jg ~~94 ~I30476 8 124 129 0 55 96 o0913 o968 1022 1077 J3i ii86 1240 129 i 1349 1404 55 8197 i45,8 53 1567 I622 676 73 785 i84o 894 1948 54 798 2oo003 2057 2112 2166 222I1 2275 2329 2384 2435 2 2492 5 799 2547 26o01 2655 2710 2764 2818 2873 2927 298 3o6 54 80oo 903090 344 3199 3253 3307 3361 34i6 3470 3524 3578 541 Sol 3633 3687 3741 3795 3849 3904 3958 4012 4o66 4120 541 802 4174 4229 4283 4337 4391 4445 4499 4553 4607 4661 54 W3 4761 4770 4824 4878 4932 486 5040 5094 5148' 5202 54 804 5256 53io 5364 54i8 5472 5 26 558@ 5634 5688 5742 54 o 5796 5850o 5904 5958 6012 6o66 6119 6173 6227 6281 54 86 6335 6389 6443 6497 655i 6604 665 6712 6766 6820 54 8 6874~ 6927 6981 7I5 7089 7143 7196 7250 7304 7358 54 7411 7465 7J 73 7626 7680 7734 7787 784 i 7895 54 809 7949 8002 8056 Io 8163 8217 8270 8324 8378 843i 5 8io 908485 8539 8592 8646 8699 8753 8807 886o 8914 8967 54 811 9021 9074 9128 9181 923 9289 9342 9396 9449 903 54 812 95561 96o10 9663 9716 9770 9823 9877 9930 9984 37 53' 83 910091 0144 0197 0251 o304 o358 0411 o464 o5 8 0571 53 814 0624 o78 0731 0784 o838 o8q9 0944 o O8 io5i Iio4 53 85 ii8 1211 1264 1317 1371 1424 1477 I530 i584 1637 53O 8 i6 169o 1743 1797 i85o 1903 1956 2009 2063 2116 2169 53 i~~~~~~~~~~~~ 2700 53 817 2222 2275 2328 238i 2435 2488 2541 2594 2647 537 8~ 8 2753, 2806 2,859 2913 I 2c66 30i1 3072 3125 3178 3231 53 89 53Bi 3841 3337 J3390 3/443 3496 3 z 62 3655 13708 [376i 53 N. J I 3 6 7 8 I 01 ii 4)5 i 14 A TABLE OP LOAJ'ITHtMS FROM 1 TO 10,000. N.T o I 2 3 4 5 6 7 8 D ii;: 820 913814 3867 3920 3973 4026 4079 4132 4184 4237 4290 53 821 4343 43(6 4449 4302 4555 4608 4660 4713 4766 4819 53 822 48721 4925 4977 5o3o 5083 536 5189 5241 5294 5347 53 823 5400 5453 55o5 5558 56ii 5664 5716 5769 5822 5875 53 824 5927 5980 6o33 6o85 638 6191 6243 6296 6349 64oi 53 825 6454 6507 6559 6612 6664 6717 6770 6822 6875 6927 53 826 6980; 7033 7083 7138 7190 7243 7295 7348 7400 7453 53 1 827 7306. 7558 7611 7663 7716 7768 7820 7873 7923 7978 52 1828 8o3o 8o83 8i35 8i88 8240 8293 8345 8397 845o 802 5 829 8555i 8607 8659 8712 8764 88i6 8869 8921 8973 9026 52 83o 919078 9130 9183 9235 9287 9340 9392 9444 9496 9549 52 831 9601 9653 9706 9758 98o10 9862 9914 9967 ",9 I 71 52 832 1920123' 0176 0228 0280 0332 o384 0436 0489 o541 0593 52 833 0645: 0697 0749 o8oj o853 0906 0958 1010 1062 1114 52i 834 1166' 1218 1270 1322 1374 1426 1478 i53o 1582 1634 52 835 i686i 1738 1790 i842 1894 1946 1998 2050 2102 2i54 5, 836 2206! 2258 2310 2362 2414 2466 2t18 2570 2622 2674 52' 837,725 2777 2829 2881 2933 2985 3037 3089 314o 3192 52 838 3244' 3296 33448 3399 345i 35o3 3555 3607 3658 37o10 52 839 4038i4 3865 3917 3969 4021 4072 4124 4176 4228 52 840 924279!' 4331 4383 4434 4486 4538 4589 4641 693 4744 52 84[ 4790, 4848 4899 4951I 5oo3 5o54 5io6 5157 5209 5261 52 842 5312 5364 5415 5467 55i8 5570 5621 5673 5725 5776 52' 843 5828 5879 5931 5982 6034 6o85 6137 688 6240 6291;5879 6i~~~~~~78862oj29 844 6342; 6394 6445 6497 6548 66oo i 665i 6702 6754 6805,1 845 6857j 6908 6959 7011 7062 71!4 7165 7216 7268 7319 31 846 7370; 7422 7473 7524 7576 727 7730 1 778I 7832 51 847 7883 7935 7986 8037 8o88 8140o 91 8242 8293 8345 51 848' 8396 8447 8498 8549 86oi 8652 8703 8754 88o5 8857 5S 849 89o8 8959 9oio 9o61 9112 9163 9215 9266 9317 9368 51 850 92941 9470 952 9572 9623 9674 9725 9776 9827 9879 51 85i 9930 998i 032i 083.I34'i85 *236 *287 338 @389 51 852 930440 0491 0542 0592 0643 0694 0745 0796 0847 0898c 51 853 0949 1000 o105i 1102 153 1204 1254 13o5 I356 1407 i 854 1458 15og i6o i6io i1661 1712 1763 1814 i865 915 5-? 855 1966 2017 2068 2118 2169 2220 2271 2322 2372 2423 51 856 2474 2524 2575 2626 2677 2727 2778 2829 2879 2930 51 857 29811 3o3 302 333 3183 3234 3285 3335 3386 3437 51 858 3487 3538 3589 3639 3690 3740 3791 3841 3892 3943 51 859 3993 4044 4094 4145 4195 4246 4296 4347 4397 4448 51 86o 482 902 493133 9386 4 498 4549 4599 4650 4700 4751 1 48o01 4852 4902 4953 50 861 5o003 5054 5104 5154 5205 5255 5306 5356 5406 5457 50 862`507 5558 56o8 5658 5709 5759 5809 586o 59io 5960 50o 863 o'iiI 6o6I 6Iii 6162 6212 6262 633 6363 643 6463 50' 864. 6514 6564 6614 6665 6715 6765 6815 6865 6916 6966 50 865 io~ 4o65oi6 6865 780161 7066 7117 7167 7217 7267 7317 7367 7418 7468 50 866 718 7568 76i8 7 668 7718 7769 7819 7869 799 7969 i 867 8019o1 8069 19ig 816 I 8219 8269 8320 8370 8420 8470 5o 868 8520' 8570 8620 8670 8720 8770 8820 8870 8920 8970 50 869 8 6920,'8930 96 0 i 869 1 90201 9070 9120 9170; 9220 I 9270 9320 9369 9419 9469 So? 27 35g96? 961g966 5 0 939519 9569 9619 9069 9719 9769 9819 9869 9918 9968 9 7I 940018 oo68 oii8 o6 0218 0267 0317 0367 0417 o0467 5o? 872 o516 o566 o6i6 o666 o0716 0765 o815 o865 o915 o964 Soi J873 j oil104 18jo103i6 873 1 014 064 1114 i63 I1213 1263 1313 I362 1412 I462 SOI 1 874 1 ii i56i i6ii i66o 6O 1809 1859 1909 1958 5oj, 875;2008 2058 2107 2157 2207 2256 2306 2355 2405 2 4S 1 So 876' 2504 2554 26o3 2653 2702 2752 280I 2851 2901o 2920 50 877 30001ooo 3049 3099 3148 3198 3247 3297 3346 3396 3445 49 878 3495 3544 3593 3643 3692 3142 3791 384I 3890 3939 491 879 3989 4038 4088 4137 4186 4236 428S 4335 4384 4433 49? 1 — - -.. o' ___ ___2 _ - _-t __ L(,Ij 01 4 8 43 8 11. L,I I i t I I2 3 A TABLE OF LOGARITHAMS FROM 1 TO 10,000. 15 N.o0 112 3 4 5 6 7 8 9 D. 88o0 9 4483 4532 458I 463i 4680 4729 4779 4828 4877 492 49 881 4976 5025 5074 5124 5173 5222 5272 5321 5370 5419 49 882 5469 55i8 5567 5616 5665 5715' 5764 58,3 5862 5912 49 883 5961 6oio 6059 6io8 6157 6207 6256 6305 6354 640o3 49 884 i 64i5 65oi 655i 66oo0 6649 6698 6747 6796 6845 689.4 49 885 6943& 6992 7041 7090 7'40 7I89 7238 i 7287 7336 7385 49 4~~~~~~~~~~~~3 7328 777 886 1 7434 7483 7532 758I 7630 [7679 7728 7777 826 7875 49 887 7924 7973 8022 868070 819 7 683 7924 ~ ~~~ 82,7o ig 8 8217 831 836.j49 I 888 8413 8462 85ii 856o 8609 8657 8706 8755 8804 8853 49 889 8902 8951 8999 9048 9097 9146 9195 9244 9292 934I 49 890 949390 9439 9488 9536 9585 9634 9683 973, 9780 9829 49 ~~~~~9536 955 96349 891 7b 9926 9975 0024 **73 012I *170 e219 267 *3i6 49 892 950365 0414 0462 o51I o56o o6o8 0657 o706 0754 oSo3 49 893 1 o85 0900 0949 o0997 46 IO95 143 1192 1240 1289 49 0997 I32JI7 894 1338 i386 1435 1483 1532 iS8o 1629 1677 1726 772 49 895 1823 1872 1920 1969 2017 2066 2114 2163 2211 2260 4 8~~~~~~~~~~67 296 27S ~ 896 2308 2356 2405 2453 2502 2550 259 2647 2696 2744 48 897 2792 2841 i 2889 2938 2986 3o34 3o83 33 3 i8 3228 48 898 3276 3325 3373 3421 3470 3518 3566 3615 3663 3711 48 899 3760 380o8 3856 39o5 3953 401 4049 4098 446 4i94 48 9o0 954243 4291 4339 4387 4435 4484 4532 4580 4628 4677 48 901 4725 4773 4821 4869 4918 49(6 501oi4 5062 51io 5i58 48 902 5207 5255 5303 5351 5399 5447 5495 5543 5592 560 48 903 5688 5736 5784 5832 5880o 5928 5976 6024 6072 6120 48 904o 6168 6216 6265 6313 636I 6409 6457 6505 6553 6601 48 905 6649 6697 6745 6793 6840 6888.6936 6984 7032 7080 48 906 7126 7176 7224 7272 7320 7368 7416 7464 7512 7 48 907 7607 7655 7703 7751 7799 7847 7894 7942 7990 8o38 48 9o8 8o86 8134 8i8r 8229' 8277 8325 8373 842 8468 85i6 48 909 8564 8612 8659 8707 8755 8803 8850 8898 8946 8994 48 910 95904, 9089 9137 9185 9232 9280 9328 9375 9423 947' 48 911 9518 9566 9614 9661 9709 9757 9804 9852 9900 997 48 912 9995 *142 90 g 138 ei85 233 e280 *328 376 423 48 913 96047' 0518 0566 o613 o066i 0709 0756 0804 oSi 0899 48 914 0946 0994 04, IoQ89 ii36 ii84 1231 1279 326 374 47 915 1421 1469 1516. i563 J6ii i658 1706 1753 18oI 1848 47 916 I 95 1943 I1990 2038 2085 2132 2180 2227 2275 2322 47 917 2369 2417 2464 2511I 2559 2606 2653 2701 2748 2795 47 918 2843 2890 2937 2985j 3032 3079 3126 3174 3221 3268 47 g919 34i6 3363 34io 3457 3504 3552 3599 3646 3693 3741 47 920 963788 3835 13882 3929 3977 4024 4o71 4,,8 4i65 4212 47 921 4260 4307 4354 440o 4448 4492 4542 4590 4637 4684 47 922 4731 4778 4825 4872 4919 4966 50oi3 5o6i 51io8 5 47 923 5202 5249 5296 5343 5390 5437 5484 5531 5578 5625 47 924 5672 5719. 5766 5813 586o 5907 5954 6ooi 6048 6095 47 925 6142 6189 6236 6283 6329 6376 6423 64 6517 6564 47 926 k6i1 6658 6705 6752 6799 6845 6892 699 6 47 689 9 9 7~ 4 927 7080 7127 7173 7220 7267 7314 7361 7408 7454 7501 47 28 7548 7595 7642 7688 7735 7782 7829 7875 7922 7969 47 929 8o016 8062 8Iog 8156 8203 82 8296 343 8390 8436 47 93o 968483 8530 8576 8623 8670 8716 8763 881o 8856 8903 47 93I 8495 8996 9043 9090 9136 9183 922 9276 9323 9362 47 932 9416J 9463 9509 9556 9602 9649 969 9742 9789 983D 47 933 9882J 9928 9975 e21 0068'D 4 Di6i 0207'254 03oc 47 933 934 97o347 0393 0440 0486 0533 0579 0626 0672 0719 0765 46 935 o3812 o858 0904 095] 0997 1o44 log 137 i 83, 1229 46 936 12761 1322 i369 415 0.962 io8 1554 60, I647 i693 46 937 14,786 1832 1879 195 1971 2018 2064 21IO 2157 46 i938 22031 2249 2295 2342 2388 243/ 2481 2527 2573 2619 46 1 939 26661 2712 i 2758 2804 28511 2897 2943 2989 1 3035 3082 46 N. o 3 4 5 6 7 8 9.. 26 [6 A TABLE OF ILOGARITHMS FROM 1 TO 10,000. N. O, - -— I94 0 9 8 17 3 4 T 6 L 7_ 8 9 1). 940 97328 374 322 3266 3313 3359 34o5 35I 3497 3543 46 941 3590 3636 3682 3728 3774 3820 3866 3913 3909 4005 46 942 4o5i 4097 4143 4189 4235 4281 4327 4374 4420 4466 46l 943 4512, 4528 4604 4650 4696 4742 4788 4834 4880 4926 46 944 49721 5o i 5o64 5,10 5i56 5202 524t8 5294 534o 5386 46 945 543 2 5478 5524 5570 56i6 5662 5707 5753 5799 5845 42 946 5891I 5937 5983 6o2q 6075 6121 6167 6212 6258 6304 46 947 635oj 6396 6442 6488 1 6533 65279 6625 6671 6717 6763 46 948 68o8 6824 69oo 6946 6992 7037 7o83 7129 712 7220 46 949 7266 7312 7358 7403 7449 7495 7541 7586 7632 7678 46 950 977724 7769 7815 7 861 76 7952 7998 8o43 8 8 46 95i 8i8i 8226 8272 8317 83631 8409 04 85oo 8546 8591 46 951 ~ ~ ~ ~ ~ ~ ~~80 85S44 ioo 8546'859, 40 I952 8637 8683 8728 8774 88I9 8865 8911 8956 9002 9047 46 953 9093 9138 9184 9230o 9275 9321 i 9366 9412 9457 9503 46 954 9548 9594 9639 9685 9730 o 977 9821 I 9867 9912 9958 46 955 980003 00oo49 0094 i014o0 oi85 023i 0276 0322 0367 0412,45 956 o458~ o5o3 o0549 0594 o64o o685 0730 o776 0821 0867 45 957 0912 0957 ioo3 iO48 1093 1139 ii84 I1229 1 275 1320 45 958 1366 1411 i I456 15o01 1547 1592 1 637 i 683 1728 1773 45 959 18 1864 1909 1954 2000'2o45 2090 I 23 2181 2226 45 959~~~~1(5 2o0o 204 960 982271j 2316 -2362 2407 i 2452 2497 2543 2588 2633 2678 45 96I 2723' 2769 2814 2859 V904 2949 2994 3o4o 3085 330o 45 962 3175 3220 3265 33io 3356 34o 3446 3491 I3536 3581 45 963 36261 3671 3716 3762 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 4032 45 966 4977 522 5067 5112 5157 5202 5247 5292 5337 5382 45 967 5426i 5471 55i6 556i 56o6 5651 5696 574I 5786 583o 45i 968 58751 5920 5965 6o010o 6055 6100 6144 689 6234 6279 45 9~~~~~~~~~~I619 I3 62792 969 6324 6369 6413 6458 65o3 6548 6593 6637 6682 6727 42 970:9867721 6817 686i 6906 6951 6996 7040 7085 7130 7'7 45 971 72191 7264 7309 7353 7398 7443 7488 7532 7577 7622 45 972 7666 771I 7756 7800 7845 7890 7934 7979 8024 8o68 45 973 8ii3 8157 8202 8247 8291 8336 838i 8422 8470 8514 45 974 8559 8604 8648 8693 8737 8782 8826 8871 8916 8960 45 975 90o5 9049 9094 9138 9183 9227 9272 9316 9361 9405 45 976 94 94349539 9583 9628 9672 9717 9761i 9806 9850 44 98 95 99 9 9983 028 ~~72 "I17 /i6i ]2o6' 250 I294 44 9 97 39 o3 03o428 0472 o5i6 o56i o6o5 o65o 0694 0738 44 979 0783 0827 0871 0916 0960 1004 1049 1093 1137 1182 44 980 991226 1270 i3iS 1359 1403 1448 1492 i536 158o 1625 44 981 1669 1273 1758 i802 1846 189o J935 1979 2023 2067 44 982 2111, 2156 2200 2244 2288 2333 277.2421 2465.2509 44 983 2554! 2598 2642 2686 2730 2774 28i9 2863 2907 2951 44 984 29951 3039 3083 3127 3172 3216 3260 33o4 3348 3392 44 985 34361 3480 3524 3568 363 3657 3701 3745 3789 3833 44 986 3877 3921 3965 4009 453 4097 414I 4185 4229 42j3 44 987 4317 4361 4405 4449 4493 4537 4581 4625 4669 4713 44! 988 4757 48oI 4845 4889 4933 4977 5021 5o65 io8 5152 44. 198 56 5240 5284 5328 5372 54i6 546o 5504 5547 5591 44 99o 995635 5679 5723 5767 58ii 5854 5898 5942 5986 6030 441 991 6074 6117J 616i 6205 6249 6293 6337 6380 6424 6468 44 992 6512 6555 6599 6643 6687 6731 6774 68i8 6862 6906 1 441 993 6949! 6993 7037 7080 7124 7168 7212 7255 7299 7343 44 994 7386 7430 7474 7517 7561 7605 7648 7692 7736 7779 44i 995 7823' 7867 7910 7954 7998 8041 8085 8129 8172 8216 44! 996 8259 83o3 8347 8390 o 8434 8477 8521 8564 8608 8652 44j 997 8695 8739 8782 8826 8869 8913 8956 9000 9043 9087 441 998 9131 9174 9218 9261 9305 9348 9392 9435 9479 9522 44j 999 9565 9609 9652 9696 9739 9783 9826 9870 993 9957 43 N. 0 2 3 4 5 6 7 8' D A TABLE OF LOGARITHMIC SINES AND TANGENTS FOR RVERY DEGREE AND MINUIE OF THE QUADRANT.,REMARK. The minutes in the left-hand column of each page, increasing downwards, belong to the degrees at the tAop; and those increasing upwards, in the right-hand 3olumn, belong to the degrees below. 18 (0' DEGREES.) A TABLE OF LOGARITHMIC M. Sine D Comine I). Tang. D. Cotang. o o.000000 ooooo000000 0000000 Infinite. 6o 6.463726 5017.17 0oooooc 0oo 6.463726 50oi7.I 7 13.536274 5 2 764756 9 93'~85 000000 00 764756 2934.83 2352444 3 9084 2082.3 oooooo.00oo 940847 2082.31 059153 57 4 7'065786 i615.17 ooooo oo 7.065786 I615.i7 12-934214 56 5 j i62696 i319.68 0000oo0 -oo 162696 319.69 837304 55 241877 1II5.75 9'999999.o01 241878 1I1.7 75 8122 54 308824 6653 999999 o 30825 96.53 691175 53 3668i6 5254 99999 366817 5254 633i83 52 9 417968 762 -63 999999 01 417970 762.63 582030 51 o10 463725 689.88 99099 01 463727 689.88 536273 5o 11 7-50518 629.81 9.999998'01 7.505120 629.81 12.494880 49 I12 5542906 579.36 999997'01 542909 579-33 47091 4 13 577668 536.41 999997'01 577672 536.42 422328 47 14 609853 499.38 999996 i 609857 499.39 3go943 46 15 6398i6 46714 999996 -0o 639820 467.I2 36oi8o 45 I6 667'45 438.89 99995 *0I 667849 438.82 332151 44 17 694173 413.72 999995 01 694179 4i3.'3 305821 43,8 718997 391.35 999994 01 71900oo4 39I.36 280997 42 19 742477 3727 999993 oi 742484 371.28 257216 41 20 764754 3535 999993 0 764761 351.36 235239 40 21 7?785943 33672 9.999992 7785951 336.73 12.214049 3 22 806146 321.75 999991 8o6i55 32176 I9384 3.j23 8 2545I il~0 123 8252451 3o8.o5 99998990 0 82546o0 308.o6 I74540o 3 9999?o 11r 24 8/3934 I224 99998 02 843944 2 5.49 I56o56 36 25 861662 2 8 999988 02 861674 2 90 38326 35 26 87865 273.i7 999988 *02 878708 273.18 121292 34 27 895o5D 263.23 999987 -02 895099 263.25 o49o 33 28 91o879 253.9 999986.02 910894 254.o 0o89106 32 9 926119 999985 02 926134 245-40 073866 3o 940842 237.33 999983.02 940858 237.35 o59142 3o 31 7.955082 22980 9'999982 02 7.955o100o 229.81 12.044900 29 32 968870 22273 999981 02 968889 222.75 031111 28 33 982233 216.08 999980 02 982253 216.0o o?747 27 34 995i?8 209.81 999979 02 995219 209.83 004781 26 36 82007727 21931 999977 02 800oo7809 203 92 II9929 25 36 o020021 98.i 999976 02 020042.33 97095 24 03i i9 1 3.02 999975 02 o3i 945 I305 96055 23 37 ~399999975 ~ 3194 38 o43501 88 oI 999973 o02 43527 I 88.o3 956473 2 39 054781 183.25 999972 02 054809 183.27 94591i 2 40 065776 78-72 999971 02 o658o6 178. 74 934194 20 4 o765o0 74 9. 999969 02 8076531 I74.44 I.923469 42 086965 170.31 999962 02 o086997 170.3 913oo003 I 43 097183 166.39 999966 02 097217 16642 902783 I i 44 o107t67 62-6 999964 o3 107202 162.68 279 45 16926 1599-o8 99963 o3 116963 159. 883o7 15 46 I26471 155.66 999961 I03 12651o i5 68 873490 14 47 i358io 152.38 999959 03 i3585i 15241 864149 i 48 144953 149.24 999958 o3 4496 149.27 85500oo4 12 49 153907 146.22 999955 153952 46.27 846048 II 50 162681 143'33 999954 I03 162727 43.36 837273 10 5i 8.171280 i40'54 9.999952 1o3 8.171328 14057 11.828672 52 179713 37.86 999950 I3 179763 137.90 820237 53 187985 135.29 999948.o3 i88o36 i 35.32 811964 54 196f22 132.80 999946.o3 ig9656 132,84 803844 6 55 204070 [30o.41 999944.o3 204126 30.44 5874 56 211895 128.o10 999902 04 2I953 I28.I4 788047 4 57 219581 125.87 999940.04 219641 125.90 78o359 3 I5 22704 12372 ~ 999938 04 227195 r23.76 772803 2 59 234557 121.64 999936.04 234621 1 216 68 765379 1I 6o 241855 119.63 999934.04 241921 __67 l758o Coi - e'- i__ __ _ _ _ __ _ _ _ _i_- - I - 1 - - - Cosine 1). Sine Cotang. D5. Tang9. M. (89 DEGREES.) SINES AND TANGENTS. (I DEGREE.) 19 Sine D. Cosine D. Tang. D3. Ci tnncg. 0 8.241855 II9.63 9.999934.04 8.24192I 119.67 II 158779 6o 249o033 I117.68 999932 -o04 249102 117.72 750898 59 2 256094 II5.8o 999929.04 256165 i5.84 5'43835 5 3 263042 113.98 999927 504 263115 1402 736885 57 269881 112i.21 999925 *o4 269956 I[,.o425 56 5 276614 iIo.o50 999922 04 276691 110o.54 723309 55 6 283243 i o8.83 999920.04 283323 Io8.87 716677 54 7 289773 107.21 999918.04 289856 10o7.26 7101144 53 8 296207 Io5.65 999915.04 296292 i10o.70 703708 52 9 302546 io4.i3 999913.o04 302634 10o4.i8 697366 51 io 308794 102.66 999910.04 308884 102.70 691116 50 II 8.3i49o4 I101.22 9.999907.04 8.315046 1o0.6 I r684954 49 12 32o'07 99.82 999905.04 321122 9987 678878 3i 327o06 98.47 999902 04 327114 98.51 672886 47 14 332924 J97' 14 J999899.0o 333025 J97.19 666975 46 15 338753 95.86 999897.oS 338856 95.90o 66i144 45 i6 3445o04 94.60 999894.05 3446io 94.65 655390 44 I 35oi8i 93.38 9998.o5 350289 93.43 649711 43 I 8 355783 921 9998?1.05 355895 92.24 644'05 42 19 3613x5 10 999885 05 361430 J9-o8 638570 41 20 366777 49.90 999882 5 366895 9.95 633,05 40, 8.372171 88.80 9.999879.o5 8.372292 88.85 ii.627708 3 21 22 377499 87.72 999876.05 377622 87.77 622378 38 23 382762 86.67 999873.05 382889 86.72 6171Ii 37 24 387962 85.64 999870.o05 388o92 85.70 611908 36 25 393101i 84.64 999867.05 393234 84.70 6o6766 35 26 398179 83.66 999864.05 398315 83.71 6oi 685 34 403199 82.71 99986.o05 4o3338 82.76 596662 33 408161 8I.77 999858.o05 408304 8.82 59696 32 29 4i3o68 8o86 999854.05 413213 80o.9 586787 3i 3o 417919 79.96 999851.o6 4i8o68 80~02 581932 30 31 8.422717 7.09 9.999848.06 8.422869 79'4 II11.577I3 1 29 32 427462 7823 999844.o06 42768 7 8.30 572382 2b 33 432156 77.40 99841.o6 432315 77.45 567685 27 34 436800 76.57 999838.o06 436962 76.63 563038 2z 35 441394 75.77 999834,o6 441o60 75.83 558440 25 36 445941 74,99 99983j.o6 44611o 75.05 553890 24 37 450440 74.22 999827 6 45o6i3 748 549387 23 454893 73.46 999823.o6 455o070 73.52 544930 22 39 4593o1 72.73 9982o 0.06 459481 72.79 54o519 21 40 463665 72.00 999816.o6 463849 72.06 53651 20 41 8.467985 7. 29 9.999812.o6 j 8.468172 71/ 35 ii.53i828 9 42 4'2263 70.60 99980.o6 472454 70.66 527546 1 ~43'~476498 69- 9990 o6, 476693 69.98 523307 17 44 48 9998i 0693 0 6 480892 5908 45 484848 5 999797.'7 48505 665 5I490 IS 46 488963 67.94 999793 807 49170 68 5 14 47 493040 67.3 99970 493250 6738 5o675o 13 48 9 99~ [.07~6.6 70 48 497078 66.69 9997 07 49793 6676 502 49 501080 66.o8 999782 07 5o1298 66.i 498702 11 o50 505045 65.48 999778.07j 0o267 65555 494733 1o 51 8.5o8974 64.89 91999774.07 8.So9r o 64-96 ii.o8oo 52 512867 64.3i 999769.07 53o8 6439 486902 53 516726 63.75 99976.07 516q6I 63.82 483039 7 54 520551 63.I9 999761.07 520790 63.26 4792I0 6 55 524343 62-64j 999757'07 524586 62.72 475414 5 56 528102 62.11 999753'07 528349 62.18 471651 4 57 531828 6i.58 999748'07 532080 6i.65 467920 3 - 535523 6x.o6 999744.0o 535779 6i'I3 464221 2 5 539186 60o.55 999740.07 539447. 60.62 460553 ~'25459 9997)99 6o 5428I9 60.04 999735.07 543084 60.12 456916 -L (',Cosine D. Sine i Cotang. ID. f Tang (88 DIEGRZBS.) 20 (2 DEGREES.) A TABLE OF LOGARITHMIC M. Sine D. Cosine D. Tang. D. Cotang.| 0 8.542819 6o-o4 9*999735.07 8*543084 6c. 12 11.4569I6 6o 1 546422 59.55 999731.07 54669I. 59.62 453309 2 54q092| 59.06 999726.0, 550268 591- I4 449732 58 3 553539 58-58 999722 o 08 553817 58166 446I83 57 4 557054 58. i 999717 *o8 557336 589 442664 6 5 560540 57-65 999713 o 560828 57.73 439172 55 6 563999 57.,9 999708 o8 56429I 5727 435709 54 7 56743,1 56.74 999704.08 567727 56.-8 432273 53 8 570836 5630 999699 i08 571137 5638 428863 52 9 5742I4 55.87 999694 08 574520 55. 5 425480 51 tO I 577566 55.44 999689 o8 1 577877 55.-2 42223 50 |I 8-58(892 55-02 9.999685 *o8 8-58I208 55.Io I414I8792 4q 12 584193 54.60 999680.o8 584514. 54.68 415486 48 13 j 587469 54-19 999675.o8 587795 54.27 412205 47 14 i 590721 53-79 999670 o8 591051 53.87 408949 46 15 593948 53.39 999665.o8 594283 53.47 405717 45 I6 597152 530oo 999660 o8 597492 53.o8 402508 44 17 600332 52.61 999655 6o8 600677 52.70 399323 43 18 6034i9 52-23 999650 o8 603839 52.32 396161 42 19 606623 5i.86 999645.0o9 606978 51.94 393022 41 20 609734 51.49 99964o.o9 6I0094 51 58 389906 40 21 8.612823 5I-I2 9.999635.09 8-613I89 5I.21 iI-3868Ix 39 22 615891 50.76 999629 *09 616262 50o.85 383738 38 23 618937 50.4I 999624.o9 6193f3 50.50 380687 37 24 621962 50o.o6 999619 0o9 622343 50 o5 377657 36 25 624965 ~ 49.72 999614 0o9 625352 49.8I 374648 35 26 627948 49.38 999608.09 628340 49 47 371660 34 27 630911 4 - o4 999603.09 6313o8 49.3 368692 33 28 633854 48-71 999597.09 634256 48.80 365744 32 29 636776 48-39 999592.09 637184 48 48 362816 31 30 639680 48-o6 999586 o09 640093 48-I6 359907 30 31 8-642563 47-75 9.999581.09 8-642982 47 84 II.3570I8 29 32; 645428 47-43 999575.09 645853 47-53 354,47 28 33 648274 47.12 999570.09 648704 47 22 351296 27 34 65iIo2 46-82 999564.09 65I537 46'9I 348463 26 35 653911 46-52 999558 I0o 654352 46 6T 345648 25 36 656702 46-22 999553 o10 657149 46.31 342851 24 37 659475 45-92 999547.Io 659928 46-02 340072 23 138 662230 45-63 999541.I0 662689 45.73 337311 22 39 664968 45-35 999535 o10 665433 45.44 334567 21 40 667689 45-o6 999529.10 668i60 45.26 331840 20 1 41 8.670393 44-79 9-999524.Ib 8.670870 44-88 I1 329130 42 673080 44-51 999518 *10 1673563 44.6i 326437 Il 43 67575i 44-24 999512 *Io 676239 44-34 32376I 17 44 678405 43.97 999506 Io 678900 44-I7 321100 16 45 68o043 43.70 999500o *10 681244 43.80 318456 15 46 683665 43-44 999493 10 684172 43-54 315828 114 47 686272 4318 9c9487 *10 686784 i 43.28 3I3216 13 48' 688863 42.92 99g9481 I0 68938I 430o3 3I06I9 12 49 691438 42-67 999475 |10 691963 42 77 308037 I' 5| 693998 42.42 999469 -I 694329 42-52 305471 10 Si 8.696543 | 42-17 9.999463.II 8.697081I 42.28 I.30219 9 5 699073 419 2 999456 *11 699617 4203 300383 53 701589 41.68 999450 II 702I39 41.79 297861 7 54 704090 41.44 999443 |II 704646 41-55 295354 6 706577 41.21 999437 I 707,40 41-32 292860 5 56 709049 40~.97 99943 1 *11 70968 4-.o8 290382 4 57 71I507 40-74 999424 |11 712083 40-85 287917 3 58 713952 | 40-. 999418 I 714534 40.62 285465 2 6 9 716383 40- 20 99941 1 II 716972 40.40 2 83028 1 j.6o 71880o 40-0 ob 999404 *11 719396 40.i7 2806o4 0 Cosine D. Sine Cotang D. Tang. M. (87 DEGuRms.) SINES AND TANGENTS. (3 DEGREES.) fl M. Sine D. Cosine D. Tang. D. Cotang. o 8.7,8800 4o.o6 9'999404 "H 8'719396 40.17 1,.280604 6o I 721204 39.84 999398.I 721806 39.95 278194 723595 39.62 999391 II 724204 39.74 275796 58 725972 39.41 999384 -II 726588 39.52 273412 57 4 728337 3'i 999378,.I 728059 39.30 271604i 56 5 73o688 389. "i6 6 7330 27 38.77 999364 12 733663 3.89 266337 54 7 735354 38.57 999357 I 2 735996 38.68 264004 53 8 737667 38.36 999350 i12 738317 38.48 261683 52 739969 38i6 999343 i 12 740626 38.27 259374 51 10o 742259 37.96 999336 I12 742922 38.07 257078 50 8I 8744536 37.76.999329.12 8745207 3787 25479 4 12 746802 37.56 999322 12 747479 37.68 252521 4 3 749055 37.37 999315 12o 749740 37.49 250260 47 14 751297 3717 999308 I2 751989 37.29 248011 46 15 753528 36-98 999301.12 754227 37.10 245773 45 i6 755747 36.79 999294 12 756453 36-92 243547 44 i; 757955 36-6i 999296 *12 758668 36.73 241332 43 75'7955 36.6, 96 24,33 760,5i 3642 999279 i12 760872 36.55 239128 42 19 762337 36.24 999272 12 763065 36.36 236935 41 20 764511 36.06 999265.12 765246 36.i8 234754 40 21 8.766675 35.88 9.999257 *.2 8.767417 36.oo II232583 3 22 768828 35.70 999250 ~3 76957 35.83 230422 38 23 1 77070 35.53 999242.13 771727 35.65 228273 37 24 773101 35.35 999235.13 773866 35.48 226134 36 25 775223 35.i8 999227.,3 775995 35.3i 224005 35 260 777333 35.oi 999220.I3 778114 35.14 221886 34 27 77934 3484 999212.13 780222 34.97 219778 33 28 781524 3467 9205 J.3 782320 34.o 217680 32 29 783605 34.51 999I97.I3 784408 34.64 215592 31 3o 785675 34.3i 999k 9.i3 786486 3447 203514 30 3i28.787736 34.18 9.999181 i 13 8-788554 34.3i 11.211446 29 32 78978k 34.02 999174.13 790613 34.15 209387 2 33 79182 33.86 999166.13 792662 3309 201338 27 34 793859 33.70 999158 -13 79470I 33.3 205299 26 35 795881 33.54 999150 13 796731 33.68 20326 25 36 797894 33.39 999,42 13 798752 33.52 20124 24 33 799897 33.2 3 99934 13 800763 33.37 199237 23 3 5799897 3 801892 33o8 999126 73 802765 33o22 197235 22 39 803876 32.93 999118.13 804758 33.07 195242 21 40 805852 32.78 99910.37 806742 32.92 193258 20 4' 8F807819 32.63 9.999I02.13 8-8o08r7 32.78 11.19I283 19 42 80)777 32.49 999094 1 8io683 32.62 189317 1 43 8II726 32.34 999086.14 812641 32.48 187359 17 44 803667 32.19 999077.14 814589 32.33 185411 6 45 81S599.32.05 999069.14 816529 32.19 183471 15 46 817522 31.gI 999061.14 818461 32.05 t8i539 14 8i036 3177 999053.14 820384 31.91 17966 13 46 8ji3z24 3i.63 999044" 14 822298 31.77 177702 12 49 823240 3i.49 999036.14 824205 3i.63 175795 I1 50 82513o 31.35 999027.14 826103 3i.5o 173897 IO 51 8.8270II 31.22 9.999019.14 8.827992 3i.36 11.172008 52 828884 3i.o8 999010 I14 829874 31.23 170126 8 53 830749 3095 999002.14 831748 3i-io 168252 7 54 832607 30'82 998993.I4 8336I3 3.96 66387 6 55 834456 30.69 998984.i4 835471 3o.83 164S29 S 56 836297 3o.56 998976.14 837321 30.70 162679 4 5, 838i o 3o.43 998967.i5 8391639 30o.7 60837 3 5 83o9S6 30.30 9989S8 0i 84o998 3o'45 159002 2 S 84i774 30.17 998950.O i 842825 30.32 157175 1 8o 843585 32.oo 998941 0i 844644 30.19 i55356 0 Cosine D. Sine - Cotang. D D. Tang.. (8 DEGREE2S.) 22 (4 DEGREES.) A TABLE OF LOGARITHIMIC M.i Sine i D. Cosine D. Tang. D. Cotang.' 8.843585 3o.o- 9.998941 i5 8.844644 30o.19 ii.i55356 6o I 845387 29~.2 998932 ~51 846455 30.07 153545 2 84183 29.0o 998923. 5 84.8260 29'9 5 ~51740 3 84897I1 29-67 998914 15 850057 292 14990 57 8575 141 9.8.i,943 571! 850751 29.55 998905 1i5 85846 29.70 I4f154 56 5, 852525!9 ~5 8~2525 29.43 998896 1.5 853628 29.58 146372 I 55 6 854291 29.31 99888.15 855403 29.46 144597 54 856049 29.19 998878 I.5 857171 29.35 142829 53 8575801 2.07 998869.15 858932 29-23,410i68 52 9 859546 28.6 998860. 5 86o686 29.Ii 139314 51i 10 861283 28.84 99885I.i5 862433 29.00 137567 5c ix 8.863oi4 28.73 9.998841.i5 8.864173 28.88 11135827 49 12 864738 28.61 998832.-i5 865906 28.77'34094 48 13 866455 28.50 998823.i6 867632 28.66 132368 47 rI14 868i658 28'3 998813 i6 86935I 28.54 I3o649 46 ~1 869868 [28.2 998804.i6 87Io064 28.43 128936 45 i6 871565 28.17 998795.I6 872770 28.32 12723o 0 44 I7 873255 28.06 998785.I6 874469 28.21 125531 43 i8 874938 27.95 998776.,6 876162 28.11 123838 42 19 876615 27.86 998766.i6 877849 28.00 122151 41 20 878285 27.73 998757.16 879529 27.89 120471 40 21 8.879949J 27.63 9'998747.i6 8.881202 27.?7 I.II8738 3q 22 881607 27.52 998738.z6 882869 27.68 1 7T1 38 23 883258 27.42 998728.i6 884530 27.58 115470 37 24 8849o3 27.31 998718 I~ 886i85 27.47 ii38i5 36 25 886542 27.21 998708 ~ i6 887833 27.37 112167 35 26 888174 27-I' 998699 % i6 889476 27.27' 1105 34 27 8898o01 27.00 998689.i6 891112 27-I7 o108888 33 28 891421 26.90 998679.16 892742 27.07 107258 132 29 893035 26.80 998669.17 894366 2697 o105634 31 30 894643 26.70 998659.17 895984 26.87 1o4oi6 3o 3i 8.896246 26.6o 9.998649.17 8.897596 26.77 I1.102404 2 32 897842 26.51 998639.17 899203 26.67 100797 2 33 899432 26.41 998629.17 9o00803 26.58 099197 27 34 901017 26.31 9986g19.7 902398 26.48 097602 26 35 902596 26.22 998609'17 9037 2638 o96oz3 25 36 904169 26.12 998599.17 905?70 26.29 094430 24 3 905736 26.03 998589'17 907147 26.20 092853 23 3 907297 25.93 998578.17 908719 26.o o091281 22 39 908853 25.84 998568'17 910285 26.oi0 o089715 21 40 910404 25'75 998558.17 911846 25.92 o8154 20 41 8.911949 25.66 9'998548.17 89'3401 25.83 1.086599 19 42 913488 25.56 998537.17 914951 25.74 085049 I8 43 915022 25-47 998527'17 916495 25.65 083505 17 44 916550 25.38 998516.I8 918034 25.56 081966 16 45 9807o3 25-29 998506.18 919568 25.47 o8o432 1 i5 946 j 99591 25.20 998495.i8 921096 25-38 078904 14 47 9211o3 25.12 998485 i8 922619 25.30 077381 (3 48 i 9226o10 25.03 998474.i8 924136 25.21 075864 12 49 924112 24'94 998464.i8 925649 25.12 074351 11 5 q 1 325609 24.86 998453.i8 927156 25.03 072844 10 5t I'92710oo 24.77 9'998442.i8 8.928658 2 I07i3 52 928587 24-89 998431.t8 930o55 24.86 o61835 I 53 930068 24.6o 99842i.18 931647 24.78 o68353 i 54 931544! 24-52 998410.i8 933134 24.70 o66866 6 55 9330o5 24.43 99839.8 934616 24.61 065384 5 56 934481 24.35 ]99838:8 936093 24.53 o63907 4 57 935942 24.27 998377 oi8 937565 24.45 062435 3 58 937398 24.19 998366.i8 939032 24.37 o6o068 2 59 938850 24.1 1 998355.i8 94o494 24.30 o595o6 I,.6o 940296 2J4.03 j 998344.8 94192 24.21 05848 0 i Cosmie PD. Sine Cotang. D. Tang. M. (85 DEGREES.) SINES AND TANGENTS. (5 DEGREES.) 28 M.l Sine D. Cosine ]D. Tang. D. Cotag. o 8-94c'296 240-o3 9.998344.19 8.941952 24.21 i1i.058048 60o I 94.738 23.94 998333 19 943404 24.13 056596 5 2 943I74 23.87 998322 I9 944852 24.05 055148 58 3 944606 23279 99831'19 946295 23.97 053705 57 4 946034 2-23.71 9983oo00 i9 947734 23.9 - 052266 56 95 97436 23-63 998289 i 19 949168 23.82 o50832 55 6 948874 23.55 998277 19 950597 23.~4 049403 54 7 950287 23.48 998266 19 952021 23.66 047979 53 8 951696 23.40 998255 I19 953441 23.6c 046559 52 9 953100oo 23.32 998243 19 954856 235S 0i45144 Si 10 954499 23.25 998232.19 956267 23.44 043733 5o II 8.955894 23.17 9.998220.19 8.957674 23.37 110io42326 4 12 957284 23.0o 998209.I 959075 23 29 040925 48 13 95s860 23-02 998197 9 960473 23.23 032527 47 14 96oo0052 22.95 998186 -9 961866 23. I4 o38134 46 15 961429 22.88 998174 o 19 963255 23807 036745 45 16 9628o01 22.8o 998163.19 o64639 23.00 o3536i 44 1 964170 522'73. 90Iu9 22.93 o33931 43 1 965534 22~66 998130 20 96 4 22.6 0326o6 42 19 966893 22.59 998128.20 968766 22.79 031234 41 20 968249 22.52 998116 20 970133 2271. 029867 40 21 8.969600 22.44 9.998104.20 8.971496 22.65 11.028504 39 22 970947 22.38 998092.20 972855 22.57 027145 38? 23 972289 22.3I 998080.20 974209 22.51 025791 37 24 973628 22.24 998068.20 975560 22.44 024440 36 25 974962 22.17 998056.20 976906 22.37 o23094 35 26 976293 22.10 998044 -20 978248 22.31 021752 34 27 977619 22.0o3 998032.20 979586 22.23 020414 33 28 978941 21'97 998020.20 980921 22.17 019079 3l 29 980259 21.9o 998oo008.20 982251 22IO 017749 31 3o 981573 2I- 83 997996 -20 983577 22-04 oi642 3o 3i 8.982883 21'77 9.997985.20 8.984899 21.97 11015101J 29 32 984189 21I70 997972.20 986217 21I.9 o03783 28 33 985491 21.63 997959.20 987532 21I84 012468 27 34 986 21.57 997947.20 988842 21.78 oxii58 26 35 988083 21.50 997935 2 9949 21. 00985 25 36 989374 2.44 997922.21 991451 21.65 008549 24 37 990660 1.38 997910.21 992750 21.58 007250 23 38 991943 21.31 997897.21 994045 21.52 005955 22 ~~~~~~~~~87 -1 994045 21 Ioo652 39 993222 21.25 997885.21 995337 21.46 004663 21 4o 994497 21'19 997872.21 996624 21.40 003376 20 4' 8.99568 21.12 9.997860.21 8.997908 21.34'11002092 19 42 997036 21z.o6 997847.21 999188 21.27 000812 18I 43 998299 21.00oo 997835.21 9.000465 21.21 10'999535 17 44 99956o 20.94 997822 -21 001oo738 21.15 998262 i6 45 9.9oo816 20.87 997809 21 0 o3007 21.09 996993 I5 46 002069 20.82 9977?7:21 004272 2I.o3 995728 14 4 oo33M8 20-' 99 21 oo05534 20.97 994466 i3 oo33 0? 9977 4 1 -97 4 00oo4563 20o-70 997771 00o6792 20o.91 993208 12 49 oo5805 20.64 997758 21 008047 2085 99i953 ii o50 007044 20.58 997745 o21 009298 20.80 990702 10 5i 9.oo8278 20.52 9.997732 21 901oio546 20.74 I0-9894S4 52 009510 2046 997719 21 0II790 20.68 988210 8 53 010737 20.40 997706 21 oi3o3 i 20.62 986969 7 54 o0962 20.34 997693.22 014268 20.56 985732 6 5 0oI3182 20o.29 99768o 22 o155o02 20.5I 984498 5 56 o0144o00 2023 997667 o22 0I6732 20.45 983268 4 57 o5603 20.17 997654 o22 I7959 20-40 982041 3 o58 6824 20.12 997641'2 o09183 20.33 980817 2 59 i8o3i 20.o6 997628 22 0oo403 20.28 979597 1 1 2000 97n'i 28.22 02604 6o 019235 2000 9976i4 2 02I62o 20.23 978380 o Csine. ( Sine a ot D. S.) (84 ]PEGREE.S.) 24 (6 DEGREES.) A tABLE OF LOGARITIIMIC I I Sin e I)D~. Clt mne - I-.{[- Tang. D-. Cotang.2000 I- 6 0 9019235 20-00 9.997614 *22 9.02i620 20-23 IO.978380 I 020435 199- 5 9976o01 22 022834 20.17 97716t 2 021632 19.89 997588 22 024044i 2011 975956 3 022825 1984 997574 22 025251 2006 974749 57 4 024016 19.78 997561 22 026455 20.00 973545 52 5 025203 19'73 997547 22 027655 i i1995 972345 55 6 026386 19.67 997534! 2 028852 19.9o 971j48 54 027567 19.62 997520 2 o3oo46 19. 85 969954 53 028744 19.57 997507.23 o3237 1979 663 52 9 029918 19.51 99743.23 032425 19.74 967575 51 10 031089 1947 9974o 0 23 033609 19.69 966391 50 11 90o32257 19.41 9.997466.23 9.03479' 1964 IO965209 49 12 033421 19.36.997452 o23 359 19.58 964031 48 13 034582 19.30 997439 Q23 037144 i1953 962856 47 14 035741 19.25 99742.23 o38316 19.48 961684 46 15 036896 19.20 997411 ~.23 039485 i9.43 96o515 45 16 038048 19.15 997397 ~23 040651 19.38 959349 44 17 039197 19.10 997383.23 o4i8i3 19.33 958187 43 8 040342 19.05 907369~ 23 042973 19.28 957027 42 19 041485 18.99 997355.23 o44i3o 19.23 955870 41 1~~ ~~~~~97045485 2 3' 9558;70 20 042625 18.94 997341.23 045284 19.18 95471I6 40 21 9.043762 i8.89 9.997327.24 9.046434 19.13 10953566 3q 22 044895 8.84 997313 -24 047582 19.0o 8 952418 3 23 046026 18.79 697299.24 048727 19 o3 951273 37 24 047154 1875 99728).24 04986 18.98 950o3 I 36 25 048279 1870 997'271.24 05oo 18.93 948092 35 26 049400 18.65 997257.24 052l44 i8.89 947856 34 2 o05059 i8.6o 997242 -24 053277 18-84 946723 33 2 o~ i63 i8.55 997228.24 054407 18.79 945593 32 29 052749 18.50 997214 -24 055535 18.74 944465 31 3o 053859 18.45 997199 ~24 056659 18.70 943341 30 31 9-054966 I8.4r 9'997185.24 9'057781 18.65 10)942219 2 32 056071 i8.36 997170.24 058goo900 i8.69 941oo 2 33 057172 i8.3i 997156.24 o6ooi6 i8.55 939984 27 34 058271 18.27 997141'24 o6i13o 18.51 938870 2(; 35 059367 18.22 997127.24 062240 18.46 937760o 25 36 060460 18.17 997112 ~24 063348 18.42 936652 24 3z o~i55i i8-0 997098'24 064453 18.37 55 2' 0; 61551 181r3' C7~ 2r 064453 r~ 935547 23 062639 i8.o8 997083.25 065556 18.33 934444 22 39 063724 18.o04 997068.25 066655 18.28 933345 21 40 064806 17.99 997053.25 067752 18.24 932248 20 41 9.065885 17.94 9.997039.25 9.068846 18.1 10.93ii54 19 42 o6696 17 97024 o25 069938 18.1 930062 18 43 068036 17 997009.25 071027 i8.,c 928973 17 44 o69~7 17'81 996994.25 072113 18-06 927887 16 45 070176 17.77 996979.25 073197 i8.o- 926803 i5 46 071242 17.72 996964.25 074278 17.97 925722 1t 4 o/2306 17.68 996949.25 075356 17.93 924644 13 073366 17.63 996934.25 076432 17.8 923568 i 49 074424 7.5 996919.25 077505 17.84 922495 i So 075480 17.5 ~ 996904 -25.078576 17.80 921424 1 9.076533 17' 50 9-996889.25 9.079644 17.76 10.920356 52 077583 17.'46 996874.25 080710 17.72 919290 53 078631 17.42 996858.25 081773 17.67 9gi8227 54 079676 17.38 996843.25 082833 17.63 917167 6 55 080719 17.33 996828.25 083891 17.S9 916109 5 56 081759 1 7.29 996812.26 084947 I755 915053 4 082797 17.22 996797.26 086oo000 1 7.5 914000o 3 j 088332 -47~a 9972 ~. 9 2'i 58 083832 17.21 996782-~ 26 087050 I I747 912950 0 59 084864 17.17 996766 26 17 9 6o 085894 17.13 9967Si.26 089144 -7.38 91 0 8 L.. c~ J —1% —---— iCoeae D. S) —- _e Cotano2. I). T ang. (83 DEGREES.) SINES AN1L TANGENTS. (7 DEGREES.) 24 I LI Sine D. Cosine D. Tang. I D. Cotang. o 9.o85894 17.-13 9.996751 I26 9-08g44 I7.38 10.9I0856 6o I o86922 I7.09 99673I 26 09018 7 7.34 909813 59 2 087947 I17.04 996720 *26 o91228 I7-30 908772 58 3 088970 17o00 996704 *26 092266 17-27 907734 57 4 089990 16 96 996688 *26 093302 I7.22 906698 56 5 010oo8 16.92 996673.26 094336 I7.19 905664 55 6 092024 I6.88 996657.26 095367 17.15 904633 54 0 o93037 I6-84 996641 *26 096395 17iI 903605 53 094047 16.80 996625 *26 097422 17.07 902578 52 9 095056 I6.76 996610 26 o98446 170o3 901554 51 1o 096062 16-73 996594 26' 099468 16-99 900532 50 I2 9 806 i7o 68 9*965996578 27 oo487 1695 I899513 4q 1 og98066 I6.65 996562.27 I0504 16190 898496 4 O3 099065 I6.6i 996546.27 102519 I6*87 897481 47 14 Iooo62 I6.57 996530.27 10 o3532 16.84 896468 46 15 101056 16-53 996514.27 104542 I6-80 895458 45 i6 102048 16.49 996498.27 io5550 16-76 894450 44 17 103037 6.-45 996482.27 io6556 16.72 893444 43 I8 104025 16.4I 996465.27 107559 6. 69 892441 42 19 IoSoio I6.38 996449.27 io856o i6.6 891440 41 20 105992 I6-34 996433.27 I09559 I6.6I 890441 40 21 9-106973 i6.30 9.9964I7.27 9 II0556 I6.58 I0.889444 39 22 10795I I6-27 996400.27 I11551 I6.54 888449 38 23 108927 i6.23 996384.27 I12543 i6.50 887457 37 24 099ggI 16-.1 996368.27 113533 16-46 886467 3 25 110873 I6.-I6 996351.27 114521 16 43 885479 35 26 11842 I6.12 996335.27 115507 I6.39 884493 34 12 I12809 i6.o8 996318.27 I16491 I6.36 8835o9 33 11 3774 i6.o5 996302.28 117472 16-32 882528 32 29 114737 r6o01 996285 28 1188452 I6.2 881548 31 30 I15698 15.97 996269.28 I119429 16-25 880571 30 3i 9.116656 I5-94 9.996252 *28 9.120404 I6-22 0.-879596 2 32 117613 15.90 996235 28 121377 i6I8 878623 2 33 II8567 15-.7 996219.28 122348 x6-I5 877652 27 34 119519 I5.83 996202.28 123317 I6.II 876683 26 35 120469 I5-80 996185.28 124284 16-o7 875716 25t 36 121417 15.76 996I68.28 125249 I6-04 874751 24 3 122362 15-73 996151.28 126211 I6-oi 873789 23 38 123306 I5.69 996134 |28 I27I72 15-97 872828 22 39 124248 i5.66 996117 |28 128130 15.94 871870 2 40 I,5187 I5-62 996io00 28 129087 I5-91 870913 20 41 9-126125 i5-59 9-996083 29 I 9300oo41 I5-87 10o869959 I1 42 127060 |5.56 996066'29 I30994 15-84 86900 6 43 127993 15-52 996049 ~29 31944 5.8i 868056 17 44 128925 15.49 996032.29 132893 15-77 867107 6 45 129854 15-45 996o05.29 133839 15.74 86616i I5 46 130781 15.42 995998.29 I34784 15-71 865216 14 47 706 15.3o 99598o.29 135726 1I567 864274 1 3 48 13263o 15.35 995963.29 136667 5.-64 863333 12 49 13355i I5-32 995946'29 137605 15-.6 862395 1I 50 134470 I5.29 995928.29 138542 I5-58 861468 10 51 9.135387 I5.25 9.995911 29 9-139476 I5.55 io0860524 52 1 1363o3 15-22 995894.29 140409 15.-5 859591 53 i372i6 I5.19 995876 |29 I41340 15-48 858660 7 54 1.38128 I5.i6 995859.29 142269 15-45 85773I 55 x39037 I512 995841.29 143196 I5.42 856804 5 56 39944 15-09 995823.29 144121 15-39 855879 4 5 140o850 1 5.0ob 995806.29 145044 15.35 854956 3 58 141754 1503 995788.29 145966 I5-32 854034 2 59 142655 15o00 995771 *29 i66885 15-29 853115 | 60 143555 14.96 995753.29 147803 15-26 852197 0 I-Cosine D. Sine I Cotang. D. J'Tn. M. (82 DEGRae.) 26 (8 DEGREES) A TABLE OF LOGARITHMIC M. Sine D. - osine D. Tang. D. Cotang. o 9. I 3555 14-96 9995753 *3o 9 147803 15-26 10.852I97 6 I I44453 14.93 995735 *3o i 148718 I5-23 851282 5q 2 145349 14.90 995717 *3o * 149632 I5-. 20 850368 58 3 146243 4-8 995699 *30 I5o544 15.17 849456 57 4 147I36 |I4.84 99568i *3o I5I454 I5-14 848546 56 5 I48026 14-8I 995664 *3o 152363 15-II 847637 55 6 I48915 1|4-78 995646 *3o I53269 i5.o8 846731 54 7 149802 14.75 995628 *30 154174 15.05 845826 53 8 i50686 I4.72 995610 *30 155077 15.02 844923 52 9 151569 14.69 99559I *30 155978 I4'99 844022 5I I0 152451 I4 66 995573 *3o 156877 14.96 843123 50 II 9I53330 14.63 9.995555.30 9-157775 14.93 1Io-842225 4 I2 154208 I4.60 995537.30 i58671 I490 841329 4 3 I55083 14.57 995519 *3o 159565 14.87 840435 47 14 155957 I4.54 995501.3i I 160457 I 4.84 839543 46 I5 i56830 14-5I 995482 *3i I6I347 14.8i 838653 45 i16 I577oo i4.48 995464.3I i62236 14.79 837764 44 17 I58569 i4.45 995446.3i 163123 I4.76 836877 43 I8 159435 14.42 995427.3i I64008 14.73 835992 42 19 i6o3oi 14.39 995409.31 164892 14- 70 835I08 41 20 I6II64 I4 —.36 995390 *3i I65774 14.67 834226 40 21 9.162025 14.33 9.995372.3i 9-.166654 I14.64 0io.833346 3 22 I62885 I4-30 995353.3I 167532 i4.6i 832468 38 23 I63743 I4.27 995334.3I 168409 14.58 83159I 37 24 i646oo00 14.24 9953i6.3i I69284 I4.55 830716 36 25 I65454 I4.22 995297 -3I 170157 14.53 829843 35 26 I66307'4I9 995278.3 171029 I4.5o0 828971 34 27 I67159 14.-6 995260.3i 171899'447 828101 33 28 I68o008 I4-I3 995241.32 172767 14.44 827233 32 29 i68856 14.10 995222.32 173634 I 4.42 826366 3I 3o I69702 14.0o7 995203.32 I74499 14.39 825501 3o 31 9. I70547 14.05 9.995184.32 9.175362 I4.36 10.824638 2 32 171389 14.02 995I65.32 I76224 14.33 823776 28 33 I72230 3.99 995146.32 177084 I4.3i 822916 27 34 I73070 3.96 995 127.32 177942 14.28 822058 26 35 173908 3-94 99510 IOJ 32 178799 14.25 821201 25 36 174744 3 I91 995089 32 1 79655 14.23 820345 24 37 175578 I3.88 995070.32 180508 14.20 819492 23 38 176411 I3-86 995051.32 181360 14.17 818640 22 39 I77242 T3-83 995032.32 18221I I4. 15 817789 21 40 178072.3.8o 9950I3 *32 183059 4.12 8i694I 21 4I 9-178900 13.77 9-994993.32 9.183907 14.09 io-816093 1 42 179726 13.74 994974.32 184752 14.07 815248 18 43 i80551 I3.72 994955.32 i85597 14-04 814403 1i 44 18I374 13-69 994935.32 186439 14-02 81356I I6 45 182i96 I3-66 994916.33 187280 13.99 812720 15 46 i83oi6 I3.64 994896.33 188120 13.96 811880 1i4 47 183834 I3.6i 994877.33 I88958 13-93 811042 13 48 i8465i i3.59 994857.33 I89794 13.9I 810206 12 49 i85466 I3-56 994838.33 190629 I3.89 8o937I I 50 I86280 |i3.53 994818.33 191462 I3-86 8o8538 I0 5 9.-I87092 |3.5i 9.994798 -33 9.192294 i3-84 10-807706 52 I87903 I3.48 994779.33 1I93124 I3.8 I 806876 8 53 I887I2 i 3.46 994759: -33 193953 13.79 806047 7 54 189519 13.43 994739.33 194780 13.76 805220 6 55 190325,3*41 994719.33 195606 13.74 804394 5 56 191130 I3.38 994700.33 196430 13.71 803507o 4 5 91933 I3.36 994680.33 197253 13.69 802747 3 1g I92734 |I3.33 |994660.33 198074 i3.66 80oI926 2 59 i93534 I3.3o 994640.33 I98894 I3.64 8oio6 6o 194332 13.28 994620 |33 I9973 I3-6I 800287 0 1 — 1- I D. Sin -I -.'Cosine D. I Sine Cotang. D. Tang. M. (81 DGREo ES.) SINES AND TANGENTS. (9 L EGREE.) 27 [ s _ine | D. Cosine D. Tang. D. Cotang. o 9 194332 i3.28 9 994620.33 91 99713 13.6i 10.800287 6c I 195129 13-26 994600.33 200529 I3.59 79947I 59! 2 195925 13*23 994580 *33 201345 356 798655 5 3 196719 13 21 994560 34 202159 I3.54 797841 57 4 197511 i3 18 994540.34. 202971 I3.52 797029 56 5 198302 I3 16 9945I9.34 203782 I3.49 796218 55 6 199091 I3 I3 994499.34 204592 3 47 795408 54 7 I99879 13'II 994479.34 205400 13.45 794600 53 8 200666 13*o8 994459.34 206207 I3.42 793793 52 9 201451 I3.o6 994438 34 207013 i3.40 792987 51 10 202234 I3*04 994418.34 207817 I3*38 792I83 50 ~I 9-.20307 13.oi 9.994397.34 9-208619 13.35 10-791381 49 12 203797 12.99 994377.34 209420 13-33?790580 48 I3 204577 I2.96 994357.34 210220 13.3 I 789780 47!4 205354 12.94 994336.34 211018 13.28 788982 46 i5 206131 I2-92 994316.34 211815 13.26 788185 45 x6 206906 I2 89 994295.34 2I2611 1324 787389 44 I 7 207679 12.87 994274 -35 213405 13.21 786595 43 208452 1285 994254.35 214198 13.I9 785802 42 19 209222 12.82 994233.35 214989 13.17 785011 4I 20 209992 12.80 994212.35 21 5780 I3I5 784220 4o 2I 9 210760 12.78 9.99419I.35 9.216568 I3.12 10-783432 39 22 211526 I1275 994171.35 2I7356 I3.o 782644 38 23 212291 12.73 994I50.35 218142 13.o8 781858 37 24 2i3035 12.7I 9941209 35 218926 I3-o5 781074 36 25 213818 12.68 994108.35 2197I10 13o3 780290 35 26 214579 I2.66 994087.35 220492 13*oi 7795o8 34 27 215338 12.64 994066.35 221272 12.99 773728 33 28 216097 2.61I 994045.35 222052 12.97 777948 32 29 216854 12-59 994024.35 222830 12-94 777I70 31 30 217609 12.57 994003.35 223606 12.92 776394 3o 3I 9-218363 12.55 9-993981.35 9-224382 12-90 10-775618 29 32 219116 I2 53 993960 *35. 225156 I2.88 774844 28 33 2I9868 12.50 993939.35 225929 12-86 774071 27 34 220618 12*48 993918.35 226700 12.84 773300 26 35 221367 12.46 993896 36 227471 I2-8I 772529 25 36 222115 12.44 993875.36 228239 I2.79 771761 24 37 222861 I2-42 993854.36 229007 I2.77 770993 23 38 223606 12.39 993832.36 229773 12.75 770227 22 39 224349 I2.37 993811.36 230539 12-73 76946I 2I 40 225092 12.35 993789.36 231302 |I271 768698 20 41 9-225833 12-33 9.993768.36 9-232065 12-69 10-767935 19 42 226573 I2-31 993746.36 232826 12-67 767I74 18 43 227311 I2-28 993725.36 233586 12-65 766414 17 44 2280o48 I2-26 993703.36 234345 12-62 765655 16 45 228784 I2.24 993681.36 235103 1260 764897 I 46 229518 12-22 993660.36 235859 12-58 764141 14 47 230252 12-20 993638.36 236614 12.56 763386 13 48 230984 12-18 993616.36 237368 12-54 762632 12 49 231714 12.16 993594.37 238120 i1252 761880 ii 50 2324441 I2-I4 993572.37 238872 1250 761.28 1 IO 51 9-233172 12-12 9.993550.37 9-239622 12-48 1:760378 9 52 233899 12-09 993528.37 240371 12-46 759629 8 53 234625 12-07 993506.37 241118 1 2-44 758882 7 54 23534 12-05 993484.37 241865 I2-42 758135 6 55 |236073| 12*03 993462.37 |242610 1240 757390 5 56 2367951 1201 993440.37 243354 1 1238 756646 4 57 237515 II-99 993418.37 244097 12-36 755903 3 58 238235 1197 993396.37 244839,12-34 7556I 2 59 238953 1i.-95 993374.37 25579 12-32 7514421 I 60 239670! 1.-93 99335I.37 246319 1 230 75368 O Cosine j ID. | Sin I Cotang. D. Tang. M. (80 DEGREES.) 28 (10 DEGREES.) A TABLE OF LOGARITHMIC M. |Sine - D. | Cosine D. Tang. D. Cotang. 0 9-239670 11.93 9.99335I.37 9.246319 12.30 10-753681 6o I 240386 IIg9 I 993329.37 247057 12.28 752943 5S 2 241101 II 89 993307.37 247794 12.26 7522o6 53 3 241814 11.87 993285.37 248530 12-24 751470 57 4 242526 ii.85 993262.37 249264 12*22 750736 56 5 243237 ii*83 993240.3 249998 12*20 750002 55 6 243947 I 181 993217.38 250710 12-18 749270 54 244656 Ii.79 993195.38 25146! 12.17 748539 53 8 245363 11.77 993172.38 252191 12.15 747809 52 9 246069 11'75 993149.38 252920 12-13 747080 5i Ie 246775 t1173 993127.38 253648 12.11 746352 50 11 9.247478 11.71 9.993,o04 38 9' 254374 12og9 o10745626 49 12 248181 11.69 993o801 38 255100 12-07 744900 48 13 248883 ii.67 993059 38 255824 12.0o 744176 47 14 249583 ii.65 993036.38 256547 12.o3 743453 46 15 250282 I1163 993013.38 257269 12-01 742731 45 16 250980 II.61 992990.38 257990 1200oo 742010 44 17 251677 I I 5 992967.38 258710o II98 741290 43 18 262373 ii.58 992944.38 259429 11.96 740571 42 19 253067 ii.56 992921.38 260146 I 194 739854 41 20 253761 ii.54 992898.38 260863 11.92 739137 41 21 9.254453 I1152 9.992875.38 9-261578 I.-90 o 738422 3: 22 255144 IIo50 992852.38 262292 89 737708 33 23 55834 1.48 992829.39 263005 11.87 736995 37 24 i 256523 I I46 -992806.39 263717 185 7 36 35 257211 11-44 992783.39 264428 Ii.83 735572 35 26 257898 II 42 992759.39 265138 II 81 734862 34 27 258583 1.41 992736.39 265847 11.79 734153 33 28 259268.39 992713.39 266555 I.78 733445 32 29 259951 jj.37 992690.39 267261 11.76 732739 31 30 260633 1.35 992666.39 267967 11I74 732o033 30 31 9-261314 1.-33 9.992643.39 9'-268671 11-72 10-731329 29 32 261994 ui-3i 992619'39 269375 11'70 730625 28 33 262673 ii.3o 992596.39 270077 11-69 729923 27 34 263351 11.28 992572.39 270779 1.67 729221 26 35 26402'7. 11.26 99254.'39 271479 i.65 728521 25 36 264703 11-24 992525.39 272178 ii'64 727822 24 37 265377 11-22 992501.39 272876 11.62 727124' 23 38 266051 1120 992478.4o 273573 1 6o 726427 22 39 266723 11.1o 992454.4o 274269 11.58 725731 21 40 267395 1i; 7 992439 4.4o 274964 11 57 725o36 20 41 9-268065 11-15 9.992406.40o 9275658 Ii.15 10o724342 1 42 268734 i Ir3 992382'40 276351 i 53 723649 I8 43 269402 11.1 992352.40 277043 115i1 722957 17 44 270069 II Io 992335.40 217734 1 50 722266 i6 45 270735 1 108 992311.40 278424 11-48 721576 15 46 271400 ii.o6 992287 ~40 279113 1.47 720887 14 47 272064 11o05 992263.40:279801 xi.45 720199 13 48 272726 1i.o3 992239.40 280488 -1143 719512 1 49 273388 11o01 992214.40.281174 1141 718826 11 50 | 74049 o1.99 992190 |40 281858 1140 718142 10 5I 9 274708 o1098 9.992166 |40 9-282542 ii-38:o'717458 52 275367 o1096 992142.40~ 283225 ii.36 716775 53 276024 10-94 992117 ~41 283907 |1135 716093 7 54 276681 o1092 99.093 *41 284588 11.33'715412 6 55 277337 10Ig1 992069.41 285268 |1131 714732 5 56 7799 89 992044.41 285947 130 714053 4 57 278644 o1087 992020 841 286624 11.28 713376 58 279297 0o.86 991996.41 287301 11-26 712699 2 59 279948 io-84 991971.41 287977 11I25 7' 2023 I 6o 20599 1 10-2 991947.41 288652 1123 711348 o.oineJD.-I Sine I _ otang. 1 i., (79 DEGREES.) SINES AND TANGENTS. (11 DEGREES.) 29 M V. sine D. Cosine D. Tanga. D. Cotang. 0 928o59o 10i82 9-991947 *41 9-288652 11 23 10 711348 6o I 281248 Io.8i 991922.'41 289326 11-22 7o10674 2 281897 10.79 991897'41 289999 11.20 71o0001 3 282544 10.77 991 i873.41 290671 II-18 7o0932 4 283190 10.76 991848 -4J 291342 11.17 708658 5 283836 10.74 091823.41 292013 i II15 707987 5 6 284480 10.72 091799.41 292632 1114 707318 J 54 235124 IO071 99177 2 293350 11I2 3o6650 i 53 285766 69 9017049 42 294017 o83 52 9 286408 IO -67 991724.42 j 294684 II.o 9 705316 51![ 287048 io.66 991699!'42 295349 11I07 704651 5o II 0'287687 io-64 9'99'674'42, 9'296ot3 11.O6;,o'-I3987! 49 o. 2 ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ 17387687 2 283326 Io.63 991649.42 296677 ii.o4 703323 j 4 13 288964 io.6i 991624 -42 297339 11.03 702601 I 47 14 289600 10.59 991599.42 298001 1101 7019), 46 15 290236 1058 991574.42 298662 11.00 70W8 45 i6 290870 io556 99149.42 299322 0o.98 700673 44 17 291504 Io.54 991524 -42 299980.0.06 700020 43 18 292137 io.53 991498'42 3o00638 10.95 (09362 42 29276899 ig 292IN 10~5Q 9914-13 *42 30I2 5 10.93 69370:) 41 20 293399 o0o50 991448.42 3019 i 10o92 698019 40 21 9.294029 Io048 9'991422 -42 9.302607 io0.90o io-67393 3 22 294658 o10.46 991397 42 303261 o-89 6967339 39 23 295286 10.45 991372.43 3o3214 10.87 696oo i 37 24 295913 10.43 991346.43 304067 io.86 695i33 36 25 296539 10.42 991321 -43 305218 Io.84 694782 35 26 297164 10.40 991295.43 305869 o0.83 69131- 34 27 297788 10-39 991270.43 3o65i Io.8i 69341i 33 28 298412 IO137 991244 43 307168 8 62832 32 29 299034 Io.36 991218.43 307815 IO.78 692185 31 30 299655 1o.34 991193.43 308463 10.77 691537 30 31 9.300276 1O.32 9'99I 67.43 9.3o909 10.o75 o1.69go89 29 32 300895 io-3I 99114'. -43 309754 O.74 690o246 28 33 30oi514 Io.29 991115.43 310o398 10-73 639622 27 34 302132 10o28 99109~0'43 31,1042 10.,71 638958 26 35 302748 o10.26 991064.43 3ii685' 10.70 6883i5 2] 36 303364 10.25 991038.43 312327 io.68 687673 24 3 303279 10.23 991o012 43 312967 o0.67 687033 23 304293 10.22 990986.43 3i36o8 io.65 686392 22 39 305207 1O 020 990960.43 314247 io.64 685753 21 40 305819 1019 990934.44 314885 io962 685i15 20 41 9.306430 10-17 9"990008'44 9'315523 io.6I o.684477 I19 42 307041 IO.i6 990882.44 316159 io.6o 683841 18 43 307650 IO.14 990855.44 316795 lo.58 683205 17 44 308259 Io.i3 990829'44 317430 10.57 682570j I 45 308867 10o.1 990803 -44 318064 Io.55 681936 15 46 309.474 Ioo010 990777.44 318697 Io.54 68i3o3 14 47 3iooo 1io.o8 990750.44 319329 10o.53 68067I 13 48 3io685 IO.O7 990724.44 319961 io.5i 68oo39 I2 49 311289 IO.05 990697.44 320292 IO.5o 679408 It 50 311893 10.04 990671.44 321222 0o.48 678778 10 51 9.312495 io.o3 9'990644.44 9.321851 10-47 I0'67814 9 52 313o097 Io-O 990618'44 1 322479 Io-45 677521.8 53 313698 10o00oo 990591'44 i 323106 110.44 676894 7 1 ~~~~~6694 ) 54 314297 9-98 990565'44 323733 io-43 676267 6 55 314897 9.97 990538'44 324358 1O.41 675642 5 56 315495 9.96 99051.45 3214983 10.40 675017 4 57 316092 9'94 9904'5.45 325607 1O039 674393 1 58 3i66 9 9.93' 990458.45 326231 IO-37 673769 2 59 317284 991 99043 i.45 326853 1io.36 673147 I 6o 317879 9.90 9904 4 -45 327475 io.35 672525 0 Cosine D. Site CotnUL. D. Manz (78 DEG:REE8.) (12 DEGREES.) A TABLE OF LOGARITHMIC Sine D. Cosine D. Tang. D. Cotang. U 9.3I7879 9.90 9-990404 *45 9.327474 io-35 io-672526 6o 1 318473 9'88 990378.45 328095 10 33 671905 5q 2i 319066 9.87 990351.45 328715 10.32 671285 5 3 319658 9.86 990324.45 329334 io.3o 670666 57 4 320249 9'.84 990297.45 329953 10.29 670047 56 5 320840 9.83 990270.45 330o70 1028 669430 55 6 32430 9 - 82 990243.45 331 87 Io026 668813 54 7 322019 9.80 990215.45 331803 10-25 668197 53 8 322607 9'79 990188.45 332418 10-24 667582 52 9 323194 9 - 77 990o6I *45 333033 10-23 666967 51 10 323780 9.76 990I34.45 333646 10o-2 666354 50 II 9.324366 9.75 9.990I07.46 9.334259 1o. Io-665741 49 12 324950 9.73 990079.46 334871 Io Iq 665129 48 13 325534 9.72 990052.46 335482 10-I7 664518 47 14 326117 9.70 990025'46 336093 IO-I6 663907 46 15 326700 9.69 989997.46 336702 io.05 663298 45 I6 327281 9.68 989970.46 3373ii io.I- 662689 44 17 327862 9.66 989942.46 337919 10-12 662081 43 I8 328442 9.65 989915.46 338527 Io-I 661473 42 19 32902I 9.64 989887.46 339133 Io0Io 660867 4i 20 329599 9.62 989860 o46 339739 Io-o8 66026I. 2I 9.330176 9.6I 9-989832.46 9.340344 Io0-07 Io659656 39 22 330753 9.60 989804.46 34o948 o1006 659052 38 23 331329 9.58 989777 -46 34 552 100o4 658448 37 24 33I9o3 9.57 989749 *47 342155 io|o3 657845 1 36 25 332478 9.56 989721'47 342757 10-02 657243 35 26 333ooi 9-54 989693 *47 343358 -oo00 656642 34 27 333624 9.53 980665'47 343958 99 656042 1 33 28 334195 9-52 989637'47 344558 9.98 655442 32 29 334766 9 50 989609'47 345157 9'97 654843 3 I 30 335337 9-49 989582'47 345755 9.96 654245 30 31 9-335906 9.48 9989553'47 9.346353 9.94 i10653647 29 32 3364752 9.46 989525'47 346949 9-93 65305i 28 33 337043 9.45 989497 *47 347545 9.92 652455 27 34 3376io 9'44 989469'47 34814I 99I | 651859 26 35 338176 9.43 98944I *47 348735 9.90 651265 25 36 338742 9-4I 1 989413 *47 349329 9.88 650671 24 37 339306,9.40 989384 47 349922 9871 650078 23 38 30434 9~3 O8928 47 3II[ 9897 65o848 23 381 339871 9.39 989356'47 35o514 9.86 649486 22 39 340434 9.37 089328 *47 35iio6 9.85 648894 21 40 340996 9.36 98930.o 47 351697 9.83 648303 20 4 9.34I558 9-35 9-989271 *47 9.352287 9.82 o10647713 I9 42 3421I9 9-34 989243'47 352876 9.8I 647124 I8 43 342679 932 989214 47 353465 9-80 646535 17 44 343239 9.3I 989186 47 35453 79 645947 45 343797 9.30 989i57 *47 354640 9-77 64536o I5 46 344355 9-29 989128.48 3-55227 9.76 644773 14 47 344912 927 989I00.48 3558J3 9.75 644187 13 48 345469 926 989071.48 356398 9-74 643602 12 49 346024 925 989042.48 356982 9.73 643018 11 50 346579 924 989014.48 357566 9-71 642434 10 5I 9.347134 9.22 9.988985 *48 9.358I49 9.7/0 10 64I85I 9 52 347687 9.2I 988956.48 35873I 9.69 641269 53 348240 9.20 988927.48 359313 9.68 640687 7 54 348792 |91 9I88898 -48 359893 9.67 640107 6 55 349343 9.17 988869.48 360474 9.66 639526 5 56 349893 9-6 J 988840.48 361053 9.65 638947 4 57 350443 9-.15 I 98811 *4 361632 9.63 638368 3 58 35o092 9-I4 988782 49 362210'9-62 6377390 2 59 35s4o 9-13 988753,49 362787 9.61 637213 i 6o 352888 9o 11 988724.49 3633641 9-60 636636 I___ C e no;_o9I. I 9_ I8 - _'49 ICosins_ i - D.I Sine Cotang. D. I Tag. Mi'. (7 7 DEGREES.) SINES ANDI TANGENTS. k 13 DEGR.EES.) 3 a.', Sine D. Cosine D. Tang. D. Cotang..~ 352o88 9-11 9988724'49 9.363364 9.60 Io.6636636 1 6o 1 352635 9o10 988695'49 363940 9-59 636obo I 353i8i 9.09 988666'49 364515 9.58 635485 3 353726 9.o0 988636'49 365090 9.57 63490o 57 4 354271 9.07 988607'49 365664 9.55 634336 56 5 354815 9.05 988578 *49 366237 9.54 633763 55 6 355358 9.04 988548'49 3668io 9.53 63319igo 54 7 355901oi 9.03 988519'49 367382 9.52 632618 3 8 356443 9-02 988489.4q 367953 9.5I 632047 52 9 356984 01 988460.4I 368524 9.50 631476 51 10 357524 899 988430'49 369094 9'49 630906 9.358064 8.98 9.988401 *49 9.369663 9.48 1o0630337 12 358603 8.97 988371 *49 370232 9.46 629768 i3 359141 8-96 988342 49 370799 9.45 629201 47 14 359678 8.95 988312.5o 371367 9'44 628633 46 15 36o215 8.93 988282.5o 371933 9.43 628067 45 i6 360752 8.92 988252.50 372499 9.42 627501 44 361287 8.91 988223.50 373064 941 626936 43 361822 8.qo 988193.50 373629 9.40 626371 42 19 362356 8.q 988163.50 374193 9.3 625807 41 20 362889 8288 988133.5o 374756 9.38 625244 40 2I 9.363422 8.8 9.988I03.50 9.375319 9.37 1o624681 3 22 363954 8.8 988073.50 375881 9.35 624119 3 23 364485 8.84 988043.50 376/44/2 9.34 62355 387 24 365oi6 8.83 988013.5o 377003 9.33 622997 36 25 365546 8.82 987983.5o 377563 9.32 6 2 2.437 35 26 366075 8.8i 987953.5o 378122 9.31 621878 34 27 366604 8,80 987922.5o 378681 9.30 6213I9 33 28 367131 8.79 987892.5o 379239 9.2 620761 32 29 367659 8.77 987862.50 379797 9.28 620203 31 3o 368185 8.76 987832.5I 380354 9.27 619646 30 31 9.368711 8.75 9.987801.5i 9.380910 9.26 io161909 2 32 369236 8.74 987771.51 381466 9.25 618534 2 33 369761 8.73 987740.5i 382020 9.24 6i,)8o 27 34 370285 8.72 987710.51 382575 9.23 617425 26 35 370808 8.71 987679 5i.383129 9.22 616871 25 36 371330 8.70 987649.5i 383682 9.21 6i63I8 24 3 371852 8.69 9876I8.5i 384234 9.20 615766 23 372373 8.67 987588.5I 384786 9.1 515214 22 39 372894 8.66 987557.51 385337 9.18 61 4663 21 40 373414 8.65 987526.51 385888 9.17 614112 20 4I 9 373933 8.64 9.987496 Si 9.386438 9.i5 Io-613562 I 42 374452 8.63 987465.51 J86987 9.14 613oi3 18 43 374970 8.62 987434.Si 387536 9.i3 612464 17 44 375487 8.6i 987403.52 388084 9.i2 61116 I6 45 376003 8-60 987372.52 388631 9.I1 611369 i5 46 376519 8.5 987341.52 389i78 9.10 610822 14 47 377035 8.5 987310o.52 389724 9.09 6io276 I3 48 377549 8-57 987279 52 390270 9.08 609730 12 49 378063 8.56 98724:52 3908,5 9.07 6085 II 50 378577 8.54 987217.52 391360 9.06 6o64o 1 Si 9.379089 8.53 9.987186.52 9399go3 9-o05 o160807 52 379601 8.52 987155.52 392447 9.04 607553 53 38oii3 8.5i 987124 52 392989 9.o3 607011 7 54 380624 8.So 987092.52 393531 9.02 606469 6 55 38ii34 8-49 98706I1 52 394073 9.01 605927 5 6 38i643 8.48 987030 52 394614 9-o00 60o5386 4 57 3821i252 8'47 986998.52 395154 8.99 604846 3 5 38266 8.46 986967 52 395694 8-98 604306 2 9 383i68 8.45 986936.S2 396233 8.97 603767 1 6 383675 8.44 986904.S2 i396771 8.96 603229 o Cosine I). Sine Cotang. D. Tang. IM. 27 (76 DEGREES.) 82 f14 DEGREES.) A T'ABLE OF LOGARITHMIC T.i Sille D. Cosine D. Tang. D. otang. 0 9.383675 8.44 9.986904.52 9. 96771 8.96 I0603229 6. I 384182 8.43 986873.53 1 397309 896 60269 2 384687 8.42 98684I 53 1 397846 895 602i54 3' 385192 8.4I 986809 53 398383 8.94 6o0617 51 4 385697 8.40 986778 53 398919 8.93 6oio8I 56 5 386201 8-39 98674o 1 3 39945 8-92 600543 55 6 386704 83 98674 3 399990 i 891 6ooo0010 i 54 6 387207 8.37 986683!53 400524 8 9 0 599476 53 387709 8.36 986651 53 401058 8 9 598942 52 9 388210 8.35 986619 153 40I591 888 59840 51 Ic 38871I 3.34 986587 53 402124 8.87 597876 50 I1 9.3892rr 833, 9.99S65555 53 ) 402656 1 8.86 10o597344 49 12 389711 8.32 956523 *53 4o3187 8.85 596813 48 1 I3 390210 8.3i 986491 53 4o371I 8.84 596282 1 47 14 390708 8.30 996439.53 5 404249 8-83 595751 46 15 3912o6 8.28 986427.53 404778 8.82 595222 45 16 391703 8.27 986395.53 405308 8-.8 594692 44 17 392199 8-26 986363.54 405836 8.80 59416 1 43 i8 392695 8-25 986331.54 406364 8.79 593636 42 19 393191 8-24 986299.54 406802 8-78 5931o8 41 20 393685 8.23 986266 -54 407419 8.77 592581 40 21 9-394179 8.22 9.986234.54 9 4o7945 8-76 10-592055 3 22 394673 8.21 986202.54 408471 8.75 591529 38 23 395166 8-20 986169 -54 408997 8.74 591003 3 24 395658 8.19 986137 54 402 8.74 590472 36 25 396 1 50 8 8 986 04 -54 41oo45 8.73 589955 35 26 396641 8.17 986072.54 410,569 8.72 58-943 I 34 27 397132 8.17 986039.54 411092 8.?7 5889o8 33 28 397621 8-16 986007 -54 41ibi5 8-70 588385 32 29 398111 8-.5 985974 -54 412137 8.69 587863 3x 30 3986oo 8-14 985942.54 412658 8.68 587342 30 31 9-399088 8-13 9.985909og.55 9-413179 8.67 1o-586821 12 32 399575 8-I2 985876.55 413699 8.66 5863oi 28 33 400062 8.-11 985843.55 414219 8-65 585781 27 34 400549 8.10 985811.55 414738 8.64 585262 26 35 40103 8.00o 985778.55 415257 8.64 584743 25 36 401520 8.oS 985745.55 415775 8.63 584225 24 37 402005 8.07 985712.55 416293 8.62 583707 23 38 402489 8.-o6 985679.55 416810 8.-6 583190 22 39 402972 8.05 985646.55 417326 8.60 582674 21 40 403455 8.04 985613.55 417842 8.59 582w58 20 41 9-403938 8-3 |9.985580.55 9-418358 8.58 I.581642 1 42 404420 8-02 985547.55 418873 8;57 5812,7 43 404901 8-oI 985514.55 419387 8-56 58o6i3 17 44 405332 8-oo 98548o.55 419901 8.55 58o009 i 1 45 405862 7.99 985447.55 420415 8-55 579585 i15 46 4o6341 7.98 985414.56 420927 8-54 379073 14 47!'97 57 563 -3~ 47 406820 7.97 985380.56 42144o 8.53 57856, -3 48 407299 7.96 985347.56 421952 8.52 578048 2 49 40o7777 7-9j 985314.56 422463 8.5i 577531 50 408254 794 985280.56 422974 8.5o 577026 10 51 9-408731 7-94 9.985247.56 9-423484 8.4 10.576516 52 409207 7-93 98523.56 423993 848 576007 53 409682 7-92 985180.56 424503 8.48 575497 I 54 410157 7 985146.56 42501 8-47 574989 55 410o632 7 0 985113.56 425519 8.46 574481 5 56 411106 7 985079 56 426027 845 573973 57 411579'.88 985042.56 426534 8-44 673466 3 68 412052 7-87 985o1I.56 427041 8-43 72959 2 59 412524 7.86 984978.56 427547 8.43 672453 1 160| 412996 7.85 984944.56 428052 842 57I948 O CoIe I D. Sine Cutang. D. (75 DEGREo S.) SINES AND TANGENTS. (15 DEGREES.) M. Sine D. Cosine D. Talng. D. Cotang. - 0 9-412996 7.85 9-984944.57 9.428052 8.42 1o-571948 60 1 413467 7.84 984910.57 428557 8.4I 571443 59 2 413938 7.83 984876.57 429062 8.40 570933 58 3 414408 7.83 984842.57 429566 8.39 570434 57 4 I 414878 7.82.984808.57 430070 8.38 J 093o 56 5 415347 7.81 984774 -5j7 430573 8.38 569427 1 55 6 415815 7.80 984740.57 431075 8.37 568925 54 46283 7'79 984706.57 431577 8.36 568423 53 48 675I 7. 78 954672.57 432079 8.35 567921 52 9 417217 7'77 984637 1.57 432580 8.34 567420 1 5 1o 417684 7-76 984603.57 433080 8.33 566920 i 50 II 9-418150 7.75 9 984569 |.57 9.43358o 8.32 I0o566420 49 12 418615 7.74' 984535.57 434080 8.32 565920 48 13 419079 7.73 984500.57 434579 8.3 565421 47 14 419544 7 73 984466.57 435078 8.30 564922 46 15 420007 7.72 984432.58 435576 8.29 564424 1 45 16 420470 7.7I 984397.58 436073 8.28 563927 44 17 420933 7.70 984363.58 436570 8-28 563430 43 18 421395 7.6q 984328.58 437067 8.27 562933 42 19 421857 7.68 98429.4.58 437563 8-26 562437 41 20 422318 7.67 984259.58 438059 8-25 561941 40 21 9 422778 7.67 9.984224.58 9.438554 8 24 o.561446 39 22 423238 7.66 984190o.58 439048 8 23 560o952 38 23 423697 7.6j 984155 1.58 439543 8 23 56o0457 37 24 424156 7.64 984120.58 440036 8 22 559964 36 25 424615 7.63 984o085 58 440529 8-2I 559471 35 26 425073 7.62 9840o50 58 441o22 8.20 558978 34 27 42553o 7.61 984015. 58 441514 8. I9 558486 33 28 425987 7.60 9g83981.58 442006 8 19 557994 32 29 426443 7.60 983946.58 442497 8.8 557503 31 |30 426899 7.-59 983911.58 442988 8.17 557012 30 31 9-427354 7.58 9.983875.58 9-443479 8.16 io.556521 29 32 427809 7 57 983840 1.59 443968 8. 6 556032 28 33 428263 7.56 983805.59 444458 8.15 5555542 27 34 428717 7.55 983770.59 44494gi7 8.14 555053 26 35 429170 7.54 983735.59 445435 8. 3 554565 25 36 429623 7. 53 983700..59 445923 8.12 554077 24 37 430075 7 52 983664 -59 44641 8. 12 553589 23 38 430527 7.52 983629.59 446898 8. - I 553102 22 39 430978 7.51 983594.59 447384 8.lo 552616 21 40 431429 7 750 983558.59 447870 8 09og 552130 20 41 9 -431879 7 49 9'983523 59 9-448356 8.09 10-551644 1 1 42 / 432329 1 7 49 983487.59 448841 8.o8.551.59 18 43 432778 7 48 983452.59 449326 8-07 55o674 17 44 433226 1 7 47 983416.59 449810 8 o06 550190 i 45 1 433675 7 /46 9833881.59 450294 8-o6 549706 10 46 434122 7.45 983345.59 450777 8o05 549223 14 143569 1744 983305.59 451260 8.04 548740 13 4 1 435"oI 6 7 44 983273.60 451743 8o3 548257 I 12 49 435462 7 -43 983238.6o 452225 8.02 547775 11 50 J435908 7 -42 983202:.6o 452706 i 80o2 547294 1 i 9.436353 17 41 9 9983166.6o 9.453187 8.o0 10o-546813 952 o436798 4 983130.6o 453668 8.00 5463)2 2 53 437242 7 40 98304 16o 454148 7 99 545852 7 54 437686 7.39 983058 6o 4Z54628 7 99 54J3'2 6 55 438129 | 7.38 i 983022.6o 455107 7.98 544393 5 56 1438572 7-37 982986 *6o 455586 7.97 544'44 4 Q. 6o 45606' 7' 6 35-493 ~57| 43g | 7-36 |982950 |6 45664 7.96 51!3 2 3 3 5 439456 7.36 982914 1.60o 456542 7'96 54158 2 59 439 -35 1 982878 i.60 4570o19 7.95,54 6o 44o338 734! 982842 -6o 457496 794 j 542504' a CosDne D Sil i Cotan D. (74 Tas aJE.) (74 D E G REEs.) 84 (16 DEGRtEES.) A TABLE OF' LOqARITHMIC M. Sine D. Cosine ID. Tar-g. D. Cotang. o 9-440338- 7.34 9.982842 *60 9-457496 7.94 10'542504 6 I 440778 7.33 982805 *60 457973 7.93 542027 59 2 441218 7.32 9827,69 *6I 458-449 7.93 5455i 58 3 441658 7 3 982733 *6I 458925 7.92 541075 57 4 442096 7.31 982696.6i 459409 7.91 54o6oo00 56 5 442535 7J 30 982660 *6i 459875 7.90 540125 55 442973 7.29 982624 *6i 4603.$9 7.'o 53965I 54 7 4434io 7 28 982587 i6I 460823 7.89 539I77 53 443847 7-27 982551 *6i 461297 7.88 538703 52 9 444284 7.27 982514 *6i 461770 7.88 538230 51 I0 444720 7.26 982477 *61 462242 7.87 537758 50 11 9445155 7.25 9.982441 *6i 9-462714 7.86 I10537286 4q 12 445590 7.24 982404 *61 463186 7.85 536814 48 I3 446025 7-23 982367 *6i 463658 7.85 536342 47 14 446459 7.23 982331 [6i 464129 7.84 535871 46 15 446893 7.22 982294.6i 464599 7.83 535401 45 I6 447326 7.21 982237 *6i 465069 7.83 534931 44 I7 447759 -.20 982220 *62 465539 7.82 534461 43 i8 448191 7.20 982183 62 466oo008 7.8I 533292 42 19 448623 7'19 982146.62;66476 7 80 533524 41 20 449054. 982109 62 466945 7.80 533055 z 40 21 9'449485 7.I7 9'982072 *62 9-467413 7'79 10'532587 39 22 449915 7.I6 982035 62 467880 7.7 532120 38 23 450345 7.I6 981998 *62 468347 7-78 531653 37 24 450775 7-15 981961 *62 468814 7.77 531i86 36 25 451204 7.I4 981924 *62 469280 7.76 530720 35 26 451632 7-13 981886 62 469746 7-75 530254 34 27 452060 7-I3 981849 *62 470211 7-75 |529789 33 28 452488 7-.2 98181! *62 47o6 J 7' 74 529324 32 29 452915 7-11 981774 *62 471141 7.73 523859 31 30 453342 7-10 981737 z62-1 471605 7.73 528395 30 3I 9-453768 7-10 9.98I699 -63 1 9-472068 7.72 10o527932 29 32 454194 7-09 98I662.63 472532 7.71 527468 28 33 4546i9 7.0 981625.63 472995 7-71 527005 27 34 455044 7.07 98i587 *63 473437 7.70 526543 26 35 455469 7-07 98I549 *63 473919 7.69 52608I 25 40 457584 7-03 981361 *63 476223 7.66 523777 20 |41 9458oo6 |7-02 9-981323.63 9-476683 7.65 10.5233I7 19 42 |458427 |701 981285.63 477142 7.65 522858 i8 43 458848 7.01 981247.63 477601 7.64 522399 17 44 459268 7-00 981209 *63 478059 7.63 521941 i6 45 459688 6-99 981171.63 478517 7.63 521483 i5 46 46oio8'6.9 981133.64 478975 7.62 521025 14 4' 460527 6.98 981095 *64 479432 7'61 520568 13 48 460946 6.97 981057.64 479889 7.61 520111 12 49, 461364 6.96 981019.64 480345 7.60 519655 11 5o 461782 6.95 980981.64 480801 7-59 519199 1o 5i 9.462199 6.95 99g809g2.64 9-481257 759 Ic-518743 52 462616 6.94 980904.64 481712 7.58 518288 8 53 463032 6.93 98o866.64 482167 7.57 517833 7 54 463448 6.93 980827 64 482621 7.57 517379 6 55 1463864 16-92 980789.64 483075 7.56 516925 15 56 464279 6'91 980750.64 483529 755 5 i06471 4 57 464694 6.9o 980712.64 483982 7.5 55 i6oi8 3 58 4651o8 6.90 980673.64 484435 7.54 51i5565 2 59 4655221 6.89 980632.64 484887 7.53 515i3 | 465935 6.88 980596.64 485339 7.53 584661 o I Cosine I D. 1 iuLe Cotanug. I D. I Tang. M. (73 DE6RIES.) SINES AND TANGENTS. (17 DEGREES.' 35 hM. Sine I D. J Cosine D. Tang. D. Cotang. 9.465935 6 88 9.980596.64 9 *485339 7.55 10o51466i 6o I 466348 6.88 980538.64 48579I 7-52 5I4209 59 2 46676I 6.87 980519.65 486242 7.51 513758 3 467173 6.86 980480 -CS5 486693 7.51 513307 57 4 467585 6.85 980442 *65 487143 7 50 5I2857 56 5 467996 6.85 980403.65 487593 7' 49 512407 55 6 468407 o 84 980364 *65 488043 749 5 1957 54 7 468817 6.83 980325 *65 488492 7.48 511508 53 8 469227 6.83 980286 *65 488941 7-47 5Iio59 52 9 469637 6.8' 980247.65 489390 7'47 51o6io 51 10 470046 6-8t 980208.65 489838 7.46 510162 50 11 9-470455 6-&)'9.98oi69.65 9-490286 7.L6 10-509714 49 12 470863 6.8 980o30.65 490733 7.45 509267 48 I3 471271 6.79 9q800oo 65 491180 7.44 508820 47 14 471679 6-78 980052.65 491627 7.44 508373 46 I5 472086 6.78 980012.65 492073 7.43 507927 45 i6 472492 6.77 979973.65 492519 7.43 507481 44 7 472898 6.76 979934.66 492965 7.42 507035 43 I8 473304 6-76 979895.66 493410 7-41 506590 42 19 473710 6.75 979855.66 493854 7.40 506146 41 20 474115 6.74 979816.66 494299 7.40 505701 40 21 9.474519 6-74 9.979776.66 9*494743 7.40 10505257 39 22 1 474923 6.73 979737.66 495186 7.39 504814 38 23 1 475327 6-72 979697.66 495630 7.38 504370 37 24 475730 6.72 979658.66 496073 7.37 503927 36 25 476133 6-71 979618.66 496515 7-37 5o3485 35 26 476536 6.70 979579.66 496957 7.36 503043 34 27 | 476938 6 - 69 } 979539 *66 497399 7.36 50260oi 33 2 477340 6.69 979499.66 49784I 7.35 502i59 32 29 477741 6-68 979459.66 498282 7.34 50I718 3I 30 478142 6-67 979420 *66 498722 7.34 501278 3o 31 9-478542 6-67 997938o 66 9.499163 7.33.500oo837 2 32 478942 6.66 979340.66 499603 7.33 5oo00397 2 33 479342 6.65 979300.67 500042 7.32 499958 27 34 47974I 6-'65 979260.67 500481 7.3I 4995I19 26 35 480I40 6.64 979220.67 500920 73I 499080 25 36 480539 6.63 979180.67 501359 730 498641 24 37 Z480o37 6.63 979140.67 50177 7.30 4982o3 23 38 481334 6.62 97900.67 502235 7-29 497765 22 39'481731 6J 6I 979059.67 502672 7-28 497328 21 40 482128 6-6I 979019.67 503109 7.28 496891 20 41 9-482525 6.6o 9.978979.67 9-503546 7-27 1O.496454 19 42 482921 6.59 978939 3 67 503982 7-27 496018 18 43 483316 6-59 978898.67 5044I8 I 7.26 495582 17 44 483712 6.58 978858.67 504854 7.25 495146 16 45 484107 6.57 978817.67 505289 7-25 494711 i5 46 484501 6.57 978777.67 505724 7-24 494276 14 47 1 484895 6.56 978736.67 50615 7.24 49384I 13 48 485289.655 978696.68 506503 7-23 493407 12 49 485682 6.55 978655.68 5070o7 7-22 492973 lI 50 486075 6-54 978615.68 507460 7-22 492540 10 51 9.486467 6-53 9.978574.68 9-507893 7-2I 10o492I107 52 486860 6.53 978533.68 508326 7.2I 491674 8 53 487251 6.52 978493 |68 508759 7*20 49I24I 7 54 487643 6-5I 978452 |68 509191 7-19 490809 6 55 48 034 6.5i 978411.68 509622 7-I9 40o378 5 56 488424 6.50 978370 68 5100oo54 7I 1 489946 4 5 48884 6.50 97832 68 510485 j 7I8 4895I51 3 58 489204 6-49 97828 |68 510916 7.17 489084 2 59 489593 6.4 978247 68 511346 7i6 488654 I 60 489982 648 978206 68 511776 7.I6 488224 0 o Cosine D. Sine D. Cotang. _ D. ira. iiIM (72 DEGREES.) 36 (18 DEGREES.) A TABLE OF LOGARITHMIC M. Sine D. Cosine D. Tang. D. Cotang. 0 9-489982 6.48 9.978206.68 9*511776 7.16 IO0488224 6o I 49037 I 6.48 978165 -68 512206 7.16 487794 5. 2 490759 6 47 978124 68 512635 715 487365 5 3 491 47 6.46 978083.69 513o64 7-I4 486936 57 4 - 491535 6.46 978042.69 513493 7'14 486507 56 5 491922 6.45 978001'69 513921 7I3 486079 55 6 492308 6-44 977959 *69 514349 7-I3 485651 54 492695 6.44 977918'69 514777 7'. 12 485223 53 493081 6.43 977877.69 515204 7.12 484796 52 9 493466 6 42 977835.69 5i563 7.1 484363 51 10 493851 6-42 977794.69 516057 7.'0 483943 5o 1, 9-494236 6-41 9.977752 69 9.5I6484 7'10 10-483516 4 12 49462I 6.4I 977711.69 51691o 7.09 483090 48 x3 495005 6.40 977669'69 517335 7.0 482665 47 14 493388 6.39 977628'69 517761 7-o8.482239 46 15 495772 6.39 977586.69 5i8i85 708 4818r 5 45 16 496154 6.38 977544.70 5186io 7.07 481390 44 17 496537 6.37 977503'70 519034 7.06 480o966 43 18 496919 6.37 97746I'70 519458 7'06 486542 42 19 497301 6.36 977419'70 519882 7.o05 48o1 18 41 20 497682 6-36 977377.70 520305 7.05 479695 40 21 9-498064 6.35 9-977335 "70 9.520728 7.04 10.479272 30 22 498444 6.34 977293.70 521151 7-03 478849 23' 498825 6*.34 9772 i.'70 521573 7.03 478427 37 24 499204 6.33 977209.70 521995 7.03 478005 36 25 499584 6.32 977167.70 522417' 7.02 477583 35 26 4.99963 6 32 977125.70 522838 7.02 477162 34 27 500342 6.3i 977083'70 523259 7.0 47674I 33 28 500721 6.3i 97704I 7o0 52368o0 7.01 476320 32 29 501099 6-3o 976999.70 524100 7.0oo 475900 31 3o 501476 6-29 976957.70 524520 6.99 475480 3o 31 9. 50o854 629 9 976374.71 9-524939 69,0 I0-475061 2 32 50223I 628 9762 71 2 6. 474641 2 33 502607 6.28 976830'.7 525778 6.98 474222 27 34 502984 6-27 976787'71 526I97 6.97 4734o3 26 35 5o336o 6-26 976745'71 526615 6.97 473385 25 36 503735 6-26 976702.71 527033 6.96 472967 24 3 5o4io 6.25 976660.71 527451 6.96 472549 23 504485 6.25 976617'71 527868 6 95 472132 22 39 504860 6-24 976574.71 528285 6.95 471715 21 40 505234. 6.23 976532.71 528702 6.94 471298 20 41 9.505608 6.23 9.976489.'7 9*529119 6.93 o10470881 19 42 505981 6-22 976446.71 529533 6.93 470465 18 43 506354 6-22 9764o4'.71 529950 6.93 470050 1 44 506727 6.-I 976361.7I 53o366 6.92 469634 16 45 507099 6 20 676318 *71 530781 6.91 469219 15 46 507471 6-20 976275'.7 531196 6.9I 468804 |4 47 507843 6-I9 976232'72 53161 6-90 468389 13 48 508214 6.19'97689.72 532025 6.o0 467975 12' 49 508585 6.Ig 976146.72 532439 6.89 46756I II 50 508956 6.-8 9761o3,72 532853 6.89 467147 10 I,509326 6.17 9.976060o'72 9533266 6.88 10-466734 Q 5. 509696 6 I6 976017'72 533679 6.88 466321 53 510oo65 6-i6 975974.72 534092 6.87 465908 7 54 o1034 6 I5 975930 72 5345o4 6.87 465496 6 55 510803 6i5 - 97582 7. 72 534916 6-86 465084 5 56 511172 6.-4 975844.72 535328 6.86 464672 4 57 i 554o 6 57 i 51i50i 1 6i13 975800.72 535739 6.85 46426I 3 58 597 63 975757.72 53615o 6.85 9 463850 1 512275 1 612 975714.72 53656i 6.84 463439 1 j 6o 512642 6&12 975670 *72 536972 6-84 463028 - Cosine - D. Sine D. -Cotang. D. Tang.. (71 DEGREES.) SINES AND TANGENTI'S. (19 DEGREES.) 37 M. Sinoe D. Cosine D.' Tang. D. Cotang. c 9.312642 6.12 9.975670.73 9.536972 6.84 1,10.463028 i 6o Sf3009og 6.i i 975627.73 537382 683 462618 5 2 513375 6.ii 975583.73 537792 6.83 46220o 3 51374I 6.io 975539 73 53820o 2 6.82 461798 57 4 514107 6.09 975496.73 5386ii 6.82 6139 56 i~~~~~68 461389 56 5 514472 6.09 975402.73 539020 6.8I 46og09 55 6 514837 6.o8 975408 73 539429 6.8I 46o51'1 54 515202 6.o8 975365.73 539537 6-80o 46oi063 53 5i5566 6.07 975321 73 540245 6.8o 459755 52 9 5i593o 6.o7 J 75277 73 54o653 6'79 459347 51 10 516294 6.06 975233.73 541o61 6'79 458939 50 II.516657 6.o5 9'975189.73 9541468 6-78 10.458532 12 517020 6.o05 975145.73 541875 6.78 458125 13 517382 6.04 975o101.73 542281 6'77 457719 47 14 517745 6.04 975057.73 542688 6'77 457312 46 iS 5181o 7 6.o3 975o013.73 543094 6.76 456906 4 i6 518468 6.o3 974963 974 543499 6'76 45650o 44 17 518829 6.02 974925 74 543905 675 456095 3 i8 519190 6.oi 974880 *74 5443o10 6'75 45569a 42 19 519551 6.oi 974836 *74 544715 6-74 455285 41 20 519911 6oo00 974792 *74 545i19 6'74 J45488, 40 2~ 9. 52027I 6.oo 9.9747,48 *74 -.545524 6.73 10o454476 22 S2o63 5.99 974703 -74 545928 6.3J 454072 23 520990 5.99 974659 *74 54633I 6.72 453663 37 24 521349 5-98 974614'74 54673 672 45326 36 25 521707 5.98 974570 *74 547138 671 452862 35 26 522o66 i 5.97 974525.74 54754o 6.71 452460 34 27 522424 5.96 974481.74 547943 6.70 452057 33 28 522781 5.96 974436'74 548345 6.70 45i655 32 29 523138 5.95 974391 I74 548747 6.69 451253 31 3o 9.5023829 5.95 974347.75 549149 6-69 45o85x 3o 32 952352 594 9.974302.75 9.54955o 668 10450450 2 J32 i 524208 5'94 974257'75 549951 66 - 450049 2 33 524564 5.93 974212.75 550352 6'67 449648 27 34 52492o 5.93 974167.75 550752 6.67 J449248 26 535 525275 5-92 974122 -75 551152 6.66 448848 20 36 j 525630 5.91 974077.75 551552 6.66 448448 24 J37 525984 5.91 974032.75 551952 6.65 448048 23 526339 5.90go 973987.'75 55235 6.65 447649 22 39 526693 5'.90o 973942.75 552750 6.65 447250 21 40 527046 5. 89 973897.75 553i49 664 446851 20 41 9.52740oo 0 5.89 9.973852.75 9.553548 6.64 10o446452 19 42 52,53 5.88 973807 75 553946 6.63 446054 J43 J5281o0j 5.88 J 97376.75 554344 6.63 445656 17 44 528458 5.87 973716.76 554741 6.62 4452::,6 45 528810 5.87 973671 -76 55539 6.62 J444861 i5 46 529161 5.86 973625 -76 555536 6.6i 444464 14 47 529513 5.86 973580.76 555933 6.6i 444067 3 48 529864 5.85 973535.76 556329 6.6o 44367 12 49 530215 5.85 973489.76 556720 6.6o 443275 11 50 530565 5-84 973444.76 5121 6.59 442379 10 Si 9.5309o 5.84 9.973398.76 9.57517 6.59 1o.442483 52 531265 5.83 973302.76 55793 6.59 442037 53 53164 5.82 973307 76 5583o8 6.58 441692 54 53i963 582~ 973261.76 558702 6.58 4412)8 55 532 12 5.8i 973215.76 559097 6.57 440o3 5 56 532661 5.8i.97369.76' 55949 6.57 44o009 4 5f 533oo9~ 5.8o 973124.76 5S9885 6.56 440110.3 bo 1 5~~~~~~~~~~~~~~~~~~~~~~~~4052 5~~~~~~~~~~~~~~~~~~~5.0 1378 6.56 439-72 533357 5~0 97'8 76 507 59 533704 5.79 973032 56673 65 439327 6o 534052 5.78 972986.77 56io66 6.35 436934 0 Cosine I D. Sine I). o ID I ~(7~0 DIEG REEB.) (70 DEGRaEES.) 8 (20 DEGREES.) A TABLE OF LOGAR1THMIC Sine 1). Cosine D. Tang. D. Cotang. o 9.534052 5'78 9.972986 77 9'561o66 6.55 1.438934 6o 5343 577 972940.77 561459 6.54 43894i 59 5 3474 53 77 972894 77 56i85i 6.54 438149 58 3 535092 5' 77 972848'77 562244 6.53 437756 57 4 535438 5.76 972802.77 562636 6.53 437364 56 5 535783 5.76 972755 *77 563028 6.53 436072 55 536129 5.75 972709'77 563419 6.52 43658i 54 7 536474 5.74 972663.77 56381ii 6.52 436189 53 8 5368i8 5'74 972617.77 564202 6.5i 435798 52 9 537163 5.73 972570.77 564592 6.5i 435408 51i'77 45I 10 537507 5673 972524.77 564983 6o 43507 o 11 9.537851 5.72 9.972478'77 9.565373 6.50 10o-434627 4 12 538194 5.72 972431.78 565763 6.49 434.237 4 13 538538 5.71 972385.78 566i53 6.49 433847 47 14 53888o 5.71 972338.78 566542 6.49 433458 46 15 539223 5.70 972291.78 566932 6.48 433o68 45 16 539565 5.70 972245.78 567320 648 432680 44 J7 539907 5'69 I; 540249 5~69 97213 78 567709 6U47 432291 43 I 540249 5 97215.78 568098 6.47 431902 *42 19 540590 5~6 972105: -78 5684?6 6-46 43i?14 41 20 ~~~~~~~~~~~~~~~~~~~~4' 20 54093, 5.68 972058.78 568873 6.46 431127 40 21 9.54272 5.67 9972o0I.78 9'56926I 6.45 io.430739 22 54,6,3 5.67 971964.78 569648 6-45 430352 3 23 541953 5.66 9719'7'78 570035 6.45 429965 37 24 542293 5.66 971870.78 570422 6.44 429578 36 25 542632 5.65 971823.78 570809 6'44 429191 35 26 542971 5.65 971776.78 571195 6.43 428805 34 27 543310 5.64 971729'79 571581 6-43 428419 33 28 543649 5.64 971682'79 571967 6-42 428033 32 5.3 97i635 i'79 715 29 54398 563 97635 79 722 642 427648 3: 30 54432 5.63 971588 79 572738 642 427262 3 31 9.544663 5-62 9.971540.79 9.573123 6.4I 10o426877 2 32 545o000oo 5.62 971493.'79 573507 6-41 426493 28 33 545338 5.6i 9714'46'79 573892 6.40 426108 27 34 545674 5.6i 971398.79 574276 6.40 425724 26 35 546o011 5.60 971351'79 574660 6.39 425340 25 36 546347 5.60 971303.79 575044 6.39 424956 24 3; 546683 5.59 971256.79 575427 6.39 424573 23 547019 5.59 971208.79 575810 6.38 424190 22 39 547354 5.58 971161.79 576193 6.38 423807 21 40 547689 5.58 971113 I79 576576 6.37 423424 20 4, 9.548024 5.57 9'97'066'8o 9'576958 6.37 10.42304i 19 42 548359 5.57 97o1018.8o 577341 6.36 422659 43 548693 5.56 970970.8o 577723 6.36 422277 17 44 549027 5.56 970922.8o 578104 6.36 421896 i6 45 549360 5.55 970874.8o 578486 6.35 42154 15 46 549603 5.55 970827.8o 578867 6.35 421133 4 47 55oo0026 5.54 970779'8o 579248 6.34 420752 3 48 550359 5.54 970731.8o 579629 6.34 420371 12 49 550692 5.53 970683.8o 580009 6-34 41999 11 50 55024 5.53 970635.80 580389 6833 419611 10 Si 9.55356 5.52 9-970586.8o 9.580769 6-33 10o419231 52 515687 5.52 970538.8o 581149 6.32 419851 53 552018 5.52 970490 8o 581528 6.32 418472 7 54.552349 5.51 970442.80 581907 6.32 418093 6 55 5526807 5.5i 970394.8o 582286 6.3 417714 5 56 553oxo 5.50 970345.8i 582665 6.3i 41733S 4 57 553341 5.52 970297.81 583043 6.3o 416957 3 58 553670 5.49 970249.8i 583422 6.3o 41/678 2 59 5540o0 5.49 970200.8i 5838oo 6.I29 41620 I 6c 554329 5.48 9752.8i, 584177 6.29 45823 o 5-4 97,52 J. 584177 6.29 j4,5~23o Cosl D. Sine D. ine Cotang. D j T.ang. J M.j (69 DEGREES.) SINES AND TANGENTS. (21 DEGREES.) 39. Sine D. Cosine D. Tang. D. Cotang. 0 9.554329 5.48 9-970152 -8I 984177 6.29 io.45823 6o 55465 5.48 970103 *8i 584555 6.29 4I5445 5q ~ 2 554987 5.47 970055 -81 584932 6.28 4I5o68 58 3 5553I5 5.47 97oo0006.81 585309 6-28 414691 57 4 555643 5.46 969957 *81 585686 6.27 414314 56 5 555971 5.46 96999 0.8I 586062 6.27 413938 55 6 556299 5.45 969860.*8i 586439 6.27 413561 5i 7 556626 5.45 96981 1.81 58681 6.26 [413185,I 8 556953 5.44 969762.81 587190 6.26 412810 52 9 557280 5.44 969714.81 587566 6.25 412434 51 10 557606 5-43 969665.81 587941 6-25 412059 5o 11 9.557932 5-43 9.9696I6.82 9.588316 6-25 10-4II684 4 12 558258 5.43 969567.82 58869i 6-24 411309 4 13 558583 5.42 969518 -82 589066 6.24 410934 47 14 558909 5-42 969469.82 589440 6.23 41o56o 46 1 15 559234 5-41 969420 *82 589814 6-2J 4101o86 45 |6 559558 5.41 969370 -82 5o00188 6-23 409812 44 17 559883 5.40 969321 82 5952 6 22 49438 3 18 560207 5.40 969272.82 590935 6.22 409o65 42 19 56o53i 5.39 969223 -82 5913o8 6.22 408692 41 20 560855 5-39 969173.82 59168i 6.21 408319 40 21 9.561I78 5.38 9.969124.82 9.592054 6.21 10o407946 3 22 5650oi 5.38 969075.82 592426 6.20 407574 3 23 561824 5-37 969025.82 592798 6.20 407202 37 24 562146 5.37 968976.82 593170 6.19 40682 36 25 562468 5.3 968926 -83 593542 6i 9 406458 35 26 562790 5.36 968877.83 593914 6.I 8 406086 34 27 56312 5.36 968827.83 594285 6.x8 405715 33 28 563433 5.35 968777.83 594656 6.i8 405344 32 29 563755 5-35 968728.83 595027 6.17 404973 31 30 564075 5.34 968678.83 595398 6.7 404602 3o 31 9.564396 5.34 9.968628.83 9.595768 6.17 10.404232 2 32 564716 5.33 968578.83 596138 6. 6 403862 28 33 565036 5.33 968528.83 596508 6.-i6 403492 27 34 565356 5.32 968479 83 596878 6.i6 403122 26 35 565676 5.32 968429.83 597247 6. I 402753 25 36 |565995 5.3i 968379.83 597616 6.I5 402384 24 37 566314 5.3i 968329.83 597985 6.15 402015 23 38 566632 5.3i 968278 83 598354 6-. 4 40I646 22 39 566951 5.30o 968228.84 598722 6. i4 401278 21 40 567269 5.3o 968178.84 599091 6.I3 400909 20 41 9.567587 5.29 9-968128.84 9.599459 6.i3 10o-400oo541 19 42 567904 5.29 968078.84 599827 6.I3 4oo00173 18 43 |568222 5.28 968027.84 600ooI94 6.-i2 399806 I7 1 44 568539 5-28 967977.84 600562 6.12 399438 i6 4) 568856 528 967927.84 600929 6ii 399071 15 |46 |569172 5.27 96787.84' 601296 6.ii 398704 14 47 569488 5.27 967826.84 601662 6.-ii 398338 13 48 |569804 5-.26 967775.84 602029 6-.xo0 397971 12 49 570120 5.26 967725.84 602395 6.i0 397605 11 50 570435 5.25 967674.84 602761 6.io 397239 10 5I 9-57075I 5.25 9-967624.84 9.603127 6-og 1o-396873 52 57o1066 5.24 967573.84 603493 6.o9 396507 8 53 571380o 5.24 967522.85 603858 6.o0 396!42 7 54 57i695 5.23 967471.85 6o4223 6.o8 395777 6 1 55 572009!5.23 96742I.85 604588 6.o8 395412 5 56 572323 |5.23 967370.85 604953 6.07 3o7 4 572636 5.22 96731 S 605317 6-07 394683 3 57295b 5.22 967268. 85 6o5682 6.o7 394318 2 59 573263 52I 967217.85 606046 6.o6 393954!6o0 573575 5.21 96716.85 6064o10 6.o6 39359o1 0o Cosine D. Sine D. Cotang. D. (68 DEGREES.) to (22 DEGREES.) A TABLE OF LOGARITHMIC I'M..' Sine D. Cosine ID. Tang. D. Cotanlg. 0 9.573575 521 9-967166.85 1 9.606410 66 o39350 I 573888 5 20 967115.85 606773 6-o6 393227 59 2 574200 5.20 967064.85 607137 605 392863 5 3 574512 5 I9 967013.85 607500 6-o5 392500 57 4 574824 5-I9 966961 85 607863 6oo4 392137 56 5 i 575136 5-IC 966910.85 608225 6-o4 391775 55 6! 575447 5-Io 966859.85 6o8588 60o4 3911412 5. 7 575758 5.18 9668o8.85 608950 6 03 39Io5 53 576069 5.17 966-56.86 6o9312 6.03 390688 52 9 576379 5.17 966705.86 609674 6-03 *390326 51 10 5766;9 5-.i6 966653 *86 610036 6-02 389964 50 11 9.576999 5-16 9.966602 86 9610o397 6.02 10.38o603 12 5773of 5-I6 966550.86 61o759 6-02 389241 i 4 13 57761 5.15 966499 -86 611120 601 388880 47 14 577927 5.15 966447.86 611480 6.oi 388520 46 15 578236 5-14 966395.86 61I84I 6.o 388159 45 16 578545 5.i4 966344 *86 612201 60oo 387799 44 17 578853 5-13 966292 *86 61256 6oo00 387439 43 18 579162 5.1I3 966240.86 612921 6.oo 387079 42 19 579470 5.13 966188.86 61328I 5.99 386719 J 41 20 579777 5.12 966136 *86 613641 5.99 386359 4o 21 9-580085 5-12 9.966085.87 9'614000 5.98 Io.386ooo 3 22 580392 5.I I 966o33.87 6I4359 5.98 385641 38 23 58o6 5 96598I.87 614718 5.98 385282 37 58IO0 5.11 965928.87 615077 5.97 384923 36 25 58I312 5.IO 965876..8 615435 5.97 384565 35 26 58z168 5.o1 965824.87 615793 5-97 384207 34 581924 5.09 965772.87 61615i 5.96 383849 33 28 582229 5.09 965720.87 616509 5.96 38349i 32 29 582535 5.09 965668.87 616867 5.96 383I33 3 30 582840 5.o8 965615.87 617224 5.95 382776 3o 31 9.583145 5.08 9.965563.87 9-617582 5.95 o10382418 32 583449 5.07 965511.87 617939 5 95 33206I 33 583754 5.07 965458.87 618295 5.94 381705 27 34 584058 5 o6 965406.87 6i8652 5.94 381348 26 35 584361 5.0o6 965353.88 619008 5.94 380992 25 36 584665 5.06 965301.88 619364 5.93 380636 2a4 37 584968 5.05 965248.88 6I972I 5.93 380279 23 38 585272 5.05 965I95.88 620076 5.93 379924 22 39 585574 5.-04 965143.88 620432 592 379568 40 585877 5-04 965090.88 620787 5-92 379213 20 41 9*586I79 5-03 9.965037.88 9.621142 5.92 10o-378858 9 42 586482 50o3 964984.88 621497 5.91 378503 8 43 586783 5-o3 964931 -88 621852 5.9I 378I48 17, 44 587085 5o02 964879.88 622207 5.90 377793 i 16 45 587386 5.o02 964826.88 62256I 5.90 377439 15 46 587688 5.0I 964773.88 622915 5-90 377085 14 47 587989 5-0.o 964719.88 623269 5.89 376731 13 48 588289 50.o 964666.89 623623 5.89 376377 12 49 588590 5.00 964613.89 623976 5.80 37624 1 1 SO 588890 5.00 964560 -89 62433o 5.88 375670,0 I 5I 9.589190 4-99 9.964507.89 9.624683 5.88 10.375317 52 589419 4-99 964454.89 625036 5.88 374964 8 53 588978; 4-99 964400.89 625388 5.87 374612 54 590088 4-98 964347.89 625741 5-87 374259 S 55 590387 1 4-98 964294.89 626093 5.87 J73907 1 5 56 5906861 497 964240.89: 626445 5.86 373555 4 5957 590984 4-97 964187.89 626797 5.86 373203 3 58 1 591282 4-97 964I33 i*9 627149 5.86 37285I 2 59 59580 4.96 96408c *89 627501 5.85 372499 1 C 591878j 4-96 91 64o026.89 I 6278'52 5.85 3721418 Cosine D. Sine ID. Cotang. D. (67 DEGRREES.) T8JNE AND TANGENTS. (23 DEGRFIS.) 41 M. Sine D. Cosine D. Tang. D. Cotang. 6 9.591878- 4-96 9.964026.89 9.627852 5.85 Io0372148 6o I 592176! 495 963972.89 628203 585 37I797 59 2 592473 4.95 963909.89 628554 5.85 371446 58 3 592770 4.95 963865.9o 628905 5.84 371095 57 4 593067 4-94 963811.9o 629255 5-84 370745 56 5 593363 4'94 963757 -90 629606 5-83 370394 55 6 593659 4.93 963704.90 629956 5-83 370044 54 7 593955 4-93 963650.9o 63o3o6 5.83 369694 53 594251 4.93 963596.9o 630656 5*83 369344 52 9 594547 4.92 963542.90 631oo5 5.82 368995 51 10 594842 4.92 963488.9o 631355 5.82 368645 50 I 9.595137 4*9 1 9.963434 g90 9.631704 5.82 Io.368296 4 12 595432 4.9I 963379.90 632053 5.8I 367947 i3 595727 4-91 963325.9o 632401 5.8i 367599 47 1 4 596021 4.90 963271 90 632750 5.8i 367250 46 I5 596315 49go 963217..90 633098 5.80 366902 45'6 596609 4.89 963163.90 633447 5.80 366553 44 17 596903 4.89 963I08 *9I 633795 5-80 366205 43 18 597196 4.-89 963054'.9 634I43 5-79 365857 42 19 597490 4.88 962999.91 63440o 5'79 365510 41 20 597783 4.88 962945.91 634838 5-79 365I62 4o 21 9.598075 4-87 9.96280o *91 9.635I85 5.78 io.364815 39 22 598368 4.87 962836'9I 635532 5.78 364468 38 23 598660 4.87 96278I *91 635879 5-78 364121 3 24 598952 4.86 962727 9I1 636226 5.77 363774 36 25 599244 4-86 962672.91 636572 5'77 363428 35 26 599536 4.85 962617 *9I 636919 5.77 36308i 34 27 599827 4.85 962562 *91 637265 5 77 362735 33 28 600oo8 4.85 962508 -91 637611 5.76 362389 32 29 600409 4-84 962453.91 637956 5.76 362044 31 30 600700 4.84 962398'92 638302 5.76 36I698 3o 3 9.6o00990 4.84 9.962343.92 9.638647 5.75 o10361353 29 32 6o028o 4.83 962288.92 638992 5.75 36ioo8 28 33 601570 4.83 962233'92 639337 5.75 360663 27 34 60o86o 4.82 962178'92 639682 5-74 3603i8 26 35 602150 4.82 962123.92 640027 5'74 359973 25 36 602439 4.82 962067.92 640371 5'74 359629 24 37 602728 4.8I 962012.92 640716 5.73 359284:23 38 603017 4.8I 96I957.92 64io6o 5.73 358940 22 39 603305 4.8I 961902'92 641404 5.73 358596 21 40 603594 4.80 961846.92 64I747 5.72 358253 20 41 9.6o3882 4-80 9.96,791'92 9.64209I 5.72 10o.357909 I9 42 604170 4'79 961735.92 642434 5.72 357566 I 43 604457 4-79 96I680o 92 642777 5.72 357223 17 44 604745 4'79 961624.93 643 20 5-7' 356880 I6 45 605032 4.78 961569.93 643463 571I 356537 15 46 65319 4'78 961513.93 643806 5-7I 356i94 14 47 605606 4'78 961458 -93 644148 5.70 355852 13 48 605892 4'77 961402.93 644490 5.70 3555Io 12 49 606179 4'77 961346'93 644832 5.70 355I68 II 50 606465 4'76 961290.93 645 74 5.69 354826 10 5I 9.60675I 4.76 9.961235.93 9.6455I6 5.69 1o.354484 52 607036 4.76 961179.93 645857 5.69 354143 53 607322 4.75 961123.93 64699 5-69 3538o01 54 607607 4.75 96Io67.93 646540 568 353460 55 607892 4474 96o1011 93 64688i 5.68 353II9 5 56 60877 474 960955.93 647222 5-68 352778 4 5 608461 4'74 960899.93 647562 5.67 352438 3 5 608745 4-73 960843'94 647903 5.67 352097 2 49 60902 4.73 960786 -94 648243 5.67 351737 6. 60o313 4.73 960730 *94 648583 5-66 35,417 0 L C oesine ID. Sine i D. Cotang_.i D. _i Tang. _ _. (66 DEGREES.) 42 (24 DEGREES.) A TABLE OF LOGARITHMIC M. Sine D. Cosine D. Tang. D. Cotang. 0 9.6093i3 4.73 9-960730.94 9.648583 5-66 Io-35I417 6a I 609597 472 960674 94 648923 566 351077 2 60980 4-72 9606I8 94 649263 5.66 350737 5 3 6IoI64 4.72 96056I'94 649602 5.66 350398 57 4 610447 471 960505 94 649942 5.65 350058 56 5 610729 4.7I 960448'94 650281 5.65 349719 55 6 6I1012 4-.7 960392.94 650620 5.65 349380 54 6II294 4.70 960335 *94 650959 5-64 349041 53 61i576 4.70 960279'94 651297 5.64 348703 52 5 6iI858 4.69 960222 *94 65i6 6. 5.64 348364 51 10 612140 4.69 960I65.94 65I974 5.63 348026 50 II 9.612421 4.6o 9-960I09.95 9.652312 5-63 Io-347688 49 12 612702 4.68 960052.95 652650 5.63 347350 48 13 612983 4.68 959995.95 652988 5.63 347012 47 14 613264 4.67 955938.95 653326 5.62 346674 46 15 6I3545 4.67 959882.95 653663 5.62 346337 45 16 613825 4.67 959825.95 654000 5.62 346000 44 17 6I4Io5 4.66 959768.95 654337 5.6I 345663 43 18 614385 4.66 959711.95 654674 5.6i 345326 42 19 614665 4-66 959654.95 655oii 5-6i 344989 41 20 614944 4-65 959596.95 655348 5-6i 344652 4o 21 9*615223 4.65 9.959539.95 9. 655684 5*60 10*344316 3 22 615502 4@65 959482.95 656020 5.60 343980 34 23 6I5781 4.64 959425.95 656356 5-60 343644 33 24 6i6060 4.64 959368.95 656692 5.59 343308 31 25 6i6338 4.64 959310.96 657028 5.59 342972 38 26 6i66i6 4.63 959253.96 657364 5.59 342636 34 27; 6I.6894 4 63 9590105.96 657699 5* 5 34230I 33 2 67I72 4.62 959I j8 96 658o34 5.58 34I966 32 29 6I7450 4-62 959081.96 658369 5.58 34I631 31 30 617727 4.62 959023.96 658704 5.58 34I296 3o 31 9.618004 4.65 9 058965.96 9.659039 5*58 I 34o0961 29 32 6I828I 4-6i 9589o8.96 659373 5.57 340627 28 33 6I8558 4,6I 95885o.96 659708 5.57 340292 27 34 6I8834 4*60 958792 496 660042 5.57 339958 26 35 6i9iio 4.60 958734.96 660376 557 33964 25 36 619386 4J60 958677 96 660710 5.56 339290 24 37 6I9662 4 59 958619 *96 66I043 5.56 33857 23 3 6I9938 4.9 958561I 96 661377 5.56 338623 2" 39 620213 4.59 9585o3.97 661 710 5.55 338290 21 40 620488 4 58 958445 *97 662043 5.55 337957 20 4I 9.620763'4.58 9.958387.97 9.662376 5.55 10-337624 19 42 621038 4.57 958329.97 662709 5.54 3372 I 18 43 621313 4.57 958271.97 663042 5.54 336958 I7 44 621587 4-57 958213.97 663375 554 336625 Ib 45 62186I 4-56 958154.97 663707 5.54 336293 15 46 622135 4.56 958096.97 664039 5.53 33596I I4 622409 4.56 958038 97 664371 5.53 335629 i3 622682 4.55 957979 97 664703 5.53 335297 12 49 622956 4.55 95792.97 6650o35 5.53 334965 II 50 623229 4.55 957863.97 665366 5.52 334634 I0 5I 9.623502 4.54 9.957804.97 9.665697 5.52 I0.334303 9 52 623774 4.54 957746.98 666029 5.52 333071 I 53 624047 4.54 957687.98 66636o 5.5 333640 7 54 6243I9 4S53 957628.98 66669I 5.5I 333309 6 55 62459I 4.53 957570.98 667021 5.51 332g79 5 56 624863 4.53 95751I 98 667352 5.5I 332648 4 57 6235 4.52 957452.98 667682 5-50 33238 3 5 625406 4.52 957303 98 668oi3 5.50 33I987 2 59 625677 4.52 95 335 98 66833 5.50 33I657 I 62594 4.5i 95j276.98 668672 5-50 335328 0 Cosine D. Sine. D. Cotang. i D. j T 3g. M. (65 DEGREES.) SINES AND TANGENTS. (25 DEGREES.) 4 il. Sine |D. Cosine- D. Tang. D. Cotang. o 9-625948 4.51 9-957275.98 9.668673 5.50 IO.33I327 6o I 626219 4. 5,1 9572I7.98 669002 5.49 330998 5 626490 4.5i 957158 *98 669332 5.49 33o668 58 3 62676o 4-50 957099 98 669661 5.49 330339 57 4 627030 4-50 957040.98 669991 5.48 330009 56 5 627300 4.50 956981 *98 670320 5-48 1 329680 55 6 627570 4'49 956921 *99 670649 5-48 32935I 54 7 627840 449 956862.99 670977 5.48 329023 53 8 628109 449 956803 *99 671306 5-'47 328694 52 9 628378 4.48 956744 *99 671634 5'47 328366 5i 1 628647 4.48 956684 *99 671963 5.47 328037 50 11 9 528916 4-47 9.956625 99 9.672291 5-47 jIo327709 49 1 j 629185 4-47 956566 *99 6726i9 5.46 327381 48 13 629453 4-47 956506 99 672947 5.46 327053 47 14 629 19 1 4-46 956447.99 673274 5.46 326726 46 I 5 629989 4*46 956387 *99 673602 5.46 326398 45 16 63025- 4-46 956327 *99 673929 5.45 326071 44 17 630524 4-46 956268 99 674257 5-45 325743 43 18 630792 4.45 956208 1*00 674584 5.45 3254I6 42 I9 63I039 4*45 956I 48 I00 674910 5.44 325090 41 20 631326 4.45 956089 I00 675237 5*44 324763 40 21 ~ 63I 53 4.44 9.956029 1*00 9-675564 5.44 10- 324436 39 2 630i85 4.44 955969 1I00 675890 5.44 3241IO 38 23 632125 4-44 955909 1*00 676216 5.43 323784 37 24 632392 4-43 955849 1.00 676543 5.43 323457 36 25 632658 4-43 955789 1.00 676869 5-43 323131 35 26 632923 4.43 955729 1.00 677194 5.43 322806 34 27 633189 4-42 955669 Ioo00 677520 5.42 322480 33 28 633454 4-42 955609 100oo 67846 5.42 322154 32 29 633719 4.42 955548 1o00 678171 5.42 321829 31 30 633984 4.4i 955488 I.00 678496 5-42 321504 30 3I 9-634249 4'4I 9.955428 IoI 9-678821 54I Io.32I179 29 32 6345I4 4'40 955368 I- o 679146 5.4I 320854 28 33 634778 4.40 955307 I-01 679471 5.41 320529 27 34 635042 4640 955247 101 679795 5.41 32020 26 35 635306 4.39 955I86 101 660120 5-40 319880 25 36 635570 4.39 955126 1-01 680444 5.40 319556 24 37 635834 4.39 955o65 I-O 680768 5.40 319232 23 38 636097 438 955005 IOI 681092 5.40 3189o8 22 39 63636o 4.38 954944 1I01 6814I6 5.39 3/884 21 |40 636623 4.38 954883 1|0I 681740 5.39 318260 20 41 9-636886 4.37 9-954823 | 0OI 9.682063| 5-3 io-317937 1 42 637148 4.37 954762 1-01 682387 5-3 37613 i 43 63741 4.37 5 I 101 682710 5-38 317290 17 44 637673 4.37 56 1 68333 38 31667 45 637935 4.36 5575 1o0 683356 5.38 316644 I5 46 638107 4.36 954518 1-02 683679 5.38 316321 14 47 63848 4-36 954457 1-02 68400I 5.37 35999 I3 48 638720 4.35 954396 102 684324 537 315676 12 49 63898I 4-35 95433 I-02 684646 5.37 35354 1i 50 639242 4.35 954274 1-02 684968 5.37 315032 10 51 9-639503 4.34 9-954213 102 9.685290 5.36 o10.34710 52 639764 4-34 954152 1-02 685612 5-36 314388 8 53 640024 4.34 954090 02 685934 5.36 314066 7 54 640284 4-33 954029 1-02 686255 536 3I3745 6 i 55 640544 -33 953968 1-02 686577 5.35 313423 5 56 640804 4-33 539o6 I-02 686898 5.35 313102 4 57 64io64 4,.32 / 3 57 |6o64 |4-32 953845 102 |687219 5-35 312781 3 58 641324.32 93783 1502 687540 5-35 312460 2 641584 4.32 953722 i -o3 687861 5.34 3I239 | 6 641842 4.3 95366o 1(3 688821 5-.34 3188 0 Cosine D 3. Sine 1..otano.. Tan. 7 (641 DEGREES.) 4:4 (26 DEGREES.) A TABLE OF LOGARITHMIC M. Sine D. Cosine D. Tang. D. Cotang. 0o 9-64I842 4-3I 9-953660 i.o3 9.688182 5.34 io03118I8 6o I 642101 4.3i 953599 I103 688502 5-34 3I1498 i5 2 642360 4-3i 953537 1*03 688823 5.34 311177 5S 3 642618 4*30 953475 i 03 689143 5.33 310857 57 4 642877 4-30 953413 I-o3 689463 5.33 3I0537 56 5 643I35 4-30 953352 i*o3 689783 5*33 310217 5| 6 643393 4.30 953290 I*o3 6 l oo3 5.33 309897 54 643650 4.29 953228 I*03 690423 5-33 309577 53 8 643908 4-29 953166 I*o3 690742 5.32 309258 52 9 644165 4.29 953104 I.o3 691062 5.32 308938 51 o10 644423 4-28 953042 i-o3 69i381 5.32 308619 50o II 9-644680 4.28 9.952980 I*o4 9*691700 5.31 o3083o00 49 12 644936 4-28 952918 1.04 692019 5.31 307981 48 I3 645193 4.-27 952855 1-04 692338 5-3i 307662 47 14 645450 4-27 952793 1I04 692656 5-3i 307344 46 15 645706 4-27 952731 1I04 692975 5*3i 307025 45 i6 645962 4-26 952669 Io04 693293 5-3o 306707 44 I7 646218 4-26 952606 1*04 693612 5*3o 3o6388 43 I8 646474 4.26 952544 1.04 693930 5-30 306070 42 19 646729 4-25 952481 1.o04 694248 5.30 305752 41 20 646984 4.25 952419 1.04 694566 5.29 305434 40 21 9 647240 4-25 9-952356 1*04 9-694883 5.29 E*305I7 39 22 647494 4-24 952294 I*04 695201 5-29 304799 38 23 647749 4-24 95223I I1 04 695518 5.29 304482 37 24 648004 4-24 952168 I.05 695836 5.29 304i64 3 25 648258 4-24 952106 Io05 696153 5-28 303847 35 26 648512 4-23 952043 1*05 696470 5-28 303530 34 2 648766 4-23 951980 Io5 696787 5.28 3032I3 33 649020 4-23 95117 Io5 697Io3 5-28 30289732 29 649274 4.22 951854 I*o5 697420 5-27 3025 80 31 30 649527 4.22 951791 IIo5 697736 5*27 302264 30 3i 9.64978I 4.22 9*951728 io5 9.698053 5.27 ID.30o194 2 32 65oo0034 422 951665 1o 698369 5-27 301631 2 33 650287 4.21 95602 Io5 698685 5. 26 3o0I 3 27 34 650539 4-21 95I539 io5 6990o1 5.26 3o.o95 26 35 650792 4-21 951476 I.o5 6993i6 5.26 3oo684 25 36 65i044 4-20 951412 1.05 699632 5.26 300368 24 37 651297 4-20 951349 I*o6 699947 5-26 300053 23 38 651549 4.20 951286 I*o6.700263 5.25 299737 22 39 651800oo 4I19 951222 I*o6 700578 5-25 299422 21 40 652052 4 19 951159 i*o6 7cLJ893 5-25 299107 20 4I 9.652304 4- 9.95Io096 i.o6 9.701208 5*24 10.298792 I9 42 652555 4.18 951032 I*o6 701523 5-24 298477 18 43 652806 4-18 950968 i*o6 701837 5*24 298163 1 7 44 653057 4-I8 950905 I*o6 702152 5-24 297848 I6 45 653308 4-I8 950841 I*o6 702466 5-24 297534 15 46 653558 4.I7 950778 i.o6 702780 5-23 297220 14 47 653808 4-17 950714 I*o6 703095 5.23 2069o5 13 48 654059 4-17 950650 1-06 7o3409 5-23 2o0391 12 49 65430 4. 1 95o586 i.o6 703723 5-23 296277 1 50 654558 44I6 950522 1.07 704036 5-22 295964 10 51 9,654808 4.16 9-950458 1.07'9.704350 5*22 10o-295650 52 655058 4-I6 950394 107 704663 5-22 295337 53 655307 4-15 950330 1I07 704977 5.22 295023 7 54 655556 41I5 950266 1.07 705290 5.22 294710 6 55 655805 4.15 950202 1-07 705603 5-21 294397 5 56 656054 4.I4 950138 1-07 705916 5-21 294084 4 5 656302 4-14 950074 I-07 706,j28 5-.21| 293772 3 65655i 4 I4 9500io 1.07 -0654 5-2I1 293459 2 59 656799 4.i3 949945 1*07 706854 5*21 293146 I 60o 657047 4-i3 94988i 1-07 707I66 5-20 292834 0 Coi ine D. _ _ _ _ _ _ _ _ _ _ _ _ _ ____.o6__ Cosine D. Sine. D. Cotang.. I Tang. M. (6 3 DEGREES.) SINES AND TANGENTS. (27 DEGREES.) 46 M. Sine D. Cosine D Tang. D. (Cotang. O 9.657047 4 13 9-949881 1.o7 9.707166 5 20 10.292834 60 I 657295 4- 3 9498I6 1 07 707478 5.20 292522 59 2 657542 4-12 949752 1 07 707790 5 20 292210 58 3 657790 4-12 949688 I o8 708102 5.20 291898 57 4 658037 4-12 949623 I o8 708414 5-I9 291586 56 5 658284 4-I2 949558 I o8 708726 5-I9 291274 55 6 658531 4- i 949494 I o8 709037 5.19 290963 54 658778 4-II 949429 1-08 709349 5.19 29065I 53 659025 4- I 949364 I o8 709660 5-I9 290340 52 9 659271 4-IO 9493oo00 o8 70997I 5.8 2o0029 51 10 659517 4'io 949235 IJo8 710282 5.I8 289718 50 11 9.659763 4'io 9'94917o I*08 97'I0593 5.I8 0o*289407 12 66ooo 4-09 949105 i o8 710904 5.I8 289096 13 h625 4' o g0 949040 I o8 711215 5.I8 288785 47 i4 66050 I 4o09 948975 I o8 711525 5.17 288475 46 15 660746 4'o6 94891o o.08'711836 5-17 288164 45 i6 660991 4.08 948845 I o8 712146 5-17 287854 44'7 661236 4'o8 948780 Io09 712456 5-17 287544 43 18 66148I 4-o8 9487I5 Io09 712766 5.16 287234 42 19 661726 4 07 948650 1.09 713076 5. 6 286924 41 20 661970 4-07 948584 og09 713386 5.16 286614 40 21 9-6622I4 4'07 9-9485I9 1.09 9'7I3696 5.I6 10-286304 22 662459 4o07 948454 I 9o 714005 5.i6 285995 23 662703 4'06 948388 109 714314 5.I5 285686 37 24 662946 4-o6 948323 I-09 714624 5.I5 285376 36 25 663 90 4-o6 948257 1I09 714933 5.I5 285067 35 26 663433 405 948192 I09 715242 5.5 284758 34 27 663677 4o05 948126 I*09 715551 5.14 284449 33 28 663920 405 948060 109 715860 5-I4 284140 32 29 664163 4o05 947995 I* O 7I6168 5-I4 283832 31 30 664406 404 947929 IIO 716477 5-I4 283523 30o 3i 9.664648 404 -947863 I-10 9.76785 5-4 10-283215 2 32 66489I 4.04 947797 1.10 717093 5.I3 282907 28 33 665i33 4o03 947731 1-Io 717401 5.I3 282399 27 34 665375 4-03 947665 1 -1 717709 5.I3 282291 26 35 665617 403 976oo I IO 7180I7 5.I3 281983 25 36 665859 4'02 947533 I.Io 71832 5.i3 281670 24 37'666Ioo 40o2 947467 10 718633 5.12 281367 23 3 666342 4'02 947401 I 10 7I8940 5.I2 281060 22 39 666583 4o02 947335 I-Io 719248 5.12 280752 21 40 666824 4-ox 947269 1-10 719555 5.12 280445 20 4' 9.667065 401 9-947203 1-IO 9-719862 5.12 10.280138 19 42 667305 4-oi 947136 I.II 720169 5-II 279831 18 43 667546 4.o0 947070 1-1I 720476 5-II 279524 17 44 667786 4oo00 947004 1 11 720783 5-II 279217 i6 45 668027 400oo 946937 1 II 721089 5.II 27II91 15 46 668267 4-oo 946871 111I 721396 5II 278604 14 47 668506 399 946804 1 11 721702 5.IO 278298 13 48 668746 399 946738 I I1 722009 5,10 27791 12 49 668986 3-99 946671 I-II 722315 5-10 277685 i 50 669225 3-99 946604 I*II 722621 5-IO 277379 Io 5 9-669464 3-98 9-946538 IJ, 9-722927 5.iO 10277073 52 6697o3 398 946471 I11 723232 5.09 276768 53 669942 398 946404 I11 723538 5.09 276462 54 67018I 3.97 946337 1III 723844 5.09 276156 | 55 670419 3 97 946270 I I2 72411/9 to9 275851 5 I 670658 3:97 946203 1I 2 724454 5-o9 275546 4 5 670896 3.97 946136 I 12 724759 5.08 275241 3 58 671134 3-96 946069 I-12 725656 5.o8 274935 2 5 1 671372 3-96 946002 1*12 725369 5-o8 274631 I 60 671609 396 945935 1I2 725674 5o8 274326 o Cosin I D. —,e D.. C.otag D. Tang. Ii. (62 DEGREES.) 46 (28 DEGREES.) A TABLE OF LOGARITHMIC M. Sine D. Cosine D. Tang. ). Cota 0 9.676o09 3.96 9.945935 1.12 19.725674 5.o8 10.274326 6o 1 671847 3.95 945 68 I.12 725979 5.o8 2740O2I 13 2 672084 3.95 945800 I.12 726284 5.07 273716 58 3 672321 3.95 945733. 1.12 726588 5.07 273412 57 4 672558 3.95 945666 I.I2 726892 5.07 2731o8 56 5 672795 3.94 945598 1.12 727197 5.07 2728o3 55 6 I 673032 3.94 94553I 1.12 727501 5.07 272492 54 673268 3.94 945464 i.i3 727805 5.o6 27219~ 53 6735o5 3.94 945396 I-I3 728109 5-o6 27189I 52 9 67374i 3.93 945328 1.13 728412 5.o6 271588 51 to 673977 3.93 94526i I.13 728716 5o6 271284 5o 9'674213 3.93 9.945193 I3 9'729020 J -o6 j0.27o98 49 12 674448 3.92 945125 i-I3 729323 5.05 270677 i 48 13 674684 3.92 945058 1-.3 729626 5.o5 27o374 4'7 14 674919 3.92 944990 1.13 729929 5.05 270071 46 15 675153 3.92 944922 1.i3 730233 505 269767 45 16 675390 3.91 944854 i.I3 73o535 5.o5 269465 44 17 675624 3.91 944786 x.i3 73o838 5.o4 269162 143 18 675859 3.9i 944718 i.i3 73II4X 5,04 [.268859 42 676094 3.9i 944650.13 73,444 5.o4 268556 41 20 676328 3.90 944582 1.14 731746 5.04 268254 40 21 9.676562 3.90 9'944514 1.14;,732048 5-o4 10o267952 39 22 676796 3.90 944446 1.14 73235I 5.o3 267649 i 38 i3 677030 3. O 944377 1.14 732653 5.oA 267347 37 24 677264 3.89 944309 1-14 732955 5.o3 267045 36 677498 3.89 944241 I.14 733257 5.o3 266743 35 2 677731 3.8 944172 1-14 733558 5-o3 266442 34 27 677964 3.88 944104 1.14 73386o 5.02 266140 i 33 28 678197 3.88 944o36 1.14 734i62, 5.02 265838 32 29 67843o 3.88 943967 1.14 734463 5.o2 265537 31 3o 678663 3.88 943899 1',4 734764 5.02 265236 3o 31 9.678895 3.87 9.94383o 1.14 9.735o66 5.02 10.264934 29 32 679128 3.87 943761 1J 14 735367 5.02 264633 28 33 679360 3.87 943693 J.iS 735668 5.oi 264332 I 27 34 679592 3.87 943624 i-i5 735969 5.oi 264o31 26 5 69864 3.86 943555 i.5S 736269 5j oi 26373. 25 35~~~~~~~~~~~~~~~~~~67' I 25 J 6oo6 6 3.86 943486 i.iS 736570 5.oi 263430 24 37 680288 3.86 943417 1.5'736871 5.oi 263129 23 68o519 3.85 943348 i.15 737171 5.00 262829 22 39 680750 3.85 943279 1.15 737471 5.00 262529 21 40 68o982 3.85 943210 ~.1S 737771 5.00 262229 20 41 9.681213 3.85 9.943141 J 15 i 9-738071 5.o0 10.261929 19 42 681443 3.84 943072 I.5 738371 5-00 261629 18 43 681674 3.84 943003 I.15 738671 4'99 261329 17 44 6819o5 3.84 942934'.i5 738971 4'99 261029 i6 45 682135 3.84 942:864 i-0 739271 499 260729 J 46 682365 3.83 942795 i.i6 739570 499 260430 14 4 682595 3.83 942726 i.i6 739870 4'99 26o03o 13 4i 682825 1 3.83 942656 i.i6 740169 4'99 25983i 12 i9 683o55 3.83 942587 i.i6 740468 4.98 259532 JI 5o 683284 3.82 942517 I.i6 740767 4.98 259233 10 96834 3.82 9942448 ii6 974066 498 I.o258934 9 683743 3.82 942378 ii6 741365 4.98 258635 8 53 683972 3.82 9423o8 i.i6 741664 4.98 258336 7 54 684201 3.8i 942239 J 1-6 741962 4'97 258o38 6 55 684430 3.8i 942169 i.i6 742261 4'97 257739 5 56 684658 3.8i 942099 11.6 742559 497 257441 57 684887 3.8o 942029 iiJ6 742858 4-97 257142 3 58 685ziS 3.8o 941959 iJ i6 743156 4'97 256844 2 685343 3.8o 941889'117 743454 4'97 256546 i J 68'571 3.8o 941819 1.17 743752 4.96 256248 0 Cosine D. Sine D. Cotang. j jD. Ta ng. (61 DEGREEES.) SINES AND TANGENTS. (29 DEGREES.) 47. Sine D. I Cosine D. Tang. I). Cotang. o 9.68557I 3-80 9-94189 1-I17 9-743752 4.96 lo-256248 6o I 685799 3.7$ 941749 I'17 744050 496 255950 59 2 686027 3.79 941679 1.17 744348 496 2556')2 5 3 686254 3.79 941609 1-17 744645 4.96 255?55 I57 4 686482 37 7 941539 I*17 744943 4.96 255057 56 5 686709 3-78 941460 117 745240 4.96 254760 55 6 686936 3.78 941398 I'17 745538 4.95 254462 54 687163 3.78 941328 I.17 745835 4.95 254165 53 687389 3-78 941258 1I.7 746132 4.95 253868 52 9 687616 3.77 941187 II7 746429 4-95 253571 5I 10 687843 3.77 94II1117 I17 746726 4.95 253274 50 I 9.688069 3'77 9.941046 ii8 9.747023 4.94 IO252977 II 12 688295 3.77 940975 I8 74739 494 25268 4 I3 688521 3.76 940905 I*I8 747616 4-94 252384 47'14 688747 3.76 940834 I.I8 747913 494 252087 46 I5 688972 3.76 940763 I.I8 748209 4-94 25179I 45 i6 689198 3.76 940693 I*I8 748505 4.93 251495 44 I7 689423 3.75 940622 I*I8 748801oi 4.93 251199 43 i8 689648 3.75 940551 I.i8 749097 4.93 250903 42 19 689873 3.75 940480 I.i8 749393 4.93 2506o7 4I 20 690o098 3.75 940409 I-I8 749689 4.93 25031I 40 21 9'690323 3.74 9.940338 1-I8 9.749985 4.93 10-250015 39 22 690548 3.74 940267 I.I8 750281 4.92 249719 3 23 690772 3.74 940196 I*I8 750576 4-92 249424 37 24 690996 3.74 940125 I*19 750872 4.92 249128 36 25 691220 3.73 940054 II9 751167 4-92 248833 35 26 691444 3.73 939982 I.I9 751462 4-92 248538 34 691668 3-73 9399II 1.19 751757 4-92 248243 33 2 69I892 3.73 939840 I'19 752052 4.9I 247948 32 29 692115 3.72 939768 I I9 752347 4.91 247653 31 3o 692339 3.72 939697 I*19 752642 4-91 247358 30 31 9.692562 3.72 9.939625 I-19 9-752937 4-9' 10o247063 29 32 692785 3-7I 939554 1-19 75323I 4-91 246769 28 33 693008 3.7i 939482 I119 753526 4-91 246474 27 34 69323I 3.71 93941o I*19 753820 4-90 246180 26 35 693453 3.71 939339 I1.9 754115 4-90 245885 25 36 693676 3.70 939267 1,20 754409 4-90 245591 24 37 693898 3-70 939i95 1-20 754703 4.90 245297 23 3 694120 3.70 939123 I.20 754997 4-90 245003 22 39 694342 3.7 93 I120 75529I 4.9~ 244709 21 40 694564 3.69 938980 120 755585 4.89 2444I5 20 41 9.694786 3.69 9.938908 120 9.755878 4.89 10-244122 19 42 695007 3.69 938836'1 -20. 756172 4.89 243828 43 695229 3.69 938763 I-20 756465 4-89 243535'7 44 695450 3.68 938691 I-20 756759 4'-89 24324I 1 45 695671 3.68 938619 I'20 757052 4-89 2429 48 15 46 695892' 3.68 938547 1-20 757345 4.88 242655 4 47 696II3 3.68 938475 1.20 757638 4.88 242362 13 4, / 696334 3.67 933402 I-21 75793I 4-88 242o69 12 49 696554 3.67 938330 1.21 758224 4.88 241776 11 o5 696775 3.67 938258 I.2I 7585I7 4.88 241483 to 5 9.696995 3.67 9.938I85 1.2I 9-7588Io 4.88 10241190 9 52 697215 3.66 938113 1.2I 759102 4.87 240898 53 | 697435 3.66 938040 1.21 759395 4.87 a40605 7 5 697654 3.66 937967 1.21 759687 4.87 240313 6 55 j $7787 / 3.66 937895 1.2I 759979 4.87 240021 5 56 I 696o9$ 3.65 937822 I.2I 760272 | 4-87 239728 4 57 698313 3.65 937749 1.21 760564 4.87 239436 3 58 698532 3.65 937676 1.21 760856 4.86 239144 2 i9 69875I 3.65 937604 1.21 761148 4-86 238852 I 60 698970 3.64 93753I 1-21 761439 4.86 238561 0 Cosine -D. I Sine D. i Cotang. I D. Tan. 28 (60 DOGRBrA.) 48 (30 DEGREES.) ATABLEOF LOGARITHMIC M. Shie D. Cosine D. Tang. D. Cotang. - 9-698970 3.64 9.937531 1 21 9'761439 4.86 1o.23856i 6o I 699189 3.64 937458 22 76173 486 238269 2 699407 3.64 937385 1.22 762023 4.86 237977 3 699626 3.64 937312 122 762314 4'86 237686 57 4 699844 3.63 937238 1.22 762606 4'85 237394 56 5 700062 3.63 937165 I122 762897 4.85 2371o3 55 6 700280 3.63 937092 1'22 763188 4.85 236812 54 700498 3.63 937019 I;22 763479 4.85 236521 53 700o716 3.63 936946 1.22 763770 4.85 236230 52 9 700933 3.62 936 72 122 76406i.85 235939 51 10 7011oi51 3.62 936799 1-22 764352 4.84 235648 50 11 9.70368 3.62 9.936725 1.22 o 764643 4.84 I0.235357 4 12 701585 3.62 936652 1.23 764933 4.84 235067 4 t3 70o802 3.6I 936578 1. 23 765224 4-84 234776 47 14 7020I9 3.6I 936505 1.23 765514 4.84 234486 46 15 702236 3.6i 93643I 1.23 765805 4.84 23,/95 45 16 702452 3.6i 936357 1.23 766095 4-84 233905 44 17 702669 3.60 936284 1I 23 766385 4.83 233615 43 i8 702885 3-6o 936210 I.23 766675 4-83 233325 42 19 7031o01 3.60 936136 1.23 766965 4.83 233035 41 o20 703317 3.6o 936062 I123 767255 483 232745 40 2! 9.703533 3.59 9.935988 1.23 9.767545 4.83 10.232455 3 22 703749 3.59 935914 1.23 767834 4.83 232166 3 23 703964 3.59 935840 1.23 768124 4-82 231876 1 37 24 704179 3.59 935766 1.24 768413 4.82 231587 36 25 704395 3.59 935692 1.24 768703 4.82 231297 35 26 7046o10 3.5 935618 1.24 768992 482 23100oo8 34 27 704825 3.58 935543 1.24 769281 4'82 230719 33 28 705040 3.58 935469 124 769570 4'82 230430o 32 29 705254 3.58 935395 1.24 769860 4-8i 230140 31 30 705469 3.57 935320 1.24 770148 4.8i 22Q852 30 3, 9.705683 3.57 9.935246 124 9'.770437 4-81 10229563 2 32 705898 3.57 93517I 1'24 770726 4.8i 229274 28 33 70611I2 3.57 935097 1-24 771015 4.81 228985 27 34 706326 3.56 935022 1.24 771303 4.8I 228697 26 35 706539 3.56 934948 1.24 771592 4.8I 228408 25 36 706753 3;56 934873 1.24 771880 4.80 228120 24 3; 706967 3.56 934798 1'25 7~2168 4'8o 227832 23 38 707180 3.55 934723 1-25 772457 480o 227543 22 39 707393 3.55 934649 125 772745 480 227255 21 40 707606 3.55 934574 1.25 773033 4.80 226967 20 41 9.707819 3.55 9.934499 1.25 9-773321 48o 10.226679 19 42 708032 3.54 934424 1.25 773608 4'79 226392 1 43 708245 3'54 934349 1.25 773896 4'79 22610o4 17 44 708458 3.54 934274 1.25 774184 4 -'79 225816 6 45 708670 3.54 934199 25 774471 479 225529 15 ~~~~~~~49./7467 ] 1'7925 225529 4 46 708882 3.53 934i23.25 774759 4'79 225241 14 709094 3.53 934048 1.25 775046 4-79 224954 12 4 709306 3.53 933973 1.25 775333 4' 7 224667 12 49 709518 3.53 933898 1.26 775621 4'78' 224379 11 50 709730 3.53 933822 I.26 775908 478 2240921 To 51 970.)4 3.52 9.933747 1.26 9.776i 5 4'78 10.223805 52 7IO153 3.52 933671.1.26 776482 4-78 223518 53 70364 3.52 933596 1'26 776769 4'78 22323 1 7 54 710575 3.52 933520 1.26 777055 4-78 222945 6 55 710786 3.51 933445 1.26 777342 4'78 222658 5 56 710997 3.5I 933369 i.26 777628 4.77 222372 4 5 711208 3.5i 933293 1.26 777915 4'77 222085 3 58 711419 3.5i 933217 I.26 778201 477 221799 2 59 711629 3.50o 933141 I.26 778487 4'77 221512 I 6o 711839 3.So 933066 1.26 778774 4-77 221226 0 _D., — l-r-J J7,, Tang. j~if~ Cosine. Sine D. Cotang.. Tag. (15 DEGRIEES.) SINES AND TANGENTS. (31 DEGREES.) 49 M. ine D. Cosine D. Tang.| D. Cotan. 0 9.711839 350o 9.933066 I126 9.778774 IO-22I22 6 1 712050 3.5o 932990 1-27 779060 4 77 220940 2 712260 3 50 932914 I27 779346 4.76 220654 3 712469 3.49 932838 I.27 779632 4.76 220368 57 4 712679 3.49 932762 1'27 779918 4.76 220082 56 5 712889 3.49 932685 1.27 780203 4.76 219797 55 6 713098 3.49 932609 1.27 780489 4.76 219511 54 7 i 713308 3-49 932533 1.27 780775 4.76 219225 53 713517 3.4 932457 1'27 781060 4.76 218940 52 9 713726 3.48 932380 I127 781346 4.75 218654 51 10 713935 3.48 932304 1'27 781631 4.75 218369 50 II 9-714I44 3.48 9.932228 1.27 9-78I9I6 4.75 IO.218084 4 12 714352 3-47 93215I 1.27 782201 4.75 217799 48 13 714561 3.47 932075 I.28 782486 4.75 217514 47 14 714769 3-47 931998 I128 782771 4.75 217229 46 15 714978 3*47 931921 I128 783056 4.75 216944 45 16 715186 3 47 O931845 I*28 78334I 4.75 216659 44 2 7I5394/ 3.436 8 1*28 783626 4.74 216374 43 7I602 3.46 13-Ib 9 I*28 783910 4-74 216090 42 19 715809 3.46 931614 I128 784195 4.74 215805 41 20 716017 3.46 931537 1-28 784479 4.74 215521 40 21 9.716224 3.45 9-931460 1.28 9-784764 474 IO.2I5236 39 22 716432 3.45 931383 1.28 785048 4 74 214952 38 23 716639 3-45 931306 I128 785332 4.73 214668 37 24 716846 3.45 931229 1.29 785616 4.73 214384 36 25 7I7053 3.45 93115* I*29 785900 4.73 214100 35 26 -17259 3-44 931075 I-29 786184 4.73 213816 34 27 7 7466 3*44 930998 1*29 786468 4.73 213532 33 28 717673 3.44 930921 1*29 786752 4-3 213248 32 29 717879 3*44 930843 1I29 787036 4.73 212964 31 30 718085 3.43 930766 1.29 787319 4.72 212681 3o 3i 9-718291 3.43 9.930688 129 9.787603 4.72 10-212397 29 32 718497 3.43 930611 I.29 787886 4.72 212114 28 33 718703 3-43 930533 1.29 788170 4.72 211830 27 34 718909 3.43 930456 1.29 788453 4.72 211547 26 35 719114 3.42 930378 1*29 788736 4.72 211264 25' 36 719320 3.42 930300 i.30 789019 4.72 210981 24 37 719525 3.42 930223 I.30 789302 4.7r 210698 23 719730 3.42 930145 I.30 789585 4.71 210415 22 39 719935 3.41 930067 I.30 789868 4.7' 210132 21 40 720140 34I1 929989 i.30 790151 471I 209849 20 4I 9-720345 3.4I1 9-929911 I.30 9.790433 4.7I 10*209567 19 42 720549 3*41 929833 i.30 790716 4.71 209284 18 43 720754 3.40 929755 I.30 790999 4.7I 209001 17 44 720958 3.40 929677 I.30 791281 4.7I 208719 i6 45 721162 3.40 929599 I.30 791563 4.70 208437 15 4i6 721366 3.40 92952I I.30 791846 4.70 208154 14 47 721570 3-40 929442 I.3O 792128 4.70 207872 13 48 721774 3.39 929364 I-31 792410 4.70 207390 12 49 721978 3.39 929286 i-31 792692 4-70 207308 I i 50 722181 3.39 929207 I-31 792974 4.70 207026 10 51i 9-722385 3.39 9-929129 I.3i 9.793256 4.70 O10 206744 9 52 722588 3.39 929050 I.3I 793538 4.69 206462 53 72279I1 3.38 928972 I.3i 7938I9 4.69 206181 7 54 722994 3 38 928893 I*3i 794101 4.69 205899 6 5 723197 3.38 928815 I.31 794383 4.69 205617 5 56 723400 3.38 928736 I*31 794664 4.69 205336 4 57 723603 3.37 928657 1-31 794945 4.69 205055 3 58 723805 3.37 928578 i.3i 795227 4-69 204773 2 59 724007 337 928499 i.-3 795508 4.68 204492 1 60 724210 3.37 928420 3I 795789 4.68 204211 0 Cosine D. Sine D.i Cotan13 D. | Tann. M. (58 DEGREES.) 50 (32 DEGREES.) A TABLE OF LOGARITHIMIC M. Sine D. Cosin D. I Tang. D. Cotang. o- 9 724210 3.37 9 928420 I.32 9-795789 4-68 lo2o4aII I 1 724412 3.37 928342 1I32 796070 4.68 203930 5o 724614 3-36 928263 I.32 796351 4-68 203649 58 3 724816 3.36 928183 1-32 796632 4.68 203368 5 4 1250!7 3.36 928104 I.32 796913 4.68 203087 5 125219 3-36 928025 I.32 797194 4.68 202806 55 6 725420 3.35 927946 1.32 797475 4-68 202525 54 7 725622 3.35 927867.32 797755 4.68 202245 53 8 725823 3-35 927787 I.32 798036 4-67 20o964 52 726024 3.35 927708 1.32 798316 4-67 20o684 j 51 I0 726225 3.35 927629 1-32 798596 4-67 201404 50 11 (,I'72642t 3.34 9. 927549 I-32 9.798877 4.67 I0201123 44 12 726626 3-34 927470 i33 799157 4.67 200843 48 I3 726827 3.34 927390 I*33 799437 4-67 200563 47 I4 727027 3.34 9273o10 I33 799717 4-67 200283 46 i5 727228 3.34 92723i I-33 99997 4-66 200003 45 i6 727428 3.33 92715I I.33;00277 4-66 199723 44 I7 727628 3.33 927071 II33 800557 4-66 199443 43 18 727828 3.33 92699I 133 8oo836 4-66 199 64 42 19 728027 3.33 926911 I.33 8oi6 4-66 198884 41 20 728227 3.33 926831 I.33 801396 4-66 198604 40 21 128427 3.32 9 926751 1.33 9-80oi675 4-66 Io. I98325 3 22 128626 3.32 926671 i.33 801955 4.66 198045 38 23 728825 3.32 926591 i33 802234 4-65 I97766 3 24 729024 3.32 926511 1.34 82513 4.65 197487 36 25 729223 3-3i1 92643i I*34 802792 4.65 197208 35 26 729422 3.3I 92635I 1'34 803072 4-65 i o96Q28 34 27 729621 3.3I 926270 1.34 8o335I 465 96649 33 28 129820 3.31 926190 I.34 803630 o 4-65 i96370 32 29 73001ooi8 3.30 926110 1-34 803908 4-65 196092 31 30 730216 3.30 926029 1-34 804187 4-65 1 95813 o 31 -730415 3-30 9925949'.34 9 -80o4466 4.64 10-195534 2 32 730613 3-30 925868 I- 4 8c4745 4-64 195255 2 33 730811 3-30 925788 I-34 805023 4-64'94977 27 34 7310o9 3.29 925707 I.34?o5302 4-64 1 94698 26 35 731206 3.29 925626 I.34 8o558o0 4.64 194420 25 36 731404 3.29 925545 i.35 805859 4-64 194141 24 37 731602 3.29 925465 i.35 806137 4.64 193863 23 38 731799 3.2 925384 i.35 806415 4.63, I 93585 22 39 731996 3.28 925303 I.35 806693 4.63 193307 2 40 732193 3.28 925222 i.35 80697I1 4.63 I93029 20 41 9'732390 3.28 9-925141 I.35 9-8o7249 4.63 l10o 192751 1 42 732587 3-28 925060'-35 807527 4.63 1 92473 18 43 732784 3.28 924979 i.35 807805 4.63 1 92195 17 44 732980 3.27 924897 x.35 808o83 4-63 191917 i6 45 733I77 3-27 924816 1.35 8o836i 4.63 191639 i5 46 733373 3-27 924735 i.36 8o8638 4.62 9Ig362 14 47 733569 3'27 924654 I.36 808916 4.62 191o84 3 48 73376 3-27 924572 i-36 809193 4.62 1 90807 12 49 733961 3-26 924491 i.36 809471 4-62 190529 11 50 734157 3.26 924409 i-36 809748 4.62 1 90252 10 5I 9-734353 3-26 9-924328 I.36 9.810025 4.62 1089975 9 52 734549 3.26 924246 i.36 81o302 4-62 1 89698 8 53 734744 3.25 924164 i.36 8io58o 4.62 1 89420 7 54 734939 3.25 924083 i.36 8o10857 462 189g43 6 55 735035 3.25 924001 I.36 81i134 4.61 i88866 5 56 73Jc | 3.5 023919.36 81410o 4'6I | 850 | 57 73P525 3-25 923837 1.36 81I687 4-6I i 883i3 3 58 735719 3*24 923755 1.37 811964 4.6i J 88036 2 59 735914 3.24 923673 1-37 812241 4-61 I87759 I 60 736109 324 92359I.37 81217 46I 87483 Cosine D. Sine Di Cotang. D.'Tang, SINES AND TANGENTS (33 DEGREES.) 61 M. Sine D. Cosine D. i Tang. D. Cotang. 0 9*7361OQ 3- 24 9.92359 I.37 98125 17 4-6I 1o*187482 6o I 736303 3 24 923509 1.37 812791 4-6I 187206 2 736498 3. 24 923427 I 37 8 3070 4.6I 186930 3 736692 3.23 923345 I.37 813347 4.60 I86653 57 4 736886 3-23 923263 I.37 813623 4.60 186377 56 5 737080 3.23 923181 I.37 813899 4.6o 8611o 55 6 737274 3.23 923098 I.37 814175 4.60 I8582 54 737467 323 923016 I.37 814452 4.60 I85548 53 8 737661 3-22 922933 1'37 814728 4.60 I85272 52 9 737855 3 22 922851 I 37 815004 4-60 I849r96 51 10 738048 3.22 922768 i 38 815279 4.60 184721 5e II 9.738241 3-22 9-922686 I.38 9.815555 4.59 o10.84445 4 12 738434 3~22 922603 I.38 81583i 4-59 184I69 13 738627 8-2I 922520 I.38 816107 4.59 I83893 47 14 738820 3 2I 922438 I.38 816382 4.59 i83618 46 15 739013 3.21 922355 I.38 8i6658 4.59 183342 45 16 739206 3-2I 922272 138 816933 4.59 183067 44 17 739398 3-2I 922189 I'38 817209 4-59 I82791 43 18 739590 3~20 922106 1.38 817484 4.59 182516 42 19 739783 3.20 922023 I.38 817759 4.5 J 182241 41 20 739975 3'20 921940 I.38 8i8035 4.58 181965 4 21 9-740167 3-20 9-921857 I139 9.818310 4.58 o. I81690 3 22 740359 3.20 921774 I.39 818585 4.58 I81415 3 23 740550 3.I9 921691 1.39 8i8860 4-58 18140 37 24 740742 3 I9 921607 I.39 819135 4.58 i8o865 36 25 740934 3.19 921524 1.39 819410 4-58 180590 35 26 741125 3"I9 921441 1-39 8I9684 4.58 i803i6 34 27 741316 3.I9 921357'139 819959 4.58 180041 33 28 741508 3 18 921274 1-39 820234 4.58 I79766 32 29 741699 39I8 921190 I 39 820508 4.57 179492 31 30 741889 3.i8 92II07 1.39 820783 4.57 179217 30 3I 9.742080 3.I8 9.921023 1.39 9.821057 4.57 10.178943 2 32 742271 3.i8 920939 1.40 821332 4.57 178668 28 33 742462 3.17 920856 1.40 821606 4.57 178394 2 34 742652 3 I7 920772 1.40 82I880 4.57 178120 35 742842 317 920688 1.40 822154 4.57 I77846 25 36 743033 3.17 920604 1.40 822429 4.57 I77571 24 37 743223 3I7 920520 1.40 822703 4-57 177297 23 38 743413 3I6: 920436 1.40 822977 4.56 177023 22 39 743602 3.I6 920352 I.40 823250 4.56 176750 21 40 743792 3.i6 920268 1.40 823524 4.56 176476 20 41 9'743982 3.I6 9'920184 i.40 9-823798 4-56 10oI76202 19 42 74417! 3 I6 920099 1.40 824072 4-56 175928 18 43 74436I 3.15 92001i.40 824345 4.56 I75655 17 44 744550 315 91993i I.41 8246i9 4.56 175381 6 45 744739. 315 919846 14' 1 824893 4.56 175107 I5 46 744928 3.5: 919762 1441 825166 4-56 174834 14 47 7 4517 3.IS 9I9677.I'41 825439| 4.55 17456i 3 48 745306 3I4'9I9593 |1'4I 8257131.4.55 174287 12 49 745494 3.14 919508 1.41 825986 455 1740I4 11 50 745683 3.I4 919424 |I41 826259 J.55 I73741 10 51 5 9745871| 3.14 9.919339 1.41 9.826532 4.55 10I173468 9 1 52 746059 3.I4 99254 I.41 826805 4.55 173I95 53 746248 33 996 827078 455 72922 54 746436 4. 3 91908 1.4I 82735' 4.51 172649 55 746624 3 i3. 910000 I-4I 827624 4 5 17237 5 56 746812 J 3.13 918915 1 42 827897 4.54 72o3 4 57 746999 3.13 918830o 1-42 828170 4.54 I7i83o 3 58 747187 3.12 918745 1I42 828442 4-54 171558 2 59 747374 3.12 918659 1.42 828715 4.54 I7128 I| 60 747562 3.12 9I8574 142 828987 4.54 1 3 Cosine D. Sine D. otg. D. (56 DEGREBE.) b2 (34 DEGREES.) A TABLE OF LOGARITHMIC Mt. 1 Sine D. Cosinre D. Tang. D. Cotang. 0 ( 9-747562 3-I2 9-918574 I,42 9-828987 4-54 I1OI7ox3 60 I 747749 3-12 918489 42 829260 4-54 170740 5 2 747936 3.12 918404 1.42 829532 4-54 170468 2 748123 3. I 918318 1.42 829805 4.54 70195 5 4 748310 3-ii I 918233 1.42 830077 4-54 169923 56 5 748497 3*I 01 8147 1.42 830349 4-53 169651 55 6 748683 3.II 918062 1.42 830621 4-53 I69379 54 748870 3.ii 917976 1*43 830893 4-53 I69107 53 749056 3.IO 9I7891 I 43 83i1 65 4-53 68835 52 9 749243 3.Io 917805 I-43 831437 4-53 i68563 51 10 749429 31io 917719 1.43 831709 4-53 168291I 5o i[ 9-749615 3-io 9-9I7634 I-43 9.83198I 4-53 10-I680I9 4 12 74980I 3.io 917548 1.43 832253 4.53 167747 48 13 749987 3o09 917462 I.43 832525 4-53 167475 14 750172 3.og 917376 I-43 832796 4-53 167204 4 I5 750358 3o09 917290 I*43 833o68 4-52 I66932 45 i6 750543 3-og 917204 1,43 833339 4-52 i6666i 44 I7 750729 3-oo 9II7I8 I*44 8336ii 4-52 i66389 43 IS 75o094 3.o8 917032 1-44 833882 4.52 I66ii8 42 I9 751o9g 3*o8 9I6946 I.44 834154 4.52 I65846 41 20 751 284 3.08 9I6859 I*44 83442 5 4-52 165575 40 21 9-751469 3.08 9-.96773 1.44 9'834696 4-52 Io.I65304 3 22 751654 3.08 9I6687 I.44 834967 4-52 165033 3 23 751839 3.o8 916600 I*44 835238 452 I64762 37 24 752023 3.o7 916514 I.44 835509 4-52 I64491 36 25 752208 3.07 916427 Ix44 835780 4-51 164220 35 26 752392 3.07 91634I 1-44 836o5i 4.5I i63949 34 27 752576 3-07 916254 I-44 836322 4-5I 163678 33 28 752760 3.07 916167 1-45 836593 4-5I 163407 32 29 752944 3.o6 91608I I.45 836864 4-5I i63I36 31 30 753128 3.0o6 915994 I-45 837134 4.5I 162866 30 31 9.7533r2 3-o6 9.9-5907 1.45 9.837405 4-5I IO.I62595 20 32 753495 3.o6 915820 oI45 837675 4-5I 162325 2 33 753679 3-o6 915733 I-45 837946 4.5I I62054 27 34 753862 3-o5 915646 I-45 8382i6 4-5I I61784 26 35 754046 3-o5 915559 I-45 838487 4-50 16i5i3 25 36 754229 3-o05 915472 I45 838757 4-50 I6I243 24 37 754412 3-o5 915385 I.45 839027 4-50 160973 23 38 754595 3-o5 915297 I-45 839297 4-50 1070 22 39 754778 3o04 915210 I-45 83956 450 I60432 21 40 754960 3.o4 915123 |I46 839838 4-50 I60162 20 41 9.755143 3*o4 9*9I5035 I*46 9*840Io8 4*50 10I59892 19 42 755326 3.o4 914948 I546 840378 4-50 159622 I 43 755508 3-04 91486o I-46 840647 4.50 I59353 17 44 755690 3-o4 914773 |I-46 840917 4-49 I59O83 1 45 755872 3.o3 914685 I*46 841187 4-49 I5883 15 46 756054 3.o3 914598 I146 84I457 4-49 I58543 14 47 756236 3.-o3 914510 I-46 841726 4-49 I58274 13 48 756418 3.0o3 914422 1-46 841996 4-49 158004 12 49 756600oo 3.03 914334 I146 842266 4-49 157734 1 50 756782 3.02 914246 1-47 842535 4-49 157465 |o 51 9-756963 3.02 9-914158 1*47 9842805 4-49 IO-157195 52 757144 3.-o 914070 9-47 843074 4-49 I56926 53 757326 3.-o02 913982 I-47 843343 4 49 I56657 7 54 757507 3.02 3894 1-47 843612 4-49 i56388 155 757688 3.01 913806 1-47 843882 4.48 I56118 5 56 |757869 3o 9 37I8 |'i7 84415I 4-48 I55849 4 5 758050 3-oz 9I3630 o-47 844420 4-48 i55580 3 5 758230 3 01 354I 147 844689 4-48 i553 2 59 1 7584Ii| 3-*oI |913453 I 47 844958? 4-48 15042 1 6o 758591 3-oi 913365 1-47 845227 4-48 154773 o Cosine - D. I Sine D. Cotan. D.' Tang.:. (55 DEGOREL3.) SINES AND TANGENTS. (35 DEGREES.) 53 Sine D. osine D. Tang. D. Cotasng. 0 9.758591 3-Io 9.9133;65 1-47 9-845227 4-48 IO-I54773 6o I 758772 3*oo 913276'*47 845496 4-48 154504 2 758952 3.00oo 913187 x 48 845764 4-48 I554236 58 3 759132 3oo 913099 1.48 846o33 4.48 153967 57 4 759312 3.oo 9*3010 9.48 846302 4.48 I53698 56 5 759492 3.oo 912922 1.48 846570 4-47 i5343o 55 6 759672 2.99 912833 1-48 846839 4.47 153161 54 7 59852 299 912744 I.48 847107 4.47 152893 53 8 76oo31 299 912655 1.48 847376 4.47 152624 52 9 760211 2.99 912566 1.48 84'7644 4.47 152356 51 10 760390o 2.99 912477 1 48 847913 4-47 152087 50 11 9,760569 2-98 9-912388 I.48 9.84818I 4-47 10-151819 49 12 760748 2-98 912299 1.49 848449 4-47 151551 48 i3 760927 2 98 912210 1.49 848717 4-47 151283 47 14 761106 2-98 912121 1 49 848986 4-47 151014 46 15 761285 2.98 912031 1.49 849254 4'47 150746 45 i6 761464 2.98 911942 I -49 849-522 4-47 150478 44 I7 761642 2.97 911853. 849 790 4'46 It0210 43 I8 761821 2.97 911763 I.49 85oo0058 4-46 I49942 42 19 761999 2.97 9x11674 I *49 850325 4-46 I49675 41 20 762177 2.97 911584 I 49 850593 4-46 149407 40 21 9.762356 2.97 9'911495 1'49 9'85086i 4.46 o-. 49139 39 22 762534 2.96 911406 I 49 851129 4.46 148871 38 23 762712 2.96 91 315 i50o 851396 4-46 148604 37 24 762889 2.96 911226 Io50 851664 4.46 148336 36 25 763067 2.96 911136 I-50 851931 4-46 148o69 35 26 763245 2.96 911046 I-50 852199 4-46 I47801 34 27 763422 2.96 910o56 I. 50 852466 4.46 147534 33 28 7636oo 2-95 910o6 -50o 852733 4.45 147267 32 29 763777 2.95 910776 I-50 853001 4-45 1469 9 I 31 30 763954 2*95 91o686 I 50 853268 4.45 146732 30 31 9.764131 2.95 9.9o0596 i-50o 9'853535 4-45 10.146465 29 32 764308 2.95 910506 I-50o 853802 4.45 14638 2 33 764485 2.94 910415 io50 854069 4.45 145931 27 34 764662 2.94 910325 1.5I 854336 4-45 I145664 26 35 764838 2.94 910235 i-51 8546o3 4.45 145397 25 36 765015 2-94 910144 1.51 854870 4.45 145130 24 37 6519i 2-94 910054 1.51 855137 4-45 I44863 23 38 765367 294 909963 I 51 855404 4.45 I44596 22 39 765544 2.93 909873 1.51 855671 4.44 144329 2I 40 765720 2.93 909782 i.5i 855938 4.44 144062 20 41 9.765896 2.93 9-gog90969I I51 9-856204 4.44 o10.143796 19 42 766072 2.93 909601 I-5I 856471 4.44 143529 18 43 766247 2.93 909510 I51 856737 4.44 143263 17 44 766423 2.93 909419 1.51 857004 4-44 142996 I6 45 766598 292 909328 1 52 857270 4.44 I42730 5 46 766774 292 909237 1.52!857537 4.44 142463 14 47 766949 2-92 909146 1-52 857803 4.44 142I97 3 48 767124 2-92 909055 1.52 858069 4-44 14I931 12 49 7673o00 292 908964 1 52 858336 4-44 141664 I 50o 767475 2.91 908873 1-52 858602 4-43 141398 1 51 9 767649 291 9-90878I I-52.. 9.858868 4.43 1o-141i32 9 52 767824 2-.9 908690 1 52 859134 4.43 140866 8 53 767999 2.91 908599 1-52 859400 4.43 140600 7 54 768173 291 908507 1 52 859666 4.43 i4o334 6 55 768348 2-.9 908416 i 53 859932 4-43 140068 5 56 768522 2.90 908324 i.53 860198 4-43 139802 4 57 68697 2-90 908233 i 53 860464 4.43 39536 3 58 768871 290 908I41 53 860730 4-43 39270 2 59 769045 290 98o49J1 1-53 860995 4.43 139005 6 769219 2.Q0 907958 I53 86126I 4.43 38739 L Cosine D. Sine D. Cotang. D. | Tang. M. (54 DEGREES.'J (36 DEGREES.) A TABLE OF LOGARITHIMIC BM. ine D). Cosine D. Ta ng. D.! tAang. 9.907 28 1.63- 9.2i 654.43 i L33739 601 097692ig 2: 90 919075?58 9~5 it-I26i 4 443 j 44 I'-,~ I9 769393 2. 907 866.53 361527 4.43 738473 2 769566 2.89 907774 i.53 861792 4.42 138:o85 3 769740 2.89 907682 i.53 86208 4.42 137942 57 4 769913 2.89 907590 i.53 862323 4.42 137677 56 5 770087 285 907498 i53 862589 4.42 137411 55 6 770260 28 907406 I 53 862854 4.42 137146 54 770433 2!8 -4- 83l 7 770433 2.88 907314 1.54 863119 4.42 13688I 53 8 770606 2.88 907222 1-54 863385 44 366i5 5a 9 770779 2.88 907129 1.54 86365o 44' 13635o i o10 770952 2988 907037 I 54 863915 4.4 136o85 50 jI 9.771125 2.88 9906945 I.54 9.864180 4.41 o0.135820 4 12 771298 2.87 906852 I. 4 864445 4.2 135555 48 13 771470 2.87 96760 54 8647o 4 42 135290 47 14 771643 2.87 906667 1.54 864975 -414 135025 46 15 771815 287 906575 154 865240 4.41 134760 45 I6 7198 2.87 906482 1 54 865505 4.4I 134495 44 16 771987 4 134495 4 772159 2.87 906389 I.55 865770 4.41 134230 43 77233I 2.86 906296 I j55 866035 4-41 133965 42 9 -7725o3 2.86 906204 I 55 8663oo 4.4i I33700 41 20 772675 2.86 90611 155 866564 4.41 133436 40 21 9.772847 2.86 9906018 1.55 9.866829 4.47 jI33 3 22 773018 2.86 90525 1.55 867094 441 132o6 23 773190 2.86 905832 1.55 867358.4 132642 37 24 773361 2.85 90573 x.55 867623 4-41 132377 36 25 773533 2.85 905646 1.55 867887 4 I4' 132113 35 6 773704 2.85 905552 1.55 868152 440 131848 34 27 773875 2.85 905459 I.55 868416 4.40 131584 33 774046 2.85 905366 io56 868680 4-40 131320- 32 29 774217 85 905272 i 56 868945 4-40 13io55 31 0 774388 284 9o5179 i.56 8692o9 4. 40 i3o994 3o 31 9.774558 84 9.90go5085 1.56 9.869473 440 o.13o52 2 32 774729 2.84 904992 1.56 869737 4.40 130263 2? 33 774899 284 9o498.56 87o0001o 4.40 I99 7 2 34 775070 2.84 904804 i.56 87o0265 4.40 12973 26 35 775240 284 904711 i-56 870529 4.40 129471 25 36 775410o 2.83 904617 1.56 870793 4.40 129207 24 3 775580 2.83 904523 i.56 87o57 4.40 1 28943 23 7757.50 2.83 904429 1.57 871321 4.40 128679 22 39 775920 2.83 904335 1.57 871585 4.40 I28415 21 40 776090.a83 904241 1.57 871849 4.39 128151 20 4' 9-776259 2.83 9-904141 1.57 9872II2 4.39 o10.127888 19 42 776429 2.82 904053 1.57- 872376 4.39 127624 i 43 776598 2.82 903959 1.5- 872640 4-39 127360 1 i 44 776768 2.82 903864 Ii 872903 4.39 2797 7 93 4-39 127097 45 776937 2.82 903770 I1.57 873167 4.39 126833 iS 46 777106 2.82 903676 1'57 873430 4'39 126570 14 47 777275 2.81 903581 1.57 873694 4.39. 12630613 4 777444 2.81 903487 1.57 873957 4.39 126043 12 49 777613 2.81 903392 1.58 874220 4.39 125780 i11 50 777781 2.81 903298 I.58 874484 4.39 125516 10 51 9.777950 2.81 99go3203 i.58 9.874747 4.39 101i25253 52 778119 2.89 go31o8 1.58 875010 4.39 124990 53 778287 2.80 903014 i.58 875273 4.38 124727 54 778455 2.80 902919 i.58 875536 4.38 124464 55 778624 2.80 902824 i.58 87580O' 4-38 124200 5 56 778792 2.80 902729 1.58 876063 4.38 123937 4 S 778960 2.80 902634 i.58 876326 4.38 1123674 3 5i 779128 2.80 902539.59 876589 4.38 I23413II,2 779295 2.79 902444 1.59 87685i 4.38 123149 I 779463 2.79 902349 I159 877114 4.38 122886J o L. Cosine DJ. S__ne_'D]. C ______ D. ______-t (53 DEGREES.') SiES AN'D TANiGENTS. (37 DEGREES.) 66 i Sine, Cosne D. Tang. j D. Cotang. 0 9.779463 2.79 9.902349 I 59 9.877114 4.38 10122886 6o I 779631 2. 79 902253 1.59 877377 4.38 122623 5 2 779798 279 902158 159 877640 4.38 122360 58 3 779966 2'79 902063 1.59 877903 4.38 I22097 57 4 7o1 33 2.79 901967 1*59 878165 4.38 121835 56 5 780300 2*78 901o 72 I.59 878428 4.38 121572 i 55 6 780467 2.78 901776 1.59 87869I 4.38 121309 54 780634 2.78 90o681.59 878953 4.37 I21047 53 780801 2.78 901585 159 879216 4.37 I20784 52 9 780968 2.78 901490 1.59 879478 4.37 120522 5i 10 781134 2.78 901394 60o 879741 4.37 120259 50 11 9-783o0I 2.77 9-901298 I.6o 9.880oo3 4.37 10o119997 4 12 781468 2.77 901202 I.60 880265 4.37 119735 48 I3 781634 2.77 90goI06 i60 880528 4.37 119472 4 14 781800 2.77 901010.60o 880793 437 119210 4 I5 781966 2-77 900914 i.60 881052 4.37 I 1948 45 i6 782132 2.77 900818 1.60 8S13!4 4.37 i 8686 44' 782298 2'76 900722 I60 881576 4.37 118424 43 782464 2.76 900626 i6o 881839 i 4-37 ii8i6i 42 I9 782630 2.76 9005290 I6o 88!01 4.37 117899 41 20 782796 76 go433 6 882363 436 117637 40 21 9.782961 2-76 9.900337 I6I 9.88262-5 4-36 o10117375 39 22 783127 2.76 900240 i.6i 882887 4.36 117113 38 23 783292 2.75 900o 44 1.6I 883148 4.36 I16852 3 24 783458 2.75 90oo047 I.6i 883410 4.36 I16590 36 25 783623 2.75 89995i i6I 883672 4.36 I16328 35 26 783788 2-7*5 89(9854 I*6I 883934 4.36 I60o66 34 27 783953 2-75 899757 i*6I 884196 4.36 I580o4 33 784118 2.75 899660 I6 884457 4.36 II5543 32 29 784282 2.7 899564 i16i 884719 4.36 115281 31 3o 784447 2-74 899467 I*62 884980 4.36 115020 3o 3i 9.784612 2.74 9.899370 I.62 9.885242 4-36 I0114758 2 32 784776 2.74 899273 I162 885503 4-36 114497 28 33 784941 2.74 899176 I162 885765 4.36 11423 27 34 7851o5 2.74 899078 I162 886026 4-36 113974 35 785269 2.73 898984 I*62 886288 4.36 I13712 25 36 785433 2-73 898884 I*62 886549 4.35 113451 24 37 785597 2.73 898787 1I62 8868io 4.35 II3190 23 3 785761 2.73 898689 I*62 887072 4.35 112928 22 39 785925 2.73 898592 I*62 887333 4.35 112667 21 40 786089 2*73 898494 i*63 887594 4-35 112406 20 41 9 786252 2.72 9.898397 I.63 9.887855 4.35 I0.112145 19 42 786416 2*72 898299 I63 8881I6 4.35 I11884 I 43 786579 2.72 898202 i I*63 888377 4.35 II1623 I7 44. 786742 2.72 898104 I-63 888639 4.35 III36I 45 786906 2-72 898006 63 888900 4. 35 1 1 I 0 46 787069 2.72 897908 I 63 889160 4.35 110840 14 47 787232 2-71 8978I0 i 63 88942I 4.35 110579 3 48 787395 2:71 897712 i63 889682 4.35 Io318 12 49 787557 2-71 897614 I-63 889943 4.35 110057 II 50 787720 2*71 897516 i.63 89o~4 4.34 Io9796 Io 51 9*787883 2*71 9.897418 -664 9.890465 4.34 Io.Io9535 9 52 78045 271 897320 I64 890725 4.34 109275 8 53;788208 27I 897222 I64 890986 4.34 o09014 7 54 788370 2.70 897123 I-64 891247 4.34 io753 6 55 78532 270 897025 I64 891507 4.34 Io8493 5 56 788.6 270 896926 i.64 891768 4.34 108232 1 4 5 7 88856 2-70 896828 I-64 892028 4-34 07972 3,89018 a270 896729 Io64 892289 4.34 107711 2 59 180 270 89663 892549 4-34 107451 60 789342 2.69 896532 i*64 692810 i 4.34 107190 0 Cosine D. | Sine ID. I Cotang. DI. Tg. —1L (52 DEG1URErJ.) I66 (38 DEGRRIES.) A TABLE OF LOGARITHMIC M. ie D. SinCosiee D. Cosi Tang. D. | (otang. 0 9'789342 2.69 9.896532 1.64 9.892810o 4.34 Io0-o07I90 6 J 789504 2.69 896433 i.65 893070 4.34 10o6930 59 2 789665 2.69 896335 i.65 893331 4.34 106669 58 3 789827 2.69 896236 i.65 893591 4.34 106409 57 4 789988 2.69 896137 i.65 893851 4.34 1 o06149 56 5 790149 2.69 896o38 I 65 894111 4.34 105889 55 6 7903IO 2.68 895939 i.65 894371 4.34 Io5629 54 7 790471 2.68 895840 I 65 894632 4.33 io5368 53 790632 2.68 895741 I.65 894892 4.33 io5io8 52 9 790793 2 68 895641 i 65 895152 4.33 104848 51 10 790954 2.68 895542 I 65 895412 4.33 io4588 50 11 9-79II15 2.68 9.895443 I 66 9-895672 4.33 Iro-14328 4 12 791275 2.67 895343 166 895932 4.-33 104068 48 I3 791436 2 6j 895244 I 66 896I92 4.33 io3808 47 14'791 56'12.67 895145 Ix66 896452 4.33 103548 46 I5 791757 2.67 895045 I.66 896712 4.33 103288 45 i6 79'9'7 2-67 894945 x.66 896971 4.33 I03029 44 17 792077 2.67 894846 I 66 897231 4.33 102769 43 I8 792237 2-66 894746 I-66 897491 4.33 102509 42 9 792397 266 894646 I*66 89775I 4.33 102249 41 20 792755 2.66 894546 i*66 89801o 4.33 10 990 4o 2' 9.792176 2.66 9.894446 I 67 9.898270 4.33 IoIo.1730 39 22 792876 2.66 894346 I 67 898530 4.33 I 01470 38 23 793035 2.66 894246 1 67 898789 4-33 II0121 37 24 793I 5 2.65 894146 1-67 899049 4.32 10oo951 36 25 793354 2.65 894046 I 67 89930 4.32 Ioo62 35 26 793514 2 65 893946 167 899568 4.32 100oo47 34 27 793673 2.65 893846 I 67 899827 4.32 IOOI,':1 33 28 793832 2.65 893745 I 67 900086 4.32 099914 32 129 7939 1 2 65 893645 I167 900346 4.32 09954 31 I 3o 794150 2.64 893544 1 67 900605 4.32 099395 3 1 3i 9.794308 2.64 9.893444 1.68 9.900864 4.32 o.o09 136 2 32 794467 2.64 893343.68 90oI24 4.32 o983876 28 33 794626 2.64 893243 1-68 901383 4.32 o0986I7 27 34 794784 2.64 893142 1-68 90o642' 4.32 o98358 26 35 794942 2.64 893041 I 68 9090 I 4.32 098099 25 36 795I0I 2.64 892940 I 68 902160o 4-32 097840 24 37 795259 2 -63 892839 I68 902419 4.32 097581 23 38 79547 2.63 892739 I.68 902679 4.32 29732I 22 39 795575 2 *.63 892638 I.68 902938 4.32 o97o62 21 2.63 6~/, 8 3 097062 21 40 795733 *2.63 892536 i*68 903197 4.3I. 096803 20 4I 9.79589I 2.63 9.8Q2435 i.699 9.903455 4.3I 10oo96545 r9 42 796049 2.63 892334 I 69 903714 4.31 o96286 iS 43 796206 2 63 892233 169 903973 4.3 096027 17 44 796364 2.62 892132 169 904232 4.3I 095768 16 45 796521 2 62 8920o30 169 904491 4.3 I 095509 15 46 796679' 2.62' 891929 I 69 904750 4.3I 095250 14 796836 2.62 891827 169 905008 4.31 o094992 13 4 796993 2 62 891726 1 69 905267 4.3i 094733 12 49 /797Io /0 2.61 89I624 I 69 905526 4-3I 094474 1 50 797307 2.6I 891523 1.70 905784 4.3i 0942I6 I1 51 9'797464 2.6i 9.891421 1.70 9.906043 4.3i 10.093957 52 797621 2.6 89I3I9 1 70 906302 4.3i o93698 8 53 797777 2 6I 891217 1.70 906560o 4.3I 093440 7 54 797934 2.6I 89iii5 1.70 906819 4.3I 093181 6 55 798209I 1.6I 8910o3 1'70 907077 4.31 092923 5 56 798247 2.6I 8~ I I. t70 907336 4.3i o92664 4 57 I 9403 2 *o 0 8:8o9 I 1.7~ 90759 4 4.3 / o92406 3 58 798560 260o 890707 1.70 907852 4.3i 092148 2 59 708716 25.60 8Ho6o5 I.7 9o08 1 4.30 091889 1 60 798;7 2-60 8j0o5o3 1 -70 908369 4.30 0o9631 o Cosine D. Sine I D. l Cotang. 1 D g. I (61 DEGRslcS.) SINES AND TANGENTS. (39 DEGREES.) 57 M.. Sine D. Cosino D. Tang. D. Cotang. o 9-798872 2.60 9.890503.I70 9 908369 4.- 3o0 io0963I 6 I 799C28 2.60 890400 1.71 908628 4] 30 091372 2 799184 2.6o 890298 1 71 908886 4-30 091114 58 3 | 99339 2.59 890195 1.71 909144 4-30 090856 5 4 799495 2-59 89oo93 1 71 909402 4 30o 00598 56 5 79965I 2.59 88 o 71 909660 4 30 090340 55 799806 2*59 88988 7 I 909918 4.30 00082 54 799962 2'59 889785 171 910177 43o 930 823 53 8o00o7 2.59 889682 1 71 910435 4.30 o89565 52 9 800272 2.58 889579 1.7 I 3 430 089307 51 10 800427 2.58 889477 I*7I 9I0931 4-30 089049 50 I 9.800582 2.58 9.889374 172 9'911209 43o0 I0-0887 4 12 800737 2.58 889271 1 72 911467 4.30 o88533 48 13 8oo092 2.58 889I68 172 911724 4.30 o88276 4 14 80Io47 2.58 889o64 1*72 9 1982 430o o88oi8 46 15 801201 2.58 888g96 172 912240 4 30 o87760 45 I6 8oi356 2.57 888858 1.72 912498 4 30 o87502 44 17 80o5II 2.57 888755 1.72 912756 430o o8244 43 8 801665 2.57 88865i 1 72 913014 4-29 08686 42 9 8oi8Ig 2.57 888548 172 9I3271 429 o86729 41 20 801973 2.57 888444 1'73 913529 4.29 o8647I 4 21 9.802128 2a57 9.888341 1.73 9.9I3787 4.29 IO0.862I3 39 22 802282 2.56 888237 1.73 914044 429 3 23 802436 2.56 888I34 1.73 914302 4.29 085698 37 24 802589 2.56 888030 1 73 914560 4.29 o85440 36 25 802743 2.56 887926 1'73 914817 4.29 o85I83 35 26 802897 2.56 887822 1*73 915075 4.29 O84925 34 27 8o0300 2.56 887718 1.73 915332 4.29 o84668 33 28 803204 2.56 887614 1.73 915590 4.29 0844i0 32 29 803357 2.55 887510 I 73 915847 4-29 084I53 31 30 8035I 2.*55 887406 I 74 916Io4 4-29 083896 3o 3I 9.803664 2.55 9.887302 I 74 9'9I6362 4.29 Io*o83638 29 32 803817 2-55 887198 1 74 916619 429 o8338I 28 33 803970 2.55 88793 174 916877 4.29 083123 2 34 804123 2.55 88698 1. 74 917134 4-29 82866 2 35 804276 2-54 886885 74 917391 4.29 082609 25 36 804428 2-54 886780 1.74 917648 4.29 082352 24 37 80458I 2.54 886676I.74 9179o5 4.29 o82o05 23 3 804734 2^54 88657I I.74 9I8 63 4.28 o81837 22 39 804886 2.54 886466 I74 9I8420 4.28 o8i580 -21 40 805039 2-54 886362 1.75 918677 4.28 08I323 20 41 9.80519 2:.54 9.886257 1.75 9.918934 4-28 io.o8io66 I9 42 805343 2.53 886152 I,75 919191 4.28 o80809 18 43 805495 2.53 886047 1.75 919448 4.28 o80552 17 44 805647 2.53 885942 1.75 9I97o5 4. 8 o80295 45 805799 2.53 885837 I.75 9 I9962 4.28 o8oo38 15 46 8059g5 2.53 885732 1x.75 9202o19 4.28 07978 14 8o6,o3 2.53 885627 175 920476 4.28 o7924 13 48 806254 2.53 885522 I 1.75 920733 4.28 079267 12 49 806406 2.52 8854i6 1.75 920990 4t'8 079010 II 5o 806557 2.52 8853ii 1 76 921247 4.28 078753 10 51 9'806709 2.52 9.885205 1.76 9.9215o03 4.28 Io0o78497 52 8o686o 2.52 885ioo 1.76 92 176o 4.28 078240 53,807011 2*.52 884994 1. 76 922017 4.-28 077983 7 54 807I63 2.52 884889.76 922274 4.28 77726 55 807314 2.52 a84783 I'76 92253o 4-28 07747~ 56 807465 2.5i 884677 x.76 922787 4.28 077213 4 5 8076I5 2.51 884572 11.76 923044 4-.28 076956 3.5o7I6~t 807766 2.51 884466 I.76 923300 4.28 076700 2 8079I 2*51 884360 1x76 923557 4.27 076443 1 80,667 25i 884254 -.77 923813 4.27 076187 0 16 I0 D. I Silne I?. ICotang. ID..Ta. j (50 DEGREES.) 58 (40 DEGREES.) A TABLE OF LOGARITHMIC M. Sine D. Cosine D. Tang. D. Cotang. 9.80806o 2.51 9.884254 1'77 9-9238I3 4.27 10 o76I87 6 808218 2I51 884148 1.77 924070 4-27 075930! 2 808368 2.51 884042 1.77 924327 4.27 075673 3 8o85i9[ 2.50 883936 1.77 92i583 4-27 075417 5 4' 808669 2.50 883829 I*77 924840 4.27 075160 56 5 808819 2.50 883723 177 925096 4-27 o74904 55 6 808969 2.50 883617 1.77 9253-2 4.27 074648 54 7 809119 2.50 883510io 77 925600 4.27 07439I 53 809269 2.50 883404 1.77 925865 4.27 074135 52 9 809419 2.49 883297 I.78 926122 427 073878 5I 10 8o9569 2.49 883I9! 1*78 926378 4-27 073622 50 i 11 98097 18 249 9883084 *78 |9926634 4.27 0*073366 4 12 809;68 2.49 88297 1.78 926890 4-27 073110 48 i3 810017 2.49 882871 1.78 927I47 4-27 072853 4 14 8ioi68 2.49 882764 1.78 927403 4.27 072597 46 15 8o0316 2.48 882657 1.78 927659 4.27 072341 45 i6 810465 2.48 882550 I 78 927915 4-27 072085 44 17 810614 2.48 882443 I.78 928171 4.27 071829 43 1 810763 2.48 882336 I-79 92842" 4-27 071573 41 19 81o0912 2.48 882229 I.79 92868i 4-27 071317 41 20 8I1061 2.48 882121 I-79 928940 4-27 071o6o 40 21 9.811210 2o 48 9J882014 1'79 9.929196 4 27 Io070804 22 811358 2.47 881907 I.79 929452 4.7 070548 3.23 811507 2.47 881799'.79 929708 4-27 070292 24 81I655 2.47 88i6 2 1.79 929964 4.26 70036 36 25 811894 2.47 881584 I*79 930220 4-26 069780 35 26 811952 2.47 881477 I.79 930475 4 26 069525 34 27 812100 2.47 881369 I.79 930731 4-26 o69269 33 28 812248 2.47 88126I I.80 930987 4.26 0690 3 32 29 8I2396 2 4 88ii53 I.80 931243 4*26 o68757 3I 30 812544 2.46 881046 I*80 931499 4.26 o685oi 3o 3I 9g812692 2.46 9.880938 i.80 9 931755 4.26 I0 o68245 29 32 812840 2*46 88o830 I80o 932010 4.26 o67990 28 33 8I2988 2.46 880722 I.80 932266 4.26 067734 27 34 8I3I35 2.46 8806I3 i80o 932522 4-26 067478 26 35 8i3283 2.46 880505 I*80 932778 4-26 067222 25 36 813430 2.45 880397 I.8o 933033 4.26 066967 24 37 8i3578 2.45 880289 I.8I 933289 4.26 0667I1 23 38 813725 2.45 88oi80 I8I 933545 4.26 066455 22 39 81 3872 2*4 880072 i8i 933800 4.26 o66200 21 40 814019 2.45 879963 I8I 934056 4.26 o65944 20 41 9-8I4I66 2.45 9q879855 1*8i 9.9363II 4.26 I0-065689 I9 42 8I43t3 2.45 879746 x.8 934567 4.26 065433 43 8I4460 2*44 87637 I*8I 934823 4.26 065177 17 81460 2*44 879529 I*8I 935078 4.26 o64922 16 45 814753 2.44 879420 oI81 935333 4.26 064667 i5 46 814900 2.44 8793II I.8i 935589 4.26 o644I I14 47 8i5o46 2*44 J 879202 182 935844 4.26 o64156 13 48 8i5193 2.44 879093 8 936100 426 063900 1 12 49 81533g 2.44 87984 i.821 936355 4.26 063645 II 50 80I5485 2*43 878875 *822 936610 4.26 o63390 10 5I. 9-81563 2-43 9.878766 1.82 %.93686 4.25 io0o63I34 52 815778 2*43 878656 i.8 937121 425 062879 53 815924 2.43 878547 I82 937376 4.25 o62624 7 54 816069 243 87848j 1.82 935763 4.25 062368 6 55 816215 2.43 878328 1.82 93188 4.25 062113 5 56 8i6361 2-43 878219 I.83 938I4 4.25 06I858 |4 5|8i507 |242 878io9 I83 93839b 1 425 061602 3 1 28I0652 |242 87799 Ix83 938653 ) 425 06I347 2 59 1i6798 2.42 877930 i *83 938908 J t o6 o I 6o 816943 2 42 877790 I.83 939163 J 4.21 0o6o083 0 _I 2 -- ICoine - D. Sine I D. I, o3g39D. (49 DIGRERBS.) SINES AND TANGENTS. (41 DEGREES.) 59 i. Sine D. Cosine D. Tang. D. Cotang. 0 98 16943 2.42 9 877780 -83 9939163 4. 25 100o60837 60 I 817088 2.42 877670 x.83 939418 4.25 060582 5 2 817233 2.42 877560 x.83 93'9673 4.25 o60327 58 3 817379 2-42 877450 i.83 939928 4.25 060072 57s 4 817524 2.41 877340 i.83 94t 183 4.-25 o59817 56 5 817668 2.41 877230 1.84 940438 4.25 059562 556 817813 2-.4 8?7720.84 940694 4.25 o59306 54 7 8i7958 241 877010 1.84 Q40949 4-25 o9o53 53 8 818io3 j 2-.4 876899 1.84 941204 4.25 058796 52 9 818247 2.41 876789.84 941458 4-25 058542 51 AO 818392 2.41 876678 1.84 941714 4-25 058286 5o II 9.818536 2-40 9.876568 I 84 9*941968 4.25 10o.58032 4 12 81868i 2-40 876457 1.84 942223 4 25 057777 48 13 818825 2.40 876347 1.84 942478 4.a5 057522 47 14 818969 2.40 876236 I 85 942733 4.25 057267 46 15 8i9113 2-40 876125 185 942988 4-25 057012 45 16 819257 2.40 876014 I 85 943243 4.25 056757 44 17 8194o0 2.-40 875904 I.85 943498 4025 o56502 43 18 819545 2.39 875793 i 85 943752 4.25 056248 42 19 819689 2.39 875682 i 85 944007 4.25 055993 41 20 819832 2.39 875571 I 85 944262 4..25 055738 4o 21 9.819976 2.39 9.875459 I.85 9.9445I7 4-25 10.055483 39 22 820120 2.39 875348 I.85 94477I 4-24 055229 3 23 820263 2.39 875237 1-85 945026 4.24 054974 3 24 820406 2.39 875126 86 945281 4-24 054719 25 820550 2.38 875014 I 86 945535 4.24 054465 35 26 820693 2.38 874903.86 945790 4-24 o542I0 O 34 27 820836 2.38 874791 I 86 946045 4-24 053955 33 2 820979 2.38 874680 x.86 946299 4.24 053701 32 29 821122 2.38 874568 I 86 946554 4-24 053446 31 30 821265 2.38 874456 J 86 946808 4-24 053192 30 3i 9.821407 2.38 9.874344 i 86 9.947063 4-24 10o052937 29 32 821550 2.38 874232.I87 947318 4-24 052682 28 33 821693 2.37 874121 1.87 947572 4.24 052428 27 34 821835 2.37 874009 1.87 947826 4-24 052174 26 33 821977 2.37 873896 I-87 948081 4-24 051919 25 36 822120 2.37 873784 1.87 948336 4:24 051664 24 37 822262 2.37 873672 I 87 948590 4-24 051410 23 38 822404 2.37 873560 1-87 948844 4.24 05II56 22 39 822546 2.37 873448 1.87 949099 4.24 050901 21 40 822688 2.36 873335 I.87 949353 4.24 050647 20 41 9.822830 2.36 9.873223 1.87 9.949607 4.24 10*o503g3 19 42 822972 2.36 873110o i88 949862 4.24 o5o,38 18 43 823114 2.36 872998 1.88 950116 4-24 049884 17 44 823255 2.36 872885 J.88 950370 4.24 049630 16 45 823397 2.36 872772 1.88 950625 4.24 049375 15 46 823539 2.36 872659 1.88 950879 4.24 049121 14 823680 2.35 872547 I-88 951133 4-24 048867 3 823821 2.35 872434 I-83 |95388 4.24 048612 12 49 823963 2.35 872321 i.88 951642 4o24 048358 II 50 824104 2.35 872208 I.88 951896 4-24 048104 10 51 9.824245 2.35 9.872095 1.89 9 952150 4-24 10o047850 52 824386 2.35 871981 1.89 952405 4-24 047595 8 53 824527 2.35 871868 1.89 952659 4-24 047341 7 54 824668 2.34 871755 1.89 952913 4.24 047087 55 824808 2.34 871641 J.89 953167 4.23 046833 5 56 824949 2.34 871528 1.89 953421 4-23 046579 4 82500o 2.34 871414 1-89 953675 4-23 046325 3 8 825i30 2.34/ 87130o 18q9 953929 4-23 46071 2 59 825371 2.34 871187 I.S9 954183 4-23 045817 I1 60 825511 1.34 871073 1.90 954437 4 -3 o45563 o Cosine D. I Sine I). Cota-. | T....L (48 iEoRIEo,.) 60 (42 DEGREES.) A TABLE OF LOGARITHMIO M. Sine D. Cosine D. Tang. ID. Cotang. 0 9.825511 234 9.871073 I.90 9954437 4.23 0 o45563 6o I 82565I 2.33 87o960 1.90 95469 I 4-23 o045309 5Q 2 825791 2.33 8708461 1 90o 954945 4-23 o45055 59 3 82593 1 2.33 870732 I.90o 955200 4.23 o 0448oo00 57 4. 826071 2.33 870618 I.90 955454 4-23 o44546 56 5 826211 2.33 870504.90o 955707 4.23 0o44293 55 6 82635i 2.33 870390o I-9o 95596 I 4-23 o44039 54 7 8264g1 2.33 870276 1.90o 956215 4.23 043785 53 J 8 66 95646 8 82663i 2.33 870oI6I I 9o 956469 4-23 o4353i 52 9 826'70 2-32 870047 1.9I 95672 4-23 043277 5x 10 8261I0 2.32 869933 I-9 9 56977 4-23 o43023 5o I I 9.827049 2.32 9-869818 1 91 9'95723I1 4-23 10.042769 44 12 827189 2.32 869704 I *9 957485 4'23 0o42515 48 13 827328 2.32 869589 I 9IJ 957739 4-23 042261 47 14 827467 2.32 869474 I 9I 957993 4'23 o42007 46 i5 827606 2.32 86936o I *9I 958246 4-23 041754 45 I 6 827745 2.32 869245 I 9 I 958500 4.23 o4I50oo 44 I7 827884 2 3 I 8691I30 I.9 958754 4-23 041246 43 i8 828023 2.3I 869o05 1 -92 95go90o8 4-23 040992 42 I 9 828I62 2'31 8689goo 1 92 959262 4-23 o40738 41 20 8283oi 2 -3I 868785 1.92 959516 4.23 o4o484 40 21 9.828439 2.3 I 9'868670 1 *92 9.959769 4-23 10o.04023I 3 22 828578 2.3I 868555 1'92 960023 4 283 o39977 38 23 828716 2. 31 868440 I.92 960277 4-23 o39723 37 24 828855 2. 30 868324 1.92 96053I 4. 23 o39469 36 25 828993 2 -30 868209 1.92 960784 4-23 o392 6 35 26 8a2913x 2.30 868093 I.92 96Io38 4 223 o38962 34 2j 829269 2. 30 867978 I'93 961291I 4.23 o03870o9 33 28 829407 2.3o 867862 I.93 961545 4.23 o38455 32 29 829545 2.30 867747 I.93 961799 4.23 03820I 31 3o0 829683 2.30 86763! I.93 962052 4.23 o37948 3o 3I 9.-829821 2 29 9'867515 I.93 9.962306 4-23 I0o.o37694 29 32 829959 2 *29 867399 I.93 962560 4.23 o37440 2 33 83oo97J 2.29 867283 I.93 962813 4.23 0o37187 27 34 83o234 2.29 867167 I.93 963067 4.23 o36933 26 35 830372 2.29 867051 I.93 963320 4.23 o36680 25 36 830509 2.29 866 35 1.94 963574 4-23 o36426 24 37 830646 2 -29 86689 1-94 963827 4.23 036I73 23 38 830784 2. 29 86673 1I94 964081 4-23 0o35919 22 39 830921 2 -28 866586 I 94 964335 4-23 o35665 21 40 83io58 2.28 866470 " 94 964588 4.-22 o35412 20 41 9.53ii95 2.28 9.866353 1-94 9.964842 4.22 160.o35x58 I 42 83I332 2. 28 866237 I *94 965095 4-22 o034905 I8 43 831469 2. 28 866120 I.94 965349 4.22 o3465i I7 44 83i6o6 2I 28 866004 I 1.95.965602 4.22 o034398 16 45 831742 2. 28 865887 1.95 965855 4.22 o34I45 I5 46 831879 2.28 865770 1.95 966105 4.22 o033891 14 47 83201 21'27 865653.95 966362 4.22 o33638 13 48 8321 52 2.*27 865536 I.95 966616 4-22 o33384 i2 49 832288 2. 27 8654g19 I 195 966869 4.22 o3313x II 50 832425 2.27 865302 1.95 967123 4 22 o32877 10 5i 9.83256J 2.27 9.865I85 I -.95 9-967376 4-22 I0o.o32624 9 52 832697 2.27 865o68 1,.95 967620 4-22 o32371 8 53 832833 2. 27 864950o 1.95 967883 4- 22 o32117 7 54 832963 2.26 864833 I.96 968136 4.22 o3x864 6 55 833 Io5 2*2.6 J8647I6 I.96 6838 4-22 o316 II 5 56 83324! 2 -26 864598 I.96 968643 4.22 o33i57 4 57 833377 2 -26 86448! I'96 968896 4I.22 o3i1o4 4 58 833512 2. 26 864363.96 969149 4.22 o3o85I 2 59 833648 2 -26 864245 I -.96 969403 4.22 o30597 I 6o 833783 2.26 864127 I*96 969656 4.22 o3o344 o ~I. o.__. Cose D. Sine ID. ICotang. D. Tang. M. (47 DEGREIES.) SINES AND TANGENTS. (43 DEGREES.) 61 IL Sine D. Cosine D. Tang. D. Cotang. o 9.833783 2.26 9.864127 I.96 9.969656 4.22 io.o3o344 6o I 833919 2 25 864010 196 969909 4.22 03o09I S 834054 2.25 863892 1'97 970162.4.22 o29838 8 3 834I89 2.25 863774 1I97 9704I6 4.22 029584 5 4 834325 2.25 863656 1.97 970669 4.22 02933I 5 834460 2.25 863538 1.97 970922 4.22 029078 55 6 834595 2.25 863419 1*97 971175 4.22 028825 54 834730 2.25 863301 I 97 971429 4.22 028571 53 834865 2.25 863183 1.97 971682 4.22 0283I8 52 9 834999 2.24 863064 1'97 97i935 4.22 028065 51 IO 835134 2-24 862946 1.98 972188 4.22 027812 50o II 9.835269 2.24 9.862827 1.98 9-972441 4.22 10027559 4 12 835403 2.24 862709 1.98 972694 4.22 027306 13 835538 2.24 862590 1.98 972948 4.22 b27052 47 14 835672 2.24 862471 1.98 973201 4.22 026799 46 15 835807 2.24 862353 1.98 973454 4.22 026546 45 16 83594I 2.24 862234 1.98 973707 4.22 026293 44 17 836075 2.23 862115 1.98 973960 4-22 026040 43 18 836200 2.23 86 996 1.98 974213 4.22 025787 42 ~9 836343 2.23 86i877 1.98 974466 4.22 025534 41 20 836477 2.23 861758 1.99 974719 4-22 02528i 40 ji 9.83661 2-23 9.86I638 I'99 9'974973 4.22 10.025027 39 22 836745 2-23 861519 1.99 975226 4.22 024774 38 23 836878 2.23 861400 I'99 975479 4.22 024521 3 24 837012 2.22 861280 1.99 975732 4.22 024268 36 25 837146 2.22 86116i 1.99 975985 4.22 024015 35 26 837279 2-22 86I041 I.99 976238 4.22 023762 34 17 837412 2.22 860922 I.99 976491 4.22 0o23509 33 8 837546 2.22 860802 I.99 976744 4.22 023256 32 [29 837679 2.22 860682 2oo00 976997 4.22 023003 31. 30 837812 2.22 860562 2.00 977250 4.22 022750 30 3I 9.837945 2.22 9.860442 2.00 9.977503 4.22 I0'022497 29 32 838078 2.21 860322 2.00 977756 4-22 022244 28 33 838211 2.21 860202 2-00 978009 4-22 021991 27 34 838344 221I 860082 2 00 978262 4.22 021738 26 35 838477 2.2I 859962 2.00 978515 4.22 021485 25 36 8386o10 22I 859842 2.00 978768 4.22 02I232 24 37 838742 22I 859721 2o01 979021 4.22 020979 23 838875 22I 859601 20oI 979274 4.22 o20726 22 39 839007 2.21 859480 2o01 979527 4.22 020473 21 40 8391gI40 2 20;859360 2*01 979780 4.22 020220 20 41 9.839272 2.20 9.859239 2.o1 9.980033 4.22 o0o019967 9 42 839404 2-20 85119g 2.01 980286 4.22 o0197I4 18 43 839536 2.20 858998 2.01 980538 4.22 019462 1 44 839668 2*20 858877 2-o01 980791 4.21 019209 Ir 45 839800 220 858756 202 981044 4.21 2I O956 o 46 839932 2.20 858635 2.02 981207 4.21 oI8703 14 4 840064 2.19 858514 2*02 981550 4.21 oI845 13 48 840196 21Ig 858393 2.02 981803 49 840328 2*19 858272 2*02 982056 4.21 017944;I 50 840459 2I19 858i51 2-02 982309 4.21 OI76g91 o 51 9.840591 2.19 9.858029 2. 02 9.982562 4.2I 10017438 52 840722 2g19 857908 2.02 98284 4.21 017186 Y 53 840854 2I19 857786 2.02 983067 4.2I oi6933 54 840985 2.19 857665 203 983320 4.21 oi668o 6 55 84III6 2*I 8I 857543 2.03 983573 4.21I o6427 5 56 841247 2.I8 857422 2.03 983826 4.2I o6174 4 5 84378 2-I8 857300 2.03 984079 4.21 01592I 3 841509 2.I8 857178 2.03 984331 4.21 oi5669 2 59 841640 2-18 857056 2.03 9C584 4.21 o154S 16 60 841771 2I8 856934 2.03 984837 5163 Cosi ne D. Sine D. Cotng. Tang. M. (46 D(EGREI.) VA (44 DEGREES.) A TABLE OF LOGARITHMIC f-_Sine D. Cosineo D. Tang. -D. Cotang. 0 9-84177I 2. 18 9.856934 2-3 9.984837 4-21 o.o-i5163 60 1 841902 2.18 85681'2 2.03 985090 4.21 014910 50 1 842033 2.18 856690 2.04 985343 4.21 o014657 5 3 842I63 2.17 856568 2.o4 985596 4.21 014404 4 842294 2.17 856446 20o4 985848 4-2I 014152 5 842424 2.17 856323 2.o4 986101oi 421 13899 55 6 842555 2-17 856201 2o04 986354 4-21 013646 54 842685 2.17 856078 2.o4 986607 4-21 013393 53 842815 2.17 855956 2o04 986860 4- 21 013144o 52 9 842946 2'17 855833 2'04 987112 4-2I 012888 51 10 843076 2.17 855711 2 05 987365 4-21 012635 50 II 9.843206 2.16 9.855588 2o05 9.987618 4-.2I 10012382 49 12 843336 2.16 855465 2.05 987871 4.21 012129 48 13 843466 2.16 855342 2.05 988123 4.21 01 877 47 14 843595 2.16 855219 2.05 988376 4.-2 011624 46 15 843725 2-.16 855096 2-05 988629 4-21 011371 45 r6 843855 2*16 854973 2*05 988882 4.21 orlli8 44 7 843984 2-16 854850 2o05 989134 4 21 oio866 43 i 844114 2.15 854727 2.06 989387 4'21 oio6i3 42 19 844243 2.15 854603 2.06 989640 4 02I o1o360 4 20 844372 2.15 854480 2-o6 989893 421 010107 40 2I 9.844502 2.15 9.854356 2-o6 9-990145 4.21 0o-oo9855 3 22 84463I 2.15 854233 2.o6 990398 4.2 o009602 38 23 844760 2-15 8541o09 2-o.06 99065 421 0o09349 3 24 844889 2.15 853986 2-o6 99090o3 4.2l 0o0o97 36 25 845018 2*I5 853862 2.o6 991156 4.21 oo8844 / 35 26 845147 2-.5 853738 2-o6 991409 4.21 008591 34 1 27 845276 2:14 853614 2.o7 991662 4-21 oo8338 33 28 845405 2'14 853490 2'07 991914 4.21 oo8o86 32 29 845533 2.14 853366 2.o7 992167 4.21 007833 31 3o 845662 2.14 853242 2-07 992420 4-21 007580 3o 3I 9-845790 2-.4 9.853118 2.07 9-992672 4.21 10-007328 29 32 845919 2.'14 852994 2.07 992925 4-2I 007075 28 33 846047 2.14 852869 2.07 993178 4.-21 006822 27 34 846I 7 25 2. 14 852745 2.07 993430 4.21 006570 1 26 35 846304 2.14 852620 2.07 993683 4.2 oo6317 25 36 846432 2.i3 852496 2o08 993936 4-2I 1 oo6064 24 37 846560 2-.3 852371 2.08 994189 4.21 oo5811 23 38 846688 2-13 852247 20o8 994441 4.21 0o5559 22 39 846816 2*.3 852122 2.08 994694 4.21 oo5306 21 40 846944 2.I3 851997 2.08 994947 4.21 005053 20 4I 9-84707' 2-13 9-851872 2-08 9995199 4-2I Io-oo48oi 19 42 847199 2.13 851747 2.08 995452 4-2I 004548 18 43 847327 2.I3 851622 2.08 995705 4-21 004295 17 44 847454 2.12 851497 2-o9 995957 4-.21 oo4o43 i6 45 847582 2*12 851372 2.09 996210 4-21 003790 15 46 847709 2.12 851246 2.o9 996463 4.21 003537 14 47 847836 21.2 851121 2.09 996715 4-21 oo3285 13 4 847964 2.12 850996 2.09 996968 4-21 003032 12 49 848091 2.12 850870 2.0o9 997221 4-2I 002779 11 50 848218 2.I2 850745 2.o9 997473 4.21 00227 10 5i 9.848345 2.-12 9.850619 2.o9 9-997726 4.21 100lo0oo2274 52 848472 2*I I 850493 2.10 997979 421 I 002021 53 848599 2- 11 85o368 2.IO 99823 I 4.-21 00 ooI769 7 54 848726 2. -11 850242 2. IO0 998484 4- 21' ooi5i6 6 55 848852 2-I I 85oii6 2.o10 998737 4.2I o001oi263 5 56 848979 2-II 849990 2.10 998989 4.21 00101 4 57 849106 2-I 849864 |2*o 999242 4- 21 ooo758 3 58 849232 2-1I 849738 2 10o 999495 4.21 ooo000505 2 59 849359 21 I 849611 2.10 999748 4.' 21 000253 60 849485 2 11 849485 2.1 10 -o000000 4.21 10'o000000 Cosine D. SiDne D. Cotang. I D. Tang.! (45 DEGREES.)