IN MEMORIAM FLOR1AN CAJOR1 PLANE AND SPHERICAL TRIGONOMETRY THE MACMILLAN COMPANY NEW YORK BOSTON CHICAGO DALLAS ATLANTA SAN FRANCISCO MACMILLAN & CO., LIMITED LONDON BOMBAY CALCUTTA MELBOURNE THE MACMILLAN CO. OF CANADA, LTD. TORONTO PLANE AND SPHERICAL TRIGONOMETRY BY LEONARD M. PASSANO ASSOCIATE PROFESSOR OF MATHEMATICS MASSACHUSETTS INSTITUTE OF TECHNOLOGY Nefo fgotfc THE MACMILLAN COMPANY 1918 All rights reserved COPYRIGHT, 1918, BY THE MACMILLAN COMPANY. Set up and electrotypcd. Published April, 1918. NorfaooU J. S. Gushing Co. Berwick & Smith Co. Norwood, Mass., U.S.A. PREFACE OF late years, in the writing of textbooks of trigonometry, a tendency to amplification has shown itself, doubtless with the idea that amplification means simplification. Unfortu- nately the amplification has spent itself upon details rather than upon principles, which latter have too often been in- adequately treated. The result has been textbooks which overlook the comparative maturity of the boys and girls who study trigonometry and which cling almost with affec- tion to the practices of the most elementary mathematics. The present text' aims to present the trigonometry in such a way as to make it interesting to students approach- ing some maturity, and so as to connect the subject, not only with the mathematics which the student has already had, but also with the mathematics which, in many cases at least, is to follow. A subject may be so burdened with detailed explanations as to become monotonous and lifeless, or, on the other hand, presented in so concise and difficult a manner as to be repellent. The present work endeavors to avoid both extremes. Full explanations are given of im- portant principles, but many simple details are left to the work of the student. The following points in the text may be noted : 1. Positive and negative angles of any magnitude and the trigonometric functions of such angles, defined by means of a system of rectangular coordinates, are taken up in the beginning of the book ; acute angles, with their functions, being mentioned as a special case. 2. Thus the basic trigonometric identities are got at once for all angles. vi PREFACE 3. The functions of 0, 90, etc., are carefully explained by the theory of limits. 4. The solution of right triangles and related problems are taken up early without the use of logarithms. 5. Logarithms are then very carefully explained and fully discussed, not so much as to their use in computation, but rather so as to clarify their meaning. 6. Right triangles are then solved by the use of loga- rithms, and the essentially approximate nature of all nu- merical results is emphasized. 7. The text next returns to trigonometric identities, giv- ing a detailed and accurate proof of the addition formulae for sines and cosines, with less detailed but sufficient expla- nation of other fundamental identities. The number of identities to be memorized is reduced to a minimum. 8. The circular measure of an angle and the inverse func- tions are then taken up, emphasis being laid upon the fact that the latter are angles. 9. There follows the solution of triangles in general. As each case is mentioned the theorems or formulas needed for its solution are derived. 10. The last subject treated in the plane trigonometry is the solution of trigonometric equations, and the fact is em- phasized that the operations are simply the solution of algebraic equations applied to a new class of quantities. 11. The lists of examples and problems are numerous and carefully chosen, many of them being taken from work in analytic geometry and calculus, though, of course, no knowledge of either of these subjects is assumed. Some of the problems are entirely new, being invented for this text, and all problems are chosen with a purpose to indicate the practical interest and value of trigonometry. 12. In the spherical trigonometry, as in the plane, the three chief aims are brevity, clarity, and simplicity ; a chapter on the Earth treated as a sphere being given to enliven an otherwise somewhat formal and lifeless subject. 13. The author has not tried to revolutionize the teaching PREFACE vii of trigonometry, believing that much that has been done in the past is good though none the less open to improvement. Such improvement has been the aim of this work. The author wishes to acknowledge the kindness of his colleagues Professor H. W. Tyler, Professor F. L. Hitch- cock, and Professor J. Lipka in reading and criticizing the manuscript of his book, and to express his thanks to Pro- fessor E. R. Hedrick, editor of the tables appended, for permission to make use of them. L. M. PASSANO. CONTENTS XBT. PAGE INTRODUCTION ......... xiii PLANE TRIGONOMETRY CHAPTER I. THE TRIGONOMETRIC FUNCTIONS OF ANY ANGLE AND IDENTICAL RELATIONS AMONG THEM . 1-13 1. Rectangular coordinates 1 2. Angles of any magnitude 1 3. Abscissa, ordinate, and distance 3 4-5. The trigonometric functions denned .... 4 6. Signs of the functions 6 7. Functions of acute angles ....... 7 8. Reciprocal functions 7 9. Tangent, sine, and cosine 8 10. Sine and cosine ........ 9 11. Tangent and secant ........ 9 12-13. Fundamental relations, collected .... 10 14. Values of the functions when one is given ... 10 CHAPTER II. IDENTICAL RELATIONS AMONG THE FUNC- TIONS OF RELATED ANGLES. THE VALUES OF THE FUNCTIONS OF CERTAIN ANGLES .... 14-27 15. Functions of negative angles 14 16. Functions of 90 a 15 17. Functions of 90 + a . . . . . . . 16 18. Functions of 180 + a 17 19. Generalization 18 20. Functions of certain angles 20 21. Functions of 30 and 60 20 22. Functions of 45 21 23. Functions of 120, 135, etc 22 24. Functions of 22 25-26. Functions of 90, etc. 25 27. Limiting values of functions 26 X CONTENTS ART. PAGE CHAPTER III. THE SOLUTION OF EIGHT TRIANGLES. LOGARITHMS AND COMPUTATION BY MEANS OF LOGARITHMS 28-44 28. Solution of right triangles 28 29. Logarithms 30 20. The common system 31 31. Mantissa and characteristic 32 32. Four computation theorems 34 33. Special properties of logarithms 36 34-35. Illustrative examples 37 CHAPTER IV. FUNDAMENTAL IDENTITIES . 45-58 36. Introductory 45 37-40. The addition formulae, sine and cosine . . . 45 41. The addition formulae, tangent and cotangent . . 50 42-43. Functions of the double angle ..... 51 44. Functions of the half angle ....... 52 45. Sum of sines ; sum of cosines ...... 53 46. Identities and equations ....... 64 CHAPTER V. THE CIRCULAR OR RADIAN MEASURE OF AN ANGLE. INVERSE TRIGONOMETRIC FUNCTIONS . 59-67 47. Circular or radian measure ...... 59 48. Inverse functions ........ 62 49. General value of an angle . . . . . . 63 CHAPTER VI. THE SOLUTION OF GENERAL TRIANGLES 68-83 50. Four cases 68 51. The law of sines 68 52. Case I ; a side and two angles ..... 69 53. Case II ; two sides and an opposite angle ... 70 54-55. Case III ; two sides and included angle ... 71 56. The law of cosines 73 57-59. Formulae from the law of cosines .... 74 66. Case IV ; three sides 76 61. Case III ; other methods of solution .... 77 62. Areas of right triangles 78 63. Areas of oblique triangles 79 CHAPTER VII. THE SOLUTION OF TRIGONOMETRIC EQUATIONS 84-93 64. Introductory 84 65. Illustrative examples 84 CONTENTS xi ART. PAGE 66. Special types of equations 86 67. Simultaneous equations 89 68. Inverse trigonometric equations 90 SPHERICAL TRIGONOMETRY CHAPTER VIII. FUNDAMENTAL RELATIONS . 94-99 69. Introductory 94 70. The law of cosines 94 71. The law of cosines ; polar triangle 97 72. The law of sines 98 CHAPTER IX. THE SOLUTION OF RIGHT SPHERICAL TRIANGLES 100-106 73. Special formulae for right triangles ..... 100 74. Special formulae ; collected. Napier's rules . . . 101 75. Rules for solution 102 76. Illustrative examples. Quadrantal triangles . . . 103 CHAPTER X. THE SOLUTION OF OBLIQUE SPHERICAL TRIANGLES 107-124 77. Six cases 107 78. Case 1 . Three sides 107 79. Case 2. Three angles 110 80. Case 3. Two sides, included angle. Case 4. Two angles, included side. Napier's analogies . . Ill 81. Illustrative examples 114 82. Case 5. Two sides, opposite angle. Case 6. Two angles, opposite side 116 83. Delambre's or Gauss's equations 120 CHAPTER XI. THE EARTH AS A SPHERE . . 125-131 84. Distances on the earth 125 85. Position and direction ; latitude and longitude; course 125 86-87. Bearings ; illustrative examples .... 126 88. Area of a triangle 129 ANSWERS 133 INTRODUCTION TRIGONOMETRY is primarily the science concerned with the measurement of plane and spherical triangles, that is, with the determination of three of the parts of such tri- angles when the numerical values of the other three parts are given. This is done by means of the six trigono- metric functions, denned in article 4 following. But these functions enter so intimately into many branches of mathe- matical and physical science not directly concerned with the measurement of angles, that their analytical properties are of fundamental importance. Analytical trigonometry, that is, the proof and use of various algebraic relations among the trigonometric functions of the same or related angles, is therefore, in modern times, of equal importance with the trigonometry which deals with triangular solutions. The same functions which enable one to solve triangles constructed in a plane suffice also for the solution of spheri- cal triangles. But the solution of triangles of which the sides are geodetic lines, that is, lines which are the shortest distances between pairs of points on the surface, on a sphe- roidal surface such as the Earth, requires the use of other functions than those needed for the solution of plane or spherical triangles. This spheroidal trigonometry is very complex, and becomes necessary only in the accurate sur- vey of very large tracts of the Earth's surface. For ordi- nary purposes of surveying and for the solution of triangles on the Earth's surface over small areas, plane and spherical trigonometry are sufficient. The study of trigonometry, as ancillary to astronomy, dates from very early times. Among the Greeks, who, xiii xiv INTRODUCTION however, were more famous as geometers than as investi- gators in other branches of mathematics, the names of Hipparchus (about 150 B.C.) and of Ptolemy (who lived in the second] century of the Christian era), both astronomers, are prominent. Hipparchus left no mathematical writings, but we are told by an ancient writer that he created the science of trigonometry. Ptolemy, making use of the inves- tigations and discoveries of Hipparchus, perfected the form of the science. The theorems of these two astronomers are still the basis of trigonometry. Ptolemy calculated a table of chords, which were used in those earliest days of the science, as we now use the sines of angles. The radius of a circle he divided into sixty equal parts. Each of these he divided again into sixty equal parts, called, in the Latin translation of his- work the Almagest, " partes minutae primae " ; and each of these in turn into sixty, called " partes minutae secundae " ; whence have come the names " minutes " and " seconds " for the subdivisions of the angular degree. Ptolemy, however, was not the first to calculate a table of chords, Hipparchus, among others, having done so previously, but he invented theorems by means of which the calculations could be more readily made. The Hindus, more skillful calculators than the Greeks, acquired the knowledge of the latter and improved upon it, notably in that they calculated tables of the half-chord, or sine, instead of the whole chord of the angle. The Arabs also were acquainted with the Almagest, and with the investigations of the Hindus. It was an Arab, Al Battani or Albategnius, who first calculated a table of what may be called cotangents, by computing the lengths of shadows of a vertical object cast by the sun at different altitudes. Another Arab invented, as a separate function, the tangent, which had previously been used only as an abbreviation of the ratio sine to cosine. Curiously enough this invention was afterwards forgotten until the tangent was re-invented in England in the fourteenth century by Bradwardine, and < INTRODUCTION XV in the fifteenth century by the German, Johannes Mtiller, called Regiomontanus, who wrote the first complete Euro- pean treatise on trigonometry. When Napier * invented logarithms, in 1614, they were at once adopted in trigonometric calculations, and the first tables of logarithmic sines and tangents were made by Edmund Gunter, an English astronomer (1581-1626). He it was who first used the names cosine, cotangent, and co- secant. During the following century the science of trigo- nometry progressed slowly, becoming more analytical in form, until, in the hands of Euler (1707-1783), it became essentially what it is at the present day. With this brief introduction to the history of trigonom- etry let us now proceed to become acquainted with that homely, perhaps, but most serviceable handmaid to so many of the arts and sciences, "... being just as great, no doubt, Useful to men, and dear to God, as they I " * John Napier, 1550-1617. '. ' PLANE TRIGONOMETRY CHAPTER I THE TRIGONOMETRIC FUNCTIONS OF ANY ANGLE, AND IDENTICAL RELATIONS AMONG THEM 1. Rectangular Coordinates. Two lines, x'x and y'y, drawn in a plane at right angles to each other, as in Fig. 1, form a system of rectangular, Cartesian coordinates. The point in which the lines intersect is called the origin ; the two lines are called the axes of coordinates. One of these, usually the horizontal line, is called the axis of abscissae, or the axis of x. The other is called the axis of ordinates, or the axis of y. We shall speak of XOY, FOX', X'OY', and Y'OX as the first, second, third, and fourth quadrants respectively. 2. Angles of any Magnitude. There are many ways in which a system of coordinates is used in mathematics. In trigonometry such a system is used primarily in defining B 1 2 PLANE TRIGONOMETRY [I, 2 the trigonometric functions, but before we proceed to do so we shall extend our ideas of angles beyond the knowledge we obtained of them in the elementary geometry. There an angle is denned by some such definition as the following : the plane figure formed by two straight lines drawn from the same point. The unit of angles is either the right angle, or -the degree, and the largest angle usually dealt with is equivalent to two right angles and is often called a straight angle. In trigonometry, on the other hand, we deal with angles of any magnitude whatever. To do so we introduce the idea of motion, of revolution. Starting from the initial position OX, Fig. 1, we may revolve the line about in the direction indicated by the arrows, stopping in any desired terminal position OPi, OP 2 , OP 3 , OP 4 , etc. In this way angles of any number of degrees whatever may be gen- erated. Thus, if we stop in the position OF, we have an angle of 90; in the position OX', 180; in the position OP 3 , 225 ; in the position OY', 270, and so on. By mak- ing one whole revolution we should arrive at an angle of 360 ; two and one half revolutions, 900; etc. Not only so, but we might revolve from the initial posi- tion OX in the opposite direction. Now oppositeness is indicated algebraically by the use of the signs plus (+) and minus ( ). So that if we agree to take the positive direc- tion of revolution counterclockwise, then clockwise will be the negative direction and we can thus generate negative angles of any magnitude whatever. Thus, Fig. 1, the angle XOP 3 is 225 if we have revolved in the positive direction, but is 135 if we have revolved in the negative direction. When an angle lies in value between and 90 it is said to be an angle in the first quadrant since its terminal side lies in the first quadrant. An angle lying in value between 90 and 180 is said to be in the second quadrant; between 180 and 270, in the third quadrant; between 270 and 360, in the fourth quadrant. I, 3] TRIGONOMETRIC FUNCTIONS EXAMPLES Construct the angles 1. 300. 3. 750. 5. - 1215. 2. -210. 4. -495. 6. 420. Add the following angles graphically : 7. 720 and 30. 10. 990 and - 60. 8. - 180 and 60. 11. - 45 and 120. 9. - 90 and - 45". 12. 135 and - 450. If A is a positive angle in the first, second, third, or fourth quad- rant respectively, add graphically 13. 450 and A. 15. 180 and -A. 14. - 270 and A. 16. - 540 and - A. 3. Abscissa, Ordinate, and Distance. Consider an angle, positive or negative, of any magnitude whatever*, XOP, of Fig. 2. From P, any point in the terminal side of this angle, drop a perpendicular upon the axis of x. The lines of the figure are named as follows : OM is called the abscissa of the point P 7 MP the ordinate, and OP the dis- tance. The abscissa OM and the ordinate MP are together called the coordinates of the point P. Note very carefully that the abscissa is always read from to M, the ordinate from M to P\ that is, in each case from the axis to the * As a matter of convenience we do not consider angles numerically greater than 360. It is obvious that the discussion applies equally well to such angles. 4 PLANE TRIGONOMETRY [I, 3 point. The distance is read from to P. Thus, for an angle in the second or third quadrant the direction of the abscissa is opposite to that of an angle in the first or fourth quadrant. For an angle in the third or fourth quadrant the direction of the ordinate is opposite to that of the ordinate of an angle in the first or second quadrant. Oppositeness in direction being distinguished as usual by difference in algebraic sign we have the following con- ventions : Tlie abscissa measured to the right of the axis of y is positive ; to the left, negative. The ordinate measured up- ward from the axis of x is positive ; downward, negative. The distance is measured from the origin outward and is taken positive. EXAMPLES 1. The abscissa of a point is 3, its ordinate 4 ; find the distance. 2. The distance of a point is 5, its ordinate 4 ; find the abscissa. 3. The ordinate of a point is 2, its distance 3 ; find the abscissa. 4. The ordinate of a point is 5, its abscissa 4 ; find the distance. 5. Prove that the square of the distance of any point is equal to the sum of the squares of the abscissa and ordinate. 6. Prove that for all points on a straight line through the origin the ratio of the ordinate to the abscissa is constant. 4. The Trigonometric Functions Defined. Let us now proceed to define the six trigonometric functions of an angle; six quantities which depend upon the angle for their values. They are the possible ratios between the various pairs of the three lines named in Art. 3. Thus, Fig. 2, the sine XOP = ordinate of P = MP ^ ^r^^ distance of P OP cosine XOP = abscissa of P =, distance of P OP tangent XOP = ornate of P = MP abscissa of P OM I, 5] TRIGONOMETRIC FUNCTIONS .' cotangent XOP = -;;--- ~ -^ ^ = abscissa of P OM ordinate of P distance of P MP' OP abscissa of P distance of P OM' OP ordinate of P MP secant XOP = cosecant XOP = Three other functions are sometimes used : The versed sine, which is unity minus the cosine ; the coversed sine, which is unity minus the sine ; the suversed sine, which is unity plus the cosine. They are relatively unimportant. 5. Trigonometric Functions are Ratios. The first thing to be noted about these functions is that, being ratios, they are independent of the actual lengths of the abscissa, M 3 M 2 M, I MO FIG. 3. ordinate, and distance. Thus, Fig. 3, the triangles OM 2 P. 2 , and OM 3 P 3 being similar, their homologous sides are proportional, so that OP, OP 2 OP 3 ' OM, Similarly the truth of the statement may be shown for the remaining functions. 6 PLANE TRIGONOMETRY [I, 6 6. Signs of the Functions. The second point to be noted is that the signs of the functions vary according .to the quadrant in which the angle lies. Thus, Fig. 2, for the angle XOP in the first quadrant the abscissa, ordinate and distance are all positive so that all the functions are positive. For the angle XOP in the second quadrant the ordinate and distance are positive, the abscissa negative. Thus we have for the angle in the second quadrant ***>?-<-=. The following table gives the signs of the functions in the four quadrants. QUAD. I II III IV QUAD. sine -f- -f cosecant cosine -f -f- secant tangent + + cotangent EXAMPLES Determine the algebraic signs of 1. cos 218. 3. sin 1100. 5. sec 315. 2. tan (-460). 4. cot (- 99). 6. esc (- 210). 7. Let the student determine, as above, the signs of the trigono- metric functions of angles in the third and fourth quadrants. I, 8] TRIGONOMETRIC FUNCTIONS 7. Functions of Acute Angles. A special set of definitions for the functions of acute angles, which are sometimes useful and should be known, follows directly as a special FIG. 4. case of the general definitions given above. Thus, Fig. 4, in which the angle XOP lies in a right triangle, _ ordinate _ opposite side distance hypotenuse v/-ki> abscissa adjacent side cos XOP = - = -^ > distance hypotenuse , abs. adj. side -, ord. opp. side , abs. adj. side ord. opp. side These definitions, it must be noted, completely agree with the more general definitions, but are applicable only to angles less than ninety degrees, since angles greater than ninety degrees cannot occur in right triangles. 8. Reciprocal Functions. Two questions would naturally suggest themselves at this point: Are the trigonometric functions of an angle related to each other in any particular 8 PLANE TRIGONOMETRY [I, 8 way? and, second, if there be a definite relation between two. given angles will the functions of those angles bear some special relation to each other? We shall proceed to answer the first of these questions affirmatively, but shall leave the discussion of the second question to a later chapter (Chap. II). Thus, if a be any angle, it follows by the definitions of the trigonometric functions that ordinate sin = YT listance ~~ tana = ord. abscissa 1 distance ~~ dist. ~~ sec n abs. ordinate _ 1 1 abscissa ~~ abs. ~~ cot a 9 ord. or, the sine and cosecant, the cosine and secant, the tangent and cotangent respectively of the same angle are reciprocals of each other. 9. Tangent, Sine and Cosine. Again, by definition, and by Art. 8, ordinate ordinate distance sin a cot a = abscissa abscissa cos a' distance 1 cos a tan a sin a sin a cos a These relations may be proved otherwise, thus, Fig. 5 : MP _. MP OP sin XOP tan XOP = T^TT = T^TT = v /^p> OM OM cos XOP ~OP I, 11] TRIGONOMETRIC FUNCTIONS QM OP _ cos XOP MP~ sin XOP' OP 9 _ ~ " y FIG. 5. 10. Sine and Cosine. Also, Fig. 5, it is obvious that MP 2 + OM 2 = OP 2 . Dividing each term by OP 2 gives OP) Whence, by definition, (sin XOP) 2 +(cos XOP) 2 = 1 or, as it is usually written, letting a = Z XOP, sin 2 a + cos 2 a = 1. 11. Tangent and Secant. Similarly, writing the first equation of Art. 10 in the form and dividing each term by OM 2 , we have (OPV_(MP\* (OM)-(OM) 10 PLANE TRIGONOMETRY [I, 11 That is (sec XOP) 2 = (tan XOPy + 1 or, sec 2 a = tan 2 a + 1. In the same way we obtain the relation csc 2 a = cot 2 a + 1. 12. Fundamental Relations. These relations, summarized below, are of great importance and must be memorized. sin a = csc a = tan a = ^^, cot a = ZZL=. cos a sin a sin 2 a + cos 2 a = 1. (3) sec 2 a = 1 4- tan 2 a, csc 2 a = 1 4- cot 2 a. (4) 13. By means of the identities of Art. 12 the value of any one of the trigonometric functions may be expressed in terms of each of the other five. Thus, by (3) 1 cos a- l tflfi n csc a sec a' cot a sin a' sina cosa <-.ota.- c 8a tan a sin a = VI cos 2 a. By (2), (1), and (4), Bina = tana- cos =^=s- tan a sec a VI 4- tan 2 a where the radical may be either plus or minus. 14. To Compute the Values of the Other Functions when One Function of an Angle is Given. By means of the relations of the preceding article if the value of any one function of an angle be given, the values of the remaining functions may be found, but a simpler method of obtaining them is illustrated by the following examples. I, 14] TRIGONOMETRIC FUNCTIONS 11 Example 1. Given sin A = -f , find the values of the remaining functions. ord. m-2 2^ -2 33* sin A = dist. m 3 The distance being always positive, the minus sign nec- essarily is taken with the ordinate. Therefore, Fig. 6, construct an angle whose ordinate is 2 and whose dis- FIG. 6. tance is 3, or any multiple (m) of 2 and 3. The third side of the right triangle is V9 4 = V5, This is the value of the abscissa and we may write the values of the six functions from the definitions. , o sn r, V5 2 -7= v5 V5 sec XOP 2 = , 12 PLANE TRIGONOMETRY [I, 14 Example 2. Given cot A = -f , find the remaining func- tions. cot A = abs. ord. m 5 m 4 5 -4' since either the abscissa or the ordinate may be negative. Construct an angle XOP 1} Fig. 7, having an abscissa 5 and an ordinate -f 4, and an angle XOP 2 having an abscissa M,-5 x x + 5 Wf y' FIG. 7. 4- 5 and an ordinate 4. In each case the distance is found to be V25 + 16 = V41, and we may write sn 4 V4l' sin XOP 2 = - V4l' sec V41 5' 5 4' V41 cos XOPo = 5 2 := cot sec o = V41 _5 4' VH esc XOP l = V41 esc XOP 2 = V41 It will be seen that the ambiguity of the two sets of values will occur in every case, no matter what function be given and no matter whether the sign of the given function be plus or minus. I, 14] TRIGONOMETRIC FUNCTIONS 13 EXAMPLES Find the values of the remaining functions, given that 1. sin a = J. 3. cot a = - 3. 5. sec a = 4. 2. cos a = - f . 4. tan a = f . 6. esc a = ^. 7. If sin x = 5, can the values of the remaining functions be found ? Why ? 8. If sec x = |, can the values of the remaining functions be found ? Why ? 9. If tan x = 4, can the values of the remaining functions be found ? Why ? " 10. Given sec a = |, find the functions of 90 a. 11. Given cot a = x, find the functions of 90 a. Prove the following relations : 12. cos a = - 19 13. cota = 20 coscc tana Vsec 2 a 1 i sin a cos a 14. seC tt = - ggg-g -- 21. . Vcsc 2 a 1 sec a = sin a. 15. sin = - 22. Vl + cot 2 a sin2 a 16. tanft = -. 23. = cos . cos a esc a 17. tan a = sin ff - . 24. tan a esc a = sec a. Vl - sin2 a 18. cot a = cos a 25. cos a- sec a = . vl cos 2 a cos ot sec a 26. sec 2 a csc 2 a = tan 2 a - cot 2 a. CHAPTER II IDENTICAL RELATIONS AMONG THE FUNCTIONS OF RELATED ANGLES. THE VALUES OF THE FUNC- TIONS OF CERTAIN ANGLES 15. Functions of Negative Angles. We shall now proceed to determine the relations which exist among the functions of two angles when those angles are related in some par- FIG. 8. ticular way. Let us consider first two angles one of which is the negative of the other, Fig. 8. Let the value of the positive angle XOP be a, and of the numerically equal negative angle XOP ' be a. On the terminal sides of these angles lay off the equal distances OP and OP', and drop perpendiculars from P and P 1 upon the axis of x. These perpendiculars will obviously cut the axis of a; in 14 II, 16] IDENTICAL RELATIONS 15 the same point, M, and the two right triangles MOP and MOP' will be congruent. Therefore, OF'=OP, OM= OM, and MP' = - MP. We then have MP' - MP Thus any function of a negative angle is equal, numeri- cally, to the same function of an equal positive angle. The algebraic sign is determined by the quadrant which a lies in when a is acute. 16. Functions of 90 a. Consider next two angles, a and 90 - , Fig. 9. Let XOP be the angle a and XOP' be 90 a. Lay off on the terminal sides of these angles the equal distances OP and OP ', and from P and P ' drop perpendiculars PM and P'M 1 upon the axis of x. Then obviously the right triangles MOP and M'OP' are congru- ent, and OF = OP, M'P ' = OM, and OM ' = MP. There- fore, 1 * By virtue of Art. 12, it is necessary to memorize only sin ( a) and cos (a). 16 PLANE TRIGONOMETRY [II, 16 S ec(90- a) = |j; = |=csc, BC ( M x / ^-a 9Q ,_ 1 p' y' y'Np FIG. 9. Thus we see that each of the functions of 90 a is equal, numerically, to the co-function of the angle a. The important case is when a is acute, and a and 90 a are complementary angles. Indeed it was because of this rela- tion that the cosine, cotangent, and cosecant received their names. They are the sine, tangent, and secant of the complementary angle. 17. Functions of 90 + a. We shall consider two more cases, limiting the discussion to acute values of a, although II, 18] IDENTICAL RELATIONS 17 the results will be equally true for any value of a whatever. In Fig. 10 let the angle XOP be a and XOP' be 90 + . On the terminal sides of these angles lay off the equal dis- tances OP and OP', and from P and P' drop perpendiculars PM and P'M ' upon the axis of x. It follows that the two FIG. 10. triangles MOP and M'OP f are congruent and that OP 7 = OP, M'P' = OM, OM' = - MP. Therefore, 18. Functions of 180 4- a. be a and XOP' be 180 + a. In Fig. 11 let the angle XOP On the terminal sides of these 18 PLANE TRIGONOMETRY [11,118 angles lay off the equal distances OP and OP, and from P and P drop perpendiculars PM and P'M' upon the axis x' M of X. Then the triangles MOP and M' OP' are congruent and OP' = OP, M'P' = - MP, OM'= - OM. Therefore, sin (180 + ) = =~ 19. Generalization. In a similar manner may be found analogous relations connecting the functions of an angle a with the functions of any integral multiple of 90 plus or minus a. Upon examining these relations we are led, by iduction, to express them in the following general rule. Any function of an even * multiple of 90 plus or minus a is the same function of the angle a. * Zero is taken as an even number, so that the rule includes the case of Art. 15. II, 19] IDENTICAL RELATIONS 19 Any function of an odd multiple of 90 plus or minus a is the co-function of the angle a. The algebraic sign of the value is determined by the quad- rant (counting in the positive direction) in which the terminal side of the angle lies when a is acute. Examples. 1. sin (720 ) = sin a, since 720 is an even multiple of 90 and the terminal side of 720 a, when a is acute, lies in the fourth quadrant. 2. cot ( 90 )= tan a, since 90 is an odd multiple of 90 and the terminal side of 90 a, when a is acute, lies in the third quadrant. 3. sec ( 180 + a) sec a, since 180 is an even multiple of 90 and the terminal side of 180 -f a, a acute, lies in the third quadrant. 4. tan 281 = tan (270 + 11) = - cot 11, or tan 281 = tan (360 - 79)= - tan 79, since I , is an j multiple of 90 and the terminal side of 281 lies in the fourth quadrant. (~z & EXAMPLES By means of a geometrical construction express each of the follow- ing as a function of a, where a is an acute angle. Check your results by the rule given above. 1. cos (270 + a}. 5. cot (270 - a). 2. sin (180 -a). 6. sec (270 - a). 3. esc (- 90 + a). 7. sin (- 180 - a). 4. tan (540 + a). 8. cos(- 270 + a). Express as a function of an acute angle & 9. sin 324. 13. sec (-537). 10. cos (-375). 14. cot 1140. 11. tan 457. 15. tan 496. 12. esc (-801 32'). 16. cos (-480). 20 PLANE TRIGONOMETRY [II, 20 20. Functions of Certain Angles. We see by the preced- ing article that the functions of angles greater than 90, and of negative angles, can be expressed in terms of the functions of angles lying between and 90. It follows that if we wish to use the trigonometric functions for com- putation or for other purposes we need find their values only for all positive acute angles. We shall not discuss the methods by means of which these values are computed in general, but shall proceed to find the values of the func- tions of certain angles which frequently occur. We shall then, in the following chapter, show how we may find the values of the functions of any angle from tables with which we are provided. We shall see, also, how the values thus found may be used in the solution of triangles ; that is, in finding the unknown parts, angles or sides, of a triangle from parts which are given. 21. Functions of 30 and 60. Let the angle XOP, Fig. 12, be an angle of 30, and from P drop a perpendicular, PM, aVT FIG. 12. upon the axis of x. Then, as we know, the angle 0PM is a 60, and if OP have the value a, MP must be equal to - and OM equal to aV3 Therefore, by definition, II, 221 IDENTICAL RELATIONS 21 tan 30 = a MP ~~2 OM a V3 V3 oV3 03f 2 V3 cos30 = = = ^T oV3 V3 MP a 2 By a similar construction, or by the relations of Art. 16, the following values may be derived : sec 60 = 2, sin 60 = ^2, tan60=V3, 2 cot 60 = V3 esc 60 = vs FIG. 13. 22. Functions of 45. Let the angle XOP, Fig. 13, be an angle of 45, and from P drop a perpendicular, PM, upon the axis of x. Then the angle 0PM is an angle of 45, and if OM have the value a, MP also will be equal to a and OP will be aV2. Therefore, by definition, 22 PLANE TRIGONOMETRY [II, 22 5 aV2 V2 OP a V2 23. Functions of Other Angles Readily Found. By sim- ilar constructions the functions of 120, 150, 135, etc., or, in general, any integral multiple of 90 plus or minus 30, 60, or 45, may be found. They may be found more readily, however, by using the rule given in Art. 19. Thus, sin 120 = sin (90 + 30) = cos 30 = . 2 or sin 120 = sin (180 - 60) = sin 60 = . EXAMPLES Find the values of the functions of 1. 120. 4. 210. 7. 300. 2. 135. 5. 225. 8. 315. 3. 150. . 6. 240. 9. 330. Prove that 10. sin 210 tan 300 = sin 120. 11. sec 315 sec 300 = sec 240 sec 225. 12. tan 210 : cos 150 = tan 150 : cos 330. 13. esc 330 sec 315 sin 225 = - sec 120. 24. Functions of Zero. Let the value of the angle XOP, Fig. 14, be represented by a, and from P, any point in the terminal side of the angle, drop a perpendicular, PM, upon the axis of x. By definition, II, 24] IDENTICAL RELATIONS 23 Now, for the sake of convenience keeping the distance OP constant in length, let the line OP approach nearer and nearer to the position OX. Then the angle a can be made,* smaller than any angle that may be assigned, however FIG. 14. small, or, as it is otherwise expressed, a will approach the limit zero. At the same time MP will approach zero as a limit, and OM will approach OP as a limit. OM Then will approach the limit zero and OP OP will approach the limit unity. Thus, as the angle approaches the limit zero (or, becomes smaller than any value that may be assigned, how- ever small) its sine approaches the limit zero (or, becomes smaller than any value that may be assigned, however small) and its cosine approaches the limit unity (or, differs from unity by a number smaller than any number that may be assigned, however small). This may be written limit sin a = 0, limit cos a = 1. o=0 OP Again, by definition, esc a = , and as a grows smaller MP Qp OP remains constant and MP grows smaller, so that becomes continually greater. Finally, when a approaches zero as a limit, MP becomes smaller than any number that * And will remain. 1*5 24 PLANE TRIGONOMETRY [II, 24 OP may be assigned, however small, and - becomes greater than any number that may be assigned, however great. OP This we express by saying that - approaches the limit infinity, or increases without limit. We may then write limit esc a = 00. a=0 Similarly it may be shown that limit tan a = 0, limit cot a = oo, limit sec a = 1. These relations are often briefly expressed, sin = 0, cosO = l tan = 0, cot 0= oo, sec = 1, esc 0= co. (6) to which there is no objection if we remember that these are merely abbreviations of the preceding statements, and that FIG. 15. means, not that we have no angle, but that" we are deal- ing with an angle which becomes smaller than any value that may be assigned, however small ; and that when this happens the sine of the angle also becomes smaller than any value that may be assigned, however small, the cotangent II, 26] IDENTICAL RELATIONS 25 becomes greater than any value that may be assigned, how- ever great, the cosine approaches the limit unity, etc. 25. Functions of 90. Let the angle XOP, Fig. 15, be represented by a, and let OJf, MP, and OP be respectively the abscissa, ordinate, and distance of P. Also, keeping the distance OP constant, let the line OP approach Y as its limiting position. Then, a approaches the limit 90, OM approaches the limit zero, MP approaches the limit OP. Therefore, limit sin a = limit - = 1. a=90 OP limit cos a = limit - = 0. a=90 OP limit tan a = limit - = oo, a=90 OM limit cot a = limit - = 0. MP OP limit sec a = limit - = oo, a=90 OM OP limit esc a = limit - = 1. a=90 MP With the same understanding as in the preceding article these may be written sin 90 = 1, tan 90 = oo, sec 90 = oo, cos 90 = 0, ' cot 90 = 0, esc 90 = 1. 26. The student should find, as in Arts. 24 and 25, the following : sin 180 = 0, tan 180 = 0, sec 180 = - 1, cos 180 = - 1, cot 180 = oo, esc 180 = oo. sin 270 = - 1, tan 270 = oo, sec 270 = oo, cos 270 = 0, cot 270 = 0, esc 270 = - 1. 26 PLANE TRIGONOMETRY [II, 27 27. Limiting Values of the Functions. We have seen (Art. 20) that all possible numerical values of the trig- onometric functions are given by angles lying between and 90. Let us now see between what limits the values of the functions lie. From the discussion and figures of articles 24 to 26 we see that the sine and cosine of an angle lie between 1 and + 1, the tangent and cotangent lie between oo and -f- oo, the secant and cosecant lie between 1 and oo or between 1 and oo. It is well to note also, for angles in the first quadrant, that as the angle increases the direct functions increase, the co-functions decrease. A very convenient and simple way to remember the range of values and the signs of the trigonometric functions is by FIG. 16. means of the unit circle, a circle with unit radius, which need not be actually drawn but merely visualized. Draw such a circle, Fig. 16, with its center at the origin of co- ordinates, and let XOP be any angle. Drop the perpendicu- lar PM upon the axis of X, and draw LQ tangent to the circle at L and meeting OP produced in Q. Then, by definition, II, 27] IDENTICAL RELATIONS 27 tan XOP = = = LQ, etc. OL 1 If now the line OP be pictured as revolving from the position OL, the sine of the angle XOP, namely MP, will be seen to increase from zero and approach unity as the angle approaches 90. The cosine, namely OM, decreases from unity to zero, and the tangent (LQ) increases without limit. Also, as the angle increases beyond 90, the direc- tions of the lines MP and OM indicate the signs of the sine and cosine. The other functions follow directly from these two by virtue of the relations of Art. 12. CHAPTER III THE SOLUTION OF RIGHT TRIANGLES. LOGARITHMS AND COMPUTATION BY MEANS OF LOGARITHMS 28. Solution of Right Triangles. With the definitions of the trigonometric functions and tables giving their nu- merical values we are now prepared to solve right tri- angles ; that is, to find the values of the unknown parts from those that are known. Two parts in addition to the right angle must be known, and one at least of these parts must be a side. We have then the general rule of procedure : Select that trigonometric func- tion which involves the two known parts and one unknown FlG - 17 ' part. The value of the un- known part can then be computed by elementary algebraic processes. Example 1. Given A = 32 16', a = 124, C = 90, find B, &, and c. See Fig. 17. Obviously B = 90 - A = 90 - 32 16' = 57 44'. Then a =124 A-32'16' : = 90 cot A = -, a A a sin A = -, c or b = a cot A. From the tables we find cot A = 1.5839. a sin A sin A = .5338. 28 Ill, 28] SOLUTION OF RIGHT TRIANGLES 29 Therefore, 6 = 124x1.5839 c= 124 .5338 = 196.4. = 232.3. Example 2. Given a = 50, 6 = 60, C = 90, find A, B, and c. In this case, A = 39 48'. B = 90 - A = 50 12'. To find c we may use either sin A = - or c 2 = a 2 -f sin A 50 =V2500 + 3600 .6402 = 78.1. = 78.1. EXAMPLES Solve the following right triangles : 1. a = 250, 3. a = .55, 5. ^4 = 59 58', A <= 36 22'. c = .70. b = 412. 2. a = 37.5, 4. B = 72 6', 6. = 24 33', 6 = 40.1. c = 502. a = 211. PROBLEMS 7. What is the height of a flagpole if at a horizontal distance of 200 feet from the foot of the pole the angle of elevation of its top is 19 28' ? 8. A rope is stretched taut from the top of a building to the ground, and is found to make an angle of 58 56' with the horizontal. If the building is 61 feet high how long is the rope ? 30 PLANE TRIGONOMETRY [III, 28 9. If a tree 74.3 feet high casts a shadow 42.6 feet long, how many degrees above the horizon is the sun ? 10. A man walking on level ground finds, at a certain point, that the angle of elevation of the top of a tower is 30. He walks directly toward the tower for a distance of 300 feet and then finds the angle of elevation of the top to be 60. What is the height of the tower ? 11. At a point, J., south of a tower the angle of elevation of the top of the tower is 60. At another point 300 feet east of A the angle of elevation is 30. What is the height of the tower ? 12. The angles of a right triangle are 42 and 48 ; the hypotenuse is 200 feet. What is the length of the perpendicular from the right angle to the hypotenuse ? 13. The height of a gable roof is 20 feet, its width 42 feet. What is the pitch of the roof ; that is, the angle it makes with the horizontal ? 14. From where I stand a tree 50 feet away has an angle of eleva- tion of 43 31'. From the same point another tree, 75 feet distant, has an angle of elevation of 32 20'. Which tree is the taller and by how much ? 29. Logarithms. The solution of right triangles as thus explained is simple in theory but may become laborious in practice because of the arithmetic computation involved. Fortunately we have in logarithms a device for simplifying such computation. The base of a system of logarithms is, in general, any arbitrarily chosen number.* In practice two systems are used : the Briggsian or common system of which the base is 10 ; and the Napierian system of which the base is e = 2.718 . The logarithm of a number to a given base (a) is the exponent of the power to which the base (a) must be raised to produce the number. Thus, if a x = m, then x is the logarithm of ra to the base a ; written x = log a m. The word power is used here in its broader sense to in- clude fractional and negative exponents. Defining frac- tional and negative exponents in such a way that the laws of exponents a m a n = a m+n ; (a m ) n = a mn hold for nega- tive numbers and fractions as well as for positive integers, * Some numbers, unity, for example, cannot be so used. Ill, 30] SOLUTION OF RIGHT TRIANGLES 31 values of x may be found to satisfy, approximately at least, such an equation as a x = &, no matter what values a and b may have. Thus, given any number, a, by raising it to a suitable power, p, and extracting a suitable root, g, of the result, we can obtain any other number, b ; that is, -^/~a p = b. But this may be written a 9 = b or a x = 6, where # = -^ , the division of p by q being carried out to any desired number of decimal places. We then call x the logarithm of b to the base a. 30. The Common System. For purposes of computation the common system, base 10, is used. Let us form a table of powers of 10 and express the relations in terms of logarithms. 10-s = .001, or log w .001 = - 3. 10-2 = .01, Iog 10 .01 10- 1 = .l, lo glo .l 100 = i 9 l oglo l = o. 10 1 =10, Iog 10 10 =1. 10 2 = 100, Iog 10 100 = 2. 10 3 =1000, Iog 10 1000 = 3,* etc. etc. This table could be extended indefinitely in either direc- tion. If we examine the table we notice that to produce a number between 1 and 10 we must raise the base 10 to a positive power between and 1 ; to produce a number be- tween 10 and 100, the exponent of the base must lie between 1 and 2 ; for a number between 100 and 1000, the exponent must lie between 2 and 3, and so on. In other words, the logarithm of a number between 1 and 10 lies between and 1, and is, therefore, a fraction, always expressed as a deci- mal. The logarithm of a number between 10 and 100 lies between 1 and 2, or is 1 plus a decimal. The logarithm of * Hereafter in this work we shall not write the base 10. Thus log 7 means Iog 10 7. In general, however, except in works on trigonometry, if no base is written, e 2.718 -"is understood. 32 PLANE TRIGONOMETRY [III, 30 a number between 100 and 1000 is 2 plus a decimal. The logarithm of a number is thus seen to consist, in general, of two parts, an integral part and a decimal part. The integral part is called the characteristic of the logarithm ; the decimal part is called the mantissa. The results of our observations may be summarized thus : NTTMBER BETWEEN CONTAINS 1 and 10 1 integral digit 10 and 100 2 integral digits 1 100 and 1000 3 integral digits 2 Whence we formulate the law : TJie characteristic of the logarithm of a number is one less than the number of digits in the integral part of the number. On the other hand, we observe from the table of this article that if a number contain no integral digits, that is, if it be purely decimal, its logarithm is negative. The characteristic in this case can be got by counting the num- ber of zeros before the first significant figure, prefixing the minus sign. It is usual, and better, however, except for special purposes, not to write the characteristic of the loga- rithm of a decimal number in the form just stated, for reasons which will now be pointed out. 31. The Mantissa. In the common system the mantissa of the logarithm of a number can be made to depend only upon the sequence of digits in the number, and be inde- pendent of the position of the decimal point. Let us assume that we know the logarithm of 1.285 to be 0.1089. It follows, multiplying successively by ten, that 10 o.io89 = L2 85, or log 1.285 = 0.1089. 10 1 - 1089 = 12.85, log 12.85 =1.1089. 10 2 - 1089 = 128.5, log 128.5 =2.1089. 10 3.io89 = 128 5, log 1285 =3.1089. 10 4.io89 = 128 5o, log 12850 = 4.1089. Ill, 31] SOLUTION OF RIGHT TRIANGLES 33 which verifies the law we have stated. If, however, we divide 10- 1089 successively by 10 we find 10 o.io89-i = 10 -o.89ii = 12 85, or log .1285 = - 0.8911. 10 -i.89ii = .01285, log .01285 = - 1.8911. 10 - 2 .89ii = .001285, log .001285 = - 2.8911. This is the true form of the logarithm of a purely decimal number, and for certain purposes this is the form which must be used.* It is obvious from the preceding discussion that the mantissa corresponding to a given sequence of digits re- mains the same as long as the sequence contains one or more integral digits, but that as soon as the sequence is a purely decimal number the mantissa changes. To obviate this difficulty and to keep the mantissa the same for a given sequence of digits regardless of the position of the decimal point, we note that the number 0.8911 may be written, without change of value, in the form 9.1089 10. We have added 10 and subtracted 10, and have therefore left the value unchanged. We may then say log .1285 = - 0.8911 = 9.1089 - 10, and if we agree to use the latter formf we see that the mantissa of the logarithm of .1285 (that is, 1089) is the same as the mantissa of the logarithm of the sequence 1285 when it contains integral digits. We may now write log 1.285 =0.1089 log .1285 = 9.1089 - 10 log 12.85 = 1.1089 log .01285 = 8.1089 - 10 and make the statement: In the common system the mantissa of a logarithm is unique for a given sequence of digits. TJie * For example, in dividing one logarithm by another. t This form, 9.1089 10, is perfectly convenient as long as the opera- tions to be performed are addition and subtraction, which are the usual operations in dealing with logarithms. D 34 PLANE TRIGONOMETRY [HI, 31 characteristic is one less than the number of integral digits. If a number be purely decimal, count the decimal point and the zeros before the first significant figure. The result sub- tracted from 10 minus 10 will be the characteristic. 32. Four Computation Theorems. The use of logarithms in computation depends upon the four following theorems : I. In any system the logarithm of a product is equal to the sum of the logarithms of its factors. To prove, log a mn s = Iog m -f log a n + + log a s. Let log d m = x then a x = m log a n = y a v = n log a s =z a* = s. Whence a* a v a 2 = a x+v+ '" + * mn s, or, by the definition of a logarithm, log tt mn -. s = x + y + + z. That is, log a mn s = Iog m + log a n + + log a s. This theorem replaces the operation of multiplication by the simpler operation of addition. II. In any system the logarithm of a quotient is equal to the logarithm of the dividend minus the logarithm of the divisor. To prove, lOg fl ~" = lgo m lgo n " n Let log a m = x then a x = m log a 7i y a y = 7i. Whence <*- _m a" 71 Ill, 32] SOLUTION OF RIGHT TRIANGLES 35 or, by the definition of a logarithm, That is, log = log ra - log tt n. tit This theorem replaces the operation of division by the simpler operation of subtraction. III. In any system the logarithm of a power of a number is equal to the exponent of the power times the logarithm of the number. To prove, log a m n = n Iog m. Let log a m = x or a* = m. Whence (a*) n = a nx = wT, or, by the definition of a logarithm, Iog m n = nx. That is log fl w" = n Iog m. This theorem replaces the operation of. involution, or successive multiplications, by the simpler operation of a single multiplication. IV. In any system the logarithm of a root of a number is equal to the quotient of the logarithm of the number by the index of the root. To prove l og(> V^ = ^. n Let log d m = x or a* = m. Whence ~\/a x = an = or, by the definition of a logarithm, n 36 PLANE TRIGONOMETRY [III, 32 That is log a ^/m = 1 ^^. n This theorem replaces the operation of evolution, or extraction of roots, by the simpler operation of division. Another theorem, important in the theory of logarithms, but of which no application is made in the study of trigonometry is the following : Proof : Let log t m = x then b 1 = ra Iog 6 a = y b y = a. L -L Whence m b x (b y ) y = a y lo ga m = * y By means of this theorem the logarithm of a number to any base can be found if the logarithms of numbers to some one base are known. Thus, assuming that logarithms to the base 10 are known, W 71 21 = lQ g' 71 ' 24 - 'g' 7t24 - 1 ' 8627 = i 2650 lo glo e log w 2.718 0.4343 As a corollary of the above theorem we have, putting 6 = m, 1 log a m =- -- log m a 33. Special Properties of Logarithms. In addition to the preceding theorems we may note the following properties of logarithms : 1. In any system the logarithm of 1 is 0. For, by the definition of zero exponent, a = 1 . Therefore, log a 1 = 0. 2. In any system the logarithm of the base is 1. For a 1 = a. Therefore, log a a = 1. Ill, 34] SOLUTION OP RIGHT TRIANGLES 37 3. In any system whose base is greater than 1 the logarithm of is oo. For, a > 1, or 00 = = = 0. a 00 Therefore, log a = oo. That is, the base of the system being greater than unity, the logarithm of a number which becomes smaller than any assigned number however small, is negative and numerically greater than any assigned number however great. 4. The cologarithm of a number is the logarithm of the reciprocal of the number. Thus, the base being 10, colog n = log - = log 1 log n, n or, colog n = log w, which may be written, colog n = (10 10) log n. Therefore, to find the cologarithm of a number to the base 10 subtract the logarithm of the number from 10 10. It may be noted that log = log m - = log m + log - = log m -f colog n. n n n Therefore we may, instead of subtracting the logarithm of a number, add its cologarithm. It is found convenient to do so in most cases. 34. The following example will illustrate the use of logarithms in making numerical computations. Example. Find the value of .0005616 x V- 424.65 (6.73) 4 x (.03194)* We note first that, with the definition of logarithms we have adopted, negative numbers have no logarithms. But 38 PLANE TRIGONOMETRY [III, 34 the numerical result of operations of multiplication and division is the same no matter what the combination of algebraic signs. We therefore find the numerical value of any expression, treating all numbers as positive, and deter- mine the algebraic sign of the result by considering the operations indicated. Thus, in the above example the factors are all positive except -\/ 424.65. Therefore, the number of which we are to extract the cube root is negative and the final result will be negative. log .0005616 = 6.7494 - 10 | log 424.65 = 0.3754 4 colog 6.73 = 6.6880 10 | colog .03194 = 1.2464 3)5.0592 - 10 log ^=8.3531 -10 ^=.02255. Therefore .0005616 x\/- 424.65 (6.73) 4 x (.03194)* NOTE. The colog 6.73 = 9.1720 10, which being mul- tiplied by four gives 36.6880 40 ; subtracting and add- ing 30 this becomes 6.6880 10, the desired form of " a number minus 10." Similarly to divide 5.0592 10 by three we first add and subtract 20. Also, in finding five sixths of the cologarithm of .03194, we first multiply by 5 and then divide by 6, in order that any error arising from inexact division by 6 may not be increased 5-fold. 35. We may now return to the problems of Art. 28 and solve them by the use of logarithms. Example 1. Given ^4 = 32 16', a = 124, (7=90, find Bj b, and c. As before, b = a cot A y c = sin A Ill, 35] SOLUTION OF RIGHT TRIANGLES 39 Therefore log b = log a + log cot A, log c = log a log sin A. log a =2.0934 log a =12.0934-10 log cot A = 0.1997 log sin A = 9.7274 - 10 log b =2.2931 logc= 2.3660 b = 196.4, c = 232.3. Example 2. Given a = 50, b = 60, C = 90. As before, tan A = - , therefore log tan ^1 = log a log b. b log a = 11.6990 - 10 log 6= 1.7782 log tan A= 9.9208-10 A = 39 48'. Also . or log c = log a log sin A. log a = 11.6990 -10 log sin A = 9.8063 - 10 logc= 1.8927 c=78.1. It must be emphasized that results obtained by logarith- mic computation are approximate. The value of the loga- rithm of a number cannot, in general, be found exactly, but only approximately to four, five, or any desired number of decimal places. TJie results of numerical computation by means of logarithms are not, in any case, correct beyond the number of decimal places in the logarithms used to make the computation. In the same way, the values of the trigono- metric functions being, in general, not exact but approxi- mate to four, five, or more decimal places, the solutions of triangles got by their use, with or without logarithms, are approximate solutions only, to the degree of accuracy of the tables used. 40 PLANE TRIGONOMETRY [III, 35 Indeed, in all but the simplest problems in applied mathematics the results are necessarily approximate, the data of a problem being themselves approximate. It is use- less to try to make results " more accurate " by using tables of logarithms or other functions carried to seven places when the data are correct only to, say, three figures. In general if data are given to three figures, three-place tables should be used ; if to seven figures, seven-place tables, etc. On the other hand, no matter to how many figures the data may be given, if we are using, say, four-place tables, the data should be used and results found to four figures only. To illustrate these points the following simple example will be worked in four ways : 1. by actual multiplication ; 2. by using four-place tables ; 3. by using five-place tables ; 4. by using seven-place tables. Example. Find the value of 123045 x 200368. 1. By actual multiplication the result is 24,654,280,560. 2. log 123045 123 1 = 5.0899 log 200368 200 | 4 = 5.3019 log product = 10.3918 product = 24,650,000,000 which agrees with the first result to four figures. 3. log 1230 1 45 = 5.09007 log 2003 | 68= 5.30183 log product = 10.39190 product = 24,655,000,000 which does not agree with the first result to the fifth figure. It will be noted that there was an accumulation of errors all in one direction. Result 3 is nearer to result 1, however, than is result 2. Ill, 35] SOLUTION OF RIGHT TRIANGLES 41 4. log 12304| 5= 5.0900640 log 20036 8= 5.3018284 log product = 10.3918924 product = 24,654,280,000 which agrees with the first result to seven figures. EXAMPLES What is the value of 1. IQlog 7.218. 2. loglO 2 - 6994 . 3. Given log 2 = 0.3010, log 3 = 0.4771, find log 12. 4. Prove 10 lo ? a + 1 = 10 a. 5. Is log 14 = log 2. log 7? Why? 6. 18^12 = 1?? Why? log 3. 3 Find the value of g log .00365 9 log 77.95 log .05312* ' log .00684* Find the value of x in the following equations : 10. logics = 3. 12. log e x = 2. 11. Iogi x=|. 13. log e x = f. 14. log x . log e x = logio 2 . 15. a logio x b logio x = a 2 6 2 . 16. log e x 3 - log e x 2 = 5. 17. \ logio z 10 - logio z 3 = 4. Express as the logarithm of a fraction : 18. log (x 2 - 6) 42 PLANE TRIGONOMETRY [HI, 35 Find the value of ate-"* b z e~ bx : 26. When x = s e_ . (a > 6) a b 27. a b (>&) Compute the values of the following : 28. log e l. 29. log, 2. 30. log, 3. By means of logarithms compute the values of : 31. Iog e 4. 32. S/.01236. 33. VI. 1193. (56.333)^ 36 99.02 34. V-. 002807. 37 (-16.65)4 38. 41 Vll.ll v'. 02983 lQ g 771 - 2 . 39. [log (.00915) ]*. 40.04 62.85 x V3111.59 42 (56.3; \/- .00986 4Q 693.08 log .00598 ) 3 x \/56.3 43. 46. 999.9 x .002008 V.08888 x 40.19 01)7 45. ( (.0002635)"^ .0001)~i V(.OOl) 3 . 44. (. J/ '485.7 x 22.01 x 11.79 \ - 55.5 X - 66.66 (5362) ~* 47. V .00298 x .00384 4 Q ( 88 ) 2 X (999) 1 v^632 x .06302 v'lOOOOOO EXAMPLES: SOLUTION OF RIGHT TRIANGLES Solve the following right triangles, and find their areas : 1. B = 24 23', 7. 6 = .2072, 13. 6 = 156.6, 6 = .02126. a = .4212. c = 856.4. 2. B = 55 45', 8. A = 82 6', 14. B = 43 46', c = 4116. 6 = .08937. a = 66650. 3. jB = 4330', 9. a = .8478, 15. B = 74 17', a = 26185.. c = 1.234. b = .00002039. 4. a = 77.38, 10. B = 60 14', 16. A = 29 56', c = 91.08. c = .007745. c = .0007814. 5. B = 76 34', 11. A = 14 53', 17. 6 = 8.243, b = 2423. a = 1353. c = 9.275. 6. A = 67 47', 12. B = 39 22', 18. B = 58 39', c = .00954. a = 121.2. c = 35.73. 19. A = 35 8', 20. 6 = 3814, a = 17270. a = 3651. Ill, 35] SOLUTION OF RIGHT TRIANGLES 43 PROBLEMS 21. A road rises 348.9 feet in a horizontal distance of one half mile. Another road rises the same height in a distance of 3019 feet along the road. Which road is the steeper and by how much ? 22. From a ship sailing due east at the rate of 7.6 miles per hour a headland bears due north at 10.35 A.M. At 12.46 P.M. the headland bears 33 west of north. How far was the headland from the ship in each position ? 23. At a distance of 502.3 feet, horizontally, from the center of a bridge the sidewalk rises at an angle of elevation of 5. The roadway, beginning 203.5 feet farther away from the center, has an angle of elevation of 4 25'. If a pedestrian and a team enter the bridge at the same moment, which will reach the center first, the man, walking 3.4 miles per hour, or the team, going 5.6 miles per hour ? 24. A flagpole 20 feet long stands on the corner of a building 143.6 feet high. Find the angle subtended by the flagpole from a point 100 feet distant from the foot of the building in a horizontal line. 25. If the radius of a circle is 835.4 feet, what is the length of the chord which subtends an arc of 45 37' ? 26. In a circle whose radius is 35.37 inches is inscribed a regular polygon of fifteen sides. Find the length of a side. 27. A tree 214.8 feet high casts a shadow 167.4 feet long. How many degrees is the sun above the horizon ? What is the time of day if the sun rose at six o'clock and will set at six o'clock ? [Assume that the sun passes through the zenith.] 28. A gable roof is 23.4 feet high and 90.6 feet broad. By how much must the height be reduced to reduce the pitch of the roof 40 per cent ? NOTE. The pitch of a roof is the angle between the slope of the roof and the horizontal line. 29. From the top of a cliff 378.6 feet above the sea, the angles of depression of a boat and a buoy, in line with the observer, are found to be 29 20' and 11 50' respectively. Is the boat or the buoy farther from the base of the cliff ? How much farther ? 30. The point B is 1249 feet due east of the point J., and the point C is 376 feet due east of B. The angle of elevation of B above A is 9 13' ; of C above B, 7 23' . A railroad runs from A to C via B. What is the increase in altitude from A to C ? 44 PLANE TRIGONOMETRY [III, 35 31. If, in Example 30, the railway could, by grading, be made to run in a straight line from A to (7, what would be the angle of eleva- tion of the new route ? 32. How much shorter would the railway of Example 31 be than the railway of Example 30 ? 33. Taking the earth as a sphere of radius 3956 miles, what is the length of the radius of the Arctic Circle, latitude 66 32' N. ? 34. Taking the earth as a sphere of radius 3956 miles, what is the latitude of a place which is 2113 miles from the earth's axis ? 35. A vessel sailing due south at a uniform rate observes at 7.15 A.M. that a lighthouse bears 70 east of south. At 8.05 A.M. the lighthouse is 12.75 miles due east from the ship. How far from the ship, and in what direction, will the lighthouse be at 9.30 A.M. ? 36. A ship sailing due south at a uniform rate observes, at 6 A.M., a lighthouse 11.25 miles away, due east. At 6.30 A.M. the lighthouse bears 17 57' north of east. What will be the bearings of the light- house from the ship at 9 A.M. ? How fast does the vessel sail ? 37. Taking the Earth as a sphere with diameter 7912 miles, what is the distance of the farthest point on the Earth's surface visible from the top of a mountain 8200 feet in height ? 38. The towns B and C lie due east from the town A, B being half- way from A to O, which are 5 miles apart. The towns 5, (7, and D are equally distant from each other. How far is D from A and in what direction ? 39. A ray of light from a source, A, strikes a mirror, 102 mm. broad, at a point two thirds of the way from the edge. The ray is then reflected to E at a perpendicular distance 25.7 mm. from the mirror. Eind the length of the path traveled by the ray. 40. From a window of a house, on a level with the bottom of a spire, the angle of elevation of the top of the spire was 41. From another window, 20.5 feet directly above the former, the like angle was 37 31'. What was the height of the spire ? 41. Having at a certain (unknown) distance measured the angle of elevation of a cliff, a surveyor walked 60 yards on a level toward the cliff. The angle of elevation from this second station was the com- plement of the former angle. The surveyor then walked 20 yards nearer the cliff, in the same line, and found the angle of elevation from the third station to be double the first angle. How high was the cliff ? CHAPTER IV FUNDAMENTAL IDENTITIES 36. In this chapter we shall discuss some of the impor- tant relations of analytical trigonometry. The number of such relations is, of course, unlimited, but there are a few, of frequent occurrence and of fundamental importance, upon which the others depend ; it is this fundamental group with which we shall now deal. Let us first observe how the need for some of the relations may arise. We have seen (Art. 27) that as the angle increases from the sine of the angle also increases. But does the sine increase at the same rate as the angle, so that, for instance, if the angle be made twice as large the sine also becomes twice as large ? This is obviously not so, for, as we have seen, the sine of 60 is not twice the sine of 30. What then are the relations, if there be any such, by which we may find the functions of twice an angle when the functions of the angle are given ? or again, is there any relation con- necting the functions of the sum of two angles with the functions of the angles separately ? Such questions as these we shall now proceed to answer. 37. The Addition Formulae. Let x and y be two acute angles, whose sum may be an angle either in the first quadrant or in the second. Construct, Fig. 18, the angle XOP equal to x and add to it the angle POQ equal to y. Then the angle XOQ is equal to x + y. From any point, A) in the terminal side of the combined angle x 4- y draw AB perpendicular to the. axis of x which is the initial side of the angle x. Then OB, BA, and OA are respectively the abscissa, ordinate, and distance of the point A and we 45 46 PLANE TRIGONOMETRY [IV, 37 may write any function of the angle x + y. But as we wish to express the functions of x + y in terms of the func- tions of x and y, we proceed to draw lines which will give us those functions. Thus, from A draw AC perpendicular to the terminal side of the angle x, and from C draw CD perpendicular to the axis of x and CE perpendicular to AB. FIG. 18. Then since AE is perpendicular to OX and AC to OP, the angle EAC is equal to the angle x, each of the angles being acute. We now have , . BA BE + EA DC.EA w(x + y)= = -- =' But these last two ratios are not functions of any of the angles in the figure. To obtain a function of x or y we must use with DC either OD or OC, and with OA either DC OC or CA. Therefore we shall multiply and divide - OA by the common line OC. Similarly with EA and OA we use CA. Thus we may write v DC OC^EA CA m(x+ y)= + .- or sin (x + y) = sin x cos y + cos x sin y. IV, 38] FUNDAMENTAL IDENTITIES In the same way N OB OD-BD OD EC ^- -- OC_EC AC* OC ' OA AC ' OA 9 47 or cos (x -f y) = cos x cos y sin x sin y. 38. The Addition Formulae (continued). Again let x and y be two acute angles where x may be either greater or less FIG. 19. than y. Construct, Fig. 19, the angle XOP equal to x and from it subtract the angle QOP equal to y. Then the angle XOQ is equal to x y. From A, any point in the termi- nal side of the combined angle, draw AB perpendicular to the axis of x which is the initial side of the angle x. Then OB, BA, and OA are, respectively, the abscissa, ordinate and distance of A. From A draw AC perpendicular to the terminal side of the angle x, and from C draw CD perpen- dicular to the axis of x and CE perpendicular to BA pro- we* * Note that in = we use AC as the positive direction of the line, AC AC therefore AC must be positive in the ratio ^ also. 48 PLANE TRIGONOMETRY [IV, 38 duced. Then since AE is perpendicular to OX and AC to OP, the angle EAC is equal to the angle x, each being acute. We now have as in Art. 37, . , , BA BE-AE DC AE **-= __ ~ OC ' OA AC ' OA\ or, sin (x y) = sin x cos y cos x sin y. OA OA OAOA = QD OC CM CA ~OC ' OA CA ' OA' or, cos (x y) = cos x cos y -f sin x sin y. 39. We have thus proved the formulae sin (x y) = sin x cos y cos x sin y, _ cos (x y) cos x cos y =F sin x sin y, for values of x and ?/ less than 90. It now remains to be proved that these relations are true for all values of x and y. This may be done by a geometric construction as in the cases given, but the following method is preferable. 40. Let x be an angle in the second quadrant and y an angle in the third quadrant. Then we may put x = 90 + a and y = 180 + &, where a and b are acute. We may now write _ _ cos (x + ?/) = cos (90 -f a + 180 + 6) = cos (270 + a + b) = sin (a + b). (Art. 19) = sin a cos b + cos a sin b. (Art. 39) But a = - 90 + x and 6 = - 180 + y. IV, 40] FUNDAMENTAL IDENTITIES 49 Therefore cos (# + ?/)= sin (- 90+#) cos (- 180 +y)+ cos (- 90+z) sin( 180 +y) = ( cos x) ( cos y) + (sin a;) ( sin y) (Art. 19) or cos (x + y) = cos a; cos y sin # sm y which is the same as the relation of Art. 39. Again, let x be an angle in the first quadrant and y an angle in the third. We may put y = 180 4- 6, where b is acute, and write sin (x y) = sin (x 180 + b) = sin (-180 + x -b) = - sin (x - b) (Art. 19) = sin a; cos b + cos x sin 6. (Art. 39) But b = - 180 + y, and therefore, sin (x y)= sin x cos( 180 + y) + cos x sin ( 180 +y) sin x( cos y) + cos x( sin y) (Art. 19) = sin x cos y cos x sin y. Thus it may be proved that the equations of Art. 39 are true for all values of x and y. The importance of these four relations, (10) of Art. 39, can hardly be over-emphasized. From, them, together with those given in Art. 12, may be derived all other trigono- metric identities. The method of so doing is shown in the following articles, and is illustrated by the following examples : Example 1. Prove the relation sin (45 + a) cos (45 - b) + cos (45 + a) sin (45 - 6) = cos (a b). 50 PLANE TRIGONOMETRY [IV, 40 This is simply a case of the first formula of (10) where x = 45 + a, y = 45 - b. We may write sin (45 + a) cos (45 - 6) + cos (45 + a) sin (45 - b) = sin (45 + a + 45 b)= sin (90 -fa b)= cos (a b). EXAMPLES Prove that 1. sin 105 + cos 105 = cos 45. 2. cos (45 - x) cos (45 + x) - sin (45 - x) sin (45 + x) = 0. 3. sin x cos (90 - x) - cos x sin (90 - x) = - cos 2 x. 4. cos (30 - 45) - cos (30 + 45) = sin 45. 5. Given sin x = f , cos y = f , find sin (x + y). 6. Given cos x = , cos y = , find cos (x y}. Given tan x = 2, tan y = 3, find 7. sin (x + y). 8. cos (x + y). 9. sin (x - y). 10. cos (x - y). 41. Tangent of a Sum. To derive an expression for the tangent of the sum or difference of two angles we proceed as follows : cos (a; y) _ sin x cos y cos a; sin y "" cos x cos y T sin x sin y sin x cos y cos # sin y __ cos # cos y cos # cos y "" cos a; cos y sin a; sin y cos a; cos y cos a; cos y or tan(*y) = ^ ^L. (ii) 1 =F tan JT tan y In a similar manner may be proved cot(xy) = cotycota; IV, 42] FUNDAMENTAL IDENTITIES 51 EXAMPLES Prove that 1 tan x 2. tan (45 + x) tan (135 + a) + 1 = 0. 3. tan (45 + x) tan (45 x) = 1. 4. Given tan a = 2, tan 6 = 4, find tan (a + 6). 5. Given sin a = , cos 6 = $, find tan (a 6). 6. Given sec a = 3, esc 6 = 4, find tan (a + 6). 7. Given tan a =|, tan 6 = ^ T , find a + 6. 8. Given sin a = f , cos 6 = f , find tan (a + 6). 9. Given sin a = f , sin b = f , find a + 6. Prove the following identities. 1Q sin (x + y) _ cot x + cot y ^ cos (x y) 1 + cot x cot y " 11. tanx-tan(x-y) = 1 + tan x tan (x ?/) 42. Functions of the Double Angle. The equations of Arts. 39 and 41 being true for all values of x and y, let us assume that y = #. Substituting x for y in the functions of the sum of two angles we obtain sin 2 x = 2 sin x cos x (13) cos 2 * = cos 2 * sin 2 x. (14) '* tan * . (15) 1 tan 2 x 1 . (16) 2 cot a 43. The student should clearly understand that the equa- tions of Art. 42 give the values of functions of twice an angle in terms of functions of the angle, no matter what the value or form of the angle may be. For example, the following relations are all true, being merely the equations of Art. 42 changed slightly in form. 52 PLANE TRIGONOMETRY [IV, 43 2 sin - cos - == sin a. 2 2i cos = cos 2 - sin 2 244 2tan( tan (2 a - -f) 44. Functions of the Half-angle. We may write the two proved relations sin 2 x + cos 2 x = 1 and cos 2 a = cos 2 x sin 2 a; in the form x t . 2 x 1 cos 2 - -|- sm 2 - = 1. cos 2 - sin 2 - = cos x. 2 2 Subtracting and adding these, we have os jr. (17) 2sin 2 ?=l -cos*. 2 cos 2 = 1 4- cos jr. 2 Dividing the last two equations one by the other we obtain , 2 1+cosJt (18) 2 1 cos x Thus we have equations which give the sine, cosine, tangent, and cotangent of one half an angle in terms of the cosine of that angle. IV, 45] FUNDAMENTAL IDENTITIES 53 EXAMPLES Given sin 6 = , find 1. sin 26. 3. tan 26. 5. cos-. 2. cos 26. 4. sin -. 6. tan |. 2 2 Prove the following identities. 7. cos 4 x sin 4 x = cos 2 x. 8. (sin x + cos x) 2 = 1 + sin 2 x. 9. tanx= Sin2x 12. tan = -**_. 1 + cos2x 2 1 + cosx 10. 2 2secx l+sin2x 11. tan = -. 14. 2 sinx 2 secx 15. tan (45 + a) + tan (45 - x) = 2 sec 2 x. 45. Sum of Sines or Cosines. By addition and subtraction of the two equations sin (x + y) = sin x cos y 4- cos x sin y, sin (x y) = sin a; cos y cos x sin y, we obtain sin (SB + y) 4- sin (a; T/) = 2 sin a; cos y, sin (SB + y) sin (SB y)= 2 cos a; sin y. If now we let x +-y = a, SB y = ft so that cc = ( + ft) and y = ^(a ft), we obtain from the last two identities sin a + sin p = 2 sin (a + P) cos| (a - P), sina-sinp = 2cos|(a+ P)sin|(a- p). Proceeding in the same way with the equations cos (a; y) = cos x cos y T sin x sin y, we obtain two more equations of importance cosa + cos p = 2cos|(a + p)cos|(a - p), i i \ * cos a - cos p = - 2 sin | (a + p) sin \ (a p). 54 PLANE TRIGONOMETRY [IV, 45 EXAMPLES Express each of the following as the algebraic sum of sines or cosines. 1. sin 6 x cos 2 x. 4. sin (x + 2 y) cos (x y). 2. cos 4 x sin 2 a; 5. sin ( 30 + x ) sin (30 - x) . 3. cos-sin. 6. cos 3 x cos (x y) . 2 2 Prove the following identities. 7. cos (30 - x) cos (60 - x) = (2 sin 2 x + V3). 8. cos 3 x sin 2 x cos 4 x sin x = cos 2 x sin x. 9. sin x cos (x 4- y) cos x sin (x y) = cos 2 x sin y. 46. Identities and Equations. It should be borne in mind that all of the equations of this chapter are identities, that is, they are true no matter what values the angles may have. We shall deal later on, in Chapter VII, with trigo- nometric equations of condition, where we shall find that not every value but only particular values of the angles involved will satisfy the equations. Also, in connection with this chapter attention should be again called to the group of fundamental identities in Art. 12. ILLUSTRATIVE EXAMPLES Example 1. Prove that sec 2 x = 1 + tan x tan 2 x. I = -^ = * = cos's = sec* a cos 2 x cos 2 x sin 2 x 1 tan 2 # 1 tan 2 a? _ 1 + tan 2 x _ ^ 2 tan 2 a; 1 tan 2 x 1 tan 2 x = 1 + tan x 2tana; = 1 + tan x tan 2 x. 1 tan 2 x By the above method we begin with sec 2 x and deduce or derive the required result. Another method of pro- cedure is as follows : IV, 46] FUNDAMENTAL IDENTITIES 55 Assume that Then sec 2 x = 1 + tan x tan 2 x. = l+ H 2tan2a; 1 tan 2 x = 1 + tan 2 a; 1 tan 2 x _ cos 2 x 4- sin 2 a; cos 2 x sin 2 aj = sec 2 #. Therefore, the original assumption is correct. ' Example 2. Prove that esc 2 x \ sec x esc 05. First method. 1 1 sec x esc x esc 2 a; = sin 2 a; 2 sin x cos a; Second method. Take the reciprocals of both members, sin 2 x = 2 sin x cos x. EXAMPLES Without the use of tables find the following : 1. Sine and cosine of 15. 3. Tangent of 15. 2. Sine and cosine of 22 30'. 4. Tangent of 22 30'. 5. Find the value of sin 3 x in terms of sin x. 6. Find the value of cos 3 x in terms of cos x. 7. Find the value of tan 3 x in terms of tan x. 8. Find the value of tan 4 x in terms of tan x. 9. Find the value of sin 4 x in terms of functions of x. 10. Find the value of cos 4 x in terms of functions of x. 11. Given sin 4 x = a, cos 4 x = 6, find sin 8 x and cos 8 x. 12. Given tan 3 x = a, find tan 6 x. 56 PLANE TRIGONOMETRY [IV, 46 Prove the following identities. 13. sin (90 -f x -f y} = cos x cos y sin x sin y. 14. sin - cos - + cos - sin - = sin \ (a + 6). 22 22 15. cos 6 a = 1 2 sin 2 3 a. .. sin 4 a -f sin 2 a _ 3 tan 2 a sin 4 a sin 2 a 1 3 tan 2 a 17 1 _ sec 2 1 + cos 2 6 ~ tan 2 e + 2 ' 18. 19. cos 5 x cos 2 x = (cos 7 x + cos 3 x). 20. sec 2 x esc 2 x = sec 2 x + esc 2 x. 21. sin 4x 22. cos 4 x sin x = \ (sin 5 x sin 3 x) . 23 sin3x 2 24. sinx= 25. sin x = 1 -f tan 2 x 2 cot x + tan x 26. tan2x= - ? - . cot x tan x 27 8Jn(x + 2y)-8in(x-. siny 28. cos (2s- y) - cos (2x + y) = 4 sin x sin y< cosy 29. 30. tan 2 - 2 esc x tan- -|- 1 = 0. 2 2 31. 1 -f tan 2 x 32. tan ^ x = esc x cot x. 33. cot - = esc x + cot x. IV, 46] FUNDAMENTAL IDENTITIES 57 2 1 + sin x + cos x ' 35 ' 2tan(x + 45) = 36. Cosx - cos3x =:tan2x. sin 3 x sin x 37. sin x (1 + tan x) + cos x (1 + cot x) = sec x + csc x. 38. cot x cot 2 x = csc 2 x. og 1 + sin 2 x _ cos x + sin x cos2x cos x sin x* 40. _ 41. Given sin x = - - = - , find the value of cos x. 2.V-1 42. Given r 2 = a 2 sin 2 0, r 2 cos + a 2 cos 2 sin _ tan3 ^ a 2 cos 2 cos 6 r 2 sin /i 43. Given r = a sec 2 - , Prove r cos + a sec 2 - tan - sin 8 99 /} - 2 - 2 - = _ cot *. Oft o a sec 2 - tan - cos 6 r sin 6 2 2 44. Given r = a (1 cos 0), Prove r cos -f a sin 2 , 3 - uin . a sin cos 6 rsmd 2 Prove the following identities. 45. cos3x + sm3x = 2cot2Xt sin x cos x 46. 1 + cos 2 x cos 2 y = 2 (sin 2 x sin 2 y -f cos 2 x cos 2 y). 47< sin 2 x cos 2 y- cos 2 x sin 2 y = ^ + ^ _ cos 2 x cos 2 y sin 2 x sin 2 y 48. sn x cos x 49. sin 2 + sin d 58 PLANE TRIGONOMETRY [IV, 46 50. (sec 20 + 1) Vsec2 0-1 = tan 2 0. gl cos + sin __ cos - sin _ 2 tan 2 0. cos sin cos 6 + sin 52. tan 2 - sec sin = tan sec 2 0. 53 1 cos x + cos y cos (x + y) _ tan^a ^ 1 + cos x cos y - cos (x + y) tan y ' 54. Given x = 3 cos 2 sin 1 Prove that y = 3 sin 2 cos 0} 2Vx 2 + y 2 = 3 sin 2 0. 55. Given x = a cos r sin 0, a _ _ 2 sin 2 y = asin0 + rcos0, V2cos20' r 2 = 2 cos 2 0, Prove, x 2 + ?/ 2 = 2 sec 2 0. CHAPTER V THE CIRCULAR OR RADIAN MEASURE OF AN ANGLE. INVERSE TRIGONOMETRIC FUNCTIONS 47. Circular or Radian Measure of an Angle. Any con- venient unit may be chosen for the measurement of angles. We have hitherto used the degree, subdivided into minutes and seconds, as the unit,* but we shall now introduce another unit called the radian, the unit angle in the circular measure of angles. The radian is an angle at the centre of a circle whose sub- tending arc is equal to the radius of the circle. FIG. 20. It is obvious that the radian is a constant angle, is the same in all circles, since the ratio of the circumference of a circle to its radius is constant. In Fig. 20 let the angle AOB be a radian, that is, let the arc AB be equal to OA, the radius of the circle. Also let the angle AOC be an angle to be measured in radians. To measure a quantity is to find its ratio to another quantity of the same kind chosen as the unit. Therefore, * It may be noted that the right angle is used as a unit in the study of geometry, partly because it is an angle easily constructed. 59 60 PLANE TRIGONOMETRY [V, 47 Circular measure AOC = radians. AOC arc AC arc AC AOB arc AB radius OA Therefore, to measure an angle in circular measure, or in other words to express the angle in radians, find the ratio of the arc subtending the angle in any circle to the radius of the circle. If we represent the angle, measured in radians, by x, the length of the arc subtending the angle by s, and the radius of the circle by r, we have the relation x = - . This is an algebraic equation involving three quantities. If any two of the quantities are known the third can be found. Thus r x Example 1. What is the radius of a circle in which an arc of 12 inches subtends an angle of \\ radians? r = - = =8 inches. x 1 Example 2. If the radius of a circle is 15 feet what length of arc subtends an angle of two-thirds of a radian ? s = 15 x |=10 feet. We know that the ratio of a semicircumference to its radius is IT = 3.1416. It follows, therefore, that the angle which is sometimes called a straight angle, and which is expressed as 180, may also be expressed as ?r radians. Thus > TT radians = 180 1 Gft 1 radian = = 57.296, approximately. 7T Also 180 = TT radians 1 = - radians. V, 47] CIRCULAR MEASURE OF AN ANGLE 61 By means of these two relations we can readily reduce a given angle from either system of measurement to the other. Example 1. Express the angle 7 radians * in degrees. TT = 180. Therefore, Example 2. Express the angle - in degrees. o Since TT = 180, = x 180 = 120. Example 3. Express the angle 110 32' 30" in circular measure. 110 32' 30" = 110f = ^|f *. Since 180=7r, therefore, 10 _ _v__ 2653 _ jr_ 2653 = 2653 TT " 180' 24 "180 24 == 4320 " An angle in circular measure is usually expressed as a multiple of IT. EXAMPLES Express the following angles in degrees. 1. 6.5 radians. 3. -. 5. 1^. 7 6 o 2ir /? Sir * -g" 4. 3.8 radians. 6 - -g- Express the following angles in circular measure. 7. 270. 9. 25 16'. 11. 208 30'. 8. 13 24'. 10. -450. 12. -98. 13. What is the ratio of a radian to a right angle ? *The name radian is often omitted. An angle written w and read O " pi over three," means " pi over three " radians, or about 1.05 radians. 62 PLANE TRIGONOMETRY [V, 47 How many right angles are there in each of the following angles ? 14. |. 15. 5^-. 16. 2*L. 17. 5 radians. 18. Through how many radians do the minute and hour hands of a clock turn in 30 minutes ? 19. Through how many radians does the minute hand of a clock turn in 35 minutes ? 20. Through how many radians does the hour hand of a clock turn in 18 minutes ? 21. The front wheel of a cart is 2 feet in diameter, the hind wheel 3 feet. Through how many radians will the hind wheel turn while the front wheel is turning through 600 ? 22. Through how many radians does the earth revolve about its axis in a week ? Is the result the same in 45 north latitude as at the equator ? 23. A wheel turns 50 revolutions per minute. Express its angular velocity in radians per second. 24. A wheel has an angular velocity of 20 radians per second. How many revolutions does it make per minute ? 25. Through how many miles will a point on the equator of the earth travel as the earth turns through 1 radians ? 26. Through how many miles will a point at 45 north latitude travel as the earth turns through one radian ? 27. The radius of a graduated quadrant is 2 feet, and the grad- uations are 5' apart. What is the distance between successive graduations ? 28. What must be the radius of a graduated quadrant if the dis- tance between graduations 5' apart is to be -fa inch ? 48. Inverse Trigonometric Functions. Let us suppose that y is the sine of the angle x. We express this briefly in mathematical symbols as y = sin x. Suppose now that we wish to make the inverse statement that x is the angle whose sine is y. To express this in mathematical symbols we write x = sin" 1 y, where, it must be noted, the minus unity is not an exponent. Having expressed our idea in symbols we next note that x depends upon y for its value, is a function of y, and we name the function the anti-sine or inverse sine. Similarly a = tan" 1 6 means that a is the angle V, 49] CIRCULAR MEASURE OF AN ANGLE 63 whose tangent is 6, and we say that a is the anti-tangent of b. In this way we have a group of six inverse trigonomet* ric functions, sin- 1 AT, 5-1 V sec" 1 x tan- 1 AT, COS'* AT, COt^Af, CSC" 1 AT. These six quantities, it must be remembered, are angles. 49. General Value of an Angle. Identities connecting the various inverse trigonometric functions exist and may be derived or proved by methods analogous to those of Chapter IV. Before taking them up, however, one important dif- FIG. 21. ference between the direct and the inverse trigonometric functions must be noted. If y = sin x, and if we give x a particular value, say 30, then y will have one and only one value, one-half. On the other hand, if x = sin" 1 y and if we give y a particular value, say -J-, then x does not have one value only but an infinite number of values, 30, 150, 390, 210 etc. This being so it is well to get an expression that will represent all the angles which have a given value of the sine, cosine, etc. Let sin x = a, a being a positive number, or x = sin~ 1 ( a) and let a be the smallest angle* which has for * That is, a if we use positive angles only ; a' if negative angles also are used. Either method may be adopted. 64 PLANE TRIGONOMETRY [V, 49 its sine the value a. By Fig. 21 we see that the possible values of x are a, TT a, 2 TT + a, 3 TT a, 4 TT -f a, , TT a, 2 TT + , 3 TT a, 4 TT + a, , which may be written in the general form x siir 1 a = mr +( l) n a (21) where n is any positive or negative integer, including zero, and a is the least angle whose sine is a. This is called the general value of the angle and a is called the principal value. Since the cosecant is the reciprocal of the sine we may . x \ x esc' 1 a = mr + (- l) n a. /r\*~*\ (22) FIG. 22. Let cos x = a or a; = cos -1 ( a) and let be the least angle whose cosine is a. By Fig. 22 we see that possible values of x are a, 2 TT a, 4 TT a, , or, in the general form x = cos" 1 a = 2 mr a (23) where TI is any positive or negative integer, including zero, and a is the least angle whose cosine is a. Since sec x = we may write for the general value, a cos x V, 49] CIRCULAR MEASURE OF AN ANGLE 65 being the principal value, x = sec' 1 a = 2 TZTT a. (24) Let tan x = a or x = tan -1 ( a), and let a be the least angle whose tangent is a. By Fig. 23 we see that pos- FIG. 23. sible values of x are a, TT -f- K) 2?r + , 3 TT TT + a, 2 TT + a, 3 7T + a, , or, for the general value, JT = tan" 1 a = mr + a (25) where w is any positive or negative integer, including zero, and a is the principal value. Giving n and a the same meaning we may write, since 1 (26) cot # tana' x = cot" 1 a = rnr + a. One need not make use of the formulae, 21-26, but may proceed as follows : Find the two smallest angles, positive or negative, which correspond to the given value of the func- tion. If we call these angles a and /3 then the complete series of angles will be given by 2 mr + a and 2 mr + p. 66 PLANE TRIGONOMETRY [V, 49 Example 1. Write the general value of cos' 1 .9205. From the tables of trigonometric functions we find a = 23. Therefore 23 7T x = cos- 1 .9205 = 2 riTT 23 = 2 180 Example 2. Write the general value of sin' 1 1. We know a = 90 = - Therefore a = sn' = ?nr -. 2i Example 3. Prove the identity 2 sin- 1 a = sin' 1 (2 a Vl - a 2 ). Let sin" 1 a = x, then sin x = a. Substituting these values in the formula to be proved we have 2 x = sin" 1 (2 sin #V1 sin 2 #), or sin 2 x = 2 sin a; Vl sin 2 x = 2 sin x cos x. Q. E. D. Or, we may proceed as follows : We know sin 2 x = 2 sin x cos x, which may be written 2 x = sin" 1 (2 sin x cos x). Let sin" 1 a = x, sin x = a and substitute : 2 sin- 1 a == sin- 1 (2 aVl a 2 ). Q. E. D. Example 4. Find the principal value of tan- 1 1 -f tan -1 . We note that this is the sum of two angles each given by the value of its tangent. We therefore write, formula (11), tan (tan- 1 + tan' 1 ) = - Therefore, tan-i + tan-ii = tan- 1 = ^ o 4 V, 49] CIRCULAR MEASURE OF AN ANGLE 67 EXAMPLES Write the general values of the following angles : 1. sin-i. 3. tan-!(- 1). 5. S ec-i(--^ r V 7. sin-i(- Y V \/3/ \ 2 / 2. cos-iQ. 4. cot- 1 -^- 6. csc-!V2. 8. cot-i(-VS). V3 Find the value of 9. sin (sin- 1 a). 11. tan (tan- 1 y}. 13. 2 cos (cos* 1 .523). 10. cos- 1 (cos x). 12. sec- 1 (sec 30). 14. cot (cot- 1 2.718). 15. sin- 1 (cos 35). / 2 \ 18. cos f tan- 1 1 + sec' 1 ~ \ 16. tan- 1 (cot 40). ' 19. 17. sin^sin-ii + cos-i-Ly 2 0. coJtan' 1 V3 + tan- 1 21. Prove that x = sec- 1 v/1 + tan 2 x. 22. Prove that tan -1 y = sec- 1 vT+1/2. Prove the following : 23. tan-i( V2 + 1) - tan- 1 (- V2 1) = 135. 24. tan- 1 V3 - tan- 1 f - \ = tan" 1 (- 3 V3). 25. tan- 1 1 - tan' 1 (_ 1) = tan' 1 2. = tan- 1 - a b a 2 - 2 62 30. sin- 1 x + cos- 1 y = tan- CHAPTER VI THE SOLUTION OF GENERAL TRIANGLES 50. Four Cases. As in the case of right triangles the solution of any triangle means the finding of the values of unknown parts from the parts that are known. Of the six parts (three angles and three sides) there must be given three, one of which at least is a side, in order that the tri- angle may be solved. Consider any triangle, Fig. 24. FIG. 24. The following cases may be enumerated : I. Given a side and two angles; say, a, A, B. II. Given two sides and the angle opposite one of them ; say, a, b, A. III. Given two sides and the included angle; say, a, b, C. IV. Given the three sides; a, b, c. 51. The Law of Sines. Cases I and II may be solved by means of the following theorem. In any triangle the sides are proportional to the sines of the opposite angles. That is, Fig. 25, a : b : c = sin A : sin B : sin C. (27) 68 VI, 52] SOLUTION OF GENERAL TRIANGLES 69 c c A D FIG. 25. Proof : In the triangle BAC draw CD perpendicular to BA. Then Therefore AC BC a DC = b sin A a sin B. Whence a : b = sin A : sin B. Similarly the theorem may be proved for the other pairs of sides and angles. 52. Case I. Given a side and two angles ; a, A, C. To find the third angle we have To find B and C we have & _ sin B c___ sin C a sin A ' a sin A ' selecting in each case that proportion, from (27), which in- volves an unknown side, b or c, and three known parts. From these two proportions we have log b = log a 4- log sin B + colog sin A log c = log a + log sin C + colog sin A 70 PLANE TRIGONOMETRY [VI, 52 Example. Given a = 412.7, A = 50 38', C = 69 13', find B * bj ' B = 180 - 119 51' = 60 9'. log a = 2.6157 log a = 2.6157 log sin B = 9.9382 log sin C = 9.9708 colog sin A = 0.1118 colog sin A = 0.1118 log b = 2.6657 log c = 2.6983 b = 463.1 c = 499.2 53. Case II. Given two sides and the angle opposite one of them; a, 6, A. We have, to find B, sin B b t sinJ. a Whence, log sin B = log sin ^4 + log b + colog a. Also, (7=180- (^ + B). Then '=* a sin .4 Whence log c = log a + log sin C -f- colog sin A. Example. Given a = 31.24, b = 49, A = 32 18', find *' ^ C ' log sin ^1 = 9.7278 log 6 = 1.6902 colog a = 8.5053 log sin B = 9.9233 B = 56 56' But since B is found from the log sine it may have two values ; namely, 56 56' and 180 - 56 56' = 123 4'. To determine which value is correct or whether both are possible we recall the theorem of geometry which states that if the given angle is acute and the side opposite is less than the other given side, then it may be possible to construct two triangles from the given parts, two sides and an oppo- VI, 54] SOLUTION OF GENERAL TRIANGLES 71 site angle. In the above example the given angle A is acute and its opposite side a is less than 6 ; there are two solutions and both values of B must be used. Figure 26 explains the case graphically. 31.24 = 32M8' Bi=5656' A =32 18' FIG. 26. b=49 a=31.24 B 2 ) Consequently there are two values of (7, namely, d = 180 - (A + B,) <7 2 = 180 - (A = 90 46' = 24 38' and two values of c, got as follows : log a = 1.4947 log a = 1.4947 log sin d = 0.0000 log sin C 2 = 9.6199 colog sin A = 0.2722 colog sin A = 0.2722 log d = 1.7669 log c 2 = 1.3868 c x = 58.46 c 2 = 24.37 If the given angle be obtuse there will be only one solu- tion. If the given angle, A, be acute and the side a be greater than the side 6, there will be one solution only. If A be acute and a be equal to the perpendicular from C to ABj there will be only one solution, a right triangle. In this case B = 90 and log sin B = 0.0000. If, A being acute, a be less than the perpendicular from C to AB, there is no solution. In this case log sin B will be greater than zero, which is impossible since sin B cannot be greater than unity. 54. Case III may be solved by means of the theorem following : In any triangle the sum of two sides is to their difference as the tangent of half the sum of the angles opposite the two sides is to the tangent of half their difference. 72 PLANE TRIGONOMETRY [VI, 54 Proof : By Art. 51 a : b = sin A : sin B. Whence, by composition and division, a 4- 6 __ sin A + sin B __ sin j- (A + jB) cos |( J. ff) a b sin .4 sin B cos (^L + 5) sin \(A B) or, a + b tan ^i + ) ~ 55. Case III.* Given two sides and the included angle; b,c,A. By Art. 54 we have C & c tan The sides & and c are known, and also l(B 4- C)= |(180 - ^)= 90 - since ^4 + 5 + C = 180. Therefore we may write tan i(J5 - (7)= ^^ - tan ^(S + C), or, log tan ^(jB-<7)=log(&-c) + colog(&+c)+ log tan Thus -| (5 C) is found, and by finding the sum and dif- ference of %(B + 0) and *(B C) the values of B and O are known. Finally, to determine a we have, as in Case I : a : b = sin A : sin B. Example. Given 6 = .06239, c = .02348, A = 110 32' ; find J5, C, a. + O)= 90 - J A = 90 - 55 16' = 34 44'. & + c = .08587 6 - c = .03891. * See also Art. 61 following. VI, 56] SOLUTION OF GENERAL TRIANGLES 73 Whence we have log (b -c)= 8.5900 colog(6 + c) = 1.0661 logtanfr(JB+ C) =9.8409 log tan KJB - (7)= 9.4970 - 0)=1726'. + C)=3444' C = 17 18'. And as We have B = 52 10' Then log b = 8.7951 log sin A = 9.9715 colog sin B = 0.1025 log a = 8.8691 a = .07398 56. The Law of Cosines. Case IV may be solved by means of the following theorem : In a triangle the square of any side is equal to the sum of the squares of the other two sides minus twice the product of those sides by the cosine of their included angle. c That is, Fig. 27, where CD is perpendicular to AB, a 2 = b 2 4- c 2 2 be cos A. We have proved the geometrical theorem a 2 = b z + c 2 - 2 c AD* But cos A = , or AD = b cos A. b * Note that in the first two triangles of Fig. 27, AD, the projection of 6, is read left to right and is positive ; in the third triangle from right to left and is negative. 74 PLANE TRIGONOMETRY [VI, 56 Therefore, a 2 = b 2 + c 2 - 2 6c cos A Q.E.D. Obviously 2 be (29) and in the same way 2 ca 2 a& so that the three angles may be found. 57. The objection to the formulae of Art. 56 is that they are not adapted to logarithmic computation. To remove this objection we proceed as follows : From (29) we have - cos . = 1 2bc 2bc 2 be 2 be or, by (17), Art. 44, 2 be Let a + b + c = 2 s. Then a - 6 + c =2(s - b), a + b - c = 2(s- c), Whence . x . _ 2(s - 6) 2(* - c) -- (30) oc Similarly ca ^ ab These formulas may be used for logarithmic computation. VI, 59] SOLUTION OF GENERAL TRIANGLES 75 58. Again, from (29), Art. 56, we have c)(b + c a) 2 be or, by (17), Art. 44, o c-a) 26c As before, letting a = 2s, this becomes or cos Similarly s(s a) ^-v^H- (31) cos i (7 = 59. These formulae also may be used for logarithmic computation, but a more convenient set is obtained by dividing the formulae of Art. 57 by those of Art. 58. We thus obtain //o fcWo W\ (32) J A comparison of these three sets of formulae (30), (31), (32), will show that for the complete solution of a triangle when the three sides are given, the first set (30), requires six different logarithms, the second set seven, the third set four. In addition to this slight advantage the tangent set, (32), gives more accurate results than the other two when the angles involved happen to be very small or very near ninety degrees. 76 PLANE TRIGONOMETRY [VI, 60 60. Case IV. We will now solve a triangle when the three sides are known. Example. Given a = 10, b = 12, c = 14 ; find A, B, C. Here 2 s = a + 6 + c = 36, so that we have 5 = 18, log s = 1.2553, colog s = 8.7447 - 10. s - a = 8, log (s - a) = 0.9031, colog (s - a) = 9.0969 - 10. s _ b = 6, log (s - b) = 0.7782, colog (s - b) = 9.2218 - 10. s - c = 4, log ( - c) = 0.6021, colog (s - c) = 9.3979 - 10. log (s-b)= 0.7782 log (s - c) = 0.6021 log ( s - c) = 0.6021 log (s - a) = 0.9031 colog s = 8.7447 colog s = 8.7447 colog Q- a) = 9.0969 colog (s- b) = 9.2218 2)19.2219 2)19.4717 .log tan | A = 9.6110, log tan B = 9.7359, |-4 = 22 13', | = 28 34', ^4 = 44 26', J5=57 8', log (- a) = 0.9031 log (s - 6) = 0.7782 colog ,s= 8.7447 cologO-c)= 9.3979 219.8239 log tan i(7=9.9120, 2 C= 78 28'! CHECK: A +B + C= 180 2'. A common method of solving Case IV is by means of an auxiliary quantity, r ".v^ ^ We may write i r i ,, r \ * \.&n (t A = , tan -5= -, tan-C = s a s b s c VI, 61] SOLUTION OF GENERAL TRIANGLES 77 In using this method log r is first found, whence the log- tangents of the three half-angles are readily obtained. 61. Case III ; Other Methods of Solution. The formulae of Art. 56 may sometimes be used to advantage in solving Case III. Example (see Art. 55). Given b = .06239, c = .02348, A = 110 32', to solve the triangle. We have a 2 = 6 2 + c 2 2 be cos A. Then a 2 = (.06239) 2 + (.02348) 2 - 2(.06239) (.02348) cos 110 32'. log b = 8.7951 log c = 8.3708 2 2 log b 2 = 7.5902 log c 2 = 6.7416 62 = .003893 c 2 = .0005516 + c 2 = .0005516 log 2 = 0.3010 -26ccos^= .001028 log b = 8.7951 a 2 = .005473 log c = 8.3708 log cos A = 9.5450 log 2 be cos A = 7.0119 2 be cos A = .001028 Then log a = $ log a 2 = 8.8691 and a = .07398. To find B and C we have the formulae of Art. 51. The above computation can in some cases be done best and quickest without the use of logarithms. Another Method of Solution for Case III, preferred by many, is as follows : From Fig. 27 we see that DC = b sin A, AD = b cos A,DB = c- AD. 78 PLANE TRIGONOMETRY [VI, 61 nc 1 Then tan J3 = , whence B is known, and (7=180 (A + B). To find a we use a = - sin Applying this method to the example above we have log b = 8.7951 log sin A = 9.9715 log 7)0=8.7666 log 7)^ = 8.6567 log tan B = 0.1099 J3=5210' log b = 8.7951 log-cos A = 9.5450 log--4Z> = 8.3401 AD = - .02188 DB ='.02348 + .02188 = .04536. log DC = 8.7666 C = 1718' log sin B = 9.8975 log a = 8.8691 a= .07398 NOTE. The fundamental importance of the law 'of sines and the law of cosines should be noted. By their use, direct or indirect, any triangle whatever may be solved. AREAS OF TRIANGLES 62. Right Triangles. Case I. Given the two legs a and 6, Fig. 28. B FIG. 28. Representing the area of the triangle by K, it is obvious 2K=ab. (33) Case II. Given the hypotenuse and an acute angle, c and A. VI, 63] SOLUTION OP GENERAL TRIANGLES 79 Then a = c sin A, b = c cos A. Whence, by (33), 2 K= c 2 sin A cos ^4 = c 2 sin or 4 jRT = c 2 sin 2 -4. Case III. Given an angle and the adjacent leg, A and b. Then a = b tan A. Whence, by (33), 2K= W tan A. Case IV. Given the hypotenuse and a leg, c and a. Then 6 2 = c 2 a 2 or b = V(c + a)(c a). Whence, by (33), 63. Oblique Triangles. Case I. Given two sides and the included angle, a,' 6, (7. In Fig. 29 the line ^4.D is perpendicular to BC. It is obvious that 2 7T= a X But Therefore, sin C= , or ZL4 = 6 sin C. b (34) 80 PLANE TRIGONOMETRY [VI, 63 Case II. Given a side and the angles, a. A, B, C. By Art. 51, 6 _ smB ^ __ a sin B a sin A sin A Whence, by (34), 2 K = fl2 8 * n -B sin C ,ggx sin .4 Case III. Given the three sides, a, &, c. The formula (34), of Case I may be written, by Art. 42, ( 13 )' 2 K= 2 ab sin -t C . cos <7. Substituting in this the values of sin -J (7 and cos \ C given in Arts. 57 and 58, we have 2 A' = 2 or JT = Vs(s - d)(s - b)(s - c). (36) EXAMPLES Solve the following triangles, in each case obtaining also the area of the triangle : 1. a B C = 1419, 29 59', 16 1'. 6. c A C = 5141, 96 3', 55 46'. 11. 6 a B = .5042, .3618, 74 43'. 2. a b B 3.384, 9.828, 109. 7. b c A = 56.2, 63.9, 71 33'. 12. a b c = .03574, .02921, .01853. 3. a b C = 302, 427, 134 29'. 8. b c C = 268.5, 282.9, 75 20'. 13. 6 a A = .2792, .2271, 65 45'. 4. a b c = 56.22, 63.91, 70.54. 9. b A C = 6.362, = 76 13', = 35 17'. 14. a A B = .01044, 26 32', 146 26'. 5. b c B = 38.65, 48.12, 34 32'. 10. a c A = 5499, 2959, 133 34'. 15. a b B = 31.49, 49.88, 44 35'. VI, 63] SOLUTION OP GENERAL TRIANGLES 81 16. c = .0357, 18. b = 4621, 20. a = 6.743, a = .0292, a = 6473, b = 3.025, 5 = 31 7'. B = 31 7'. c = 4.271. 17. a = 32.15, 19. 6 = .4312, 21. c = .01825, 6 = 67.54, c = .8901, 6 = .02893, A = 28 26'. A - 29 55'. B = 83 30'. PROBLEMS IN THE SOLUTION OF TRIANGLES 22. A man owns a triangular lot on the corner of two streets which do not intersect at right angles. The frontage- on one street is 300 feet, on the other 250 feet. The back line of the lot is 350 feet long. If he buys land to add 275 feet to the 300-foot frontage, by how much is his lot increased in size ? 23. A man owns a triangular lot on the corner of two streets which intersect at an angle of 62. The frontage on one street is 200 feet, on the other 150 feet. If the land is worth one dollar a square foot and the man has $ 1200 with which to increase the size of his lot, by how much can he lengthen the 150-foot frontage ? 24. The perimeter of a triangle is 100 feet, and the perpendicular from the vertex C to the base AB is 30 feet. The angle A is 50. Find the length of the base AB. 25 . What is the perpendicular height of a hill which is known to rise 72 feet for every 100 feet of length of its slope, if the angle of elevation of the hilltop from a point 100 yards from the base of the hill is 31 ? 26. From where I stand, 50 feet from the bank of a stream, the angles of depression of the near and far banks of the stream are respectively 15 37' and 6 24'. How wide is the stream ? How far am I above the level of the stream ? 27. A man 5 feet 6 inches tall, standing on a bluff 40 feet high, measures the angles of depression of the near and far shores of a bay. The angles are 46 52' and 5 3' respectively. How wide is the bay ? 28. A man 5 feet tall, standing on the edge of a pond, finds the angle of elevation of the top of a tree on the other bank to be 44 26'. The angle of depression of the reflection of the treetop is 60 47'. Find the height of the tree. The reflection of an object appears as far below the surface as the object is above the surface. 29. The frontage on the beach (AB) of a quadrangular lot ABCD cannot be measured. The sides BC, CD, and DA are found to be 236, 155 and 105 feet respectively. The angles DJ.Cand DBG are 32 20' and 29 50' respectively. Find the length of AB. G 82 PLANE TRIGONOMETRY [VI, 63 30. The bases of a trapezoid are 48.25 and 94.75 feet. The angles at^the ends of the longer base are 63 52' and 70 55'. Find^the lengths of the other two sides. 31. Two sides of a triangle are 8.53 and 7.41. The difference between the angles opposite these sides is 18 23'. Solve the triangle. 32. The area of a triangle is 979 square feet. The angle A is 56 22' and the side b is 44.80 feet. Solve the triangle. 33. Two sides of a parallelogram are 8005 and 5008. The included angle is 60 53'. Find the lengths of the diagonals. 34. The diagonals of a quadrangular field ABCD intersect at O at an angle of 78 3'. The lines AO, BO, CO, and DO are 27.5, 31.8, 58.5 and 63.2 feet respectively. What is the area of the field ? 35. Two sides of a triangle are b = 302 and c 40.8. Find the angle A so that the triangle may have the same area as the triangle whose sides are 62, 51 and 30. If b were 30.2 and c were 40.8 could A be found ? Why ? 36. Two vessels start from the same point and sail, one northeast at the rate of 6 miles per hour, and the other east. 30 south at the rate of 8 miles per hour. How far apart will the ships be after 2| hours ? 37. A submarine in submerging drifts back 5 feet for every 20 feet it sinks. After the submarine has sunk vertically 300 yards, at what angle must a torpedo be shot from a cruiser one mile away to hit the submarine, if the latter drifts away from the cruiser ? 38. A post 6 feet high casts a shadow 10 feet long. What is the length of a flagpole that casts a shadow 60 feet long if the pole makes an angle of 82 with the horizontal on the side away from the sun ? 39. In problem 38 find the length of the flagpole if the angle made with the horizontal is 82 on the side towards the sun. 40. Two yachts begin a race by sailing from a point A, along the windward leg of the course in the direction northeast until they reach a buoy B. They then sail before the wind, east 32 south, until they reach a point C, 5 miles east along a straight coast from A. The first yacht sails to windward 5 miles per hour, and before the wind 6.5 miles per hour ; the second 5.8 miles per hour to windward and 6 miles before the wind. Which yacht wins the race and by how much? 41. A triangular beach lot has a frontage on the sea of 100 yards. The boundary lines running from the beach make, on the inner side of the lot, angles of 60 and 50 respectively with the shore line. How must a line be drawn from the middle point of the shore line to form two equal lots ? VI, 63] SOLUTION OF GENERAL TRIANGLES 83 42. A point A, on the south bank of a river 1.5 miles broad and flowing due east is to be connected by bridge and road' with a town U, 3 miles back in a straight line from the north bank of the river. It is found that the bridge can be built to a point, C, on the farther bank lying north 22 west from J., or to D lying north 41 east from A. The town 1} lies north 12 east from A. If the bridge costs $ 2000 per mile to build and the road $ 500 per mile, which route is the more w economical and by how much ? (See Fig.) 43. The distances of a point (7, on the far side of a river from two points A and B on the near side, are to be found but can not be directly measured. In the direction CA a distance / \ AD = 150 feet is measured, and in the direction CB a distance BE = 250 feet. The distance from A to B is 279.5 feet, and by measurement it is found that BD = 315.8 feet, DE = 498.7 feet. How far is C from A and B ? (See Fig.) 44. The distance from ^ a point, A, on the coast to // a lighthouse, i, is to be B found. A straight line is run from A along the coast, and on the line two points, B and (7, are taken from which the lighthouse is visible. By measurement it is found that AB = 236.7 feet, BC = 215.9 feet, the angle ABL = 142 37', the / angle ACL = 76 14'. How far is the lighthouse / from A? (See Fig.) A 45. On the north side of a river lie two points A and B both of which can be seen from (7, and from no other point, on the south side of the stream. From a point D, whose dis- tance from C is 425.3 feet, A and C are sighted. It is found that the angle ADC = 37 15', and the angle A CD = 42 35'. From another point whose distance from C is 405.4 feet, and from which B and C are visible, the angles CEB = 53 15', and ECB = 58 5' are measured. The angle ACB is also measured and found to be 65 11'. What is the distance from AtoB? (See Fig.) CHAPTER VII THE SOLUTION OF TRIGONOMETRIC EQUATIONS 64. The trigonometric equations hitherto dealt with have been identical equations ; equations, that is, true for any values of the variables involved. We shall now deal with trigonometric equations which are not identities, and shall examine the methods by which such equations are solved. No methods applicable to all such equations can be given, but methods applicable to several important classes will be discussed. In general it may be said that all such equa- tions are algebraic in form, the one difference being that now the unknown quantities are the trigonometric func- tions, sine, tangent, etc., or, occasionally, the inverse func- tions. Therefore, all methods applicable to the solution of algebraic equations are applicable to the solution of trigo- nometric equations. Moreover, in the case of trigonometric equations we have the various fundamental identities, treated in former chapters, which being true for all values of the variables involved can be used in connection with any equation whose solution is desired. 65. For example, given the equation 2 sin 2 x cos 2 x + | = to find the value of x. In form this is an algebraic, quadratic equation in two unknowns, sin x and cos x. To find the values of two un- knowns we must have two consistent and independent equations. But we also know that cos 2 x = 1 sin 2 x. Therefore, our equation may be written 84 VII, 65] TRIGONOMETRIC EQUATIONS 85 2 sin 2 x -(1 sin 2 x) = , 3 sin 2 x = |, whence sin # = -, and it* = sin" 1 ( ). The principal values of x are, therefore, - and the 6 general values are or we may proceed thus : Given 2 sin 2 a* cos 2 x = J. We know sin 2 # + cos 2 x = 1. Adding the two equations, 3 sin 2 x = }, etc. Example 2. Solve the equation cos x V3 sin x + 1 = 0. For sin x substitute Vl cos 2 x. Then cos x V3 Vl cos 2 x + 1 = 0, cos a; + 1 = V3 Vl cos 2 x, cos 2 x + 2 cos a; + 1 = 3 3 cos 2 #, 2 cos 2 a; -j- cos x 1 = 0, a quadratic equation in cos x whose roots are cos x = 1 or i. Therefore, a = cos- 1 (- 1)= 2 WTT -h TT =(2 n -f- I)TT, and x = cos" 1 ( - ]= 2 WTT - V2 3 86 PLANE TRIGONOMETRY [VII, 65 These roots, as in every case, must be tested by substitution in the original equation. It is found that 2 n-n- - does not 3 satisfy the equation, while the other two values do. The roots are, therefore, x=(2n + V)7r and 2mr +- 3 Another method of solving the last equation is as follows: Given cos x V3 sin x = 1. Divide by 2, But Therefore we may write, sin cos cos x sin - sin x = - Whence x + - = cosY- -"\ = 2 mr ?-?, 3 \ 2J 3 and x = (2 n - I)TT or 2 mr + - 3 Note that the two general solutions (2 n + I)TT and (2 n !)TT are identical since each represents any odd multiple of TT. 66. Special Types of Equations. This last solution is an example of the type of equation 1. a cos x + b sin x = c, c C'Ocosa = BTA' 2 + CM 72 - 2 '4' - CM' cos A, or 2'O- C'Ocosa = WO 2 - WA' 2 + (70* - C'A* 2 + 2 B'A' C'A' cos A But B'OA' and C'OA' are right triangles, and therefore, We then have B'O C'Ocosa = 04' 2 + B'A' C"^' cos A, or 04' 04' ,B'4' 0'4' But f) A' cos J304 = cos c, -^p = cos 40C = cos 6, , BO CO = sin 504 = sin c, - = sin 40C = sin b. BO CO Hence cos a = cos 6 cos c 4- sin 6 sin c cos 4, which is the law of cosines. 96 SPHERICAL TRIGONOMETRY [VIII, 70 In the above demonstration the sides b and c were taken less than 90 in order that the construction of the right tri- angles B'OA' and C'OA' might be possible. The resulting theorem, however, is true in all cases. Let us assume 90 < b < 180 and 90 < c < 180. Then, Fig. 33, pro- duce the arcs AB and AC to meet in A', thus forming a lune. In the triangle A'BC, b' and c' are less than 90. FIG. 34. The law of cosines is, therefore, true for the triangle A'BC, so that, since A' = A, cos a = cos b' cos c' -f- sin b' sin c' cos A. But b 1 = 180 -b and c' = 180-c. Whence cos a = cos (180 b) cos (180 c) + sin (180 - b) sin (180 c) cos A, or, cos a = cos b cos c -f- sin b sin c cos A. Q.E.D. Again, let b < 90 and 90 < c < 180. Produce the arcs BA and BC, Fig. 34, to meet in B', thus forming a lune. Then, in the triangle AB'C, b < 90 and c f < 90, and, therefore, cos a' = cos 6 cos c' + sin 6 sin c' cos CAB'. But a' = 180 - a, c'= 180 - c, and CAB' = 180 - A. Hence cos (180 a) =cos b cos (180 - c) + sin b sin (180 - c) cos (180 - A), VIII, 71] FUNDAMENTAL RELATIONS 97 or, cos a = cos b cos c -f sin b sin c cos -4, which proves the law of cosines for all cases. We thus have the three fundamental equations, the law of cosines : cos a = cos b cos c + sin b sin c cos A, cos 6 = cos c cos a -f- sin c sin a cos JB, (37) cos c = cos a cos b -f- sin a sin & cos C, by means of which any spherical triangle may be solved. For example, given a = 60, b = 70, A = 65. We have cos 60 = cos 70 cos c + sin 70 sin c cos 65, r> .500 = .342 cos c + .940 sin c x .423, .342 cos c + .398 sin c = .500, .342 cos c + .398 sin c .500 V(.342) 2 +(.398) 2 " V(.342) 2 + (.398)2 ' .342 , .398 . .500 __cosc + sine- , .651 cos c + .758 sin c = .952, .651 = cos 49.4, .758 = sin 49.4. Therefore, cos (c- 49.4) = .952, c - 49.4 = cos- 1 .952 = 18.2 and c = 67.6. Similarly the other parts may be found. The equations are not, however, adapted to logarithmic computation, so that for practical use, as will presently be shown, they must be transformed in various ways. 71. Law of Cosines Applied to the Polar Triangle. The law of cosines, being true for any triangle, is true for the polar triangle of ABC. Therefore, denoting the six parts n 98 SPHERICAL TRIGONOMETRY [VIII, 71 of the polar triangle by the same letters accented, we have cos a' = cos b' cos c' -f- sin &' sin c' cos A'. But a ' = 180 -A,b' = 180 - B, c' = 180 - C, A' = 180 - a. Whence, cos (180 -A)= cos (180- B) cos (180- (7) -f sin (180- B) sin (180 - <7) cos (180 - a), or cos A = cos B cos (7 sin B sin (7 cos a, so that the truth of the three following equations is obvious : cos A = cos B cos C +- sin B sin C cos a, cos B = cos (7 cos .4 -h sin (7 sin A cos 6, (38) cos (7 = cos ^4 cos B + sin ^4 sin B cos c. 72. Law of Sines. Another theorem of importance in the solution of spherical triangles, known as the law of sines, is as follows : In any spherical triangle the sines of the sides are proportional to the sines of the opposite angles. That is, sin a : sinb: sin c sin A : sin B: sin C. (39) From equations (37) we have, A cos a cos b cos c COS A = : : . sin b sin c Whence 1 - cos 2 A = 1 - ( CQS - cos 6 cos c) 2 sin 2 b sin 2 c or, . 2 . _ sin 2 b sin 2 c (cos a cos b cos c) 2 sin 2 b sin 2 c _ (1 cos 2 6)(1 cos 2 c) (cos a cos 6 cos c) 2 sin 2 b sin 2 c _ 1 cos 2 6 cos 2 c cos 2 a + 2 cos a cos 6 cos c sin 2 & sin 2 c VIII, 72] FUNDAMENTAL RELATIONS 99 Whence, sin 2 A _ 1 cos 2 a cos 2 6 cos 2 c + 2 cos a cos 6 cos c sin 2 a sin 2 a sin 2 b sin 2 c or, sin A _ VI cos 2 a cos 2 6 cos 2 c+2 cos a cos 6 cos c sin a sin a sin 6 sin c where the positive sign is taken because A and a are each less than 180. The right-hand member of this expression is symmetric in a, 6, and c, so that if we started with cos B or cos C instead of with cos A, the final result for the right- hand member would be identical with that written above. Therefore, obviously, we have sin A _ sin B __ sin C sin a sin b sin c ' the law of sines which was to be proved. CHAPTER IX THE SOLUTION OF RIGHT SPHERICAL TRIANGLES 73. Special Formulae for Right Triangles. If we let C be the right angle in a right spherical triangle, and put C = 90 in the third equation of (37), we have cos c = cos a cos b. (40) The third equation of (38) gives = cos A cos B + sin A sin B cos c, or, cos c = cot A cot B. (41) The first two equations of (38) give cos A = sin B cos a, cos B sin A cos b. Using the proportions of (39) when C = 90, we have sin A _ sin j? __ 1 sin a sin 6 sin c Whence, sin a = sin A sin c, (43) sin b = sin B sin c. From (42) by (43) we have or, by (40) cos A = sin B cos a = . cos a, sine A sin 6 cos c tan cos A = sm c cos b tan c 100 IX, 74] RIGHT SPHERICAL TRIANGLES 101 Similarly, (44) tanc By (43), (44), and (40), sng sin A sin c sin a tan c tan A = - -; = z r = -: - 7 r cos A tan 6 sin c tan b tanc sin a sin a tan a tan b cos c tan & cos a cos 6 sin b In the same way, (46) sin a 74. The formulae (40) to (45) may be assembled, in a slightly different form, as follows : sin a n sin b sin A = sin 5 = sin c sin c tan c tan c (46) sin b sin a cos A = cos a sin B. cos B = cos b sin 4. cos c = cos a cos &. cos c = cot ^4 cot B. A device, known as Napier's Rules, was formulated by Napier to facilitate the remembering of the above formulae. Let us take for the Jive parts of a right triangle the sides a and 6, and the complements of A, B, and c. These five parts, Fig. 35, arrange themselves so that each is a middle to two adjacent parts and a middle to two opposite parts. NAPIER'S RULES state I. The sine of the middle part equals the product of the tangents of the adjacent parts. 102 SPHERICAL TRIGONOMETRY [IX, 74 II. The sine of the middle part equals the product of the cosines of the opposite parts. co-B co-A By applying these rules to the various parts all the formulae of (46) may be obtained. Thus, for example, sin (co-^1) = tan b tan (co-c). That is, sin (90 - A) = tan 6 tan (90 - c), or, cos A = tan b cot c = 5- tanc 75. Rules for Solution. In a right triangle, the right angle being always known, only two other parts need be known to solve the triangle. To solve a right triangle by means of the formulae (46) we have, therefore, the general rule : Select that equation which involves the two known parts and one unknown part. The algebraic signs of the functions must be carefully noted in order to determine the sign of the resulting func- tion and thereby the angle. If the part to be found is got from a cosine, tangent, or cotangent there is no ambiguity, for if these functions are plus the part will have a value less than 90. If they are minus the part will have for its value the supplement of the angle found from the tables of trigonometric functions. On the other hand, if the unknown part is determined by a sine, the sine being positive for all angles between and 180, the value may be either that got from the tables or its IX, 76] RIGHT SPHERICAL TRIANGLES 103 supplement. In general both solutions must be used unless the ambiguity can be removed by the following laws : 1. If the sides adjacent to the right angle are in the same quadrant, the hypotenuse is less than 90 ; if they are in dif- ferent quadrants, the hypotenuse is greater than 90. 2. An angle and its opposite side are in the same quadrant. PROOF OF LAW 1. By (40) cos c = cos a cos b. Let a<90 and b < 90. Then cosa = , cos& = , and cos c = ( )( ) = -f . Therefore, c < 90. Again, let a ^90, and b ^90. Then cos a = , cos& = =F, cose -, and c>90. PROOF OF LAW 2. By (45) sin b = tan^L Since sin b is necessarily positive, it follows that tan a and tan A are both plus or both minus. Therefore a and A are each less than 90 or each greater than 90. 76. The solution of right triangles is illustrated by the following examples : Example 1. Given A = 33 50', b = 108, find B, a, and c. From the formulae (46) we select tan a A tan b tan A = -r - , cos A = , cos B = cos 6 sin A. sin b tan c or, tan a = tan A sin 6, cot c = cos A cot b, cos B = sin A cos b. + log tan A = 9.8263 +log cos A = 9.9194 +log sin b = 9.9782 "log cot b =9.5118 + log tan a = 9.8045 "log cot c' = 9.4312 c' = 74 54' a = 32 31' c =105 6' 104 SPHERICAL TRIGONOMETRY [IX, 76 4 log sin A = 9.7457 -log cos b = 9.4900 -log cos B' = 9.2357 B' = 80 6' B = 99 54' To check the results we select a formula involving the three parts to be found ; a, c, and B. Thus cos B = JJ?, tanc log cos B = log tan a -f- log cot c 9.2357 =9.8045 + 9.4312 CHECK. Example 2. Given a = 47 30', c = 120 20', find A, B, and 6. We have sin a T. tan a ', cos c , cos5 = , cos b = . sin c tan c cos a log sin a =9.8676 + log tan a =0.0379 ~log cos c =9.7033 log sine =9.9361 "log tanc =0.2327 + log cos a =9.8297 log sin A =9.9315 -log cos '=9.8052 -log cos &'=9.8736 '=50 19' &'=41 38' A = 5840' B = 129 41' b = 138 22 CHECK. cos B = cos 6 sin A log cos B = log cos b + log sin A 9.8052 = 9.8736 + 9.9315 = 9.8051. NOTE. The value of A less than 90 is taken by virtue of law 2, Art. 75. Example 3. Given B = 105 59', b = 128 33', find A, o, and c. We have, cos B = cos b sin A, tan sn a sin c IX, 76] RIGHT SPHERICAL TRIANGLES 105 or sin a = tan b cot B, , , --. cos b sm jB -log cos B =9.4399 ~log cot B= 9.4570 log sin b =9.8932 -log cos b =9.7946 -log tan b =0.0986 log sin 5=9.9828 log sin A = 9.6453 log sin a = 9.5556 log sin c = 9.9104 ^ 1= 26 14' ai=21 4' Cl =54 27' A 2 = 153 46' a 2 = 158 56' c 2 = 125 33' CHECK. sin a = sin A sin c. log sin a = log sin A -f log sin c 9.5556 = 9.6453 + 9.9104 = 9.5557. By law 2 both sets of values must be used ; but by law 1 the acute value c x belongs with the obtuse values of A and a, the obtuse value c 2 with the acute values of A and a. Thus the two solutions are : 1. A = 26 14', a = 21 4', c = 125 33'. 2. .4 = 153 46', a = 158 56', c = 54 27'. NOTE. A quadrantal spherical triangle is one which has a side equal to a quadrant. The polar triangle of a quad- rantal triangle is right. Therefore, to solve a quadrantal triangle solve its polar triangle and take the supplements of the parts thus found. EXAMPLES Solve the following triangles in which C = 90. 1. A = 40 13', 5. a =165 19', 9. a = 144 1', a = 26 25'. c = 46 50'. 6 = 123 6'. 2. B = 8315', 6. 6 =40 49', 10. 4 = 69 17', 6 = 76 46' . c = 135 40'. B = 51 46'. 3. B = 110 50', 7. a = 21 18', 11. ^ = 137 18', 6 = 118 30'. 6 = 49 55'. B = 119 30'. 4. 6 = 127 36', 8. a = 78 32', 12. A = 71 46', c = 94 52'. b = 132 25'. B = 148 3'. 106 SPHERICAL TRIGONOMETRY [IX, 76 13. A = 20 34', 19. A = 98 17', 25. A 97 24', c = 23 18'. a = 143 8'. a = 103 D 12'. 14. B = 97 36', 20. a = 172 28', 26. b = 164 10', c = 96 31'. c 124 39'. c = 133 60'. 15. A = 100 38', 21. a = 4 54', 27. b = 34 3', c = 51 44'. b = 169 27'. a = 54 26'. 16. B = 59 54', 22. A = 76 17', 28. A 156 30', a = 6 60'. B = 144 1'. B = 104 50'. 17. B = 47 34', 23. B = 82 43', 29. A = 165 1', a = 144 24'. c = 99 26'. c = 50 30'. 18. A = 102 49', 24. TJ 99 47', 30. B = 37 56', b = 10 19'. a 26 43'. a =r 157 12'. Solve the following quadrantal spherical triangles (c = 90) : 31. ^1 = 30 12', 33. 6 = 51 33', 35. A = 159 20', a = 72 29'. C = 25 48'. a = 136 30'. 32. = 118 16', 34. ^ = 141 13', 36. b = 18 41', a =137 57'. C = 49 35'. - A = 39 24'. CHAPTER X THE SOLUTION OF OBLIQUE SPHERICAL TRIANGLES 77. Six Cases may be enumerated in the solution of oblique spherical triangles. 1. Given the three sides, a, b, c. 2. Given the three angles, A, B, C. 3. Given two sides and the included angle, a, b, C. 4. Given two angles and the included side, A, B, c. 5. Given two sides and the angle opposite one of them, a, b,A. 6. Given two angles and the side opposite one of them, A, B, a. We shall proceed to consider these cases in the order named. 78. Case 1. Given the three sides. The law of cosines is sufficient to solve this case, but the equations are not adapted to logarithmic computation. We therefore de- velop them as follows : We have proved the formula cos^l 2L =\/ ^ 1 -h cos A By the law of cosines ^_ cos a cos 6 cos c sin b sin c 107 108 SPHERICAL TRIGONOMETRY [X, 78 Whence 1 cos A _ sin b sin c + cos b cos c cos a 1 -f- cos A sin b sin c cos b cos c -f- cos a _ cos (6 c) cos a _ cos a cos (b c) ~ cos a cos (b -f c) cos (6 + c) cos a But cos a cos (b c) = 2 sin | (a + b c) sin (a b -f- c), cos(6 + c) cosa = 2 sin 1 (a + 6 + c) sin^(6 + c a). Hence c)sin|(a 1 + cos .4 sin ^(a + 6 + c) sin ^(b-\- c a) Let a + 6 + c = 2s; then a + 6 c = 2 (s c), a 6 + c = 2(s 6), and 6 + c- a = 2(s a). Therefore, 1 cos A _ sin (s b) sin (s c) 1 + cos A sin s sin (p a) tan^=.--. (47) Similarly, sin (-*) We may write sin (s a) sin (s b) sin (s c) sin (s a) * sin s or, putting /sm (s a) sin (s b) sin (s c) sins (48) sin (s c) X, 78] OBLIQUE SPHERICAL TRIANGLES 109 Either set of formulae (47) or (48) may be used in the solu- tion of Case 1. If a check is desired in the solution the law of sines may be so used. Thus, since sin A _ sin B _ sin C sin a sin b sin c ' it follows that log sin A log sin a = log sin B log sin b = log sin C log sin c. It must be remembered, however, that results may check and still be incorrect. If they do not check they are wrong ; if they check they may be right, or may be wrong, since errors may compensate each other. It is important to check one's work, but far more important to learn, by care- ful attention, to work accurately. Example. Given a = 103, 6 = 53, c = 61, find A, B, and C. Using the formulae (48) we find 8 = i (103 + 53 + 61) = 108 30', s-a = 530', s-6 = 5530 r , s-c = 4730'. log sin (s a)= 8.9816 log sin (s - b) = 9.9160 log sin (s c) = 9.8676 log esc s = 0.0230 2)18.7882 log&= 9.3941 log k = 9.3941 log k = 9.3941 log sin (sa) = 8.9816 log sin (s b) = 9.9160 log tan A = 0.4125 log tan | B = 9.4781 % A = 68 51', I B = 16 44', A = 137 42', B = 33 28', 110 SPHERICAL TRIGONOMETRY [X, 78 log k = 9.3941 log sin (s c) = 9.8676 log tan C = 9.5265 iC=1835', C = 37 10'. CHECK. log sin A = 9.8280 log sin B = 9.7415 log sin C = 9.7811 log sin a = 9.9887 log sin b = 9.9023 log sin c = 9.9418 9.8393 9.8392 9.8393 79. Case 2. Given the three angles. This case may be solved by the same formulae that are used in Case 1, by making use of the principle of polar triangles. Thus, using accented letters to represent the corresponding parts of the polar triangle, we have a' = 180 - A, b' = 180 - B, c' = 180 C. Knowing the sides a', &', c', we can find the angles A', B', C', as in Art. 78. Then the sides of the original triangle will be a = 180 -A',b = 180 - B', c = 180 - C'. Example. Given A = 123, B = 43, C = 64, find a, 6, c. Here a' = 180 - A = 57, b' = 137, c' = 116, 8 = K57 + 137 + 116) = 155, s-a' = 98, s - b' = 18, s - c' = 39. log sin (s - a') = 9.9958 log sin (s - b') = 9.4900 log sin (s - c') = 9.7989 log esc s = 0.3741 2)19.6588 logA:= 9.8294 log k = 9.8294 log k = 9.8294 log sin (s - a') = 9.9958 log sin (s - b') = 9.4900 log tan \ A' = 9.8336 log tan \ E' = 0.3394 A' = 34 17', I B' = 65 24 r , A' = 68 34', B' = 130 48', X, 80] OBLIQUE SPHERICAL TRIANGLES 111 log Jc = 9.8294 log sin (s - cQ = 9.7989 log tan | (7 = 0.0305 I a = 47 i', C'=942'. Therefore a = 111 26', 6 = 49 12', c = 85 58'. CHECK. log sin A = 9.9236 log sin B = 9.8338 log sin C = 9.9537 log sin a = 9.9689 log sin b = 9.8791 log sin c = 9.9989 9.9547 9.9547 9.9548 NOTE. Using the law of cosines as stated in (38), .Art. 71, whence cos B cos C 4- cos A cos a = sin B sin (7 and proceeding as in Art. 78, the following formulae may be got : / cos (S B) cos (S C) or cos (8 A) cos (8 B) cos (8 - O) = where and similar formulse for cot ^ b and cot % c. These formu- lae are simple and convenient, but it is unnecessary to bur- den the memory with them. 80. Cases 3 and 4, two sides and the included angle, two angles and the included side, are solved by means of Napier's Analogies, which we shall proceed to derive. From (48), Art. 78, we may write A ^ B * tan ^- tan ^ = 2 2 sin(s-a) sin(s-6)' 112 SPHERICAL TRIGONOMETRY [X, 80 or, since , 2 _ sin (s - a) sin (s b) sin (s c) sins sin sin 2 2 _ sin (s - c) "2"' ~^~ sins cos cos 2 2 Whence, sin - sin _ 1 _ _2 _ 2 _ 1 _ sin (s - c) cos^cos*' sins 2 2 or, cos cos -- sin sin 22 22 _ sins-sin(s-c) eosfcosf = That is, cos i (A + B) _ 2 cos |(2 s c) sin | c .4 J5 sins COS C S ~ Whence, since 2s c = a-|-6-|-c c = a + 6, cos J (^ + B) _ 2 cos i (a 4- 6) sin i c ,.~ ^|^f = S5T ^' Also, from (a) above, sin sin -, , _2 _ 2 _ sm(s-c) += which being transformed in the same manner gives cos i (A B) __ 2 sin|(q + 6) cos j-c "~Z 5~ : sins cos cos 2 2 X, 80] OBLIQUE SPHERICAL TRIANGLES 113 Dividing (ft) by (y) we have cosiQi + B) = tan|c (49) cos|(4-5) tan(a+&) Again, from (48) Art. 78 we may write tan _ 2 _ sin (s 6) ^~ S~sin(-a)' 2 sin ^ cos * 2 2 _ sin (s Whence, sin cos 2 2 sin( S - 6 ) cos sn sin (s a) sin cos cos sin 2 2 _ 2 2 _ sin (s b) sin (s a) ^4 . 5 sin (s a) cos sin 2 2 Using the upper signs, sin 1 (A + B) _ 2sini (2s a 6) cos -*- (a - 6) cos | sin | = sin ( s - tt ) ^ g i n j c CQS k (& &) . /sj\ sin (s a) Using the lower signs, sin %(A - B) __ 2 cos |(2s - a - 6) sin j-(a - 6) sin (8 -a) cos sn 2 cos | c sin ^ (a b) t ,* sin (s a) 114 SPHERICAL TRIGONOMETRY [X, 80 Dividing (8) by (c), sin -, (A + B) tan c - ; = - -. (50) sin g ( A B) tan ^ (a b) Applying (49) and (50) to the polar triangles we obtain cos i (A 1 + B') = tan 1 c' cos ^ (A' B') tan \ (a f + 6') ' sin ^ (A' + #') = tan j- c' sin i ( J.' 5') tan 1 (a' &') Remembering that A' = 180 - a, a' = 180 - A, etc. these become cos I (a + ft) cot ^ C cos I (a &) tan | (A -f- (51) ~ (^) The formulae (49), (50), (51), and (52), called Napier's Analogies because of their similarity to formula (28) of the plane trigonometry, can obviously be written in other forms by the cyclical interchange of the letters. 81. Case 3. Example. Given a = 100 30', b = 40 20', (7=46 40', find A, B, c. Napier's analogies (51) and (52) may be written _. cos i (a - 6) ; sm which will determine A and B. Then to find c we may use either (49) or (50). The latter may be written sm %(A B) X, 81] OBLIQUE SPHERICAL TRIANGLES 115 We have (a 4- b) = 70 25', \ (a - 6) = 30 5', C = 23 20'. log cos (a - 6) = 9.9371 log sin (a - 6) = 9.7001 log cot \ C = 0.3652 log cot | C = 0.3652 log seel (a + 6) = 0.4748 log esc % (a + 6) = 0.0259 log tan 1(^1 + )= 0.7771 log tan (A-B) = 0.0912 i (J. + B) = 80 31', (A - B) = 50 59'. Whence A = 131 30', jB = 29 32'. log sin %(A + B)= 9.9940 log tan i (a -6) =9.7629 log esc $(A-B)= 0.1096 log tan i c = 9.8665 |c = 3620 f , c = 7240'. The signs are all plus in the above computation. Case 4. Example. Given B = 110 40', C = 100 36', a = 76 38', find 6, c, A Napier's analogies (49) and (50) may be written cos iB-C tan 1 a sin 5- 6* tan which will determine 6 and c. To find A either (51) or (52) may be used. The latter is sin i (6 c) Here | (5 + C) = 105 38', $(B - C) = 5 2', 1 a = 38 19'. log cos i(B 0) = 9.9983 + log sin 1 (B - 0) = 8.9432 + log tan i a = 9.8977 +log tan 1 a = 9.8977 log sec (B+C) =0.5695 + log esc | (B + 0) = 0.0164 -log tan (& + c) = 0.4655 + log tan | (6 - c) = 8.8573 180 - $(b + c) = 71 6', |(6 - c) = 4 7'. 116 SPHERICAL TRIGONOMETRY [X, 81 Whence b =113 1', c = 104 47'. log + sin i (6 + c)= 9.9759 log + tan ( - O) = 8.9449 log + csc i (6 -c) = 1.1440 log + cot i ^1 = 0.0648 i ^ == 40 44', .4 =81 28'. Note that the algebraic signs are not all plus, and that the quadrant in which the angle lies is determined by the sign in the case of the tangent, cotangent, or cosine. 82. Cases 5 and 6, two sides and an opposite angle or two angles and an opposite side, may be solved by the law of sines together with Napier's analogies. Thus, if a, b and A are given, we may write sin b sin A sin B = sm a which, however, does not determine B unambiguously, since B may be either acute or obtuse. In this case, indeed, there may be two solutions, one solution, or none. We know, however, that if two sides (or angles) of a spherical triangle are unequal the angles (or sides) opposite are un- equal, and the greater angle (or side) lies opposite the greater side (or angle). These theorems enable us to deter- mine which values of the angle (or side) are possible. Thus ifb^a, then only values of B which are | Qrea ^ er \ ^ ian A are possible; both values of B may be so, or only one value. If the sine of B is greater than unity ; that is, if log sin B is positive, no solution is possible. These same con- siderations obviously apply to case 6 also. Another method of removing the ambiguity of Cases 5 and 6 is as follows : Two angles are of the same species when they are both acute or both obtuse. Also, since each side and angle of a spherical triangle is less than 180, we X, 82] OBLIQUE SPHERICAL TRIANGLES 117 see that $(A + B) and | (a + b) are each less than 180 ; while %(A B) and 1 c are each less than 90. It follows, in Napier's first analogy, cos i (A + B) _ tan^c cos i (A - B) ~~ tan |(a + b) ' that tan ^ c and cos ^ ( A B) are both positive. Then cos -i ( A + B) and tan \ (a + b) must have the same algebraic sign, and, therefore, 1 (A + B) and |(a + 6) are o/ fte same species. Thus, when a and b are given and A or 5 is to be found, if i(a + 6) ^ 90 m^s aZso %(A+B)^ 90 ; and the values of A or 5 must be so chosen as to satisfy this condition. Having thus found B, say, (whether there be two values or only one) we may complete the solution of the triangle by the use of Napier's analogies. Example 1. Given a = 46 30', b = 30 20', B = 36 40', solve the triangle. We have . sin a sin B sin A = sm b log sin a = 9.8606 log sin B = 9.7761 log esc b = 0.2967 log sin A = 9.9334 ^1 = 59 4' or 120 56' = ^'. Here a > 6, and, therefore, must A> B. This is true of both values of A found, so that there are two possible solu- tions of the triangle. To find C and c we may use (52) and (50). cot C= sin JO + 6) tan (A-B) 2 sina-& , c = 2 sin |(^1 -J5) 118 SPHERICAL TRIGONOMETRY [X, 82 We have First solution Second solution .1 (a + 6) = 38 25,' ' |(a + b) = 38 25', |(a-&)=86', K-&) =8 5', (4 + -B) = 47 52/ %(A' + B) = 78 48', l(A -B)= 11 12'. i(^' - B) = 42 8'. log sin |(a + 6) = 9.7934 or 9.7934 log tan $(A -B)= 9.2966 9.9565 logcsc^(a-&) =0.8519 0.8519 log cot - = 9.9419 or 0.6018 ^=48 49', Y = 14 3 '> (7 =97 38'. O' = 286'. log sin %(A + B) = 9.8701 or 9.9916 log tan (a - 6) = 9.1524 9.1524 log esc %(A- B)= 0.7117 0.1734 log tan =9.7342 or 9.3174 -=28 28', - = 11 44', c=5656'. c' = 2338'. The two complete solutions are, therefore, ^1=59 4', or 120 56', C = 97 38', 28 6', c = 5656'. 23 28'. Example 2. Given a = 126, c = 70, A = 56, solve the triangle. Using the formula sin A sin c . ^ sm C sin a X, 82] OBLIQUE SPHERICAL TRIANGLES 119 we have log sin A = 9.9186 log sin c = 9.9730 log esc a = 0.0920 log sin C = 9.9836 C = 74 20' or 105 40'. But since a > c, must A > C. Therefore, there is no sohtr tion. Otherwise thus : (a + c) = 98, i (A + C) = 65 10' or 80 50', which are not of the same species. Example 3. Given A = 84, C = 19, a = 28, solve the triangle. Using the law of sines, sin c = ; - - sm^. log sin C= 9.5126 log sin a = 9.6716 log esc ^4 = 0.0024 log sin c = 9.1866 c = 8 50' or 171 10 f . But since C < A, must c < a, and the second value is impossible. To find b use (50). log sin (A + C)= 9.8935 log tan i (a - c) = 9.2275 log esc fr(^-Q = 0.2698 log tan i 6 = 9.3908 |6 = 1349 r , 6 = 27 38'. To find B we may use (52), which has the advantage of giving an unambiguous result, or the law of sines. Selecting the latter method we have log sin (7= 9.5126 log sin b = 9.6663 log esc c = 0.8137 log sin B = 9.9926 B = 79 27' or 100 33'. 120 SPHERICAL TRIGONOMETRY [X, 82 But since b < a, must B < A, and the second value is im- possible. The complete solution is, therefore, c = 8 50', b = 27 38', B = 79 27'. 83. Delambre's Analogies or Gauss's Equations. Using the law of cosines we may write cos a cos b cos c COS A = : : sin b sin c Whence 1 - cos A = 2 sin' 1 A = (c S 6 COS c + sinfe sin c )~ cos a sm 6 sin c or, 9 A cos(6 c) cos a_2 sini(a-h6 c)sini(a 5+c) L Sin = ; ; ; . ** * 2 sm b sin c - sin b sm c That is, sm ^l _ /sin (s b) sin (s c) 2 ^ sin 6 sin c _B f* with similar formulae for sin and sin 2i In the same manner, by adding unity to each side of the first equation of this article, may be obtained formulae of which the type is cos- = J singsin ( g -). 2 ^ sin b sin c From these obviously follows . A B _ sin(s-6) sin s sin (s - c) Dill - OUO ; Al r 2 2 sin c ^ sm a sm b _ sin (s b) C_ (a) sine 2 In the same way we obtain 2 sm c X, 83] OBLIQUE SPHERICAL TRIANGLES 121 Adding (a) and (/3), sin 4 cos * + cos ^ sin | = sin(.-a)+sin(.-6) cog <7 2222 sine 2 Whence sin 4- c cos 4 (a 6) (7 sin 4 M + B) = - -^ cos -^ sm^ccos^c 2' or cos i c 2 Similarly may be obtained sA - = coa sin, III cos - c 2 cos 1(A-B) = TV* T- "/ sin ^ IV sm i c 2 which are the analogies or equations sought. These im- portant equations may be conveniently used in the solution of Cases 3 and 4 of oblique triangles. Example. Given a = 132 47', b = 59 50', C = 56 28', solve the triangle. We have (a + 6) = 96 19', i(a - 6) = 36 29', 1(7 = 28 14'. log sin i (a + b) = 9.9973, log sin (a - 6) = 9.7742, log cos i (a + 6) = 9.0414, log cos (a - b) = 9.9053, log sin - = 9.6749, log cos - = 9.9450. 2 2 From equations II and IV, log Jsin i c sm$(A - B) \ = 9.7742 + 9.9450 = 9.7192, log { sin | c cos %(A-B)\= 9.9973 + 9.6749 = 9.6722. 122 SPHERICAL TRIGONOMETRY [X, 83 Whence log tan $(A-B) = 0.0470, From I and III, log+ { cos i c sin J (A + JB) j = 9.9053 + 9.9450 = 9.8503, log- \GOS i c cos (4 + ) } = 9.0414 + 9.6749 = 8.7163. Whence log- tan %(A + B) = 1.1340. 180 -%(A + B) = 85 48', | (A + 5) = 94 12'. Therefore, .4 = 142 18', = 46 6'. Also log sin $(A B) = 9.8718. Therefore, log jsin 1 c sin %(A B)\ = 9.7192 _ log sin i(A-B)= 9.8718 log sin i c = 9.8474 |c = 4444', c = 8928'. Possibility of Solution by Inspection of Data. Before at- tempting the solution of a spherical triangle it may be desirable to determine whether the triangle is possible with the given data. Case 1. Given the three sides. The triangle is always possible if the sum of the sides is less than 360, and if no one side is greater than the sum of the other two. This follows at once from well-known geometrical theorems. Case 2. Given the three angles. This- case can be read- ily tested by the criteria of Case 1 applied to the polar triangle. For example, the triangle A == 78, B = 100, C = 160 is impossible because the sides of the polar tri- angle, a' = 102, V = 80, c' = 20, are such that a' > 6' + c'. Case 3, given two sides and the included angle, and Case 4, two angles and the included side, are always possible. Cases 5 and 6 have been discussed in Art. 82. X, 83] OBLIQUE SPHERICAL TRIANGLES 123 EXAMPLES 1. a 68 25', 14. a - 111 20', 27. A 132, b =. 71 11', c = 41 30', B = 139 50', c = 56 57'. C = 26 10'. b = 127 10'. 2. a = 100 8', 15. A 159 1', 28. A 79, b = 50 2', C = 36, B = 40, c = 60 6'. a = 9 5'. c = 108. 3. A = 51 59', 16. A = 25 20', 29. a 40, B = 83 55', C = 153 30', b = 118 21', C = 58 54'. a = 73 33'. A = 29 25'. 4. A = 142 33', 17. B i= 142 30', 30. C 148, B = 27 53', C = 71 20', B = 22 20', C 32 27'. c = 39 35'. c == 136. 5. b = 42 10', 18. A =: 110 5', 31. a 114, c 96 11', B = 123 20', c 148, A = 110 5'. b = 126 55'. C = 135 V. 6. a 146, 19. a = 59 34', 32. A 73, c = 69 20', b = 136 11', B - 81 50', B = 125 10'. c = 150 2'. a = 122 47'. 7. a - 90 50', 20. a = 109 24', 33. B 61 40', c = 117 50', c = 81 50', C = 140 15', B = 120 6'. A =3 107 40'. c 150 25'. 8. B = 41 6', 21. a = 99 50', 34. a = 125 16', C = 122 10', c = 64 10', b = 151 37', a = 37. A = 96 13'. c = 75 55'. 9. A 135, 22. A 35 31', 35. A 60 40', C = 50 50', B = 24 43', C = 105, b = 68 50'. C = 138 25'. a = 64 30'. 10. A = 147 30', 23. A - 31 20', 36. a = 55 5', C = 163 10', C = 122 40', c = 138 5', b = 76 25'. b = 40 40'. A = 42 28'. 11. a 29 2', 24. a = 120 45', 37. B =. 116 6', b = 14 3', c = 70 25', C = 73 50', A = 49 5'. B = 50 16'. c = 80. 12. b 98, 25. A - 120 21', 38. B = 134, c = 36, B = 130 22', C = 51, C = 163. C = 140 7'. a = 70 20'. 13. a = 132, 26. c & 109 20', 39. b 108, b = 56, b = 80 20', c = 40, A = 116 18'. C = 106 50'. C = :39. 124 SPHERICAL TRIGONOMETRY [X, 83 40. a = 58 20', 42. A = 70 5', 44. A = 115, 6 = 138 5', B = 122, B = 60, c = 116 3'. C = 95 4'. C = 135. 41. a = 61, 43. flf = 60, 45. a = 150, c = 97, 6 = 120, 6 = 160, 5 = 110. c = 50. B = 10. 46. a = 112 30', 47. ^1 = 20 30', 6 = 108 40', B = 32 30', c = 14010'. = 124 30'. CHAPTER XI THE EARTH AS A SPHERE 84. Distances on the Earth. As we remarked in the intro- ductory chapter, plane trigonometry is sufficient for the survey of small areas. For larger areas and in navigation, except in the most refined work, the Earth is treated as a sphere and we make use of the principles of spherical trigo- nometry already enunciated. The shortest distance between two points on the Earth is the arc of a great circle joining them. If we know the number of degrees in that arc we can compute its length by the formula s = xr (Art. 47), where s is the length of arc, x the angle in circular measure, and r the radius of the sphere ; in this case 3960 miles, the radius of the Earth. Example. Find the length of an arc of 26 on the Earth's surface. 26 '= Therefore, s =^ x 3960 = 12 . ?? . 3960 = 1798 miles, yo yo 7 It is convenient to compute and remember the number of miles in one degree of arc for the purpose of finding lengths of arcs. s = -?- x 3960 = 69.1 miles, approximately. JLoU 85. Position and direction. The position of a point on the Earth is determined by its latitude and longitude; that is, by the number of degrees the point lies north or south of the equator, and the number of degrees east or west of a 125 126 SPHERICAL TRIGONOMETRY [XI, 85 great circle, through the Earth's axis, chosen as a reference line. We shall use the great circle, or meridian, through Greenwich. A point moving along a great circle of the Earth, unless that circle be a meridian or the equator, is constantly changing its direction, or course. For example, Fig. 36, at A the compass points north along AN, and a ship at A is FIG. 36. sailing, say, due west. When the ship has reached B the compass points north along BN and the ship is sailing west 30 south. On the other hand if a ship sailed constantly on a course, say, west 30 south it would move around the Earth in a spiral approaching continually nearer to the South Pole. 86. Bearings. To illustrate the use of spherical trigo- nometry in determining positions, directions, and distances on the Earth's surface, consider, Fig. 36, a ship sailing from C to A along the great circle CBA. The lines NC, NB, NA y and NG are meridians, the last being the meridian of Greenwich. Suppose the latitude and longitude of C are 44 40' N., 63 35' W. ; of A, 53 24' K, 3 4' W. The longitude of G, obviously, is 0. The positions of the points C and A being thus known, let us find the directions (called of A from C and of C from A, and the distance XI, 86] THE EARTH AS A SPHERE 127 from C to A. From the meaning of longitude we have 34:', GNC = 6335', whence a = ANC = 60 31'. Also, by the meaning of latitude, = go - 53 24' = 36 36' ; CN= 90 - 44 40' = 45 20'. We therefore have, in the spherical triangle CNA, CN = a = 45 20', AN= c = 36 36' and the included angle a = 60 31', which is case 3 in the solution of spherical triangles. The data : (a + c) = 40 58', i (a - c) = 4 22', | a = 30 15.5'. log cos i (a - c) = 9.9987 log sin 1 (a - c) = 8.8816 log cot - = 0.2340 log cot * = 0.2340 2 2i log seel (a + c) =0.1220 log esc |(a + c) = 0.1834 log tan $(A+C)= 0.3547 log tan 1 (A - C) = 9.2990 (A + C) = 66 10' i (A - C) = 11 16'. Whence A = 77 26'. C = 54 54'. Therefore the bearings of C from A are N. 77 26' W. ; of A from (7, K 54 54' E. To find the side CA = x we have log sin i (>1+ C) = 9.9613 log tan i (a - c) = 8.8829 log esc %(A - C) = 0.7092 log tan -=9.5534 a =19 47.5', x = 39 35'. Therefore length CA = 39.6 x 69.1 miles = 2736 miles. If only the distance sailed is required it is simpler to use the law of cosines. Thus, cos x = cos c cos a -f sin c sin a cos a. 128 SPHERICAL TRIGONOMETRY [XI, 86 log cos c = 9.9046 log sin c = 9.7754 log cos a = 9.8469 log sin a = 9.8520 9.7515 log cos = 9.6921 number = .5643 9.3195 _ .2087 number = .2087 cos x = 0.7730, x = 39 23' and the distance = 39.4 x 69.1 = 2723 miles. 87. The course of the ship at C would be N. 54 54' E. To show how the ship's course changes as it sails along CA let us find the course as the ship crosses the meridian 38 W. at the point B, Fig. 35. In the triangle NCB we have b=CN = 45 20', C = NCB = 54 54', N= CNB = 63 35' - 38 = 25 35' ; that is, two angles and the included side. i(C+ N) = 40 14.5', (C-N) = 14 39.5', b = 22 40'. log cos (C-N) = 9.9856 log sin | (<7 - N) = 9.4033 log tan i b = 9.6208 log tan | b = 9.6208 = 0.1173 log esc $(C + N) = 0.1897 log tan (c + n) = 9.7237 log tan | (c - n) = 9.2138 (c + ) = 27 53.5' (c - n) = 9 17.5'. Whence c = BN=37 11', n = CB = 18 36'. The latitude of is 90 - BN = 52 49' N., and the dis- tance sailed is CB = 18.6 x 69.1 miles = 1285 miles. To find the angle B = CBN we have log sin i (c + n) = 9.6700 log tan i (C - N) = 9.4176 log esc i (c - n) = 0.7919 log cot % B = 9.8795 \ B = 52 51', B = 105 42', and NBA = 180 - B = 74 18'. Therefore the ship's course at J5 (the bearing of A from B) is N. 74 18' E. XI, 88] THE EARTH AS A SPHERE 129 88. The Area of a Spherical Triangle may be found as fol- lows : The theorem has been proved that the area of a spherical triangle is equal to its spherical excess (the excess of the sum of its angles over two right angles) times the area of the tri-rectangular triangle; it being understood that the right angle is the unit of angles. Thus, using A to represent the area of a triangle whose angles (in degrees) are A, B, and C ; and noting that the tri-rectangular triangle is one eighth of the surface of the sphere ; we have A + B+C-ISQ 4 irr* = (A + B + C - 180)in* 90 8 180 Example. Given A = 105, B = 80, C = 95, and taking r = 3960 miles, the radius of the earth, A = (105 + 80 + 95" - 180) / 3960 N 2 = 5?r(3960) 2 180 9 log 5 = 0.6990 log TT = 0.4971 2 log r = 7.1954 colog 9 = 9.0458 - 10 log A = 7.4373 and A = 27,370,000 square miles. TABLE OF LATITUDE AND LONGITUDE Baltimore 39 17' K, 76 37' W. Boston 42 21' N., 71 4' W. Chicago 41 53'*K, 87 38' W. Greenwich 51 29' N., 0W. Honolulu 21 18' N., 157 55' W. Liverpool 53 24' K., 3 4' W. New York 40 43' K, 74 W. Pernambuco 8S., 34 W. Rio de Janeiro 22 54' S., 43 10' W. San Francisco 37 48' K, 122 24' W. Washington 38 54' N., 77 3' W. K 130 SPHERICAL TRIGONOMETRY [XI, 88 EXAMPLES In the following problems assume that one can travel directly along the arc of a great circle between the points named. 1. A ship sails from Baltimore to Boston. Find the course of the ship as she leaves Baltimore, her course as she crosses the meridian of New York, and the entire distance she sails. What are the bearings of Baltimore from Boston, and of Boston from Baltimore ? 2. Find the course at Liverpool, the course at 55 W., and the total distance sailed by a ship going from Liverpool to New York. What are the bearings of these cities from each other ? 3. A ship sails from Baltimore to Rio de Janeiro. She sails first to a point off Pernambuco in latitude 8 S., longitude 34 W., and from there to Rio. How far does she sail, and what is her course off Pernambuco ? NOTB. In the Southern Hemisphere latitudes are taken as algebra- ically negative. Use the north-polar distances of places as sides in solving triangles. 4. In problem 3 what course will the ship be sailing after she has gone 1000 miles ? What will be her latitude and longitude at that point ? 5. How far is the Washington Observatory from the Greenwich Observatory ? What are the bearings of the two observatories from each other ? 6. A ship sails from Boston on a course East 12 South. At what distance would she be sailing due East ? What are her latitude and longitude at that instant ? 7. A ship sails northwest from San Francisco. What would be the highest latitude she would reach? What would be the ship's longitude at that instant ? 8. Find the number of square miles in the triangle whose vertices are Baltimore, Boston, and Chicago. 9. A ship sails from Honolulu to San Francisco. Find the entire distance sailed, and the course of the ship when she has gone halfway. 10. An aeroplane sails from Washington to Chicago along a great circle arc one mile above the surface of the Earth. In what time is the flight made at a rate of 75 miles per hour ? XI, 88] THE EARTH AS A SPHERE 131 11. Find the number of square miles in the triangle whose vertices are Baltimore, New York, and Chicago. 12. Find the face and edge angles of a regular triangular pyramid. 13. What is the latitude of three points on the Earth equally dis- tant from each other and from the North pole ? 14. Each face of a triangular pyramid is a triangle whose sides are 3, 4, and 5 respectively. Find the face and edge angles of the pyramid. ANSWERS 1. 6 = 340 c = 422 4. 6 = 478 a =154 7. 70.7ft. 12. 99.5ft. 8. 1.912 17. 100 CHAPTER III Art. 28 2. A = 43 5 c = 54.9 5. a = 713 c = 823 8. 71.2ft. 9. 60 10' 3. ^ = 51 47' b = .433 6. b = 96.4 c = 232 10. 260ft. 11. 212ft. 13. 43 36' 14. Heights equal. Art. 35 9. - .874 14. c 2 15. 10+* 16. e 5 20. 21. 22. 1 23. Impossible. 2(log e a-log e 6) a b 27. 29. .6931 30. 1.099 31. 1.386 32. .2312 33. 1.029 34. -.3088 35.4.408 36.238.2 37. - 358300 38. .07212 39. Impossible. 40. -312.1 41. -1747 42. 57090 43. .00003162 44. .03728 45. 100 46. 3.241 47. .001347 48. 1142 49. 2448 SOLUTIONS OF RIGHT TRIANGLES 1. a = .04691 2. a = 2316 c = .05151 b = 3402 IT = .0004988 K = 3,941,000 5. a = 578.8 c = 2491 K= 701,000 8. a = .6441 c = .6503 b = 48.04 #=1859 7. A = 63 48' c = .4694 #=.04364 #=.02879 133 3. b = 24850 c = 36100 K= 325, 400,000 6. a = .00883 b = .003607 K= .00001593 9. J. = 4324' 6 = .8966 #=.3801 134 ANSWERS 10. a = . 003845 11. 6 = 5091 12. 6 = 99.43 b = .006723 c = 5268 c = 156.8 K = .00001293 K = 3,444,000 # = 6030 13. A - 79 28' 14. 6 = 63,840 15. a = .000005737 a = 842 c = 92,280 c = .00002118 K = 65,900 K = 2,128,000,000 K = .0000000000585 16. a = .0003899 17. A = 27 17' 18. a = 18.59 6 = .0006772 a = 4.252 6 = 30.51 1T = .000,000,1321 K= 17.53 JT= 283.7 19. 6 = 24,540 20. ^ = 43 45' c = 30,010 c = 5280 # = 211 ,900,000 K - 6,970,000 21. First steeper by 54'. 22. 24.7 mi: and 29.5 mi. 23. Team by 15 seconds. 24. 3 25' 25. 648ft. 26. 14.7 in. 27. 9 hr. 28 min. A.M. or 2 hr. 32 min. P.M. 28. Reduced by 10.1 ft. 29. Buoy farther by 1133 ft. 30. Increase in altitude 251.4 ft. 31. 8 45'. 32. 1 ft. shorter. 33. 1575 mi. 34. 57 43' N. or S. 35. N: 58 15' E. 15 mi. 36. E. 62 46' N. 7.29 mi. per hour. 37. 112.5 mi. 38. E. 80 N. or S. 4.33 mi. 39. 127.9 mm. 40. 155.1ft. 41. 74.17yd. V2 2 CHAPTER IV Art. 40 5. 9. 9 4x/7 20 V2 a 1 2V30 12 10 7V ^ 10 10 Art. 41 4 5 J=9V3 j= 8\/2 6 32\/29\/15 5 7 7. One value, 45 8. - 2 7 S 9. Two values, 90 Art. 44 5. 1 6. (32V2) V6 ANSWERS 135 18. 23. 25. CHAPTER V Art. 47 Minute hand, TT 19. 20. 21. 22. 14 T 6 20 9 Hour hand, radians per second. 24. 191 revolutions per minute. 3 6934 mi. 26. 1978V2 mi. 27. .035 in. 28. 6.7ft. Art. 49 17. V2 + V6 4 18 V6-V2 19 512 20 -LO. -LC7. 4 3 CHAPTER VI 1. 6 = 986 c - 544.3 # = 193,000 2. A = 19 3. = 52 c = 8.19 # = 13.1 A = 18 39' B = 26 52' c = 673.9 #=45,990 4. B = 59 18' C = 71 36' #=1705 5. A = 100 35' or 10 21' C = 4453' 135 7' a = 67.02 12.25 # = 914 167.1 6. a = 6184 6 = 2937 #=7,510,000 7. 5 = 49 8' 8. C = 59 19' a = 70.48 #=1703 A = 37 58' B = 66' 42' a = 179.9 #=23,370 9. a = 6.64 c = 3.95 #=12.21 10. = 23 29' 11. C = 22 57' b = 3024 #=3,243,000 C = 6128' c = .4592 #=.0802 12. A = 94 16' B = 54 36' C = 31 8' #= .0002699 13. No solution. 14. b = .01292 c = .002861 #=.00000826 15. A = 26 19' C = 109 6' c = 67.14 #=742 16. b = .0185 17. A = 54 40' C = 94 13' # = .000002694 5 = 90 c = 59.39 #=955 136 ANSWERS 18. J. = 4623' or 133 37' 19. # = 22 37' C = 102 30' 15 16' C = 127 28' c = 8730 2354 a = .5593 K = 14,600,000 3,938,000 K = .0958 20. A =134 22' 21. 4 = 57 41' B = 18 42' C = 38 49' C - 26 56' a = .02461 K= 4.622 K=. 0002232 22. 33,695 sq. ft. 23. 15 ft. 24. 35.6 ft. 25. 428 ft. 26. Width 74.6 ft. ; height above stream, 14 ft. 27. 472 ft. 28. 17.1ft. 29. 109ft. 30. 61.93 and 58.81 ft. 31. A = 61 43' = 80 7' C = 38 10' c = 5.20 32. B = 53 26' a = 46.45 ft. c = 52.48 ft. 33. 11,320 and 7082. 34. 3997 sq. ft. 35. A = 7 5' ; no. 36. 21.7 mi. 37. 9 17' with the surface of the water. 38. 33.5 ft. 39. 40 ft. 40. Second yacht by 1 min. 12 sec. 41. At 82 33' with shore on the side towards the 60 angle. 42. A-C-B\)j $560. 43. AC =152.1 ft. BC = 319.4 ft. 44. 441ft. 45. 336.9ft. CHAPTER VII 1. riTT 2. WTT | 3. 2 nir ^ 4. 2 wr + cos-' 5. 2 iw 6. rnr 37 " 2 T17T + COS" 1 2 T17T o 7. 2nT- 8. iiT+ 9. 2n?r TT riTT TT 36 28 10. 2 riTr coB- =v 11. nv. 12. ?ITT 2nir ANSWERS 137 13. 16. mr . TT ~3~ 12* 14. 4 T + ^ 15. 18. TIT + tan-i f 19. 21. mr TT 2 8 4 20. 2ri7r^, riTTitan-i^ 4 ' 22. = 2 mr - 23. = n?r, 2 nir 2 ^1 JL r = 2 ' 2 2 4 25. x = y = = 0, * , n even av/3 26. * = + ir*=^P nodd 5 27. fl = r = 28. x = + - 5vf 2 3 [. e= . r=0, n even 2 32. 35. 39. 47. 6 = mr + - 4 r = 2>/2 33. = 7* =: 36. 4 40. 1 44 "" 2' 48. = i 34. V 2V2 v /c 07 I v 2 'i _ 52 + 6x/r 38. 1 or '7 42 1 i 14 0, V3 ka - y -i 41 1 T/3 V t or 2 2"~ ' 3 -^ 45. 0. 6 _T 7T 4q ~2' 6 49< X 46 = i sin ~ 1 ( ; 2/ = V2V5-2 138 Kf\ ANSWERS , S T KI " 50 ,2-8>/6 O\J. z: in 37r 8 10 10 CHAPTER IX 1. B 58 30' 6 = 35 59' c 43 33' 121 30' 144 1' 136 27' 2. A = 30 53' a = 30 13' c = 78 35' 149 7' 149 47' 101 25' 3. A 48 11' a = 44 29' c = 109 52' 131 49' 135 31' 70 8' 4. A 83 39' 5. A = 159 39' 6. A = 147 34' B = 127 20' B = 104 14' B = 66 3' a = 82 1' b = 136 a = 157 26' 7. A 27 8. A = 81 29' 9. A = 139 5' B 73 B = 131 50' Tt 110 57' c 53 8' c = 97 42' c = 63 47' 10. a =3 49 26' 11. a = 147 37' 12. a = 53 45' b 43 58' b = 136 32' b = 153 17' c = 62 5' c = 52 11' c = 121 53' 13. a 3 7 59' 14. a = 49 11' 15. o = 129 30' b - 21 58' b = 100 6 = 166 50' B = 70 59' A = 49 37' B = 163 8' 16. A = 30 47' 17. A = 126 53' 18. a = 141 47' b 11 36' b = 32 29' c = 140 37' c =: 13 26' c = 133 18' B = 16 25' 19. B 10 23' 6 = 6 16' c = 142 41' 169 37' 173 44' 37 19' 20. A 170 50' 21. A = 25 5' 22. a = 66 12' B 84 45' B = 114 38' b = 146 25' b = 55 1' c = 168 23' c = 109 39' 23. a 142 40' 24. A = 28 19' b 78 7' b = 110 59' A = 142 3' c = 108 39' 25. b 33 37' c = 101 B = 34 20' 146 23' 79 145 40' 26. A 74 12' 27. A = 68 11' 28. a = 161 32' B 157 47' B = 39 43' b = 129 57' a 43 57' c = 61 11' c = 52 28' ANSWERS 139 31. 32. 35. 29. a = 168 30' 30. b = 130 29' B = 99 40' b = 20 23' B = 10 35' 159 37' 169 25' JL = 14132' 33. ^ = 163 16' C = 111 46' B = 19 55' b = 108 29' a = 138 36' B = 23 26' C = 30 51' 156 34' 149 9' 36. B = 12 7' C = 139 5' a = 75 40' A = 124 32' 6 = 16 48' c = 151 57' C = 34. a = b = B = b = 148 10' 31 50' 124 38' 46 49' 33 43' 50 50' 129 10' CHAPTER X 1. 4..= 76 2. A 138 18' 3. 90 14. A = 36 42' B = 160 32' b = 148 44' c = 138 32' 143 18' 38 52' 78 c = 146 42' 15. No solution 16. c 90 17. a = 138 34' b = 18 15' b = 155 50' B = 8 2' A = 100 16' 18. a = 63 59' c = 156 10' c = 155 2' 116 1' 72 54' 87 36' 19. A = 110 4' 20. 6 _ 115 19' 21. B = 96 16' B = 131 2' B 114 2' C = 65 14' C = 147 2' C 90 b = 99 52' 140 ANSWERS 22. a 61 2' 23. B = 37 30' 24. 6 = 69 46' b 39 2' a = 33 49' A = 67 37' c = 92 2' c = 64 15' G = 25 17' 25. a = 90 58' 26. a = 120 29' b 118 A = 119 3' c = 132 2' B = 90 27. a - 66 40' c = 160 54' C = 164 38' 113 20' 102 2' 127 42' 28. C = 109 58' a 96 42' b = 40 34' 29. B = 42 15' C = 160 10' c = 153 38' 137 45' 49 50' 89 56' 30. A = 44 54' 31. No solution. 32. No solution a 112 14' b = 29 53' 33. A = 89 36' a = 138 54' 6 = 42 48' 26 48' 25 28' 137 12' 34. A = 142 35. B = 58 6' 36. 6 = 96 24' B = 159 6 = 61 31' B = 54 54' C = 133 c = 90 C = 146 38' 37. A 72 54' 38. A = 51 16' a 78 30' b = 119 47' b = 112 57' c = 69 41' 39. a 129 44' A = 131 8' B = 68 36' 95 40' 76 58' 111 24' 40. A = 70 44' 41. 6 = 110 50' 42. a = 62 42'. B - 132 12' A = 61 35' 6 = 126 44' C 94 54' C = 86 27' c = 109 42' 2. CHAPTER XI Course at Baltimore : E. 37 55' N. Course at meridian of New York : E. 36 14' N. Bearings Boston from Baltimore : E. 37 55' N. Bearings Baltimore from Boston : W. 34 17' S. Distance sailed : 359 mi. Course at Liverpool : W. 14 51' N. Course at 55 W. : W. 26 56' S. Bearings New York from Liverpool : W. 14 51' N. Bearings Liverpool from New York : E. 40 31' N. Distance sailed : 3303 mi. ANSWERS 141 3. Course at Pernambuco : S. 36 35' E. arrives. Course at Pernambuco : S. 29 33' W. departs. Distance sailed : 5141 mi. 4. Course : S. 42 32' E. Position : 29 12' N., 64 1' W. 5. Bearings Greenwich from Washington : E. 40 41' N. Bearings Washington from Greenwich : W. 18 33' N. Distance : 3669 mi. 6. Distance : 11,550 mi. Position : 43 42' S., 91 25' E. 7. Position : 66 2' N., 153 54' W. 8. 117,700 sq. mi. 9. Distance : 2398 mi. Course : E. 29 6' N. 10. 8 hours, nearly. 11. 35,580 sq. mi. 12. Face angle, 60 ; edge angle, 70 32'. 13. 1928'S. 14. Face : 36 52', 53 8', 90. Edge : 180, 0, 0. Printed in the United States of America. HE following pages contain advertisements of a few of the Macmillan books on kindred subjects. Descriptive Geometiy BY ERVIN KENISON Associate Professor of Drawing and Descriptive Geometry AND HARRY C. BRADLEY Assistant Professor of Drawing and Descriptive Geometry in the Massachusetts Institute of Technology Edited by Professor E. R. HEDRICK The point of view maintained in this text is that of a draftsman. Mathematical formulae and analytic computa- tions have been almost entirely suppressed. The method of attack throughout is intended to be that which shall most clearly present the actual conditions in space. Wherever experience has shown that a simple plan and elevation are not sufficient for this purpose, additional views or pro- jections have been introduced freely, corresponding to the actual drafting practice of making as many side views or cross sections as may be needed. The amount of ground covered by this book is that which is considered sufficient to enable the student to begin the study of the technical drawings of any line of engineering or architecture. It is not intended to be a complete treatise on descriptive geometry. Detailed exposition of such branches as shades and shadows, perspective, stereographic projection, axonometry, the solution of spherical triangles, etc., will not be found here, but the student is prepared to take up any of these subjects. THE MACMILLAN COMPANY Publishers 64-66 Fifth Avenue New Tork College Algebra BY E. B. SKINNER Associate Professor of Mathematics in the University of Wisconsin Edited by Professor E. R. HEDRICK Cloth, i2mo, Illustrated, $1.50 The essentials of a course in algebra are put into the smallest possible compass consistent with clearness of presentation to elementary students. Enough material in the way of examples is provided to enable the student to acquire some skill in manipulation. The practical side of the work is emphasized by giving much space to the detailed study of the functions that are most important in science and in applied mathematics. The student is encouraged to think of his algebra in a concrete way by the giving of geometrical interpretations as far as possible. The way is prepared not only for a clear comprehension of the imme- diately practical side of algebra, but also for a natural and easy transition to the Analytical Geometry and the Cal- culus. THE MACMILLAN COMPANY Publishers 64-66 Fifth Avenue New Tork Elementary Mathematical Analysis BY JOHN WESLEY YOUNG Professor of Mathematics in Dartmouth College AND FRANK MILLET MORGAN Assistant Professor of Mathematics in Dartmouth College Edited by EARLE RAYMOND HEDRICK Professor of Mathematics in the University of Missouri $2.60, Cloth, 1 2mo, 54.2 pages A textbook for the freshman year in colleges, universities, and technical schools, giving a unified treatment of the essentials of trigonometry, college algebra, and analytic geometry, and introducing the student to the fundamental conceptions of calculus. The various subjects are unified by the great centralizing theme of functionality, so that each subject, without losing its fundamental character, is shown clearly in its relation- ship to the others, and to mathematics as a whole. More emphasis is placed on insight and understanding of fundamental conceptions and modes of thought; less emphasis on algebraic technique and facility of manipula- tion. Due recognition is given to the cultural motive for the study of mathematics and to the disciplinary value. The text presupposes only the usual entrance require- ments in elementary algebra and plane geometry. THE MACMILLAN COMPANY Publishers 64-66 Fifth Avenue New York Mathematics for Students of Agriculture and General Science BY ALFRED MONROE KENYON Professor of Mathematics AND W. V. LOVITT Assistant Professor of Mathematics at Purdue University Edited by Professor E. R. HEDRICK Cloth, i2mo, $2.00 This book is designed as a text in Freshman mathematics for students specializing in agriculture, chemistry and physics in colleges and in technical schools. The selection of topics has been determined by the definite needs of these students. Since this course is to constitute the entire mathematical equipment of some students, some chapters have been inserted which have seldom been available to Freshmen, for example, the chapters on annuities, averages and correlation, and the exposition of Mendel's Law in the chapter on the binomial expansion. Particular attention has been given to the illustrative examples and figures, and to the grading of the problems in the lists. Exercises constitute about one-fifth of the text and contain a wealth of material. Four place mathematical tables printed at the end of the text have been selected and arranged for practical .use and are adapted to the requirements of the examples and exercises in the book. THE MACMILLAN COMPANY Publishers 64-66 Fifth Avenue New York A First Course in Higher Algebra BY HELEN A. MERRILL Professor of Mathematics in Wellesley College AND CLARA SMITH Associate Professor of Mathematics in Wellesley College Cloth, ismo, $1.50 At this time when combination courses are receiving so much attention, a book which introduces notions of the calculus early should be valuable. The present volume, an elementary text in higher algebra, is based entirely on the theory of limits, which is treated as rigorously as an elemen- tary text allows. It is intended for use in required Freshman mathematics courses, and it seems especially well adapted for this purpose. Its most admirable features are : 1. It gives the student the ability to use the derivative in a course in analytic geometry, enabling him to derive equations of tangents, polars, etc., in the best possible way. 2. The sense of familiarity which the student gains with the beginnings of the calculus makes for ease and rapidity in that study. 3. He is equipped earlier than is customary to use mathematics in science courses. THE MACMILLAN COMPANY Publishers 64-66 Fifth Avenue New York 14 DAY USE RETURN TO DESK FROM WHICH BORROWED LOAN DEPT. This book is due on the last date stamped below, or on the date to which renewed. Renewed books are subject to immediate recall. 290cr 60M.ro ?* VfTO L'U OtC J9 15811 - 110r.t'62MH r?nT^o i o r\CLw LJ t-i-^ OPT r? IQfi? UUi / i r c.. mJZ ^ > V REC'D LD FEB 5'64-HAI LD 21A-50m-4,'60 CA9562slO)476B General Library University of California Berkeley Q.A53\ P35 . < THE UNIVERSITY OF CALIFORNIA LIBRARY