LIPRARY UNIVERSITY Of V CALIFORNIA PLANE AND SPHERICAL TRIGONOMETRY THE MARSH AND ASHTON MATHEMATICAL SERIES. BY WALTER R. MARSH, HEAD MASTER PINGRY SCHOOL, ELIZABETH, N.J. AND CHARLES H. ASHTON. INSTRUCTOR IN MATHEMATICS, HARVARD UNIVERSITY. The series will include text-books in * ELEMENTARY ALGEBRA, COLLEGE ALGEBRA, PLANE AND SOLID GEOMETRY, PLANE AND SPHERICAL TRIGONOMETRY, PLANE AND SOLID ANALYTIC GEOMETRY. PLANE AND SPHERICAL TRIGONOMETRY AN ELEMENTARY TEXT-BOOK BY CHARLES H. ASHTON, A.M. ASSISTANT PROFESSOR OF MATHEMATICS IN THE UNIVERSITY OF KANSAS AND WALTER R. MARSH, A.B. HEAD MASTER PINGRY SCHOOL, ELIZABETH, N.J. NEW YORK CHARLES SCRIBNER'S SONS 1908 COPYRIGHT, 1902, BY CHARLES SCRIBNER'S SONS PREFACE THERE have been two distinct methods of presenting the subject of Trigonometry. In one the treatment is purely geometrical, no attention being paid to the direc- tion of lines. Here the algebraic signs of the functions are assigned in a purely arbitrary manner, and none of the proofs hold except for the particular figure. When this method is employed, the result must be shown to be general by some algebraic process. In the second method all lines have direction as well as magnitude, and the proofs are given in such a form that they are general and hold for every possible figure. This would seem to be the logical method of developing the subject ; for Trigonometry is the connecting link between elementary Geometry and those subjects in which Algebra and Geometry are combined in such a way that the directed line must be used constantly. This second method has been employed in this work, which is in- tended as a text-book for a fifty hour course in high schools and the ordinary first year classes in college, and is, therefore, made as elementary as possible. All matter not required for such a course has been excluded. We have attempted to avoid the usual mistake of mix- ing the two methods mentioned above. Two distinct proofs of the Addition Theorem are given. The first employs the method of projection, the formulas for which have been simplified by employing the French symbol for yi PREFACE an angle between two directed lines; the second is a geometric proof, similar to the one usually given, which has also been made perfectly general. The second dem- onstration may be preferred in a shorter course. Carefully selected exercises are given at the end of almost every article. In solving triangles, the natural functions are used where the solution may be obtained easily without the aid of logarithms. This method has been followed because, where logarithms are used exclu- sively, the student often does not know the meaning of the operation he is performing. The subject of trigono- metric equations, which is usually accorded little atten- tion, is here given in a separate chapter, for it is thought that in no other way can the student acquire so good a knowledge of general principles or so great skill in applying the formulas. AUGUST, 1902, CONTENTS PAET I PLANE TRIGONOMETRY CHAPTER I FUNCTIONS OF ACUTE ANGLES ART. PAGE 1. Definitions 1 2. Functions of complementary angles 5 3. Variation of the functions as the angle varies . . .6 4. Functions of 45, 30, and 60 7 5. Relations between the functions ...... 9 6. Trigonometric identities ....... 13 CHAPTER II RIGHT TRIANGLES 7. Solution of right triangles 16 8. Solution by the aid of logarithms 18 9. Heights and distances 19 CHAPTER III FUNCTIONS OF ANY ANGLE 10. Directed lines 24 11. Theorem 25 12. Angles 26 13. The measure of angles 27 14. Addition of directed angles . . . . ' . .29 vii viii CONTENTS ART. PAGE 15. Angles between directed lines ...... 30 16. Functions of angles of any magnitude ..... 31 17. Algebraic signs of the functions : 33 18. Functions of the quadrantal angles 35 19. Line values of the functions 38 20. Variations of the functions 39 CHAPTER IV RELATIONS BETWEEN THE FUNCTIONS 21. Relations between the functions of any angle ... 45 22. Functions of a, - a, TT a, - TT a . . . . 46 23. Reduction of the functions of any angle to functions of an angle less than ........ 50 24. Projection . 52 25. Projection of a broken line 53 26. Projections on the axes of any line through the origin . 54 27. Projections of any line on the axes . . . . . .54 28. Functions of the sum and difference of two angles . . 56 29. Second method of finding sin (a + /?) and cos (a + (3) . 59 30. Tan (a 0) 62 31. Functions of twice an angle 64 32. Functions of half an angle 66 33. Sum and difference of the sines and of the cosines of two angles 69 CHAPTER V INVERSE FUNCTIONS AND TRIGONOMETRIC EQUATIONS 34. General values . . - 72 35. Inverse trigonometric functions 76 36. Solution of trigonometric equations 79 37. Trigonometric equations 81 38. Trigonometric equations 83 CONTENTS ix CHAPTER VI OBLIQUE TRIANGLES ART. PAGB 39. Introduction . . 89 40. Law of the sines . . . . ' 89 41. Law of the cosines ......... 90 42. Law of the tangents ........ 92 43. General form of the law of the sines and the law of the cosines ' 92 44. Solution of oblique triangles . . . . . . .95 45. Case I. Given two angles and one side . . . .95 46. Case II. Given two sides and the included angle . . 97 47. Solution by logarithms . . 98 48. Case III. Given two sides and the angle opposite one of them 99 49. Case IV. Given the three sides . . . . . .103 50. Solution by logarithms . 104 51. Area of an oblique triangle 107 PART II SPHERICAL TRIGONOMETRY CHAPTER VII RIGHT AND QUADRANTAL TRIANGLES 52. Introduction 117 53. Right triangles 119 54. Napier's rules of circular parts . . . . . . 122 55. Solution of right triangles 123 56. Quadrantal triangles 126 CHAPTER VIII OBLIQUE TRIANGLES 57. Law of the sines . . . . ' 128 58. Law of the cosines 129 X CONTENTS ART. PAGE 59. Transformation of formulas 130 60. Napier's analogies ......... 132 61. Solution of oblique spherical triangles 134 62. Case I. Given the three sides 135 63. Case II. Given the three angles 136 64. Case III. Given two sides and the included angle . . 137 65. Case IV. Given two angles and the included side . . 138 66. Case V. Given two sides and the angle opposite one of them 139 67. Case VI. Given two angles and the side opposite one of them . 141 PLANE AND SPHERICAL TRIGONOMETRY PART I PLANE TRIGONOMETRY CHAPTER I FUNCTIONS OF ACUTE ANGLES 1. Definitions. Let ABO (Fig. 1) be a right triangle, right-angled at C\ and let A'B' 0' be any second right triangle which is similar to the first. From the definition of similar triangles it follows that Z A = /.A', and OB AB O'B' AO AC A'B' AB A'B' and OB C'B' AO AC' These five equations between the angles and the ratios of the sides are true for any pair of similar right tri- angles. Bat any two right triangles are sim- ilar, if an acute angle of the one is equal to an acute angle of the other, or if any pair of homolo- gous sides are propor- tional. Hence if any one of these five equations hold, the remaining four also hold and the triangles are similar. 1 PLANE TRIGONOMETRY [Cn. I, 1 It follows then that in every triangle equiangular with ABO the ratios of the sides are the same as in ABO, and that in every right triangle not equiangular with ABO these ratios are not the same as in ABO; and, conversely, if any right triangle has any one of these ratios equal to the corresponding ratio of ABO, the triangle must be equiangular with ABO. The ratio, then, of any two sides of a right triangle is a number which is entirely independent of the lengths of those sides, and, if the angle A is fixed, each of these ratios has a deter- minate value which is dif- ferent from the value of the corresponding ratio A ^X* O A' C' f or an y other angle. FlG>1 ' These three ratios of the sides of the triangle may then be spoken of as functions of either of the acute angles of the triangle. (One quantity, y, is said to be a function of another, x, when y has a determinate value, or values, for every value given to x.) It has been found convenient to give names to these ratios as follows : If A is either of the acute angles of a right triangle (see Fig. 1), sine of A = sin . cosine tangent AB hypotenuse A AO adiacent leg of A = cos A = - ^ = J - -> AB hypotenuse of A = ten A=<= p , posite } e S AC adjacent leg CH. I, 1] FUNCTIONS OF ACUTE ANGLES 3 The abbreviations in the second column should always be read the same as the first column. Names are also given to the reciprocals of these ratios as follows : ,. ,, A AB hypotenuse cosecant of A = esc A = = ^ . - , CJD opposite leg f A A AB hypotenuse secant of A = see A = - = ^ * AC adjacent leg '. A AC adjacent leg" cotangent of A = cot A = J : =-. CB opposite leg From these definitions it will be seen at once that esc A = , sec A = -, and cot A = sin A cos A tan A The last three ratios are used much less frequently than the first three, and will usually be studied as the recip- rocals of the first set. It is in this way that the student is advised to memorize their definitions. There are also two other functions which are occasion- ally used. They are defined by the equations, versed sine of A = vers A = 1 cos A. coversed sine of A = covers A = 1 sin A. The student should note that these eight quantities are all abstract numbers^ since they are the ratios of two lines. They are called the trigonometric functions of the angle A. PLANE TRIGONOMETRY [On. I, 1 EXERCISE I In the following problems, C is the right angle of the right triangle ABC, and the small letters a, b, c are used to represent the lengths of the sides opposite the corresponding angles j that is, c represents the hypotenuse, a and b the legs. 1. Find all the functions of the angle A, if (a) a= 5, 6 = 12. (c) a= 8, c = 17. (6) 6= 4, c= 5. (d) a =10, c=15. 2. What are the functions of B in the same triangles ? 3. What relations do you notice between the functions of A and B ? 4. Construct the angle whose (a) sine is f, (b) tangent is 5, ^y (d) cotangent is 2, (c) cosine is J, -J >> Y \ (e) secant is 3. 5. Draw, on a large sheet of paper, a line AC 10 in. long. At C erect a perpendicular, and at A lay off (with a protractor) angles of 10, 20, etc. Measure carefully the sides of the triangles thus formed, and from these measurements determine (to two decimal places) the functions of the angles given in the following table : A sin A cos A tan A cot A 10 .1737 .9848 .1763 5.6713 20 .3420 .9397 .3640 2.7475 30 .5000 .8660 .5774 1.7321 40 .6428 .7660 .8391 1.1918 50 .7660 .6428 1.1918 .8391 60 .8660 .5000 1.7321 .5774 70 .9397 .3420 2.7475 .3640 80 .9848 .1737 5.6713 .1763 6. By the aid of the table given above find the legs of a right triangle, if A = 40 and c = 12. CH. I, 2] FUNCTIONS OF ACUTE ANGLES 5 SOLUTION. By definition, a b sin A = - . and cos A = - c' c Hence a = c sin A = 12 sin 40 = 12 x .6428 = 7.7136, and b = c cos A = 12 cos 40 = 12 x .7660 = 9.192. 7. Find c,ifA = 30 and a = 8. 8. Find A,iia = 171 and c = 500. 9. Find A, a, and c, if B = 50 and 6 6. 10. Find .4, 5, and c, if a = 91 and b = 250. li th e cosine .AB) nearly unity, the secant (reciprocal of the cosine) a little more than unity, while the cotangent and cosecant (recip- rocals of the tangent and sine) are very large. As the angle increases, both BC and AB increase and approach equality. Hence the sine, tangent, and secant increase, while the cosine, cotangent, and cosecant de- crease. As the angle A approaches 90, AB and CB are very large and nearly equal. Hence the sine approaches unity, and the tangent and secant increase indefinitely. The cosecant decreases toward unity, while the cosine and cotangent decrease indefinitely. It appears, then, that for angles less than 90 the sine and cosine are less than unity, secant and cosecant are greater than unity, tangent and cotangent may have any positive value. 4. Functions of 45, 30, and 60. (a) Functions of 45. In the right triangle ABO (Fig. 4) let A O= OB = a. Then angle A = angle B= 45, and AB = Vo" 2 + a 2 = a V2~. It follows at once from the defini- tions of the functions that sin 45 = cos 45 = -? = \ V2, tan 45 = cot 45 = - = 1, a sec 45 = esc 45 = = V2. PLANE TRIGONOMETRY [Cii. I, 4 (b) Functions of 30 and 60. Let each side of the equilateral triangle ABD (Fig. 5) be represented by 2 a. Bisect the angle B by the line BC. Then this line bisects the base AD and is perpendicular to it. A right triangle AB C is thus formed, in which angle A = 60 and angle =30. Also AB= 2 a, AC= a, and BC= V-4 a 2 - a 2 = a VS. It follows at once from the definitions of the functions that D FIG. 5. sin 60 = cos 30 = V3 9 ' cos 60 = sin 30 = ^- = |, tan 60 = cot 30 a cot 60 = tan 30 = --^ = i VS, sec 60 = esc 30 = esc 60 = sec 30 = EXERCISE III Find the numerical value of 1. 2 sin 30 cos 30 cot 60. 2. tan 2 60 + 2 tan 2 45. NOTE. tan 2 60 is equivalent to (tan GO ) 2 , or the square of the tangent of 60. CH. I, 5] FUNCTIONS OF ACUTE ANGLES 3. tan 3 45 + 4 cos 3 60. 4. 4 cos 2 45 + tan 2 60 + 3 sec 2 30 5. sin 60 cos 30 + cos 60 sin 30 6. sin 2 45 + cos 2 60 - sin 2 30. Prove tluit PAO 2 tan 30 = 8. sin 60 = 2 sin 30 cos 30. 9. cos 60 = 1 -2 sin 2 30. 10. cos 2 60 : cos 2 45 : cos 2 30 = 1 : 2 : 3. 5. Relations between the functions. It appeared at once from the definitions of the six trigonometric func- tions that they were not all independent ; for three of them were defined as the reciprocals of the other three. From these definitions we have the following relations : sin A esc A = 1 9 [1] cos .4 sec .4 = 1, [2] * hi s /I' 1 We shall now show that there are other relations be- tween these functions ; that, intact, each depends on the others, so that, if one is giveii, all the others may be found. In the right triangle ABO, Dividing by Alf, we have A& or siii 2 ^! + cos 2 ^ = 1. [4] F IG . G. 10 PLANE TRIGONOMETRY [Cn. I, 5 Again dividing the same equation by AC 2 , we have - + 1 ,= AC' 2 or 1 + tan 2 A = sec 2 A. [5] Dividing the same equation by wehave 1 or l+cot 2 ^ = csc 2 ^. [6] Dividing sin A by cos A, we have sin A ~AE CB AB But tanJ. = A C Hence Since cot A = cos^l tan^l cot A=^. [8] These formulas should be carefully memorized by the student, as they are in constant use throughout the study of the subject. The student should also be per- fectly familiar with the following forms, which may be obtained easily from the formulas given above : CH. I, 5] FUNCTIONS OF ACUTE ANGLES 11 sin. A = VI cos 2 A. cos A = Vl sin 2 ^., sec A = Vl -h tan 2 A, esc A = Vl + cot 2 A, tan J. = Vsec 2 A 1, cot J. = VcscM 1. One of the simplest applications of these formulas is in finding the remaining functions, when any one function of an angle is given. For example, let sin A = |-. Then cos A = Vl-sin 2 J. = Vl - J = -| V3. cos^L 1V3 cot ^1 = __ = -^ = V3, tan^l sec ^ = = -^ = * V3, cosJ. 1.V3 i esc A = - - = - = 2. sin A Again, let tan A = 5. Then cot ^4 = sec ^4. = Vl + tan 2 A = V26, in^ = _J_= 5 V26, esc ^4. sec A 12 PLANE TRIGONOMETRY [Cn. I, 5 EXERCISE IV Find all the other functions of A, if 1. sinJ. = f. 6. esc ^4 = 10. 2. cos A = i. 7. sin -4 = -J-}. 3. tan J. = 3. 8. cos A = f. 4. cot A = 6. 9. tan ^4 = -|. 5. sec A = 2. 10. sec A = a. 11. Find all the other functions of 45 from the fact that tan 45 = 1. 12. Find all the other functions of GO from the fact that sec 60 = 2. 13. Find all the other functions of 30 from the fact that cot30 = V3. 14. Find all the other functions of A in terms of (a) sin A, (b) cos A, (c) tan A, (d) sec A. 15. Transform each of the following expressions into an- other form which shall contain sin A only : (a) ^AeofA+^A cot A . cos A cos A sm A cos 2 A cot A (c) sec 2 A + cos 2 A - tanl4 cot 2 A. v 16. Transform each of the following expressions into an- other form which shall contain cos A only : (a) sin A cos A tan A cot A. (b} sec A 1 4. tan A cos A sin A (c) sin 2 A + cos 2 A tan 2 A cot 2 -4. . CH. I, 6] FUNCTIONS OF ACUTE ANGLES 13 17. Transform each of the following expressions into an- other form which shall contain tan A only : (a) sin A cos A + cos A tan A -f sin A cot A. /7N A A 1 sin A cos A (b) sin A cot A 1 + sin A cos A (c) sec A esc A + cot A. 6. Trigonometric identities. Another important use of the formulas obtained in Art. 5 is in proving the identity of certain trigonometric expressions. The student should here make himself familiar with the distinction between identities and conditional equations, such as usually occur in algebra. An identity is an equa- tion in which the two members are equal for every possible value of the- variables in it ; while a conditional equation holds only for certain values of these variables. For example, 2x + 3 x = 5 x and' sin 2 x + cos 2 x 1 are identi- ties ; while x 2 2 x = 5 and sin x cos x are conditional equations. We are at present concerned only with the proof of identities. The method to be employed is to change the form of one of the members of the equation, by the applica- tion of the formulas of Art. 5, until it has been made to assume the form of the other member. Skill in choosing the proper formulas to accomplish this result with the least labor is acquired only after considerable practice ; but the form which we wish to obtain will soon suggest to the student which formulas he should use. The following examples illustrate the mode of pro- cedure : 14 PLANE TRIGONOMETRY [Cn. I, 6 EXAMPLE 1. Prove that (1 cos 2 A) sec 2 A = tan 2 A. If we replace 1 cos 2 A by its equal sin 2 ^L, and sec 2 A 1 2 A by -, the first member becomes S1I \ J , which is evi- cos^-4 cos 2 JL dently equal to tan 2 A. EXAMPLE 2. Prove that esc 2 A tan 2 A - 1 = tan 2 A. Here it is probably simplest to express the first member in terms of the sine and cosine. This is often advisable. It becomes 1 sin 2 J. i 1 2/1 1 = _ 1 = sec 2 A 1 = tan 2 A. sin 2 A cos 2 A cos 2 A EXAMPLE 3. Prove that sec A tan A sin A = cos A. As in the previous example, express the first member in terms of the sine and cosine. It becomes 1 _ sin A . M _ 1 sin 2 A _ cos 2 A _ . cos A cos A co A cos A EXERCISE V Prove the following identities : 1 . cos A tan A = sin A. 2. sin A sec A = tan A. 3. cos A esc A = cot A. 4. sin A sec A cot ^1 = 1. 5. sin 2 ^4 sec 2 ^4 = sec? A 1. ^ sin J. cos^l _ ^ esc A sec yl 7. COS^ CH. I, 0] FUNCTIONS OF ACUTE ANGLES 15 8. tan 2 ^4 sin 2 .4 = sin 4 .4 see 2 .4. 9. cot 2 A cos 2 A = cot 2 A cos 2 A. # 10. tan A + cot A = sec A esc A. 11. sin 4 A cos 4 .4 = sin 2 A cos 2 .4. 12. cos A sin A tan A + cot J 13. - = sin A cos A. 14 15. sin.4 1 tan A 1 cot A 16. sec 2 A esc 2 .4 = tan 2 A-\- cot 2 .4 + 2. ' 17. sin A (tan A 1 ) cos A (cot ^4 1) = sec A esc A. ^ 18. V 19. ^ 1 + sin A 20. (sin A + cos ^4) (tan A + cot ^4) = sec A + esc A 2 1 . tan 2 yl cot 2 A = sec 2 A esc 2 J. (sin 2 J. cos 2 A) . 22. (1 + cot ^4 esc ^4) (1 + tan J. + sec A) = 2. 23. 2 vers ^4 + cos 2 A = 1 + vers 2 A. 24. cot 4 ^4+ cot 2 A = esc 4 ^4 -cse 2 A. 25 . (sin ^4 + esc ^4) 2 + (cos A -f sec ^4) 2 = tan 2 A + cot 2 ^1 + 7 CHAPTER II RIGHT TRIANGLES 7. Solution of right triangles. A right triangle may be solved (that is, the unknown parts may be found) when any two parts, one of which is a side, are given. It is convenient to letter the triangle as in Fig. 7, the small letters a, &, c representing the lengths of the sides opposite the corresponding angles. When either A or B is known, the other angle may be deter- mined by subtracting the given angle from 90. When any two of the sides are known, the third side may be found by the aid of the equation a 2 + b 2 = ft. But when it is necessary to find the angles, having given two sides, or to find the other sides, having given one side and an angle, the trigonometric functions must be intro- duced. The unknown parts may then be found by the aid of the definitions of the functions, v b FIG. 7. sin A = cos 1 ? = i C J cos A = sin 1 c # tanA = cot J 5 =6 16 CH. II, 7] RIGHT TRIANGLES 17 Choose that one of these formulas tvhicJi contains the two given parts and one unknown part, and solve for the un- known part; its numerical value may then be found by the aid of a table of trigonometric functions. If necessary, repeat the operation with one of the other formulas to find the remaining parts. Each of the unknown parts should be determined directly from the given parts, without using the results of a previous operation. The accuracy of the work may then be tested by determining one of the given parts from the parts just found. The different cases which may arise might be separated, and the particular formulas to be used in each indicated ; but it is thought best to let the student determine the best method of solving each problem, only giving a few typical examples. EXERCISE VI 2. Given c = 16 and A = 26 15' ; find the remaining parts. To obtain a, use sm^. = -, or a = c sin A = 16 x .44229 = 7.0766. c To obtain b, use cos^l = -, or b = c cos A = 16 x .89687 = 14.35. C The accuracy of these results may be tested by determining whether they satisfy the equation a 2 -f b 2 = c 2 . The angle B is the complement of A, or 63 45'. 2. Given a = 8 and b = 12 ; find the remaining parts. Here c may be found at once from the equation c 2 = a 2 + b' 2 . From this equation, c = V64 + 144 = 14.44. 18 PLANE TRIGONOMETRY [Cn. II, 8 To obtain A, use tan A = - = = .66666. 6 12 From this, by the aid of the tables, A = 33 1'. B = 90 A = 56* 19'. These results may be tested by determining whether they satisfy either the equation for sin A or cos A. Find the remaining parts in each of the following problems : 3. a = 3, 6 = 4. 7. 6 = 13, c =85. S 4. a = 9, 6 = 40. y 8. a = 12, A = 34 42'. 5. a = 12, c = 13. ^9. c=15, = 23 34'. I. a = 60, c = 61. /xlO. 6 = 17, A = 13 52'. 8. Solution by the aid of logarithms. The student should now make himself familiar with the use of loga- rithms by reading the explanation of the tables, and by doing some of the problems given there. Logarithms are not of great use in solving right triangles, since they do not greatly decrease the labor involved ; but it is best to become familiar with their use in solving simple problems. EXERCISE VII 1. Given a = 12.73 and c = 43.18 ; find the remaining parts. sin A = - - sin B = -, or 6 c sin B. c c Hence log sin A= log a log c. Hence log6 = logc-J-logsiiiJ3. log a = 1.10483 log c = 1.63528 log c = 1.63528 log sin = 9.98026 log sin A = 9.46955 log 6 = 1.61554 .4 = 17 8' 47". 6 = 41.26. = 90 -^1 = 72 51 '13". CH. II, 9] RIGHT TRIANGLES 19 2. Given ^4=43 48' and I =67.92 ; find the remaining parts. A cos A = -, or c = c cos A tan -4 = or a=btanA. Hence log c = log b log cos A. Hence log a = log b -f- log tan A. log 6 = 1.83200 log b = 1.83200 log cos A = 9.85839 log tan A = 9.98180 log c = 1.97361 log a = 1.81380 c = 94.10. a = 65.13. Find the remaining parts in each of the following problems : 3. = 5678, 6 = 6789. ' 4. a = 2222, c = 3333. 5. a = .4545, c=.5454. , 6. = 4567, 4 = 23 52'. 7. c = . 8765, = 27 25'. 8. a = 206.14, .4 = 24 24'. 9. c = 2.383, 5 = 32 42'. 10. a = 1758, 6 = 1312.7. 11. a = .581, 6 = 13.5. 12. 6 = 75.84, A = S732'. 13. 6 = 8.4, c = 14. 14. a = 795, 6 = 164. 15. c=543.3,5=1720'35". 16. a = 1.456, yl=326'42". 17. a = .0065, c = .0094. 18. c = 765, .4=84 16'. 19. c = 1000,^l = 75 . 20. a = 1.006, c = 1.06. 9. Heights and distances. One of the applications of Trigonometry is in finding the height of objects or the distance between points, when that height or distance can- not be easily measured. This is accomplished by measur- ing other easily accessible lines and certain angles, and computing the desired lines from these data. It is necessary to explain a few terms used in such prob- lems. Let A and B be two points not in the same horizon- tal plane, and let A C and BD FIG. 8. 20 PLANE TRIGONOMETRY [Cn. II, 9 be horizontal lines through these points. If B is observed from J., the angle CAB is called the angle of elevation of B from A. If A is observed from B, the angle DBA is called the angle of depression of A from B. Again, let AB be any object, and let Q be the point of ob- servation. The angle ACB is spoken of as the angle which AB subtends at C. The earlier problems of the following exercise give simple right triangles, and can be solved by the student without further assistance. EXERCISE VIII 1. A vertical pole 13 ft. high casts a shadow 21 ft. long. What is the angle of elevation of the sun at this moment ? 2. How high is the tower which casts a shadow 125 ft. long, when the angle of elevation of the sun is 28 20' ? 3. What is the angle of elevation of the top of a tower 150 ft. high at a point in the same horizontal plane as its foot and 200 ft. distant ? 4. From the top of a tower 100 ft. high, the angle of de- pression of an object in the same horizontal plane as its foot is found to be 31. How far is the object from the tower ? 5. The angle of elevation of the top of a tower from a point in the same horizontal plane and 57 ft. from its foot is found to be 22 14'. How high is the tower ? 6. A building 95 ft. high stands on the bank of a river. The angle of elevation of the top of the building from the opposite bank of the river is found to be 25 10'. Find the breadth of the river. CH. II, 9J EIGHT TRIANGLES 21 7. The two equal legs of an isosceles triangle are each 10 ft. and the opposite side is 6 ft. Find the three angles. NOTE. Draw the altitude of the triangle. Then there will be two equal right triangles formed, in which the base is 3 ft. and the hypotenuse is 10 ft. 8. From a point directly in front of the middle of a build- ing and 100 ft. distant, the length of the building is found to subtend an aogle of 34 15'. How long is the building? 9. Two trees stand directly opposite each other on a straight road 80 ft. wide. From a point in the centre of the road the line joining their trunks subtends an angle of 5 28'. How far is the point from the trees ? 10. A circular balloon 10 yd. in diameter is noted by an observer to subtend an angle of 40'. At the same time the angle of elevation of its apparent lowest point is 50 10'. Find, approximately, the height of the balloon. 11. A. flagstaff 20 ft. long stands on the corner of a building 150 ft. high. Find the angle subtended by the flagstaff at a point 100 ft. from the foundation of the corner. ' 12. A strip of river bank is straight. It is 300 ft. long and it subtends a right angle at a point on the opposite shore. The a;igle between a line drawn from the point to one end of the strip and the perpendicular from the point to the strip is 15. Find the width of the river. 13. A ladder 30 ft. long leans against a house on one side of a street making an angle of 60 with the street. On turning the ladder about its foot till the top touches the house on the opposite side, the angle is found to be 30. Find the width of the street. 14. To find the height of a chimney a distance of 125 ft. is measured from its base. From the point thus reached the angle of elevation of the top of the chimney is found to be 48 25'. What is the height of the chimney ? 22 PLANE TRIGONOMETRY [Cn. II, 9 15. From the top of a telegraph pole 35 ft. tall a wire 50 ft. long is stretched to the ground. Find the angle which the wire makes with the ground. '16. A man lies on the ground with his eye to the edge of a well and, looking into the well, he sees the reflection of the opposite edge in the water. The direction in which he looks makes with the vertical an angle of 17. The well is 6 ft. broad. Find the distance from the edge of the well to the surface of the water. NOTE. The angle of reflection equals the angle of incidence. " IT. From one point 6t oblservaQcm^cne angle of elevation) of the top of a building is found to be 35. The observer walks 100 ft. directly away from the building in the same horizontal plane and then finds the angle of elevation of the top of the building to be 25. Find the height of the building. SOLUTION. Let AB (Fig. 10) represent the face of the building, and let C be the first point of observation and D the second point. Then .405 = 35, CDB = 25, and CD = 100 ft. Draw CE perpendicular to BD. Then in the right triangle ODE, FIG. 10. CE= CD sin 25, = 100 sin 25. In the right triangle BCE, the angle CBE = 10. CE 100 sin 25 Hence In the right triangle sin 10 C sin 10 100 sin 25 sin 35 sin 10 CH. II, 9] RIGHT TRIANGLES 23 Applying logarithms, log 100 = 2.00000 log sin 25= 9.62595 log sin 35 = 9.75859 11.38454 log sin 10= 9.23967 log AB= 2.14487 AB = 139.6. SECOND SOLUTION. -The following is another method of solution in which natural functions are used. In the right triangle ABC. - = tan 35. y In the right triangle ABD, -^ = tan 25. y + 100 Solving this pair of equations for aj, we have x = 100 tan 35 tan 25 = 100 x .70021 x .46631 " tan 35 -tan 25 .2339 Applying logarithms, log 100 = 2.00000 log .70021= 9.84522 log .46631= 9.66868 11.51390 log .2339 = 9.36903 log AB= 2.14487 AB = 139.6. 18. The shadow of a tower standing on a level plane is found to be 60 ft. longer when the sun's altitude is 30 than when it is 45. Prove that the height of the tower is 30(1 + V3). NOTE. Use second method. 19. Find the height of a chimney if the angle of elevation of its top changes from 31 to 40 on walking toward it 80 ft. 'in a horizontal line through its base. 20. From the top of a cliff 100 ft. high the angles of de- pression of two buoys, which are in the same vertical plane as the observer, are found to be 5 and 15. Find the distance between the buoys. CHAPTER III FUNCTIONS OF ANY ANGLE 10. Directed lines. If a point moves from A to B in a straight line, we shall say that it generates the line AB ; if it moves from B to A, it generates the line BA. The position from which the generating point starts is called the initial point of the line ; the point where it stops, the terminal point. In our study of Geometry, AB and BA meant the same thing, the line joining A and B without regard to its direction. But we shall now find it convenient to distinguish between AB and BA by calling one of them positive and the other negative. When either direction along a line has been O ^ >X chosen as the positive FlG 1L direction (as OX in Fig. 11), then all dis- tances measured along this line, or any line parallel to it, in this direction, shall be represented by positive numbers, and those in the opposite direction by negative numbers. In Fig. 11, if OX is chosen as the positive direction, AB and MN are positive lines, while CB is a negative line. The measures of AB and MN must, therefore, be positive numbers ; but CB must be represented by a negative number. 24 CH. Ill, 11] FUNCTIONS OF ANY ANGLE 25 The lines which we shall use in our further study will all be directed lines, unless the opposite is expressly stated, and we shall be concerned not so much with the lines themselves as with the measure of those lines. We shall therefore find it convenient to use the symbol AB to represent " the measure of the line AB " (its absolute magnitude with its proper sign attached) ; while if we wish to speak of the line itself, we shall write "the line AB." Since the lines AB and BA are equal in magnitude but opposite in direction, AB = - BA. 11. THEOREM. If A, B, and O are any three points on a straight line, ^^ + BC = AC. When the three points are situated as in Fig. 11, the theorem is evident, since all the numbers are positive, and the measure of AC _ ^ equals the sum of the A. _ O" JS measures of AB and BC. In Fig. 12, for the same reason as above AB = AC+CB. But, since we are dealing with numbers, we may treat this as an ordinary equation. Hence AC=AB-CB A FIG. 13. . " In Fig. 13, CA + AB = OB. Hence -AC + AB= - BC, or AC = AB + BC. 26 PLANE TRIGONOMETRY [Cii. Ill, 12 Let the student place the points in other positions, and show that the theorem holds in all cases. This theorem may be readily extended into the fol- lowing : If A, B, J, K are any number of points on a line, + BC+ ...IJ+ JK = AK. 12. Angles. In the previous chapters only angles less than 90 have been considered ; but the student is already familiar, in his study of plane geometry, with angles be- tween 90 and 180. In the further study of mathematics it is convenient to regard an angle as formed by a straight line revolv- ing in the plane about some point , t / , in the line. If a line starts from \u \s the position OA and revolves in a * \ / \ \ y / fixed plane about the point into V 4 "^ " / . . . the position OB, it is said to gener- ate the angle A OB. The position from which the moving line starts is called the initial side of the angle ; the position where it stops, the terminal side. We shall regard the amount of such rotation as the measure of the angle. In this sense there is no restriction on the size of an angle, since there is no limit to the possible amount of rotation of the mov- ing line ; after performing a complete revolution in either direction, it may continue to rotate as many times as we please, generating angles of any magnitude in either direction. Since there are two directions in which the moving line may be made to rotate, it is found convenient to distinguish On. Ill, 13] FUNCTIONS OF ANY ANGLE 27 between them by using positive and negative signs, just as in algebra it is found convenient to attach signs to the ordinary arithmetic number. There is no special reason for choosing either direction rather than the other as positive ; but the usual convention is to regard as positive an angle formed by a line revolving in the direction op- posite to the direction of rotation of the hands of a clock ; the clockwise direction of rotation is then negative. In reading an angle in the ordinary way a letter on the initial line is read first ; for example, A OB means the angle formed by a line in rotating from OA to OB, while BOA means the angle formed by a line in rotating from OB to OA. This method of reading an angle is evidently ambiguous, since there is an indefinite number of positive and negative angles, all of which must be read A OB. For the rotating line, starting from OA, may make any number of complete revolutions in either the positive or negative direction and then continue to the position OB, and any one of these angles must still be read A OB. Such angles which have the same initial and terminal sides are called congruent angles. It will be seen later that it is usually unnecessary to distinguish between congruent angles, since their trigonometric functions will be found to be the same; but we shall understand that the smallest of the congruent angles is meant unless another angle is indi- cated by an arrow in the figure. 13. The measure of angles. The angular unit of meas- ure with which the student is familiar is the degree, or J^th part of one right angle. When we wish to measure an angle, we say that it contains a certain number of these 28 PLANE TRIGONOMETRY [Cn. Ill, 13 units. This system is convenient for numerical problems in the solution of triangles, etc. But there is another unit which is almost universally used in higher mathe- matics. On the circumference of any circle lay off an arc AB equal in length to the radius, and join its ex- tremities to the centre of the circle. Since the ratio of a circumference to its radius is constant (and equal to 2 TT), the arc AB is always the same fractional part of a complete circumference, namely Then, since angles at the cen- tre of a circle are proportional to the arcs which they subtend, the angle AOB is the same fractional part of four right angles, and is, therefore, constant. FlG> 15 ' This angle A OB is the unit angle of circular measure, and is sometimes called a radian. The circular measure of any angle is the ratio of that angle to this unit angle, or the number of times the given angle con- tains the unit angle. It is usually written without the name of the unit. Since a complete revolution, or four right angles, has been shown to contain 2 TT radians, the circular measure of a right angle is ^; of an angle of 60, ^; of 45, ^. e t c . If a is the circular measure of any angle A OM and r is the radius of the circle, AOM arc^Of krcJLBf _ ~ r Hence the circular measure of any angle is equal to the arc it subtends divided by the radius, and, conversely, the CH. Ill, 14] FUNCTIONS OF ANY ANGLE 29 length of any arc equals the radius multiplied by the number expressing the circular measure of the angle. Or arc AM= ar. Since there are 2 TT radians in a complete revolution, or 360, orno one radian = 57 17' 45", approximately. 2 7T We have seen that it is convenient to call angles formed in the counterclockwise direction of rotation- positive ; the circular measures of such angles are, therefore, ex- pressed by positive numbers, ^, -^, etc. ; while the cir- cular measures of angles formed in the clockwise direction of rotation are negative numbers, , ^ , etc. EXERCISE IX 1 . Express in circular measure the angles 15, 30, 40, 120, 250, 300. 2. Express in degrees, minutes, and seconds the angles -, ** OK 7T Z 7T O 7T 7T 7T 7T K 9' T' ~6~' 3. In a circle whose radius is 10 inches, what is the length of an arc which subtends at the centre of the circle an angle of 3?T 7T 3 7T r, T' 6' IT' 4. In. a circle whose radius is 5 inches, what is the circular measure of an angle at the centre which subtends an arc of 10 inches ? 14. Addition of directed angles. If the moving line starts from OA (in any one of these figures) and rotates first through the angle A OB, and then through the angle BOO, it is evident that the position 00 which the line finally reaches is the same as if, starting from 30 PLANE TRIGONOMETRY [Cn. Ill, 15 OA, it had rotated through the single angle AGO. The angle A 00 is called the sum of the angles A OB and BOO. That is Z.AOB + Z.BOO = Z.AOQ. By a process similar to that used in Art. 11, it may be shown that the measure of the sum of any number of angles is equal to the sum of their measures, or that the above equation holds when, instead of the angle, we mean, in each case, the measure of the angle. It will not be necessary to distinguish further between an angle and its measure. 15. Angles between directed lines. We shall have occasion to speak of the angle from the positive direc- tion of one line to the positive direction of another line, and it is convenient to have a symbol to represent it, since the ordinary method of reading an angle is not suffi- cient. We shall adopt the following notation : {AB, MN) shall indicate the angle from the positive direction of AB to the positive direction of MN. (Sometimes it is more convenient to use single letters to represent the lines, as a and 6, when the symbol (a, 6) will be used for the same purpose.) In this symbol it is entirely immaterial whether we write AB or BA, MN or NM ; since it is always to indicate the angle between their positive directions without reference to the way they are read. CH. Ill, 16] FUNCTIONS OF ANY ANGLE 31 Here again, as in the ordinary method of read- ing an angle, the symbol is ambiguous, since it may represent any one of the congruent angles ; but the smallest of these will be understood unless an- other is indicated in the figure. If the arrows indicate the positive di- rections of the lines, in Fig. 17 (a) (AB, MN} = AOM, a positive acute angle ; while in Fig. 17 (li) (AB, MN) = AON, a negative obtuse angle. 16. Functions of angles of any magnitude. We must now define the trigonometric functions of an angle of any magnitude. For this purpose, let the plane be divided into four quadrants by a pair of indefinite lines perpen- dicular to each other, one horizontal, X'X, called the /-axis, and the other vertical, Y f Y, called the X-axis. Let the positive direction of the JT-axis be from left to right, and let the positive direction of the JT-axis be up- ward. Then all lines drawn parallel to these axes must have the same positive directions ; that is, a line drawn to the right or upward is positive, while lines drawn to the left or downward are negative. The positive direction of any line not parallel to one of the axes will be determined by other conventions. The point where the axes cross is called the origin. The quadrants are numbered as in the figure. 32 PLANE TRIGONOMETRY fCn. Ill, 16 'II X For the purpose of defining the trigonometric functions of any angle, let the angle be placed with its vertex at the origin and with its initial line coincident with OX, the positive segment of the Jf-axis. It is called an angle in the first, sec- ond, third, or fourth quad- rants, according to the position of its terminal -x IV Y FIG. 18. 7T line; that is, angles from to are angles in the first 2 quadrant, ^ to TT in the second quadrant, etc. The positive direction of segments measured along the terminal line of an angle will always be from the origin along that terminal line. For example, if we are considering the angle XOA, OA is positive and OK is negative ; while if we are consider- ing the angle XOK, OK is positive and OA is negative. From any point P on the terminal line of the angle XOA drop the perpendicular MP to the Jf-axis. Then the trigonometric func- tions of the angle XOA are defined as follows : sin XOA = cos XOA = tan XOA = MP OP' OM. OP' MP OM FIG. 19. CH. Ill, 17] FUNCTIONS OF ANY ANGLE 33 The secant, cosecant, and the cotangent are defined as the reciprocals of the cosine, sine, and tangent. In each of these definitions the direction as well as the magnitude of the lines is to be considered. The point P may be taken as any point either on the positive segment, OA, of the terminal line, or on OK, the negative segment of that line. For the similarity of the triangles shows that the ratios are the same numeri- cally for all positions of P '; and the signs of the ratios are also unchanged by a change of P from the positive segment OA to the negative segment OK, since this change simply reverses the signs of MP, OP, and OM. From these definitions it appears that the value of any function is the same for all congruent angles, since we are concerned only with the positions of the initial and ter- minal lines. The functions of Z XOA and Z XOK, which differ by TT, are numerically equal, but may differ in sign. 17. Algebraic signs of the functions. These definitions, when applied to angles less than , will be seen to agree with the definitions in Chap. I. In the first quadrant all the functions are positive, v since OM, MP, and OP are all positive. But in the other quadrants the signs of some of the func- v > m M \i/ M M tions will be seen to be negative. It will be sim- plest always to place P on the positive segment FIG. 20. X 34 PLANE TRIGONOMETRY [Cn. Ill, 17 of the terminal line of the angle, so that OP shall always be positive, and it will only be necessary to consider the signs of OM and MP. In the second quadrant MP is positive and OM is nega- tive. The sine and cosecant of angles in the second quad- rant are, therefore, positive and all the other functions negative. In the third quadrant both MP and OM are negative, so that all the functions of angles in the third quadrant are negative except the tangent and the cotangent. In the fourth quadrant OM is positive and MP negative. The cosine and secant of angles in the fourth quadrant are, therefore, positive and all the other functions negative. ~ X The student may find Fig. 21 use- ful in remembering the signs of the . functions in the different quadrants. FlQ 21 The function written at the head of each arrow is positive in the quad- rants through which the arrow passes. EXERCISE X '1 Determine the signs of the functions of the following angles: 100, 200, 300, 400, 500, 000, 700, -50, -15() c . -350 %, In which quadrant must an angle lie, if (a) its sine and cosine are negative, (&) its cosine and tangent are negative, (c) its sine is positive and its tangent is negative, (d) its cosine is negative and its tangent is positive ? CH. Ill, 18] FUNCTIONS OF ANY ANGLE 35 3. In which quadrants may an angle lie, if its cosine and secant are negative ? 4. For what angles in each quadrant are the absolute values of the sine and cosine the same ? For which of these angles are they also alike in sign ? 5. Determine the limits for x between which sin x + cos x is positive. 18. Functions of the quadrantal angles. The angles 0, , TT, - , and 2 TT, which have their terminal lines coinci- dent with one of the axes, are called quadrantal angles. It is necessary to consider the definitions of the functions of these angles separately ; for in some cases these defini- tions will be found to have no meaning. Let XOP be a small angle and let it decrease. If the length of OP remains fixed, MP will decrease indefinitely, and the length of OM will approach that of OP. The sine M FIG. 22. cosine (MP\ J evidently decrease indefinitely and ap- proach zero as a limit ; while the ( ) approaches unity as a limit. Then sin = 0, cos = 1, and tan = 0. Also, since sec A = -, sec = 1. ^^ cos ^ But when we consider the other reciprocal functions, the cosecant and cotangent, we meet with a difficulty, since the reciprocal of zero does not exist. There is, then, no cosecant or cotangent of a zero angle. For small values of 36 PLANE TRIGONOMETRY [Cii. Ill, 18 the angle XOP, the cosecant -73 and the cotangent / OM\ f - J are large, and they continue to increase without limit as XOP approaches zero. This fact is usually represented by writing esc = oo , and cot = oc . But these equations must be distinctly understood to be abbre- viations of the statement that, as an angle approaches zero, its cotangent and cosecant increase indefinitely ; they must never be understood to mean that the cotangent and cosecant of a zero angle exists. It is in this sense only that the symbol oo will be used throughout this work. If, in the expression -, x is made x to decrease indefinitely, the value of - will increase indefi- x nitely. This may be abbreviated into - = oo , which "1 should be read - increases indefinitely, as x approaches x zero." This is sometimes abbreviated still further into \ = oo . But this must never be interpreted as an ordinary equation, in which one member is equal to the other. It has no meaning whatever except as an abbreviation of the sentence above. When it is necessary to express the fact that the nega- tive values of a variable increase numerically without limit, the symbol oc will be used. When the angle XOP increases toward , OM ap- proaches zero and MP approaches equality with OP. Then the sine (-775) an( i the cosecant - approach CH. Ill, 18] FUNCTIONS OF ANY ANGLE 37 unity ; the cosine f - j and the cotangent ( ) ap- \OP/ fMP\ \MP J proach zero; while the tangent (-) and the secant fOP\ f j increase indefinitely. Then ^ = 0, cot^=0, esc- = 1, tan x and sec x = 00, = 00. O M FIG. 23. FIG 24. It must be here noted that, if x is an angle in the second quadrant (as XOP in Fig. 24), OM is negative, and hence the tangent and secant are negative. If x is now made to decrease toward , the negative values of these functions 2i increase numerically without limit. This will be found to be true in every case where any trigonometric function becomes infinite. When the angle is made to approach this value from one side, the function has positive values which increase indefinitely, while if the angle is made to approach this value from the opposite side, the function has negative values which increase numerically without limit. 7T This fact is usually expressed by writing tan = QD, 2 38 PLANE TRIGONOMETRY [Cii. Ill, 19 But the student must keep constantly in mind the fact that this is only an abbreviation for the statement made above. Let the student show in the same manner that the func- tions of the other quadrarital angles are as follows : sin TT = 0, cos TT = 1, tan TT = 0, sec TT = 1, esc TT = GO, cot-TT oo. sin TT = 1, cos TT = 0, tan TT = oo, sec TT = oo, 19. Line values of the functions. The trigonometric functions have been denned as the ratios of lines to each other, and are, therefore, abstract numbers. But, by the aid of a circle drawn about the origin with unit radius, lines may be found which will represent, in magnitude and direction, the values and signs of the various func- tions. This method of representing the functions will be found to be an aid in remembering the changes in sign and value of the func- Y tions. T Construct a circle with its centre at the origin, having as radius a line which we shall use as the unit of length. Drop a per- pendicular from the point P where the ter- minal line of the angle meets the circle. Then CH. Ill, 20] FUNCTIONS OF ANY ANGLE 39 smx, or - , is represented by MP both in magnitude and sign, for the denominator OP is always positive unity. In like manner, OM will always represent cos x. To represent tan x, it is necessary to erect a perpen- dicular at A, in order that the denominator of the ratio AT shall be the unit of measure. Then tan x = , and AT OA will represent tan x. If the angle is in the second or third quadrant, the line which is to represent the tangent must still be drawn at A to meet the terminal line of the angle, produced in the negative direction ; for, if it is drawn at A', the denominator of the fraction would be OA, which is equal to 1, and cannot, therefore, be used as the unit of length. The other functions may also be represented by lines, but it seems best to think of them as the reciprocals of the three given above. Thus, the secant of any angle has the same sign as the cosine, and in magnitude is the reciprocal of the cosine. 20. Variations of the functions. The changes in value of the functions were partially considered in Art. 18. But the student will find it much easier to remember the changes in both sign and value by making use of the line values discussed in the previous section. The sine. The line MP (Fig. 26) which represents the sine of XOP is seen, as the angle increases, to increase from to 1 in the first quadrant ; to decrease from 1 to in the second quadrant ; to continue to decrease from to 1 in the third quadrant ; and to increase from 1 to 40 PLANE TRIGONOMETRY [Cn. Ill, 20 P, in the fourth quadrant. This variation in value and sign may be represented by the aid of a curve as follows: Let y = sin #, where the angle x is expressed in circular measure. Then a graph of this equation may be formed just as is done in the case of an ordinary alge- braic equation. Using a pair of perpendicular axes, lay off from along OX various val- ues of #, and at the points thus determined erect perpendiculars, whose lengths are the corresponding values of y. For example, take any convenient distance, as OA, to represent an angle of , and at A erect a perpen- 2 dicular of unit length to represent the fact that sin = 1. At B, midway between and A, erect a perpendicular equal to -|-V2, or sin , etc. Do this for various values of x between and 2 TT, and then pass a smooth curve through their extremities. This curve will form a picture of all CH. Ill, 20] FUNCTIONS OF ANY ANGLE 41 the changes in value and sign as the angle changes from to 2 TT. It will be seen that the curve crosses the axis at x = TT, since sin TT = ; that it remains below the axis from x = TT to x = 2 ?r, forming a curve like that formed above the axis from x= to # = TT; and that this curve will be repeated indefinitely both to the right and left, if x is allowed to take on all possible values. The horizontal distance which represents a radian, and the vertical distance which represents unity are usually chosen equal. The cosine. The l^ine OM (Fig. 26), which represents the cosine of XOP, is seen, as the angle increases, to decrease from 1 to in the first quadrant ; to decrease FIG. 28. from to 1 in the second quadrant; to increase from 1 to in the third quadrant; and to continue to increase from t/ 1 in the fourth quadrant. Let the student show that this variation is represented by the figure given above. The form of this curve is seen to be the same as that of the sine curve, but it is moved along the axis a distance ^ to the left. 2 The tangent. It has been shown that the tangent is always represented by a tangent to the unit circle drawn from A. to meet the moving radius produced. Using the 42 PLANE TRIGONOMETRY [Cn. Ill, 20 symbol 00 in the sense explained in Art. 18, it is easily seen that, in the first quadrant, AT increases from to + oo ; in the second quad- rant, AT 2 increases from oo to ; in the third rp quadrant, AT 3 increases X from to +00 ; and in the fourth quadrant, AT increases from oo to 0. If we let y = tan x, the variation of y will be represented by a curve starting at (Fig. 30) and going upward indefinitely, as x approaches . Erect at the point x = a line MN Q TT TT TT X FIG. 30. perpendicular to the axis. For every value of x less than 7T - there are finite values of the tangent, growing indefi- OH. Ill, 20] FUNCTIONS OF ANY ANGLE 43 nitely large as x approaches . But when x = , there ' 2 is no tangent. The curve, therefore, goes upward indefi* nitely at the left of MN, never touching or crossing this line, but constantly approaching it. Such a line is called an asymptote of the curve. When x is slightly greater than ^, the tangent has a large negative value; it in- creases toward as x approaches TT ; and increases indefi- nitely as x approaches |TT. This branch of the curve has, then, the two lines MN and US as asymptotes, and crosses the axis at the point X = TT. When x increases from |-7r, the tangent again starts with a large negative value and passes through the same changes as before. If x is allowed to take all positive and negative values, there will be a series of such branches, just alike, at intervals of TT from each other. The cotangent. Since the cotangent is the reciprocal of the tangent, it decreases from co to in the first quadrant; decreases from to GO in the second quad- rant; decreases from -f- GO to in the third quadrant; and decreases from to oo in the fourth quadrant. The secant. Since the secant is the reciprocal of the cosine, it increases from 1 to + GO in the first quadrant; increases from GO to 1 in the second quadrant; de- creases from 1 to GO in the third quadrant; and decreases from + GO to 1 in the fourth quadrant. The cosecant. Since the cosecant is the reciprocal of the sine, it decreases from +00 to 1 in the first quadrant; increases from 1 to -j-oo in the second quadrant; increases 44 PLANE TRIGONOMETRY [Cn. Ill, 20 from GO to 1 in the third quadrant ; and decreases from 1 to -co in the fourth quadrant. EXERCISE XI 1. How many angles less than '360 have their cosine equal to | ? In which quadrants do they lie ' 2. How many angles less than 720 have their tangents equal to 5 ? In which quadrants do they lie '! 3. Are there two angles less than 180 which have the same sine? the same cosine? the same tangent? 4. Construct the curve which represents the change in value and sign of the cotangent. 5. Construct the curve which represents the change in value and sign of the secant ; the cosecant. CHAPTER IV RELATIONS BETWEEN THE FUNCTIONS 21. Relations between the functions of any angle. The relations between the functions of an acute angle which were proved in Art. 5 may v easily be shown to hold for all angles. In the triangle MOP, it is always true that x- M Y FIG. 31. By dividing successively by OP 2 , OM 2 , and MP 2 , we obtain, as in Art. 5, sin 2 a + cos 2 a = 1, 1 -f- tan 2 = sec 2 a, 1 -f cot 2 a = esc 2 a. It is also apparent from their definitions that MP MP< OP sin a cos a tan a = - = : = , and that cot = -: OM OM cos a sin a OP 45 46 PLANE TRIGONOMETRY [Cn. IV, 22 EXERCISE XII 1. If sin# = i, find the possible values of the other func- tions of x. 2. If cos x = i, and sin x is negative, find the other functions of x. 3. If tan x = 3, and x is an angle in the fourth quadrant, Hi id the other functions of x. 4. If sec x = 4, find the possible values of the other functions >> x. 5. If secA = , find the possible values of the other functions of A. ,^2 nj2 6. If sin^L equal -^ ^, find the values of cos A and cot A x -\-y 7. If sin , mr -f- 2 mn -+- 2 ?i 2 22. Functions of a, a, IT a, -IT a. In this 2 2 article we shall determine the values of the functions of a, , etc., in terms of functions of a. In proving 2 these relations, it is necessary to consider four cases ac- cording to the quadrant in which ex, lies ; but the student should first go over the demonstration in the simplest case where a is an angle in the first quadrant, and then assure himself that the demonstration applies to all values of . The formulas will be obtained for the sine, cosine, and tangent only ; the three reciprocal functions are seldom used and, if needed, formulas for them may be obtained easily from those given. CH. IV, 22] RELATIONS BETWEEN FUNCTIONS (a} Functions of <*. 47 Y * p \ 1 A'' s\ x x' \ " N V c | M M / f X s~ N P' > P / Y' X X (a) (6) FIG. 32. (c) In any of these figures, let XOP =, and XOP 1 = a. Drop a perpendicular from P on the Jf-axis and continue it to meet OP f . In either figure, the right triangles MOP and MOP 1 are equal, since OM is common and the geo- metrical angles MOP and MOP 1 are equal. In every case OM is identical for the two angles, OP 1 = OP, and MP' = -MP. MP r MP Hence sin (- a) = -^^^ = ^ = - sin a, OP 1 cos ( ) = tan ( ) = OP' OP OM OP OM OM = cos a, = tan . Here XOP may be the angle indicated by the arrow, or it may be any one of the positive or negative congruent angles which are read in the same way. We must then understand by XOP' that one of the congruent angles which is equal to XOP. For example, XOP may be the negative angle formed by revolving from OX to OP in the negative direction. Then XOP' is the positive angle formed by revolving from OX to OP'. 48 PLANE TRIGONOMETRY [Cn. IV, 22 The above demonstration is therefore general, and a may be any angle, positive or negative. These state- ments apply also to the demonstrations which follow. Functions of ^ . jL (6) FIG. 33. (c) (d) 7T In any of these figures, let XOP = a and XOP f = -| +. If OP' is taken equal to OP, the triangles .ftfOP and M'OP 1 are equal ; for the geometrical angles MOP and M'P'O are equal, having their sides perpendicular. Then the sides of these triangles are equal in magnitude, but, when their signs are considered, OM' = MP, and M'P' = OM. OM 7T , M'P f Hence sm[-+ = OP = cos - MP = cot a. Since the demonstration just given holds for all values of ce, we may replace a by a. CH. IV, 22] RELATIONS BETWEEN FUNCTIONS 49 7T Hence sin [ ) =s cos ( a) = cos a, cos ( ^ CM = sin ( a) = sin a, tan ( - a ) = cot ( a) = cot a. \2 / Functions of TT a. M Y (O) M P FIG. 34. Y' <<9 111 either of the given figures let XOP = a, and ' = 7r + a. If OP' is taken equal to OP, the tri- angles If OP and Jf'OP' are equal, and M'P' = -MP, and OM'=-OM. M' P' MP Hence sin (TT + a) = -^^ = ^ = - sin .a, OP' OP , Olf ' - OM cos (TT -f a) = = ^7 = cos OP' Olf' OP OM = tan a. Replacing a by a, we have sin (TT ) = sin ( a) = sin a, cos (TT a) = cos ()= cos a, tan (TT a) = tan ()= tan a. 37" 50 PLANE TRIGONOMETRY [Cn. IV, 23 (d) Functions of | TT a. Let the student construct figures for the other values of a and show that sin (| TT + a) = cos os, * cos (f TT + ) = sin oc, tan (|- TT -}- ) = cot a Also that sin (| TT ) = cos , cos (|^ TT ) = sin a, tan (| TT ) = cot a. (e) Functions of 2 TT . The functions of 2 TT cc are equal to the same functions of #, since these angles are congruent. For the same reason the functions of 2 HTT + a (where n is any integer and a is any angle, positive or negative) are equal to the same functions of . 23. Reduction of the functions of any angle to functions of an angle less than . It appears from the results of the previous article that the functions of any angle may be obtained in terms of the functions of an angle less than, or equal to, 45. This may be done by the aid of the formulas there derived ; but since the functions of angles in the first quadrant are all positive, it is best to use the following simple rule, which is easily derived from the formulas of the previous article : If 180 or 360 is subtracted from a given angle, or if the given angle is subtracted from 180 or 360 (so as to obtain CH. IV, 23] RELATIONS BETWEEN FUNCTIONS 51 in either case an acute angle), the functions of the resulting angle will be numerically equal to the same named functions of the given angle ; while if the given angle is combined in the same way ivith 90 or 270, the functions of the resulting angle will be numerically equal to the co-named functions of the given angle. In any case, attach to the result the proper sign of the function of the given angle, according to the quadrant in which it lies. ' For example, to obtain sin 290. This is equal numeri- cally to cos 20 ; but since 290 is an angle in the fourth quadrant and the sine is negative in that quadrant, sin 290 = - cos 20. Any multiple of 360 may be added to, or subtracted from, an angle without changing the value or sign of any of its functions. EXERCISE XIII 1. Express each of the following functions in terms of func- tions of angles not greater than 45. (a) sin 100, (e) sin -110, (i) cos 395T~ (6) cos 245, (/) cos -125, (j) tan 560, (c) tan 310, (g) tan -335, (&) sin ITT, sec 190, (A) esc -25, (I) COS|TT. 2. Find all the functions of each of the following angles. (See Art. 4.) (a) 120, (e) 225, (i) 330, (b) 135, (/)240, (j) -30, (c) 150, (g) 300, (fc) -45, (d) 210, (7i) 315, (I) -60. 52 PLANE TRIGONOMETRY [Cn. IV, 24 v 3. Prove geometrically the formulas for the functions of 37T ^ a \, (TT a), and j a J in terms of functions of a. 4. From the formulas for the functions of a and ( --f a\ derive algebraically the formulas for the functions of (?r a) and ^ J. SUGGESTION. sin (TT a) = sin - 4- ( - a ) = cos ( - a = sin a. / \ 5. Obtain the functions of Y a j in terms of the functions of a. 6. Obtain the functions of (a TT) in terms of the functions, of a. 24*. Projection. The projection of a point on a line is the foot of the perpendicular dropped from the point to the line. The projection of one line on another is the locus of the projections of its points, or the distance measured along the second line from the projection of the initial point of the first line to the pro- jection of its terminal point. - In Fig. 36, CD is the pro- c FIQ ^ D jection of AB on OX. The direction as well as the mag- nitude of CD must be considered, and CD will be positive or negative according as it is drawn in the positive or negative direction of OX. Throughout the present work the projection will always be upon two perpendicular axes, X'X and Y' Y. CH. IV, 25] RELATIONS BETWEEN FUNCTIONS 53 When we wish to speak of the projection of a line, as AB, on the JT-axis, it will be written proj x AB ; on the I^-axis, projy AB. These symbols will always mean the distance measured along the axes from the projection of A to the projection of B, and will be positive or negative according as they are drawn to the right or upward, or to the left or downward. V 25*. Projection of a broken line. The sum of the pro- jections on any axis of any series of lines, AB, BC, CD, etc., in which the initial point of each line is joined to the terminal point of the preceding line, is equal to the projec- tion on the same axis of AD, the line which joins the initial point of the first line with the terminal point of the last line. In Fig. 37, ab = proj^ AB, bc=p?oj x BC, etc. But b y Art U ' ab + be + cd = ad. Hence pro].,. AB + proj^ BO + proj x CD = proj x AD. If the terminal D of any such broken line coincides with the initial point A, ad = 0, and we see that' the projection of any closed contour is zero. The lines AB, BO, etc., are directed lines, but it is not necessary that their positive direction should be from A to B, etc. The measure of some of the lines as read may be positive and of others negative without affecting the truth of the theorem. PLANE TRIGONOMETRY [CH. IV, 26 26*. Projections on the axes of any line through the origin. The definitions of the functions may be given very concisely by the aid of projection. Let P be any point on either the positive or negative segment of the terminal side of the angle XOA. The angle (OJT, OP) is then the same as the angle XOA. Also OM = proj^ OP, and MP = proj ?/ OP. Substituting these expressions in the definitions of the sine and cosine given in Art. 16, they become M , and FIG. 38. OP) = OP Clearing of fractions, we have proj^OP - OP cos(OJf, OP), and pro } y OP = OP sin ( OX, OP) . The student must remember that in these formulas proj^OP and proj^OP are the measures of the distances from the projection of to the projection of P, the signs being determined by the directions of the axes ; that OP is the measure of the line OP ; and that ( OX, OP) is the angle from the positive direction of OX to the positive direction of OP. 27*. Projections of any line on the axes. We shall now proceed to obtain the more general formulas for the projection of any line, AB, on the axes. Through A, which may be either the positive or nega- tive extremity of the line AB, draw a pair of axes, AX' CH. IV, 27J RELATIONS BETWEEN FUNCTIONS and AY', parallel to the given axes. The pro- jections of AB on these new axes are evidently equal in magnitude and direction to its projec- tions on the original axes, OX and OY '; and the angles made by AB with these axes are the same as those made by it with the original axes. Then, from the formulas obtained in the previous article, = AB cos (OX, AB), [9] = ABsin (OJT, AB). [10] FIG. 39. EXERCISE XIV 1. What are the projections on the axes of a line 5 inches long which makes an angle with the X-axis of 30 ? of 100 ? of 200 ? 2. If proj a! AB = 3, and proj y AB = 4, find the length of AB and the angle (OX, AB). 3. If A B = 10, and proj^JB = - 3, find the angle (OX, AB). 4. Describe an equilateral triangle (each side being 10 inches in length) by going from A to B, then to C, and back to A. What is the angle (AB, BC) ? (AB,AC)? (BC,AB)? What is the projection of AC on AB? of BC on AB? of CA on BC? 5. In the triangle of problem 4 drop a perpendicular CD from C on AB. Let its positive direction be from D to C. What is the projection of BC on this perpendicular ? of CA ? of AB ? What is the projection of DC on BC? on AC? 56 PLANE TRIGONOMETRY [CH. IV, 28 28*. Functions of the sum and difference of two angles. We shall now proceed to find the functions of the sum and difference of two angles in terms of the functions of those angles. Let the first angle a be placed in the posi- tion XOA, and let the second angle /3 (=AOB} be added to a, making + (3 ( = XOB) . From any point P of the FIG. 40. terminal line OB of /3, drop a perpendicular KP on OA. The positive direction of all lines in the figure except KP is fixed by our former conventions ; for we have agreed that the positive direction of the initial and terminal lines of an angle shall be from the vertex along those lines. To determine the positive direction of KP, draw from a line OR, making with OA a positive angle Then OR bears the same relation to OA that OY bears to OX; and for any angle having OA as its initial line, the posi- tive direction of KP, a perpendicular to OA, must be taken parallel to OR. From Art. 26 we have But by Art. 25 the projection of OP on the !F-axis is CH. IV, 28] RELATIONS BETWEEN FUNCTIONS 5,7 equal to the sum of the projections of OK and KP on the same axis. Hence or by [10], = - [ <9JTsin( OX, OK) + KP sin (OX, KP)~] . The angle (OX, OK) is seen at once to "be a, and since the positive direction of KP is the same as that of OR, (OX, KP) = XOR = + - Hence sin ( OX, KP) = cos . d K JfT* Noting that = cos /3 and = sin (3, we have sin (a + P) = sin a cos p + cos a sin p. [11] In like manner, Here cos (OX, OK) = cos , and cos ( OX, KP) = cos fa + J = sin a. Hence cos (a + p) = cos a cos p - sin a sin p. [12] 58 PLANE TRIGONOMETRY [On. IV, 28 Since the magnitudes and directions of all the lines and angles in the figure have been carefully considered in this demonstration, it holds for all cases. But in order that the student may assure himself of this fact, let him draw various figures according to the same directions, using the angles of different magnitudes (for example, see Fig. 42), and go through this demonstration carefully, making sure that every statement made above applies to every figure. Y A Let him also take note that the demonstration applies letter for letter when fi is a negative angle (see Fig. 43). We may, therefore, replace ft by fi in formulas [11] and [12] and obtain the following formulas for the sine and cosine of the difference of two angles : sin [ + ( /3)] = sin a cos ( /3) + cos a sin ( /3), sin (a p) sin a cos p cos a sin p, cos [ + ( /3)] = cos a cos ( fi) sin a sin ( /3), cos (a - p) = cos a cos p + sin a sin p. [14] CH. IV, 29] RELATIONS BETWEEN FUNCTIONS 59 29 . Second method of finding sin (a + p ) and cos (a + p) . The following method of finding the formulas for sin( + ) and cos ( + /8) avoids the use of projection and may be pre- ferred by some teachers. Let Z XOA = , and ZAOB=/3. Then ZX OB = a -h @. Through any point P of the line OB draw PM perpendicular to the X-axis and PTTper- pendicular to the line OA. Through K draw .BT^V per- pendicular to the Jf-axis and LK parallel to the same axis. FIG. 44. Prolong LK toward the right to S. Then Z SKA = a and Z SKP = a + 90. Then sin MP = NK LP OP OP OP' NK OK OK OP LP KP KP ' OP But NK OK = sin XOA = sin ce, = cos/3, = sin SKP = sin( + 90) = cos a, OP = sin/3. Hence sin (a + p) = sin a cos p + cos a sin p, pi] 60 PLANE TRIGONOMETRY [Cn. IV, 29 AI Also But . OM ON , NM COB( + 0) = = +^ , = ON OK KL OK' OP KP ON -- = cos XOA = cos a, OK = cos ACS = cos KP OP' ~ - cos = cos (a -f 90) = - sin a, KP OP = sin AOB = sin Hence COS (a + P) = COS a COS P - Sill a sin p. In Fig. 44 not only are a and /3 acute, but their sum a + /3 is also acute. But the proof applies without change B to Fig. 45, in which a -f /3 is obtuse. The only difficulty which the student is likely to meet is the equation OM=ON+NM, _ which is used in finding cos ( + /3) ; but this is seen to hold in Fig. 45, when the direction as well as the magnitudes of the lines is considered. By a very slight change, this proof may be made per- fectly general, so that it will apply to all values of a and /3. Through draw OR, making a positive right angle with OA, and having its positive direction from toward R. Let KP, which is parallel to OR, have the same positive direction as OR. If, in the above demonstration, we replace Z.SKP by N FIG. 45. CIL IV, 29] RELATIONS BETWEEN FUNCTIONS Gl T, KP), we shall have taken account of the directions of all lines and angles, and the demonstration will, there- fore, be perfectly general, holding for all values of a and ft, both positive and negative. For example, consider Fig. 46, in which a and /3 are both obtuse. The construction is the same as above. From P, any point of the terminal line of ft, draw PM perpendicular to the X-axis and PK perpen- dicular to the line OA, produced through the origin . Through K draw KN perpendicular to the X-axis and LK parallel to the same axis, prolong- ~r ^7' t i c* FIG. 46. ing LK to the right to S. Draw OR, making a positive right angle with OA. Let the student now follow the demonstration on page 59. The first statements are evident. In considering the functions of **, since K is on the terminal line extended through the origin, OK is negative, but by definition sin a= , and cos a= In determining the functions OK OK of ft, turn the book so that OR points upward. Then it KP OK is seen that, by definition, sin ft = -, and cos ft = In determining the functions of (SK, KP), think of K as a new origin and LS as a new X-axis. Then KL KP sin (tfJT, KP) = , and cos (SK, KP) = 62 PLANE TRIGONOMETRY [Cn. IV, 30 The student should have no difficulty in seeing that all the other equations hold for this figure. Let him also construct other figures according to the same directions and go through the demonstration carefully, making sure that every statement made above applies to every figure. Since this demonstration holds for all values of a and /3, we may replace ft by ft and obtain the following formulas for the sine and cosine of the difference of two angles : sin [a + ( /3)] = sin cos (/&) + cos a sin ( y#), sin (a - P) = sin a cos p - cos a sin p, [13] cos [<* + ( /3)] = cos acos( /3) sin a sin( /?), cos (a - p) = cos a cos p + sin a sin p. [14] 30. tan(ap). Since ten we have tanQ + cos* sin cos a cos p sin a sin /3 But this may be expressed entirely in terms of the tan- gent by dividing both numerator and denominator by cos a cos /3. This gives In like manner, let the student show that tan (a _ p) tan a -tan P [1G] 1 +tanatanp CH. IV, 30] RELATIONS BETWEEN FUNCTIONS 63 EXERCISE XV 1. Find sin 75 from the functions of 30 and 45. SOLUTION. sin 75 = sin (30 + 45), = sin 30 cos 45 + cos 30 sin 45. = J(V2+V6). 2. Find cos 75. 3. Find tan 75. 4. Find sin 15, cos 15, tan 15. 5. If sin a = ^ and sin ft = \, find sin (a + ft), when a and ft are both acute :) when they are both obtuse. 6. By the aid of formulas [11] to [14], prove the various formulas for the functions of (^ + a ) (TT ) ? etc. (See Art. 22.) 7. Find sin( SOLUTION. Replacing ft in formula [11] by ft -f- y, we have sin [ + (/3 + y)] = sin cos (ft -f- y) + cos a sin (ft + y) = sin a (cos J3 cos y sin ft sin y) + cos a (sin /? cos y 4- cos /? sin y) = sin a cos /? cos y + cos a sin ft cos y + cos a cos /2 sin y sin a sin ft sin y. 8. Find cos (a -f- y). 9. Find tan ( -f y). Transform the first member into the second. 10. sin (a 4- ft) sin ( /?) = sin 2 sin 2 /?. 11. cos (a + /?) cos (a ft) = cos 2 a sin 2 ft. 64 PLANE TRIGONOMETRY [Cn. IV, 31 12. cos g-)cosg-0)-sing-) sing-/?) = sin ( -f 0). 13. cos (a -f ft) cos a + sin a sin (a -f /?) = cos ft. tan tan 6 / 14 . = tan ( fl) tan a. cot + tan ft 31. Functions of twice an angle. If in formulas [11] and [12] we place /8 = , they become sin ( + )= sin cos a + cos sin a, or sin 2 a = 2 sin a cos a, [17] and cos (a -|- ) = cos a cos sin a sin a, or cos 2 a = cos 2 a - siii 2 a. [18, a] By replacing cos 2 a by 1 siii 2 , [18, a] becomes cos2a^l-2sin 2 a. [18, 6] Also, by replacing sin 2 a by 1 cos 2 , it becomes cos2a = 2cos 2 a- 1. [18, c] These three forms for cos 2 a should be remembered, and the one most convenient for the purpose in hand should be used. If we place ft = a in formula [15], it becomes f tan a + tan a tan + 1 tan a tan a Q tilH fL r--< f\-^ or tai,2a= - V. [19] (JH. IV, 31] RELATIONS BETWEEN FUNCTIONS 65 The functions of 3 a, 4 a, etc., may be found in a similar manner by letting ft = 2 , etc. These equations express the relations between the func- tions of any angle and the functions of twice that angle. It is not necessary that the angle in the first member should be expressed as 2 a and that in the second as a. It is only necessary that the angle in the first member should be twice that in the second. We may then replace 2 a by a, if at the same time we replace a by $ . These formulas may then be written sin a = 2 sin ^ a cos ^ a, [20] cos a = cos 2 * a - sin 2 \ a, [21, a] = l-2sin2ia, [21, ft] = 2cos 2 |a-l, [21, c] tana- [22] EXERCISE XVI 1. Find sin 2 a, if cos a = i-. V%. Find cos 2 a, if tan a = 3. \s 3. Find tan 2 a, if sin a = .6. 4. Find sin 3 a. SOLUTION, sin 3 a = sin (a -f 2 a) = sin a cos 2 + cos a sin 2 a, = sin a (1 2 sin 2 ) -f cos a 2 sin a cos a, = sin a (1 2 sin 2 ) -f- 2 sin (1 sinV), = 3 sin a 4 sin 3 a. 66 PLANE TRIGONOMETRY [On. IV, 32 *"" 5. Find cos 3 a and tan 3 . 6. Find sin 4 a, cos 4 a, and tan 4 a. 7. Find sin 5 , and cos 5 . Transform the first member into the second. Is* 8. 2 cot 2 = cot a tan rc. 9. 2 esc 2 a tan + cot a. /^ 10. esc 2 + cot 2 = cot . 11. - ) - - = esc 2 l-tan 2 [--a V 4 _ _ cos 4- si n cos sin <* , i & . - ; --- ; - = > tan & cc. cos sin a cos + sin a 2 cot rc 13. tan 2 a = cor 1 14 . 2 CQt 2 = (1 + tan a) cot a. 1 tan a 1 sin2 cos 2 a L-16. tan [- + ] + tan [ - aW 2 sec 2 4 4 32. Functions of half an angle. It has been shown in the previous article that cos a = 1 2 sin 2 1 a. Solving this for sin | a, we have [23] Cn. IV, 32] KELATIONS BETWEEN FUNCTIONS 67 Also from formula [21, c], cos a = 2 cos 2 a 1. Solving for cos J , we have i /I + cos ct ro j -\ COS-a = \-^ [/4J Dividing [23] by [24], we have [25, a] - I - sin a (1 + cos a) 2 1 + cos a These three forms for tan | should all be remembered, and the one most convenient for the purpose in hand should be used. It must be noted that the sign before the radicals does not mean that it has two values for any given angle a, but that it is impossible to determine in gen- eral what sign should be used. In any particular case, determine the quadrant in which | a lies, and affix the proper sign for each function. The ambiguous sign is omitted before the last two forms, since 1 cos a and 1 -f- cos are always positive, and tan | may easily be shown always to have the same sign as sin a. It is sometimes convenient to express the functions of a in terms of the functions of 2 a. This may be done 68 PLANE TRIGONOMETRY [Cn. IV, 32 by replacing | a by a, and a by 2 a in the above formulas. They then become ^, [26] /I + cos 2 a [27] [28, a] [28, 6] T28. /-I V 2 . /I COS 2 a * 1 + COS 2 a' 1 COS 2 a sin 2 a : sin 2 a EXERCISE XVII 1. Find sin | a and cos-|, if sin a = .6; find tan|, if tan a = 4. 2. Determine the functions of 22 30' from the functions of 45. 3. Determine the functions of 15 from the functions of 30 ; compare the results with those obtained in Exercise XV, problem 4. 4. Determine the functions of 165 from the functions of 330. 5. Obtain the formula for tan la by solving formula [22]. Transform the first member into the second. 6. sin J a -}- cos i a = Vl + sin a. y 7. sin i a cos ^ a = Vl sin a. 8. tan(7 + )=eQ + tana. U CH. IV, 33] RELATIONS BETWEEN FUNCTIONS 69 V 2 tan- 1-tan 2 " 10. - =sin. 11. - 1 + tan 2 " l + tan a " 12. tan - + cot t j = 2 esc . 13. 1 + cot a cot " = esc a cot -. y 14. tan i x + 2 sin 2 1 a; cot x = sin ce. \/ 15. tan 3 1 a? (1 + cot 2 1 x) 3 = 8 esc 3 x. 33. Sum and difference of the sines and of the cosines of two angles. In Art. 28, it has been showji that sin ( + /3) = sin a cos /3 + cos a sin /3, sin ( )= sin ct cos /3 cos sin /?, cos ( + /3) = cos a cos /3 sin a sin /3, cos Q /3) = cos a cos /3 + sin a sin /3. From these, by addition and subtraction, we have sin (a + /3) -f- sin ( /3) = 2 sin cos /3, sin (a + ff) sin (a /9) = 2 cos a sin yS, cos ( + /3) + cos (a /3) = 2 cos a cos /?, cos (a -*~ /?) cos ( yS) = 2 sin a sin /3. If we let + = . A /v _L Qin s. rt 10. sin 4 a. + sin 75 - sin 15 _ V3 cos 75 + cos 15 3 ' CH. IV, 33] RELATIONS BETWEEN FUNCTIONS 71 EXERCISE XIX Prove the following identities : 1 - tan a tan ft 2. (sin + cos a) 2 = 1 + si n 2 a. 3. cos 4 sin 4 a = cos 2 a. 4. tan 3 -a tan = 2 sin a sec 3 a. 5. cot a 2 cot 2 tan #. ar/"- 4- P\ - 2 csc 2/3 + sec/?. - 7. sin - cos - = 1 sin a. 8. sin [ - 4- a } sin [ ^ a ) = \ cos 2 a. V 4 J V 4 J 9.14- tan a tan - sec a. . tan ( 4- a ] tan ( - a } = 2 tan 2 a. V 4 / V 4 / f 11. sin 3 a 4- sin 2 a sin a = 4 sin a cos " cos - 22 IT 12. sec 2 a(l 4- sec 2 a) = 2 sec 2 a. . esc a 2 cot 2 a cos = 2 sin a. 14. cos 6 a = 32 cos 6 a 48 cos 4 a + 18 cos 2 a 1. sin a esc 2 a 14- tan 2 a 1 4- esc 2 a (1 4- tan ) 2 . (cos 2 4- cos 2 ft) 2 + (sin 2 a 4- sin 2 /3) 2 = 4 cos 2 (a cos cos . sin | a = (1 -f- 2 cos a)-yi cos a 2 20. 3 sin 2 a sin 6 a = 32 sin 3 a cos 3 CHAPTER V INVERSE FUNCTIONS AND TRIGONOMETRIC EQUATIONS (2n+l)7T T (2 n + '/n) TT , 34. General values. If sin x = -, x = ^, or ^, or any of the other angles which have the same terminal lines. There are, then, an indefinite number of angles which satisfy this equation, or any similar one. It is convenient to have general formulas to express all angles which have the same sine, cosine, or tangent, and we shall now pro- ceed to find such formulas. We shall first obtain general expressions for all angles which have their terminal lines along one of the axes. Let n be any integer, positive, negative, or zero. Then all angles which C have their terminal lines in coin- cidence with the initial line OX are represented by 2 nir ; for this expression represents the series of angles 0, 2 TT, 4?r, etc., and 2?r, 4?r, etc., which evidently contains all the angles which have their terminal lines in coincidence with OX. In like manner (2n + l)7r represents all the angles which have their terminal lines in coincidence with OX' ; for TT is the smallest of these angles and the addition or subtraction of any multiple of 2 TT will evidently give the same terminal line. 72 Y' FIG. 47. CH. V, 34] INVERSE FUNCTIONS 73 Y FIG. 48. Let the student show that the angles which have their terminal lines along OY are represented by (2n + J)TT; and along OY' by (2w->7r. If OP is the terminal line of any angle , all the angles (positive and negative) which have OP as their x '_; i^ ** y ^ terminal line are represented by 2 WTT + . All angles which differ from each other by any multiple of 2 TT have the same terminal line, and hence the same functions. This fact is expressed by saying that they are periodic functions, having a period Of 27T. Let the student show that the period of the tangent is TT. All angles ivMcTi have the same sine or cosecant as a may be expressed by 2 mr + and (2 n + 1) TT . For it was shown in Art. 22 that sin a = sin (TT a) . Then all angles which have the same terminal lines as a and ?r a are expressed by 2 nir + , and 2 n-rr + (TT a) or (2 n + 1) TT a. Hence the p theorem. This result may also be shown easily from Fig. 49, where a is taken as an angle in the first quadrant. Similar figures may be drawn for any value of a. Since esc a = , these are also the formulas for all sin a angles which have the same cosecant. o Y FIG. 49. 74 PLANE TRIGONOMETRY [Cn. V, 34 EXAMPLE 1. Give general expressions for , if sin a The smallest positive value of a is evidently Then the general expressions for a are 2 nir + \ TT and (2 n -f 1) TT - \ TT, (2 n + 1) TT and (2 ^ + f ) TT. These are seen to represent the series of angles or -TT, TT, |TT, J TT, TT, etc., and |TT, J TT, ---^-Tr, ---\ 5 -7r, etc. EXAMPLE 2. Give general expressions for a, if sin a =-f The smallest positive value of a is ^ ?r. Then the gen- eral expressions for a are 2 mr + I TT and (2 n + 1) TT - -J TT, or (2/i + I) TT and (2 w - -J) TT. angles which have the same cosine or secant as ct may be expressed by 2 mr a. For it was shown in Art. 22 that cos a = cos ( a). Then all angles which have the same terminal lines as a and a are expressed by 2 mr + and 2 nir a ; or in one formula, 2 mr a. Hence ~ the theorem. This result may also be shown X- FIG. 50. easily by the aid of Fig. 50. CH. V, 34] INVERSE FUNCTIONS 75 EXAMPLE. Give the general expression for , if cos a = J. The smallest value of a is evidently | TT. Then the general expression for a is 2 UTT f TT, or (2*f>r. J.ZZ angles which have the same tangent or cotangent as a may be expressed by mr -f # For it was shown in Art. 22 that tana = tan(?r+ a). All angles which have the same terminal lines as a and 7r+ are expressed by 2n7r + a and 2njr + (TT + a), or (2 M -f- 1) TT + a. But since 2 w and 2^ + 1 include all integers, these two formulas may be written as the one, TITT + a. Hence the theorem. This result may also be shown by the aid of Fig. 51, since it is evident that all angles which have the same tangent have their terminal lines in the same line through the origin. We shall evidently reach one or the other of these FlG - 51 - terminal lines when we add any multiple of TT to a. EXERCISE XX Give general expressions for x, if 1. sinx 0. 8. sec# = l. 2. cos x = 0. 9. sec x = 1. 3. tan x = 0. 10. tan x = 1. 4. sina? = l. 11. tan# = 1. 5. sin 05 = 1. 12. sin x = - 6. cos 05 = 1. 13. sin# = 7. cos x = 1. 14. cos x = 7G PLANE TRIGONOMETRY [Cn. V, 35 15. cos# = jV2. ^X~23. cos-aj=l. 16. cos x \ V2. 24. sin x = 1. 17. sin x = ^ V3. 25. tan x = 2. 4^18. eosa? = JV3. 26. cosx = -J. ^19. tan 2 a? = i. 27. sec a; = 10. ^20. cos-a = -J. 28. esc 2 a; = 2. 21. tan a = 2 -VI 29. tan 2 a? = 3. 22. secx = 2. 30. tan 2 a; = 7 4 V3. 35. Inverse trigonometric functions. Throughout all the previous work the trigonometric ratios have been con- sidered as functions of the angle ; but it is also possible to think of the angle as a function of the trigonometric ratios. For this purpose, if y sin #, it is convenient to have a short method of writing the fact that " x is an angle whose sine is y" The usual method employed in this country is x = s'm~ l y, which may be read "a; equals anti-sine?/," or "inverse-sine y"; but the student must remember that this is only an abbreviation for the longer statement "a;, is an angle whose sine is ?/." With the same meaning, we use x = cos" 1 y, x = tan" 1 y, x sec" 1 y, etc. These are called inverse trigonometric or inverse circular functions. We have seen that when an angle is given, its trigo- nometric functions are determined. For example, if y = cos a;, y has a single determinate value for every given value of x. But if a value is given to y, it has been shown in the previous article that x has an indefinite number of values. Thus, if y = -J-, x = cos" 1 \ ? = 2 nir ^ o We then may use x = cos" 1 1 and x = 2 mr as different CH. V, 35] INVERSE FUNCTIONS 77 methods of expressing the same idea. But in working with inverse trigonometric functions we shall understand that the least positive angle of the set is meant, unless the contrary is stated. This least positive value is called the principal value of the function. With this meaning we may prove the equality of such expressions as sin" 1 x and cos" 1 Vl x 2 . But these two expressions do not represent the same series of angles, as may readily be seen by giving x a numerical value. If x = , Vl - x 2 = I V3, and sin- 1 a; =30, 150, 390, 510, etc., while cos- 1 Vr^? = 30, 330, 390, 690, etc. EXAMPLE 1. Prove that sin" 1 x = cos" 1 Vl x 2 . Let y=*sin OJ. Then x = s\ny. From this we see that cos y = Vl x 2 . Hence y = cos" 1 Vl x 2 , and sin" 1 x = cos" 1 Vl x 2 . EXAMPLE 2. Prove that tan" 1 x + tan- 1 y= tan- 1 ^ 1-xy Let tan" 1 x = z and tan ~ l y = w. Then x = tan z and y = tan w. T) , N tan z + tan w x + y But tan (z + w) = = y . 1 tan z tan w \ xy Hence z + w tan" 1 ? ^ , \xy and tan" 1 x + tan" 1 y = tan" 1 ^ ^ . 78 PLANE TRIGONOMETRY [Cn. V, 35 1 r 2 EXAMPLE 3. Prove that 2 tan -1 x = cos" 1 '- Let tan" 1 x = y. Then x = tan ?/. "1 9 We wish to show that 2 y = cos" 1 ~ x - But cos 2 y = cos 2 '?/ sin 2 ?/, _ 1 tan 2 ?/ ~ 1 + tan 2 y 1 - Hence 2 / = cos" 1 ^ "I 9 and 2 tan" 1 x = cos" 1 - - EXERCISE XXI Prove the following identities for the principal values of the inverse functions. 1. sin (cos- 1 1) = 1 V3. 3. tan (2 tan" 1 ) = f . 2 . tan (sin- 1 if) = -L 2 . 4. tan ( ta.n- 1 if) = |. 5 . tan (tan- 1 J tan- 1 1) = i. 6. cos" 1 x = sin" 1 Vl x 2 . 8. 2 cos- 1 x = sin- 1 (2 a; Vl - x 2 ). 9. tan^i-f tan- 1 i = -. 10. 3 tan- 1 x = tan- 1 CH. V, 30] TRIGONOMETRIC EQUATIONS 79 11. tan(2tan- 1 a)=-^- . I -a 2 2i a 14. sin (sin- 1 a + sin- 1 6) = a Vl - 6* + 6 Vl - a 2 . 15. tan- 1 1 + tan- 1 1 + tan- 1 1 + tan- 1 f = - 36. Solution of trigonometric equations. In all the previous work, except Art. 34, the equations with which we have dealt have been true for all values of the angles, and the student has been asked to prove this fact. There is another important class of problems in which the student is given an equation which is true only for cer- tain values of the angle, and he is asked to determine the values for which it is true. In other words, he is asked to " solve the equation." The first step toward this end is usually to transform the given equation into one which contains a single trigonometric function of a single angle. This function may then be looked upon as the unknown quantity and its value may be obtained by the algebraic solution of the equation for this unknown. If the equa- tion can be reduced to either a simple or a quadratic equation, the solution may be obtained by elementary methods, and only such equations will be considered here. The following problems illustrate the method of pro- cedure in the simpler cases, where the equation contains functions of a single angle only. 80 PLANE TRIGONOMETRY [Cn. V, 36 EXAMPLE 1. Solve the equation sin x + esc x=2. Since esc x = - , the equation becomes sinx sin x -\ -- = 2. Clearing, sin 2 x 2 sin x -f- 1 = 0. Solving, sin x = 1, x = sin" 1 1, *=! All values of # are then represented by EXAMPLE 2. Solve the equation sin # = tan 2 a?. Since tana; = 5HLE, the equation becomes COS X sin x = cosa; Clearing, sin # cos 2 a; = sin 2 x, or sin x (cos 2 # sin x) = 0. Hence sin x = 0, and cos 2 # sin # = 0. The first of these equations gives at once x = 0, or TT, = 717T. CH. V, 37] TRIGONOMETRIC EQUATIONS 81 The second equation, cos 2 x sin x= 0, may be written 1 sin 2 x sin x 0. Solving as an affected quadratic in sin a?, we have sin x = ~ l * V ^= 0.61803, or -1.61803. The second of these values is evidently impossible, since sin a; cannot be numerically greater than 1. Im- possible solutions of this nature are of frequent occurrence in solving trigonometric equations. This value of sin a; satisfies the original equation, but since there is no angle whose sine is 1.61805, it does not give a possible solu- tion of that equation. From the other value of sin a;, we have x = sin" 1 .61803 = 38 10' 21". There will also be an angle in the second quadrant, 180 - 38 10' 21". The general answers are then nir, 2nir + 38 10' 21", and (2 n + I)TT - 38 10' 21". EXERCISE XXII 1 . cos 2 x = sin 2 x. 6. sec 2 a? = 4 tan x. 2. 2 cos x = V3 cot x. 7. tan 2 x sec x = 1. 3. tan x -f cot x = 2. 8. tan 2 x + esc 2 x = 3. 4. sec x + 2 cos x = 3. 9. 4 tan x cot x = 3. 5. sinic = tan#. 10. tan 2 x + cot 2 x = *-. 37. There are many equations in which it is convenient to introduce expressions containing radicals when we attempt to write the equation in terms of a single func- tion. It will then be necessary to square the equation in the solution; and in doing this an ambiguity will be 82 PLANE TRIGONOMETRY [Cn. V, 37 introduced, and the solutions of the resulting equation will contain not only the solutions of the original equa- tion, but also the solutions of the equation obtained by changing the sign before the radical. It then becomes necessary, as in any radical equation, to determine, by actual substitution, which of the results satisfy the given equation. This difficulty may often be avoided by using formulas which do not contain radicals, but it is not always convenient to do this. EXAMPLE. Solve the equation V3 sin x cos x = 1. Replacing smx by Vl cos 2 2-, V3 Vl cos?x = 1 4- cos a;. Squaring, 3 (1 cos 2 #) =1 + 2 cosx + cos 2 #. Uniting, 4 cos 2 x + 2 cos x 2 = 0. Solving, cos#=|, or 1. , 7T x = , or TT. o If cos x J, sin x | V3, and the equation is evi- dently satisfied if sin x = -f- \ A/3, but not if sin x = | A/3. The solution ~ must, therefore, be discarded. It is o easily seen that the equation is satisfied when x = IT. The general solutions are, then, (2 n + J)TT, and (2 n-\-V)ir. SECOND SOLUTION. This difficulty may be avoided in any problem which is in the form a sin x + b cos x = c by the following device : Divide the equation through by Va 2 + b 2 . Then a may always be expressed as Va 2 + V b the sine of some angle, and ~ as the cosine of CH. V, 38] TRIGONOMETRIC EQUATIONS 83 the same angle. In this problem Va 2 + 6 2 = 2. Dividing V3 . 1 1 by 2, Substituting = sin^, and -= cos^, we have L O 2 O sin sinx cos cosx = -, or cos(^-f#)= -- . 2 \3 / 2 Hence 7r, and z=(2ra-l)7r, or EXERCISE XXIII 1. sin aj cos a? = 0. **B. cot x tan x= since + cos x. 2. sinic+ cosic = 1. If fl. cos cc + tan x = sec a?. 3. sin a? cos x = V|. ^"8. esc x = 1 + cot x. 4. esc a/* cot x = V3. 9. tan a? + sec x = VS. ^5. ^ cos x J sin x = ^. 10. 5 sin a; -f- 2 cos x = 5. 38. In the previous problems only functions of x have occurred. If the equation contains functions of multiples of #, as 2#, 3#, J#, etc., these may all be replaced by their values in terms of functions of x, and the equation solved as in the previous problems. But many equations may be solved more readily by various devices, and some may be solved by these devices which, if expressed in functions of #, would give equations of the third or higher degree, which the student could not solve. There is an excellent chance for the display of the ingenuity of the student in discovering methods for shortening the work in many of these problems. A few of these are illustrated in the following examples. 84 PLANE TRIGONOMETRY [Cn. V, 38 EXAMPLE 1. Solve the equation sin 3 x + sin 2 x + sin x = 0. By formula [29, a], sin 3 x + sin a; = 2 sin 2 a; cos x. The given equation may then be written sin 2 x = 2 sin 2 a; cos #. From which sin 2 a; = 0, and cos x = J. Hence 2 z = WTT, # = (2 n f ) TT. * = T* The values of x less than 360 are seen to be 90, 120, 180, 240, 270. EXAMPLE 2. Solve the equation esc x cot x = V3. By substitution this becomes or COS X sin x sin x 1 cos x sin a; = V3, VS. This reduces, by formula [25, 5], to tan |- # = VS. Hence J # = nir 4- ? o and a;=(2 + f)7r. The only value of x less than 360 is seen to be 120 C OIL V, 38] TRIGONOMETRIC EQUATIONS 85 EXAMPLE 3. Solve the equation cos 3 x = cos 2x. From Art. 34, cos 2 x = cos (2 mr 2 x). Hence 3 x = 2 nir 2 #, and # = 2 WTT, and 5 x = 2 WTT, or a? = | HTT. The values of ^ less than 360 are seen to be 0, 72, 144, 216, 288. EXAMPLE 4. Solve the equation 2 sin x sin 3 x = 1. By formula [29, d~] this becomes cos 2 x cos 4 a; = 1. By formula [18, 5, there is one con- struction. Fig. 56 (). O o (6) (0 FIG. 56. If A is acute, and a 5; otherwise tnere will be no construction. Fig. 56 (d). Many of these results appear also in the trigonometric solution of the triangle. From the law of the sines, sin B = b sin A Since sin B cannot be greater than 1, we see at once that there will be no solution if a < b sin A. Also that there will be only one solution (.#=90) when a = bsmA. When a > b sin A, there will be apparently two values of B, one acute and the other obtuse, which are supple- mentary. But both of these apparent values are not always possible ; for, if a > 6, plane geometry tells us, that A > B, and hence B cannot be obtuse. Again, if A is obtuse, B must be acute, and there can be only one solution. Bat here there will be no solution when a 90, and a>b, there is one solution. When A > 90, and a a > 6 sin A, there two solutions. sin B = , or log sin B = log b -f log sin A log a. ci log b= 0.90309 log sin A= 9.80807 10.71116 log a = 0.77815 log sin B= 9.93301 Hence B= 58 59' 15", and ' = 121 0'45". C= 180 -(A + B) = 81 0'45", C" = 180 - (A + B 1 ) = 18 59' 15". _ a sin (7 f _ a sin (7' sin yl sin A loga= 0.77815 loga= 0.77815 log sin C = J&99464 log sin (7 ' = 9.51236 1O77279 10.29051 log sin A = _9.80807 log sin A = 9.80807 logc= 0.96472 logc'= 0.48244 c= 9.2198. c'?= 3.037. 2. a = 77.04, 6 = 91.06, 5 = 519 r 6". 3. a = 80, 6 = 401, 5 = 84 16' 31". 4. a = 319, c = 481, A = 41 32' 40 ;; . 5. a = 695, 6 = 345, B = 21 14' 25". CH. VI, 49] OBLIQUE TRIANGLES 103 6. a = 4.32, 6 = 7.61, B = 59 14'. 7. a = 704, 6 = 302, B = 25 14' 13". 8. a = 49, 6 = 45, = 17 41' 9". . a = 242, 6 = 767, E = 36 53' 2". 10. c = 1042, 6 = 55.8, = 32 22' 42". 49. CASE IV. Given the three sides. If we solve the equation a 2 = 6 2 -f- c 2 2 &G J cos ^4. for cos -4, ^ _ we nave cos A = - . 2 be The corresponding formulas for the other angles are C _ cos B = - , and cos C = , . 2 ca 2 ab When the sides of a triangle are expressed by small numbers, the angles may be found easily from these formulas, with the aid of a table of cosines. Each angle should be found in this way, and the accuracy of the result tested by adding the three angles. EXERCISE XXX 1. Find the three angles, if a = 8, 6 = 5, c = 7. SOLUTION. cos A = 25 + 49 ~ 64 = .1429. A = 81 47'. 70 cos B = 64 + 49 f ~ 25 = .7857. B = 38 13'. 11.Z 0=60. 80 Adding, A + B 4- C = 180, and the solution is accurate as far as minutes ; but since we are using a four-place table, we cannot determine the seconds. With a five-place table, the seconds may be determined with fair accuracy ; but even with 104 PLANE TRIGONOMETRY |Cn. VI, 50 a five-place table the results may often differ a number of seconds from the true value. Find the three angles, if 2. a = 13, b = 8, c = 15. 3. a = 26, b = 31, c.= 21. 4. a = 25, 6 = 26, c = 27. 5. a = 17, 6 = 20, c = 27. 50. Solution by logarithms. When the sides of the triangle are expressed by large numbers, this method is long, since the formulas are not adapted to the use of logarithms. They may, however, be changed into a dif- ferent form, to which logarithms may be applied. From formula [23], sin A = 1 cos A. Substituting in this the value of cos A obtained in Art. 49, 1- sin A = 2 be J * be O- ft + c)(a+b-c) 4 be For convenience, let a + b + c = 2 s. Then a b + 6 2(* - 5), and a -f b - c = 2(s c). Substituting these values, CH. VI, 50] OBLIQUE TRIANGLES 105 In like manner, 2J C cos A = Either of these formulas might be used for solving tri- angles, when the three sides are given, arid are well adapted to the use of logarithms. But for values of ^ A near , the first is inaccurate (since the sine of such angles changes very slowly) and the same is true of the second for very small values of |- A. It is, therefore, best to obtain the formula for tan J A, which may be used for all angles, 8 a If we let J(* -*X*- s this formula becomes [85, a] 106 PLANE TRIGONOMETRY [Cii. VI, 50 Since r is not changed by interchanging the letters, the corresponding formulas for the other angles are ^\C = ^. [35, e] These formulas are evidently well adapted to the use of logarithms, and, since the tangent varies rapidly for angles of any magnitude, they may be used in all cases. Each of the angles should be found by these formulas, and the accuracy of the work tested by adding the three values,, With a five-place table the sum should not differ by more than a few seconds from 180. EXERCISE XXXI 1. Find the angles of the triangle, if a = 15.47, b = 17.39, c = 22.88. SOLUTION. . a = 15.47 b = 17.39 c = 22.88 log (s - log (s - log (s - a) = 1.09342 b) = 1.02036 c) = 0.69810 2 s = 55.74 s = 27.87 s - a - 12.40 s - b = 10.48 sc= 4.99 logs log r- 2.81188 = 1.44514 = 1.36674 log r = 0.68337 Subtracting log (s a), log (s b), and log (s c) successively from log r, we have log tan A = 9.58995, or A = 42 30' 44", log tan i B = 9.66301, or B = 49 25' 49", log tan i C = 9.98527, or C = 88 3' 27". 180 00 00. CH. VI, 51] OBLIQUE TRIANGLES 107 Find the three angles, if 2. a = 17, b = 113, c = 120. 3. a = 3359.4, b = 4216.3, c = 4098.7. 4. d -3.9009, 6 = 2.7147, c = 3.0012. ^X 5. a = 289, 6 = 601, c = 712. 6. a = 354.4, 6 = 277.9, c = 401.3. 7. a = 5.134, 6 = 7.268, c = 9.313. 8. a = 0.099, 6 = 0.101, c = 0.158. 9. a = 33.112, 6 = 44.224, c = 55.336. 51. Area of an oblique triangle. Let K denote the area of the triangle ABC. Draw CD perpendicular to AB. G h D e D FIG. 57 (a). c B FIG. 57 (b). Then K But in either figure CD = a sin B. Hence = ~ ac sin B. [36] In like manner it may be shown that K= | ab sin C = J be sin A. Or, the area of any triangle is equal to one-half of the product of any two sides and the sine of the included angle. 108 PLANE TRIGONOMETRY [CH. VI, 51 When the three sides of a triangle are given, the area may be obtained as follows : By formula [20] sin B=Z sin J- B cos | B. Hence K ac sin | B cos ^B. Substituting in this expression the values of sin | B and cos B obtained in Art. 50, we have K= -- - . * ac K = Vs(s-a)(s- &)(s-c). [37] When any other three parts are given, find either two sides and the included angle or the three sides, and apply one of the above formulas. EXERCISE XXXII Find the area of each of the following triangles : 1. a = 5, 6 = 6, (7=78 9'. 2. a = 45.34, c = 56.45, B = 100 10'. 3. 6 = .1001, c = .3204, A = 30 33' 25". 4. a = 14, 6 = 14, c = 15. ^ 5. a =.39, b = .8, c=.89. ^6. a = 56, b = 90, c = 106. v7. a = 318, 6 = 181, ^1 = 64 58'. 8.6 = 34.51, c = 183.94, A = 23 53' 17". < 9. a = .7845, 6 = .07859, (7= 120 43' 50". 10. a = 23, A = 76 53' 25", B = 13 29' 15". CH. VI, 51] OBLIQUE TRIANGLES 109 EXERCISE XXXIII. /-I. The diagonals of a parallelogram are 73 and 95, and they cross each other at an angle of 35 28'. Find the sides and angles of the parallelogram. 2. At one point of observation the horizontal angle sub- tended by a round fort is 4 35'. On going 500 ft. directly toward the fort, it is found to subtend an angle of 6. Find the diameter of the fort. ^- 3. The parallel sides of a trapezoid are 16 and 23ft. The angles at the extremities of the longer side are 35 54' and 76 20 '. Find the non-parallel sides. 4. A tower stands at the foot of an inclined plane whose inclination to the horizon is 9; a line 100 ft. in length is measured straight up the inclined plane from the foot of the tower, and at the upper extremity of this line the tower sub- tends an angle of 54. Find the height of the tower. 5. Looking out of a window, with his eye at the height of 15 ft. above the roadway, an observer finds that the angle of elevation of the top of a telegraph pole is 17 18' 35", and the angle of depression of its foot is 8 32' 15". Find the height of the pole and its distance from the observer. 6. A and B are two points, 200 yards apart, on the bank of a river, and C is a point on the opposite bank. The angles ABC and BAC are respectively 54 30' and 65 30'. Find the breadth of the river. 7. A ship sails due east past two headlands which are two miles apart and bear in a line due south. Half an hour later one of the headlands bears 15 south of west and the other 30. What is the rate of the ship ? 8. What is the approximate distance at which a boy must hold a coin one inch in diameter from his eye to conceal the moon, if its apparent angular diameter is half a degree ? 110 PLANE TRIGONOMETRY [Cn. VI, 51 9. A balloon rises vertically at a horizontal distance of 3000 yards from an observer, who finds the angle of elevation to be 15 when he first sights the balloon. When he again measures the angle he finds it to be 30. Through what dis- tance has the balloon risen between the two observations ? 10. A pole 10 ft. high stands upright in the ground. The angle of elevation of the top of a tree from the foot of the pole is 32 27', while the angle of elevation of the top of the pole from the foot of the tree is 22 44'. Find the distance between the pole and the tree, also the height of the tree. 11. A telegraph pole stands on the bank of a stream. Its angle of elevation from a point directly opposite on the other bank is 36 53', and from a point 60 ft. from the bank in a straight line with the first point and the pole, the angle is 16 42'. Find the width of the stream and the height of the telegraph pole. 12. A church stands on the bank of a river. From the opposite side of the river the angle of elevation of the top of the spire is found to be 57 25'. The observer moves back 200 ft. in a direct line with the foot of the spire and there finds the angle of elevation to be 48 30'. Find the width of the river. 13. A man wishing to determine roughly the length of a pond finds that a line joining two stakes, driven one at each end of the pond, runs N.W. He then takes 150 paces from one of the stakes toward the N.E., turns and takes 200 paces to the other stake. What is the length of the pond, if one of the man's paces is 2-J- ft. ? What is the angle through which the man turns ? 14. The angle of elevation of a steeple is 71 34', when the observer's eye is on a level with the bottom. From a window 25 ft. above the place where the observer stands the angle of elevation is 69 27'. Find the observer's distance from the steeple and the height of the steeple. Cn. VI, 51] OBLIQUE TRIANGLES 111 15. A man at a station B at the foot of a mountain observes the elevation of the summit A to be 50. He then walks one mile toward the summit up an incline, making an angle of 30 with the horizon to a point C, and observes the angle ACB to be 150. Find the height of the mountain. 16. A tower I ft. high stands in a plane. The angles of depression from the top of the tower of two objects lying in the plane in a direct line from the foot of the tower are for the nearer a and for the more remote /3. Find the distance between the objects. 17. Two inaccessible objects P and Q lie in a horizontal plane. To find the distance PQ a base line AB of 500 yards is measured in the plane. At its extremities A and B, the follow- ing angles are measured: ZBAQ = 3612', ZQAP=5046', Z ABP = 43 22', and PBQ = 72 9'. What is the distance PQ ? 18. There is a tower on the top of a hill. From a point in the plane on which the hill stands the angle of elevation of the base of the tower is 37, and of the top of the tower 50. From another point straight away from the hill in a line through the first point and 200 ft. from that point the angle of elevation of the top of the tower is 31 22'. Find the height of the tower. . An obelisk stands on a hill whose slope is uniform. A man measured from the foot of the obelisk a distance of 32 ft. directly down the hill and found the angle between the incline and the top of the obelisk to be 45 ; after measuring down- ward an additional distance of 68 ft., the angle found in the same manner was 2147 f . What is the height of the obelisk and the inclination of the hillside ? 20. An observer in a ship sees two rocks, A and B, in the same straight line N. 25 E. He then sails northwest for 4 miles, and observes A to bear due east and B northeast of his new position. Find the distance from A to B. 21. The circular basin of a fountain subtends an angle of 25 at a distance of 44 ft. from the edge of the basin, measured on a diameter produced. Find the radius of the basin. ;^ , / V 112 PLANE TRIGONOMETRY [Cn. VI, 51 22. From the top of a vertical tower whose height is 100 ft., the angle of depression of an object is observed to be 60, and from the base to be 30; show that the vertical height of the base of the tower above the object is 50 ft. 23. A flagstaff 40 ft. tall stands on a castle wall. At a horizontal distance of 20 ft. from the foot of the wall the flagstaff subtends an angle of 15. Find the height of the wall. 24. The angle of elevation of a tower at a distance of 20 yards from its foot is three times as great as the angle of elevation 100 yards from the same point. Show that the height of the tower is ~~= ft. 25. Two parallel chords of a circle, lying on the same side of the centre, subtend respectively 72 and 144 at the centre. Show that the distance between the chords is half the radius of the circle. 26. A person standing due south of a lighthouse observes that his shadow, cast by the light at the top, is 24 ft. long ; on walking 100 yards due east, he finds his shadow to be 30 ft. Supposing him to be 6 ft. high, find the height of the light from the ground. 27. A man 5 ft. tall stands on the edge of a pond. The angle of elevation' of a tree on the opposite bank is 45 and the angle of depression of its reflection is 60. Find the height of the tree. NOTE. The reflection of the top of the tree appears as far below the surface of the water as the top of the tree is above the water. 28. One end of a pole rests on the ground and the other end touches the top of a window. When the lower end of the pole is moved away 16 ft. farther from the wall, the top rests on the sill of the window. If the first angle the pole makes with the ground is 71 25' and the second 48 35', find the length of the window. CH. VI, 51] OBLIQUE TRIANGLES 113 29. Two captive balloons are floating at equal heights in calm air. A man standing in the straight line between their points of attachment finds the angle of elevation of the nearer balloon to be tan" 1 1. He then walks a distance of 240 ft. at right angles to the straight line joining the points of attach- ment and finds the angle of elevation of the same balloon to be tan" 1 -|, and that of the other tan" 1 -/-$. Find the height of the balloons and the distance between their centres. 30. Wishing to find the inclination of a roadway rising from a level park, a man walked 100 ft. up the incline and observed the angle of depression of an object in the park to be 30. After walking up the plane 100 ft. farther, the angle of depression of the same object was 45. Show that the angle of inclination is cot- 1 (2 - 31. Two persons stand facing each other on opposite sides of a pool. They are of such heights that the eye of one is 5 ft. above the ground and that of the other 6 ft. When the line of vision of each makes an angle of 54 with vertical, the reflection of the eye of either is visible to the eye of the other. What is the width of the pool ? 32. Viewed from the S.E. corner of a room the N.E. corner has an elevation of and the S.W. corner of <. Find the elevation of the N.W. corner, also the angles the diagonals make with the edges of the room. 33. The angle of elevation of a tower from a point A due south is , and from a point B due west of the first station it is ft. If the distance, AB, between the two stations is b, show that the height of the tower is 6 sin a sin V sin (a + ft) sin (a - ft) FlG . 58 . 114 PLANE TRIGONOMETRY [Cn. VI, 51 34. A man walked up an inclined plane a feet and observed the angle of depression of an object in a horizontal plane to be a. When he had walked a distance of 2 a ft. further up the incline the angle of depression was 2 a. Show that the angle the incline makes with the horizontal plane is 1 4 sin- a 35. A man walks along the bank, AB, of a straight stream and at A observes the greatest angle subtended by two objects, P and Q, on the opposite side to be 75 ; he then walks a dis- "tance of 300 ft. to B and finds that the objects are in a straight line, which makes an angle of 15 with the bank. Find the distance between the objects. NOTE. The point on the bank where the greatest angle, subtended by the objects, is made, will be the point of tangency of a circle pass- ing through them. Z.BPA = Z.QAB = can be expressed in terms of 15, 75, and 90. 36. A person on the top of a tower observes the angles of depression of two objects in the plane on which the tower stands to be 60 and 30. He knows the distance between the two objects to be 500 ft. The angle subtended at his eye by ,the line joining the two objects is 30. Find the height of the tower. 37. From two stations a ft. apart a balloon is observed. At one of the stations the horizontal angle between the balloon and the other station is y, and the angle of elevation of the balloon is a; at the other station the corresponding angles are 8 and ft. Show that the height of the balloon is a sin y tan (3 sin (8 + y) Also show that tan a sin 8 = tan ft sin y. CH. VI, 51] OBLIQUE TRIANGLES 115 38. An inclined plane AB, of length I, has a vertical rod, PQ, fastened to its surface at such a distance from its foot that the upper end of the rod and the top of the plane are in the same straight line with a point D at a distance I from the foot of the plane. The angles subtended by the rod and < the part of the plane below are each equal to a. Find the distance at the foot of the rod from the lower end of the plane, also the length of the rod. A FIG. 60. PART II SPHERICAL TRiaONOMETRY CHAPTER VII RIGHT AND QUADRANTAL TRIANGLES 52. A spherical triangle is a portion of the surface of a sphere bounded by the arcs of three great circles which intersect. Spherical Trigonometry is concerned with the relations of the sides and angles of a spherical triangle, and the computation of the unknown parts when any three parts are given. In the following treatment a knowledge of Solid Geometry is presupposed, but it is thought best to begin with a statement of the definitions and theorems which are most important for our subject. Let ABO be a spherical triangle on a sphere whose centre is at 0. Join to the vertices A, B, and (7, and pass planes through and the sides of the triangle. These sides are measured in degrees, and their measures are, therefore, equal to the measures of the correspond- ing plane angles A OB, BOO, 00 A. 117 118 SPHERICAL TRIGONOMETRY [Cn. VII, 52 A spherical angle is equal to the angle between tan- gents to the sides of the angle, drawn at the vertex. Hence the angle A is equal to the angle formed by draw- ing, in the faces A OB and AOC, any two lines perpen- dicular to OA at the same point. A spherical angle may also be measured by the arc of a great circle having the vertex of the angle as a pole and intercepted between its sides. We shall restrict our study of triangles to those in which each of the angles and sides is less than 180. The sum of the three sides may have any value less than 360, while the sum of the angles must lie between 180 and 540. A triangle may contain one, two, or three right angles, or each of the angles may be greater than 90. If each of the angles is a right angle, each of the sides is a quad- rant, and it is called a tri-rectangular triangle. Polar triangles are so related that the vertices of each are the poles of the corresponding sides of the other. Each angle of either triangle is the supplement of the side lying opposite it in its polar triangle. When one or more of the sides of a spherical triangle are quadrants, it is called a quadrantal triangle. EXERCISE XXXIV 1. Prove that, if a triangle has three right angles, it is its own polar. 2. Prove that the polar of a right triangle is a quadrantal triangle. 3. Prove that, if a triangle has two right angles, the sides of the polar triangle opposite these are quadrants, and that the third side measures the third angle of the given triangle. CH.VII,53] RIGHT AND QUADEANTAL TRIANGLES 119 4. Prove that, if one side of a triangle is a quadrant, either of the other sides and the angle opposite it are either both less or both greater than 90. 5. Prove that, if a triangle has one right angle, each of its remaining angles is in the same quadrant as the side oppo- site it. 6. Prove that, if the sides about the right angle of a right spherical triangle are in the same quadrant, the hypotenuse is less than 90; while if they are in different quadrants the hypotenuse is greater than 90. 7. Prove that, if one of the sides of a right triangle is equal to the opposite angle, the remaining parts are each equal to 90. 8. The angles of a triangle are 80, 75, and 105. Find the number of degrees in each side of the polar triangle, and, if the radius of the sphere be 90 ft., compute their lengths in feet. 9. The sides of a spherical triangle are 70, 80, and 110. Find the angles of its polar triangle. 10. In an equiangular spherical triangle each of the angles is 120. Find the value of each side of the polar triangle. If the angles increase to 180 each, state the limits of the tri- angle and its polar. 53. Right triangles. Let the spherical triangle ABO (Fig. 2) be right-angled at (7, and let each of the other parts be less than 90. Pass planes through the sides of the triangle and 0, the centre of the sphere. Repre- sent the measure of the sides opposite A, B, and by the corresponding small letters a, 5, c. Then these are also the measures of the corresponding angles at 0. That is, Z.BOO=a, ZCOA = b, ZAOB=c. Through B pass a plane BED perpendicular to OA. Then EB and ED are perpendicular to OA, and the 120 SPHERICAL TRIGONOMETRY [Cn. VII, 53 angle BED is the measure of the spherical angle A. Also BD is perpendicular to the plane AOC, since it is the line of intersection of two planes which are perpendicu- lar to that plane. Then the triangles BOE, BOD, DOE, and BDE are all right triangles. FIG. 2. Then COS G = OE OB Also OE = OD cos b = OB cos a cos b. Hence Again Hence cos c = cos a cos b. 1 a EB OB sin c sin c sin a = sine sin A. Interchanging a and b, A and J5, sin 6 = sin c sin J5. Again cos J. tan 0^ tan c tan or cos A = tan 6 cot c. Also cos .B = tan a cot c. Formula [4] may be written sin b cos c cos A = But sin cos 5 sin c = sin .#, and [2] [3] [5] cos 6 = cos a. CH.VII,53] RIGHT AND QUADRANTAL TRIANGLES 121 Hence cos A = cos a sin B. [6] Also cos B = cos b sin A. [7] Substituting the values of cos a and cos b obtained from these last equations in formula [1], we have cos c = cot A cot B. [8] A . . , ED DBcotA cc& A Again sin b = = = OD DB cot a cot a Hence sin b = tan a cot A. [9] Also sin a = tan b cot B. [10] In deriving these formulas we have used a triangle in which none of the parts is greater than 90 ; but they may be easily shown to hold for any right triangle. Let ABO (Fig. 3) be a right tri- A. angle in which a < 90 and I > 90. Then by Problem 6, Exercise xxxiv, c> 90. Con- tinue the sides AB and A O until they meet at A f . A' BO is a right triangle in which each of the five parts is less than 90. Since each of the arcs ABA' and AC A' is a semicir- cumference, A'B = 180 - sin a. log sin a = 9.56695 log tan a = 9.59872 log s'mA = 9.82693 log cot 4 = 10.04298 log sin c = 9.74002 log sin b = 9.64170 c = 33 20' 15", b = 25 59' 28", or = 146 39' 45". or = 154 0' 32". log cos^4= 9.86991 log cos a = 9.96823 log sin 5= 9.90168 S = 52 53', or = 127 1'. These values must be combined according to the laws stated in Examples 5 and 6 in Exercise xxxiv. In this problem all the values less than 90 form one solution and those greater than 90, the other. This will be true only in case the given parts are less than 90. 126 SPHERICAL TRIGONOMETRY [Cti. VII, 56 EXERCISE XXXV Find the remaining parts of the right spherical triangle in wnich 1. a = 9 45' 19", b = 12 16' 42". 2. a = 28 26' 56", b = 29 37' 36". 3. c = 41 5' 6", A = 41 32' 38". 4. a = 12 16' 42", B= 79 29 '45". 5. A = 15 38' 6", B = 80 14' 41." 6. b = 48 27' 22", c = 56 15' 43". 7. B = 56 15' 43", c = 58 40' 13". 8. A = 47 37' 21", B = 61 33' 4". 9. a = 48 27' 22", c = 64 9' 43". 10. b = 74 21' 54", A = 38 57' 12". 11. a = 35, ^4 = 61. 12. b = 75 45', B = 65 38'. 13. b = 105 30', c = 80 25'. 14. b = 98 35', A = 47 38'. 15. b = 79 35', B = 80 25' 20". 56. Quadrantal triangles. If one side of a spherical triangle is a quadrant, the angle opposite that side in the polar triangle is a right angle. From the two given parts of a quadraiital triangle two parts of its polar triangle may be obtained. This right triangle may then be solved and, from these solutions, the unknown parts of the quadraiital triangle may be obtained. On. VII, 56] RIGHT AND QUADRANTAL TRIANGLES 127 EXERCISE XXXVI Find the remaining parts of the spherical triangle in which c = 90 and 1. C= 163 53' 38", A = 169 29' 45". 2. C = 141 2' 48", B = 142 5' 54". 3. B = 140 2' 56", a = 163 53' 38". 4. A = 148 40' 13", b = 127 54' 6". 5. a = 138 54' 54", b = 100 30' 15". 6. a = 65, A = 48 35'. 7. B = 50 38' 20", b = 75 37' 30". 8. C = 70 30' 28", b = 128 35' 12". CHAPTER VIII OBLIQUE TRIANGLES 57. Law of the sines. In the oblique spherical triangle ABC draw the arc CD perpendicular to AB, forming the two right spherical triangles ADC and CDB. Then by formula [2], sin DC = sin a sin B, and sin DC = sin b sin A. Hence sin a sin B = sin b sin A, sin a sin A FIG. 7. or In like manner and sin b sin />' sin & _ sin B t sin c sin C sin c_sin C sin a sin A In this proof we have used a figure in which D falls between A and B\ but all of the statements will be seen to be true for Fig. 8, in which D falls without AB, if we note that, in this case, sin DBC = sin (TT J5) = sin B. 128 CH. VIII, 58] OBLIQUE TRIANGLES 129 58. Law of the cosines. Let ABC be a spherical tri- angle in which two of the sides, b and ' sin c' cos ^4/, cos b f = cos r cos a r + sin ] cos C = - cos A cos B + sin ^1 sin B cos c. [13, c] 59. Formulas [12] and [13] are unsuited to logarith- mic computation ; but by a transformation very similar to that used on page 104 of the Plane Trigonometry, we may obtain from them formulas which are well adapted to the use of logarithms. From [12, a], cos a cos b cos c cos A = sin b sin c CH. VIII, 59] OBLIQUE TRIANGLES 131 But sin J A = V" , [231, Pt. 1, sin o sin c cos a -}- cos o cos c 2 sin 5 sin c foogft-.Q-ooaa fe ^ pt ^ V ci n n em / Z Sill Sill g = /sinj_(a_ sin 6 sin 'i(''-0i(-a)''i"(-'0 sin s tan H = ^(~^> [H] tan l*= s -nr^)> t 14 '*] tan ; c =^cf^)- C 14 '<1 Let the student obtain in a similar manner the following formulas from formula [13] : tan | a = K cos (8 - A) 9 [15, a] tan|& = JKcos(S--B), [15,5] tan I c = K cos (S - C), [15, c] in which 2 $ = A + B + C 7 , and - cos 8 cos (tf - A) cos (8 - B) cos (# - (7) 60. Napier^s analogies. Dividing [14, a] by [14, 5], we have sin ^ A cos j B _ sin (s 5) cos JJ. sin ! .B sin (s a) By composition and division, sin j- ^4. cos ^B -\- cos j J. sin \ B _ sin (s b) + sin (s a) sin J A cos J -S cos J A sin J j5 sin (s 6) sin (s ) Applying [11], [13], [29, a], [29,5], Pt. I, we have [16] tanc tan^(a-ft) CH. VIII, 60] OBLIQUE TRIANGLES 133 Multiplying [14, a] by [14, 5], we have sin I A sin \ B _ _ lfi_ _ _ sin (s c) ^ cos J A cos ^ B sin ( a) sin (s 6) sin s By inversion ; then by division and composition, cos | A cos j- B sin j- J. sin -j J? _ sin s sin (s c) cos J ^4. cos J -B + sin J ^4. sin ^ 5 sin s -f- sin (s c) Applying [12], [14], [29, 6], [29, a], Pt. I, we have cos|(^ + B)^ ton|c Let the student obtain in a similar manner the follow- ing formulas from [15, a] and [15, 6] : sin (a -b) cos J (a + 6) cot|c [18] [19] Let the student also obtain from each of the above four formulas two others by interchanging the letters. EXERCISE XXXVII 1. Show that J (a + &) and |- (A + -B) may have any value less than 180 ; but that each -of the other angles used in Napier's analogies must be less than 90. 2. Show by the aid of formulas [17] and [19] that A + B is less than, equal to, or greater than 180, according as a -f- b is less than, equal to, or .greater than 180; and the converse. 3. Prove that in any spherical triangle each angle is greater than the difference between 180 and the sum of the other two angles. 134 SPHERICAL TRIGONOMETRY [On. VIII, 61 4. Deduce from Napier's Rules for the right-angled spheri- cal triangle the relation sin a _ sin A sin b sin B 5. Show what changes occur in formulas [11, a, &, c], (1) when C = 90, (2) when c = 90, (3) when B=C= 90. 6. If a perpendicular be dropped from vertex Oof an oblique spherical triangle upon the opposite side AB, to meet it at X, prove by Napier's Rules that, disregarding signs, sin AX _ cot A sin BX~ cotB' 7. By means of Napier's Rules, derive formulas for finding in an oblique spherical triangle the parts required in the follow- ing cases : (a) Given A, C, a, required b. (b) Given B, C, c, required b. 8. A certain point X on a sphere is joined to three points P, Q, R on an arc of a great circle. By application of the law of cosines to the triangles formed and by reduction, deduce the formula, sin PQ cos RX+ sin QR cos PX sin PR cos QX= 0. 61. Solution of oblique spherical triangles. When any three parts of a spherical triangle are given, the remaining parts may be determined by the aid of the seven formulas, [11] and [14] to [19]. Six cases may occur. There may be given 1. the three sides, 2. the three angles, 3. two sides and the included angle, 4. two angles and the included side, 5. two sides and the angle opposite one of them, 6. two angles and the side opposite one of them. CH. VIII, 62] OBLIQUE TRIANGLES 135 There will be one determinate solution in each of the first four cases, but in 5 and 6 there may be two solutions. 62. CASE I. Given the three sides cr, 6, and c. The angles may be determined by formula [14]. tan \A = , tan 4 B = , 01 " fat a) sm (s b) tan 1 = - sin (s k sn EXAMPLE 1. Find the three angles of the spherical triangle in which a = 16 6' 22", b = 52 5' 54", and c = 61 33' 4". s = 64 52' 40", log sin ()= 9.87627 log sin (s - b) = 9.34478 log sin (- tan 4 ( O A) = -; f-7 -- ^ cot J- .B, sin I (c* + a) and cos c + <) i- ( c _ ) = 19 40 r 12", and J (c + a) = 59 49' 33". log sin J (= 9.52712 log cos J ( sin a, there is no solution. But if sin b sin A < sin a, there are sometimes two solutions. After the two values of B have been obtained, the number of solutions may be determined from the fact that the greater side is opposite the greater angle. It will also be necessary to see that the theorem of Problem 2, Exer- cise xxxvii, is satisfied. The remaining parts, c and C, may now be found from formulas [16] and [18], or from [17] and [19]. 140 SPHERICAL TRIGONOMETRY [Cn. VIII, 06 EXAMPLE 1. Find the remaining parts of tlie spherical triangle in which a = 103 10', b = 120 12', B = 131 40'. . ., sin a sin B sin A = - -- sin o log sin a = 9.98843 log sin B = 9.87334 19.86177 log sin b = 9.93665 log sin -4= 9.92512 A = 57 18' 45", or 122 41' 15". Both of these values of A will be seen to satisfy the conditions stated above. There are, therefore, two solu- tions. Using the first value of A, we shall proceed to find the corresponding values of c and 0. J (a + 5) =111 41', (5-a) = 8 31', |- (A + B)= 94 29' 22", \ (B-A) = 37 10' 37". Since -| (A + B) is near 90, more accurate results will be obtained by using formulas [17] and [19], which contain the cosine. cot l (7= cos + a tan \ (B + X). cos \(p a) log cos %(B + A)= 8.89363 log cos J (b + a) = 9.56759 log tan J (5 4- a) = 10.40054 log tan | (B + A}= 11.10504 19^29417 20.67263 log cos $(B-A)= 9.90133 log cos J (5 -a) = 9.99518 log tan l c = 9.39284 log cot -J- = 10.67745 c = 27 45' 26". C= 23 44' 14". CH. VIII, 67] OBLIQUE TRIANGLES 141 In each of the above formulas two factors in the second member are negative. The first members are, therefore, positive, and the acute values of | c and ^ O must be chosen. Using the second value of A we find, by the aid of the same formulas, *= 113 28' 14", 0= 127 32' 56". EXERCISE XLII Find the remaining parts of the spherical triangle in which 1. a = 16 6' 22", c = 525'54", A = 15 38' 7". 2. Z> = 38 57' 12", c= 56 15' 43", = 47 37' 21". 3. 6 = 50 2' 56", c = 56 52' 23", = 64 9' 43". 4.6 = 28 35' 30", c = 30 28' 15", B = 85 38' 40". 5. S = 869', a = 72 18' 15", 6 = 71 54' 15". 6. J. = 12035'28", b = 98 48' 24", a = 105 30' 30". 7. ^ = 103 28' 12", 6 = 20 25' 35", a = 28 58' 25". Show that the following triangle is impossible ; also find a value for c such that B shall be equal to 90. 8. 0=98 35' 28", b = 70 35' 24", c = 50 28' 22". 67. CASE VI. Given two angles and the side opposite one of them. If J., 5, and a are given, b may be found from formula fill. . n - . 7 sin U sin a sin = : sin JL The number of solutions may be determined as in Case V. The remaining parts c and O may now be found as in Case 5, by formulas [16] and [18], or by [17] and [19 j. The logarithmic work is very similar to that of Case V. 142 SPHERICAL TRIGONOMETRY [Cn. VIII, 67 EXERCISE XLIII Find the remaining parts of the spherical triangle in which 1. b = 56 15' 43", = 38 57' 12", C= 138 54' 54". 2. b = 48 20', A = 76 50', B = 59 48'. 3. A = 70 30' 28", a = 45 28' 32", B = 60 20' 32'-'. 4. .4 = 78 47' 20", a = 63 49' 10", C = 80 25' 30". 5. c = 112 49' 24", (7= 152 49' 27.5", A = 29 42' 13.7". 6. C=S 48' 48", c = 85 26' 45", 5 = 23 49' 15". 7. A = 57 48' 23", B = 120 38' 27", a = 48 25' 20". 8. J. = 7028', a = 80 25' 40", C=12528'. REVIEW EXERCISE 1. If a median be drawn from the vertex C of a spherical triangle to the opposite side c, and the parts of the angle adjacent to sides a and b of the triangle be named a and ft respectively, show that 5^ = sin ft sin a 2. The city of Quito is situated nearly on the equator and its longitude is 78 50' west of Greenwich. The latitude of Greenwich is 51 28'. Find the distance from Greenwich to Quito (on the arc of a great circle). Assume the radius of the earth to be 4000 miles. 3. In a spherical triangle whose sides are 48,* 57, and 65, respectively, a median is drawn to the side whose length is 48 from the opposite vertex. Find the length of the median and also the parts into which it divides the angle. 4. If the values of two sides of a spherical triangle are and ft and the angle between them n, find the length of the perpendicular upon the third side from the vertex of the angle n. CH. VIII, 67] OBLIQUE TRIANGLES 143 5. If a lune whose angle is a be drawn upon a sphere whose radius is c feet, and an arc of a great circle be drawn to intersect at equal angles the sides of the lune, making the part of the arc so included /?, find the distance from each vertex of the lune to the transverse arc in feet. 6. A ship starting from a point on the equator in longitude 130 West sailed for 3 days, arriving at a point whose latitude is 20 North and longitude 150 West. What was its rate per hour, allowing the radius of the earth to be 4000 miles ? 7. The sides of a spherical triangle enclosing an angle of 75 are respectively 60 and 54. Find the length of the bisector of the angle and the angles it makes with the base. 8. There is a regular tetrahedron each of whose face angles is 60. Find the angle between any two faces. NOTE. Suppose one vertex of the tetrahedron to be at the centre of a sphere whose radius is an edge of the tetrahedron. The other three vertices of tho solid will determine upon the surface of the sphere the vertices of a spherical triangle whose sides are measured by the face angles of the tetrahedron. 9. If the face angle at the vertex of a regular four-sided pyramid is 50, find the angle between any two lateral faces. 10. Find the area of a spherical triangle whose sides are 45 26', 53 44', 68 46', respectively, on a sphere whose radius is 10 ft. N OTE ._ The formula for the area of a spherical triangle is: area = """ ^ , where E denotes the excess in degrees of the sum of -LoU the angles of the triangle over 180. This excess may be found when the three sides of the triangle are given by THuilier's Formula, tan I E = Vtan \ s tan \ (s - a) tan \ (.s - b) tan (* - c )> in which a, &, and c denote the sides of the triangle, and s, as usual, the half sum of the sides. 144 SPHERICAL TRIGONOMETRY [Cn. VIII, 67 11. In a sphere of radius 12 is a spherical pyramid whose base is a spherical triangle of which the sides are 85, 65, and 120. The vertex of the pyramid being at the centre of the sphere, find its volume. 12. If a line makes an angle with its projection on a plane passing through one end of the line and if the projec- tion makes an angle < with a second line drawn in a plane which interse'cts the first plane in the line of the projection at an angle of 30, show that the angle between the first line and the second is cos" 1 \ (2 cos cos < sin sin ). 13. A flight of stone steps faces due south. A rod rests with one end on a step and leans against the edge of the step above, in a plane perpendicular to the steps. At noon the horizontal part of the shadow is marked on the step and also the vertical part. If the rod makes an angle with the step upon which its foot rests, show that t hours after noon the angle the horizontal part of the shadow makes with its posi- tion at noon may be determined by the equation tan x = sin 6 tan t, and the angle the vertical part of the shadow makes with its position at noon, by the equation tan?/ = cos tan t. NOTE. This example illustrates the principle of both the hori- zontal and the vertical sun dial. 6 represents the latitude of the place. Let the lower end of the rod be the centre of a sphere whose surface is pierced by the rod and its two horizontal shadows in three points which are the vertices of a spherical right triangle. By means of this triangle the first relation may be proved. It should be remembered that each hour of time corresponds to 15. 14. If the longitude of New York is 40 43', find the angles which the shadows the sun would cast at three o'clock P.M. upon a dial constructed according to the principle of Ex. 13, would make with the shadows cast at noon. ANSWERS ANSWERS Exercise I, page 4 1. (a) sin A = ^, cos A = if, tan A = T \, sec A = Jjj, esc J. = J/, cot .4 = -^, vers .4 = T ^, covers A = T 8 ^. (6) sin -4 = |, cos A = f , tan ^4 f , sec A = , esc J. = |, cot A = f . (c) sin ^4 = T 8 r , cos A = Jf , tan ^4 = T %, sec J. = J, esc ^4 y , cot ^4 = - 1 /. (d) sin -4 = |, cos ^4 = | V5, tan ^1 =i f V5, sec ^1 = f V5, esc ^1 = f , cot A= JV6. 7. c = 16. 8. ^1 = 20. 9. ^1 = 40, a = 5.0346, c = 7.83. 10. A = 20, ^ = 70, c = 266. 11. B = 20, c = 29.24, a = 27.475. 12. .4 = 70, a = 18.794, b = 6.84. Exercise II, page 6 1. cos 20, sin 5, esc 27, cot 33 12', tan 4'. 2. x = 45. 3. x = 30. 4. x = 15. Exercise III, page 8 1. J. 2. 5. 3. f. 4. 9. 5. 1. 6. \. Exercise IV, page 12 1. cos A = I \/7, tan .4 = f V7, sec J. = f V7, esc ^4 = f , cot ^1 = \/7. 2. sin ^ = | V6, tan ^1 = 2 \/6, sec A = 5, esc ^1 = & V6, cot .4 = & V6. 3. sin -4 = ^yVlO, cos ^4 = 3^ VlO, sec^l=VlO, csc^l = ^VlO, cot -4 = |. 4. sin J_= ^V A/37, cos A = & A/37, tan -4 = |, sec JL = J V3T, esc A = V3T. 5. See Art. 4. 147 148 6. cot A 7. 8. 9. cot A ANSWERS sin A = r V, cos A = & VTT, tan A = -fa VTT, sec A = VTT, cos A = j\, tan A = *, sec A = - 1 /, esc -4 = |f, cot .4. = sin A = $, tan .4 = f , sec A f , esc ^4 = , cot ^4 = f . sin 4 = , cos A = -, tan .4 = v a 2 1, esc A = 14. 15. (a) 00 fa*\ cos .4 tan A cot A sec .4 esc A sin J. cos -4 cot A sec .4 sin 7 .4 = Vl - sin 2 A, sin .4 (6) sin^l = Vl -cos 2 .4, Vl - cos 2 A Vl - sin 2 .4 cos A cos A Vl sin 2 .4 _ 1 Vl - cos 2 A 1 esc A 1 1 sin A tan A (cZ) sin A = cos A Vl - cos 2 A Vsec 2 A 1 Vl + tan 2 A 1 sec .4 1 Vl 4- tan 2 A 1 tan^l = cot A = esc A = 4- sin 2 .4-1 sec A Vsec 2 A 1, 1 Vsec 2 A - 1 sec A tan A = Vl + tan 2 .4, Vl 4- tan 2 A tan A -2sin 5 ,44-2sm3.4 Vsec 2 A- I sin 2 ^1- sin 4 1 - sin 2 A - sin 3 A 1 - sin 2 A 00 16. (a) cos^lVl-cos 2 A (6) 00 cos 1 sin 2 A 4- sin 4 A 1 -sin 2 A ,, 1 - 3 cos 2 A 4 -> cos 4 A cos 4 ^4 - cos 2 A 1 l-tan^l + tnn'M Vl 4- tan 2 A 1 4- tan J. 4 tan- A (tan A - 1) Vl + tan 2 .4 tan .4 ANSWERS 149 Exercise VI, page 18 3. .4=36 52', 7?=53 8', c=5. 4. ^1=12 41', 5=77 19', c=41. 5. ,4=67 23', J?=2237', 6 = 5. 7. -4=81 12', S= 8 48', a=84. 8. #=5518', 6 = 17.83, c = 21.08. 9. ^1=66 26', a = 13.75, 6 = 5.997. 6. .4=79 37', 5=10 23', 6 = 11. 10. 5=76 8', a=4.197, c=17.51. Exercise VII, page 19 3. ^ = 39 54' 28", c = 8850.6. 4. A = 41 48' 35", 6 = 2484.3. 5. .4 = 56 26' 27", 6 = 0.3015. 6. 6 = 10322, c = 11287. 7. a =0.778, 6 = 0.4036. 8. 6 = 454.43, c = 499. 9. a = 2.005, 6 = 1.287. 10. A = 53 15' 6", c = 2194. 11. ^1 = 2 27' 52", c = 13.48. 12. a = 1760.5, c = 1762.2. 13. ^1 = 53 7' 48", a = 11.2. 14. ^1 = 78 20' 39", c = 811.74. 15. a = 518.61, 6 = 161.95. 16. 6 =24.187, c = 24.23. 17. A = 43 44' 51", 6 = 0.00679. 18. a = 761.17, 6 = 76.42. 19. a = 965.93, 6 = 258.82. 20. .4 = 71 38', 6 = 0.334. 1. 31 45' 33". 2. 67.4ft. 3. 36 52' 12". 4. 166.43ft. 5. 23.3ft. 6. 202.2ft. Exercise VIII, page 20 7. 34 54' 54", 13. 40.98ft. 72 32' 33", 72 32' 33". 14. 140.88 ft. 8. 61.6ft. 15. 44 25' 37". 9. 838.8 ft. 16. 9.81 ft. 10. 660yd. 19. 169.3ft. 11. 3 13' 29". 20. 769.8ft. 12. 75ft. . Exercise IX, page 29 1. ^TT, !-, 2^ | r> n^ !.. 2 . 36, 20, 120, 150, 900. 3. J*yf7r, 15 TT. 4. 2. Exercise XII, page 46 1. cos x = f V6~, tan cc = T V VG~> sec x = r 5 2 V6, esc x = 5, 2. sin a; = - f V2, tan x = - 2 \/2, sec x = 3, cscx = - f\/2", cotse =- JV2. 3. sinx = -j\VTo, cosx = fjyVIO, secx=VlO, cot x = - . 150 ANSWERS 4. sinx = iVl5, coso: = |, tana; = Vl5, cscx = j^VlS, cotx = ^VTs. 5. sin^l = ^ cos^ = x/x2 ~ y2 , tan^ = ^ , csc^ = ?, x Vx 2 - v a ^ Exercise XIV, page 55 1. |V3, f; -0.868,4.924; -4.698, -1.71. 2. 5, - 53 8'; or - 5, 126 52'. . 3. 197 27' 28", or - 17 27' 28". 4. If the triangle is described in the positive direction of rotation, the angles are 120, - 120, - 120, 5 ; - 5 ; - 5. 5. 5V3; -5V3; 0; 7 ; - 7J. Exercise XV, page 63 2. i(V6->/2). 3. 2>V3._ 4. -*-( V6 - V2), i(V6 + V2), 2- V~3. 5. i V3 + i V2 ; - J- Vl - |V2. 8. cos a cos j3 cos 7 + cos a sin /3 sin 7 sin a sin /8 cos 7+ sin a cos j8 sin 7. g tan ct tan ft + tan 7 + tan a tan ft tan 7 1 + tan tan /3 tan a tan 7 + tan /3 tan 7 Exercise XVI, page 65 1. iV3. 2. -f. 3. 3.43. 5. cos3<* = 4cos3<*-3cos a. tan 3 g = 3 tan " ~ . 1-3 tan 2 6. sin 4 ct = 8 sin ex, cos 3 4 sin a cos a. cos 4 a = 8 cos 4 a - 8 cos 2 a + 1. tan4<*= 4tan-4tantt m 1-6 tan 2 a + tan 4 a 7. sin 5 a = 5 sin a 20 sin 3 a + 16 sin 5 , cos 5 a = 5 cos a 20 cos 3 a + 16 cos 5 a. Exercise XVII, page 68 1. 0.316, 0.9487 ; 0.78. 2. sin2230'=^v2_V2, cos 22 30' =2+ V2, tan2230 f = V2-1. 4. sin 165 =J(v / 0-v / 2), cos 165= -J(V2+ V6), tan 165= V3-2. ANSWERS 151 Exercise XX, page 75 1. mr. 2. O-fi)7r. 3. nir. 4. (2n + ^)ir. 5. (2rc-|)7r. 6. 2 nir. 7. (2n+l)7r. 8. 2 n?r. 9. (2n+l)ir. 10. O + ^)TT. 11. (n + f)ir. 12. (2n-i)ir, (2n-f)ir. 13. (2n + )ir, (2 n + f )TT. 14. (2ni>. 15. (2nJ>. 16. (2n|)r. Exercise XXII, page 81 1. 45, 135, 225, 315 ; mr -. 4 2. 60, 90, 120, 270; 2 nir -, 2 nir +, (2n + l)7r- ; 2 o 3 3. 45, 225 ; mr + -. 4. 0, 60, 300 ; 2 nir, 2 mr 5. 0, 180; WTT. o 6. 15, 75, 195, 255; mr + -^ mr + ~ 7. 60, 180, 300 ; 2 mr -, (2 n + 1) TT. o 8. 45, 135, 225, 315 ; mr -. 4 9. 45, 165 58', 225, 345 58' ; mr + -, mr + tan-i(- ). 10. 30, 60, 120, 150, 210, 240, 300, 330; mr -, nir -. 6 3 Exercise XXIII, page 83 1. 45, 225 ; mr + - 3. 285, 345 ; 2 mr - , 2 WTT - . 2. 0, 90 ; 2 WTT, 2 WTT + -. 4. 120 ; 2 WTT -f . 2 3 5. 24 27', 261 49' ; n 360 - 36 52' 61 19'. 6. 27 58', 135, 242 2', 315; mr + , Jsin- 1 (2V2 - 2). 7. 0, 90, 180; nir, 2wir+. 8. 0, 90; 2 nir, 2 nir +2:. 9. 30, 270 ; 2 WTT + -, 2 WTT - - 6 2 10. 46 24', 90 ; 2 mr + -, n 360 + 46 24'. ' 2 152 ANSWERS Exercise XXIV, page 86 1. 90, 270; nir+f- 2. 51 19', 180, 308 41'; (2 n + !)JT, cos-if. 3. 0, 180; mr. 4. 0, 60, 120, 180, 240, 300; mr, (n + |)TT - 6 5. 0, 7, 37i, 971, 127|, 180, etc.; mr, 6. 30, 60, 90, 120, 150, 210, 240, 270, 300, 330; 7. 45, 60, 120, 135, 225, 240, 300, 315; mr-, mr -. 8. 0, 180; mr. 9. 18, 162, 234, 306; sin-i ~ 1 V5 . 10. 18, 54, 90, 126, 162, etc. ; 11. 45, 90, 135, 225, 270, 315 ; mr -, mr + -. 4 2 12. 22 6', 67 54' ; sin-i | (5 - Vl3). 13. 0, 65 4', 252 45' ; 2 mr, 14. 45, 67|, 90, 1571, 225, 247|, 270, 337|; mr + 1, W7T 37T T + ~F' 15. 22 30', 112 30' ; nir + - . 16. 60, 90, 120, 240, 270, 300 ; mr , WTT + ^. 3 2 Exercise XXV, page 87 1. f. 5. \/6. 7. -2 -V3. 9. i. 11. f. 13. 14. 0, 45, 180, 225 ; nir, mr + ^. 18. 22|, 112 J, 202|, 292^; \ I mr +-Y 23. (1) (np + n + p - 1) (np - n - p - 1) sec A sec 5 ; (2) esc A esc B csc 2 .B-l- 27. 18, 54, 126, 162, 198, 234, 306, 342; sin-i (V5 1). ANSWERS 153 Exercise XXVI, page 96 2. C = 54 18', 6 = 3.317, c = 3.925. 3. A = 123 12', a = 23.63, c = 20.51. 4. B =25 12', c = 227.7, 6 = 157.4. 5. C = 35 4', b = 577.3, c = 468.9. 7. O = 87 32' 5", 6 = 17.632, c = 21.746. 8. ^1 = 38 21' 47", a = 13.509, b = 17.632. 9. A = 29 25' 18", b = 2675.9, c = 3674. 10. B = 67 27' 33", b = 77.08, c = 79.06. 11. B - 100 22' 45", a = 1337.2, 6 = 1758.9. 12. ^4 = 139 21' 42", a = 100, c = 63.15. Exercise XXVII, page 97 2. c = 8.9. 3. a = 13. 4. c = 2. 5. b = V3. Exercise XXVIII, page 99 2. A = 12 22', B = 149 15', c = 34.37. 3. A = 64 19' 28", B = 42 24' 22", c = 612.06. 4. A = 84 12' 33", C = 45 46' 59", b = 0.5591. 5. # = 37 48' 5", C = 42 11' 55", a = 0.0117. 6. ^ = 33 5' 18", C r =410'42", 6 = 96.42. 7. 4 = 31 50' 20", .B = 50 4' 25", c = 3139.9. 8. A = 133 51' 34", 5=11 59' 10", c = 2479.2. 9. A = 70 22' 38", B = 21 24' 42", c = 33.787. 10. .4 = 72 40' 41", # = 158'1", c = 15.272. Exercise XXIX, page 102 2. A = 41 13' 0", C = 87 37' 54", c = 116.82. 3. 4 = 11 26' 58", C = 84 16' 31", c = 401. 4. B = 48 27' 20", C = 90, b = 360. 5. A = 46 52' 10", C = 111 53' 25", c = 883.65. A 1 = 133 7' 50", C' = 25 37' 45", fe = 411.92. 6. A = 29 11' 39", = 91 34' 21", c = 8.853. 7. ^1 = 83 40', C=715'47", c = 670.1. 4' = 96 20', C" = 58 25' 47", c' = 603.5. 8. -4 = 19 19' 3", (7 =142 59' 48", c = 89.15. 4' = 160 40' 57", C" = 1 37' 54", c' = 4.218. 9. 4 = 10 54' 58", C =132 12', c = 946.68. 10. A = 57 37' 18", C = 90, a = 88. 154 ANSWERS Exercise XXX, page 103 2. .4 = 60, # = 32 12', C7 = 8748'. 3. A = 56 7', .# 81 47', C = 42 6'. 4. .4 = 56 15', 5 = 59 51', C y = 6354'. 5. ^ = 38 57', = 47 41', C =93 22'. Exercise XXXI, page 106 2. ^1 = 7 37' 42", B = 61 55' 38", C = 110 26 '40". 3. ^1 = 47 38', _B = 681'6", C = 64 20' 54". 4. A = 85 55' 7", 5 = 43 57' 33", C = 50 7' 20". 5. J. = 23 32' 12", B = 56 8' 42", C = 100 19' 6". 6. .4 = 59 39' 30", # = 42 35' 20", C = 77 45' 10". 7. A = 33 15' 39", B = 50 56', C= 95 48' 21". 8. ^ = 37 22' 19", B = 38 15' 41", C =104 22'. 9. A - 36 45' 14", B = 53 3' 8", C = 90 11' 38". Exercise XXXII, page 108 1. 14.68. 3. 0.00815. 5. 0.156. 7. 28621. 9. 0.0265. 2. 1259.6. 4. 88.66. 6. 2520. 8. 1285.3. 10. 63.34. Exercise XXXIII, page 109 1. 27.65, 80.08 ; 65 19' 58", 14. 75.13 ft. ; 225.4 ft. 114 40' 2". 15. 11646ft. 2. 169.45ft. 16. Z(cot|8-cota). 3. 4.43ft.; 7.35ft. 17. 753.1yd. 4. 114.41 ft. 18. 91.772 ft. 5. 46.14 ft. ; 99.92 ft. 19. 47.168 ft. ; 16 20'. 6. 171.08yd. 20. 8. 574 miles. 7. 13 miles per hour, nearly. 21. 12.15ft. 8. 114.6 in. 23. 20 V3 ft. 9. 2000 (2 V3- 3) yd. 26. 106ft. 10. 23.87 ft. ; 15.18 ft. 27. 5(2 + V3) ft. 11. 39.97 ft. ; 29.99 ft. . 28. 9.24ft. 12. 520.44 ft. 29. 180 ft. ; 500 ft. 13. 330.72ft. ; 138 35' 30". 31. 15.14ft. 32. tan- 1 tan e tan ^ ; tan- 1 (tan cot sec 0) ; Vtan 2 -f tan 2 tan- 1 (tan cot sec 0) ; tan" 1 (cot cot Vtan 2 + tan 2 0) . 35. 50V6ft. 3g I . 2Zsin2 36. 250V3ft. ANSWERS 155 Exercise XXXIV, page 118 8. 100, 105, 75; 157.08, 164.934, 117.81. 9. 110, 100, 70. 10. 00 ; a great circle and its pole. Exercise XXXV, page 126 1. c =15 38' 6", .4 = 38 57 '12", J3 = 525'54". 2. c=409'21", A = 47 37' 21", .B = 50 2' 56". 3. a= 25 50' 17", b = 33 7' 37", B = 56 15' 43". 4. & = 48 54' 54", c = 50 2' 56", A = 16 6' 22". 5. a = 12 16' 42", & = 51 2' 48", c = 52 5' 54". . 6. a = 33 7' 37", A = 41 5' 6", B = 64 9' 43". 7. a = 42 22' 39", 6 = 45 15' 43", .4 = 52 5' 54". 8. a = 39 57' 4", & = 49 50' 39", c = 60 22' 24" . 9. 6 = 48 54' 54", A = 56 15' 43", B = 56 52' 23". 10. a = 37 54' 6", c = 77 43' 18", B = 80 14' 41". 11. c= 40 58' 50", 6 = 22 50' 19", .S = 36 17' 17", c = 139 1' 10' , 6= 157 9' 41", B = 143 42' 43". 12. Impossible. 13. A = 127 30' 11", a = 128 32', B = 102 14' 30". 14. 5 = 96 19' 51.6", c = 95 48 '28", a = 47 18' 44". 15. a = 66 37', c = 85 52', ^1 = 66 57' 48". a = 113 23' , c = 94 8', A = 113 2' 12". Exercise XXXVI, page 127 1. B= 167 43' 18", a = 138 54' 54", b = 129 57' 4". 2. A= 170 14' 41", a = 164 21' 54", & = 102 16' 42". 3. C = 138 54' 54", A = 169 29' 45", & = 102 16' 42". 4. C= 138 15' 43", B = 146 15' 43", a = 132 22' 39". 5. C= 102 16' 42", A = 140 2' 56", B = 106 6' 22". 6. B = 31 54' 40", C = 55 50' 7", ft = 39 42' 23". B = 148 5' 20", C = 124 9' 53", & = 140 17' 37". 7. A = 18 12' 24", C = 52 57' 12", a = 23 2' 44". A = 161 47' 36", C = 127 2' 48", a = 156 57' 16". 8. B = 132 31' 45", A = 60 25' 39", a = 67 18' 38". Exercise XXXVIII, page 136 1. A = 20 35' 37", B = 36 10' 39", C - 129 57' 4". 2. A = 43 2' 7", 5 = 47 37' 21", C = 102 16' 42". 3. .A = 45 26' 42", B = 47 37' 21", C = 102 16' 42". 4. A - 52 6 ( 54", B = 56 52' 23", C = 88 42' 27". 156 ANSWERS 5. A = 50 2' 56", B = 58 40' 13", C = 105 6' 41". 6. A = 52 5' 54", B = 68 12' 58", C = 113 53' 57". 7. A = 52 5' 54", B = 56 52' 23", C = 138 5' 42". 8. A = 84 26' 48", B = 108 24' 54", C = 56 28' 32". 9. A = 126 18' 42", B = 119 42' 8", C = 111 51' 42". 10. A = 67 9' 28", B = 67 9' 28", C = 74 27' 56". Exercise XXXIX, page 136 1. a = 36 10' 39", b = 40 9' 21", c = 50 13' 58". 2. a = 38 57' 12", b = 55 D 1' 2", c = 56 15' 43". 3. a = 39 20' 24", b = 41 5' 6", c = 60 22' 24". 4. a = 50 2' 56", 6 = 52 5' 54", c = 99 57' 42". Exercise XL, page 138 l.o = 20 32' 33", B = 52 5' 54", C = 123 7' 37". 2. c = 60 22' 24", J. = 26 40' 20", B = 38 57' 12". 3. 6 = 56 15' 43", A = 21 34' 28", C = 132 22' 39". 4. c = 60 22' 24", ^ = 47 37' 21", B = 77 39' 26". 5. C = 21 41' 25", B = 178 0' 29", a = 50 33' 38". 6. C = 129 53' 4", ,4 = 29 47' 28", 6 = 71 45' 15". 7. B = 84 11' 45", C = 72 59' 41", a = 93 58' 18". 8. B = 161 46' 32", A = 157 58' 8", c = 85 19' 46". Exercise XLI, page 139 1. a = 56 52' 23", b = 19 29' 45", C= 119 15' 56". 2. b = 52 5' 54", c = 63 21' 53'', A = 58 40' 13". 3. a = 44 44' 17", c = 8014'4i", = 52 5' 54". 4. b = 60 22' 24", c = 79 39' 38", A = 38 57' 12". 5. a = 40 12' 34", b = 53 38' 28", C=829'. 6. ' o = 112 25' 37", c = 59 19' 25", B = 87 14'. 7. b = 108 20' 51", c = 108 58' 5", A = 53 53' 6". 8. o = 108 32' 10", b = 88 35' 18", C = 121 47' 14", Exercise XLII, page 141 1. 6 = 40 32' 33", B = 39 9' 35", (7 = 129 57' 4". 6 = 61 33' 4", B = 121 19' 47", C = 50 2' 56". 2. a = 35 30' 24", A = 43 2' 7", (7 = 102 16' 42". a = 55 1' 2", A = 74 18' 19", C = 77 43' 18". 3. a = 21 7' 35", A = 25 26' 16", O = 100 30' 15' . a = 46 059", 4 = 47 39' 0", C = 79 29' 25". ANSWERS 157 4. Impossible. 5. A = 90, C = 12 29' 4", c = 11 53' 42". 6. B = 61 59', c = 13 38' 16", C = 12 9' 24 '. B = 118 1', c = 132 29' 46", C .= 138 48' 6". 7. -B = 44"28' 46", c = 16 35' 58", C = 34 59' 24". 8. c = 68 50' 36". Exercise XLIII, page 142 1. a = 5 21' 59", c = 60 22' 24", A = 4 3' 15". a = 77 43' 18", c = 119 37' 37", A = 47 37' 21". 2. A= 57 18' 43", C = 66 31' 42", c = 52 27' 4". ^1 = 122 41' 17", C = 152 14' 42", c = 156 15' 54". 3. b = 41 5' 17", c = 42 55' 48", C - 64 14'. 4. c = 64 26' 20", 6 =. 40 48' 50", B = 45 35' 50". c =. 115 33' 40", b = 176 34' 16", B = 176 15' 4" , 5. a = 90, & = 25 57' 12", B = 12 28' 38". 6. Impossible. 7. & = 49 30' 48", c = 2 5' 26", (7=2 21' 54". b = 130 29' 12", c = 118 5' 56", C= 94 4' 4". 8. c = 121 33' 9", b = 77 39' 20", B = 69 0' 18", Review Exercise 2. 5799.8 miles. 3. median = 58 3' 15". angles = 26 12' 29" and 28 31' 6". _ sin n tan a _ \ Vsin 2 n + (tan a - tan ft cos n) 2 J* tan - cot 5. . 180 6. 27 miles (nearly). 7. bisector = 50 35' 21". angles = 84 34' and 95 26'. 8. 70 31' 43". 10. 42.43ft. 9. 87 15' 2". 11. 904.7808. 14. 33 7' 2"; 37 9' 37". vq >\ 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. jANIS CZ UtC o LD 21A-60m-2,'67 (H241slO)476B General Library University of California Berkeley vJ*W* YB 17136 lLi? E ELE Y LIBRARIES *'4