^'i IN MEMORIAM FLORIAN CAJORI "if^l^yU ^ da^p -^, ELEMENTS OF TEIGONOMETKY WITH TABLES BY HERBERT C. WHITAKER, Ph.D. CENTRAL MANUAIi TRAINING SCHOOL PHILADKLPHLA., PENNSYLVANIA PHILADELPHIA D. ANSON PARTRIDGE 1898 Copyright, 1898, By H. C. WHITAKEK. Nortoooti ?3w8S J. S. Gushing & Co. — Berwick Sc Smith Norwood Mass. U.S.A. PREFACE. The introduction and first five chapters of this book have been prepared for the use of beginners in the subject. The second appendix has been added for those intending to take up work in higher departments of Mathematics. The subject of logarithms is usually considered in treatises on algebra ; but for the convenience of those not familiar with the theory of this subject, the first appendix has been added. To aid in a clear understanding of each process in the various investigations, the aim has been to closely associate with every equation a definite meaning with reference to a diagram. The tables used are in general as precise as the data usually obtained by students in engineering, physics, or chemistry. It is assumed that a few seven-place tables are accessible to a class ; and it is recommended that some class- room work be done with seven-place tables, especially in the solution of spherical triangles. It is hoped that the answers are free from error. The numerical computations were made with care by the author, but the results have-not been verified by any one else. Any corrections or suggestions relating to the book will be thankfully received. H. C. WHITAKER. Philadelphia, May, 1898. Digitized by the Internet Archive in 2007 with funding from IVIicrosoft Corporation http://www.archive.org/details/elementsoftrigonOOwhitrich CONTENTS. Table of natural values of functions i Table of circular measure of angles vii Table of hyperbolic functions vii INTRODUCTION. § 1. Meaning of geometric operations 1 § 2. Negative segments and angles 3 § 3. Products and quotients of segments 5 § 4. Homogeneity of trigonometric equations .... 6 § 5. Significant figures of numbers 7 § 6. Approximate values of numbers 7 § 7. Contractions in arithmetical operations .... 7 CHAPTER I. Functions of Angles. § 8. Generation of an angle 11 § 9. The four quadrants 12 § 10. The six ratios 14 § 11. Relations between functions and segments ... 17 § 12. Device for memorizing relations 19 § 13. Absolute values and algebraic signs of functions . . 20 § 14. Functions of 180° ± x, of 90° + x, and of - x . . . 22 § 15. The curve of each function 24 § 16. Limiting values of functions 26 § 17. The number of angles corresponding to a value of a function 26 § 18. Functions oi\x and 2 x 28 § 19. Functions of particular angles 29 CONTENTS. CHAPTER II. Explanation of the Tables. PAGK § 20. The table of natural values of functions .... 31 § 21. Principles of logarithms 37 § 22. Use of the table of logarithms 41 § 23. The table of logarithms of functions 45 CHAPTER III. Right Triangles. § 24. Problems are indeterminate, ambiguous, determinate, or impossible § 25. Relations of the parts of a right triangle § 26. Solutions of triangles using numbers. § 27. Applications § 28. Area of a right triangle § 29. Solutions of triangles using logarithms S 30. Problems 46 47 49 51 55 56 58 CHAPTER IV. The Isosceles Triangle and the Regular Polygon. § 31. Formulse for isosceles triangles . ... . . .60 § 32. Problems in isosceles triangles 61 § 33. Formulse for regular polygons 63 § 34. Problems in regular polygons 65 CHAPTER V. Oblique Triangles. § 35. Notation 66 § 36. Methods of solution 66 § 37. Solutions using numbers 70 §.38. Cot .5 = 0^^^^ -cot J. 74 ^ b CONTENTS. PAGK § 39. The area of a triangle 76 § 40. Solutions using logarithms 77 § 41. Applications with solutions 81 § 42. Problems 87 CHAPTER VI. Pboperties of Triangles. § 43. The altitudes, bisectors, and medians .... 94 § 44. The incircle and the excircles 96 § 45. The circumcircle 98 §46. The sum and difference of two sides and two angles . . 98 § 47. General method of solution, given three segments . . 100 § 48. Points of intersection of the circumcircle . . . .100 § 49. The nine-points circle 102 § 50. Abscissae and ordinates of points 104 § 51. Problems 105 § 62. Relations between the angles of a triangle . . . 108 CHAPTER VII. Spherical Triangles. 53. Tlie general spherical triangle 110 54! Formulae for right triangles 112 55. Device for rapidly obtaining formulae . . . .114 56. Solution of right triangles 115 57. Polar, quadrantal, and isosceles triangles ; the regular polygon 117 58. Oblique spherical triangles 120 59. The incircle and the excircles 120 60. The incircle of the polar triangle and the circumcircle . 121 61. Formulae for oblique triangles 122 62. Methods of solution 122 63. Solutions . . 123 64. A desired part directly in terms of given parts . . 126 65. Applications to geodesy and astronomy . . . .128 66. Area of a spherical triangle 132 CONTENTS. APPENDIX I. Theory of Logarithms. § 67. Arithmetic and geometric series § 68. Napierian logarithms treated geometrically § 69. Present systems treated geometrically § 70. Natural logarithms § 71. The number corresponding to a given logarithm § 72. A logarithm is an exponent of a base § 73. The logarithm corresponding to a given number § 74. The logarithmic curve S 75. Problems 133 134 136 137 138 140 141 143 144 APPENDIX II. GONIOMETRY, COMPLEX QUANTITIES, HYPERBOLIC FUNCTIONS. § 76. Functions of the sum or difference of angles . . . 144 § 77. Trigonometric equations ; the quadratic equation . . 146 § 78. The circular measure of an angle 149 § 79. Arc functions 150 § 80. Complex Quantities : addition, subtraction, multiplication, division, powers and roots 152 § 81. Functions of multiple angles 159 § 82. Exponential values of sin 6 and cos d . . . . 159 § 83. Hyperbolic Functions : the equilateral hyperbola . . 160 § 84. Definitions of hyperbolic functions 161 § 85. Exponential values of hyperbolic functions . . . 162 § 86. General relations 163 § 87. Longitude of w 164 § 88. Solution of the cubic equation 164 Problems for Examination 167 Answers to Problems 177 Table of Logarithms of Numbers viii Table of Logarithms of Functions x Table of Formula xvi Table of Constants xvi A TABLE OF THE NATURAL VALUES OF THE TRIGONOMETRIC FUNCTIONS OF ANGLES FROM 0° TO 90° FOR EVERY 10' A TABLE FOR CONVERTING THE SEXAGESIMAL MEASURE OF AN ANGLE INTO CIRCULAR MEASURE A TABLE FOR CONVERTING CIRCULAR FUNCTIONS INTO HYPERBOLIC FUNCTIONS NATURAL VALUES OF FUNCTIONS. X sinjif tanjf secAf CSCJf cotjr cosjr 0° .0 00000 00000 1.00 00 Infinite. Infinite. L.OOOO 90° 10' 02909 02909 00 343.775 343.774 00 50' 20' 05818 05818 00 171.888 171.885 00 40' 30' 08727 08727 00 114.593 114.589 00 30' 40' 11635 11636 01 85.946 85.940 .99 99 20' 50' 14544 14545 01 68.757 68.750 99 10' 1° .0 1745 .0 1746 1.00 02 57.299 57.290 .99 98 89° 10' 2036 2036 02 49.114 49.104 98 50' 20' 2327 2328 03 42.976 42.964 97 40' 30' 2618 2619 03 38.202 38.188 97 30' 40' 2908 2910 04 34.382 34.368 96 20' 50' 3199 3201 05 31.258 31.242 95 10' 2° .0 3490 .0 3492 1.0006 28.654 28.636 .99 94 88° 10' 3781 3783 07 26.451 26.432 93 50' 20' 4071 4075 08 24.562 24.542 92 40' 30' 4362 4366 10 22.926 22.904 90 30' 40' 4653 4658 11 21.494 21.470 89 20' 50' 4943 4949 12 20.230 20.206 88 10' 3° .0 5234 .0 5241 1.00 14 19.107 19.081 .99 86 87° 10' 5524 5533 15 18.103 18.075 85 50' 20' 5814 5824 17 17.198 17.169 83 40' 30' 6105 6116 19 16.380 16.350 81 30' 40' 6395 6408 21 15.637 15.605 80 20' 50' 6685 6700 22 14.958 14.924 78 10' 40 .0 6976 .0 6993 1.00 24 14.336 14.301 .99 76 86° 10' 7266 7285 26 13.763 13.727 74 50' 20' 7556 7578 29 13.235 13.197 71 40' 30' 7846 7870 31 12.745 12.706 69 30' 40' 8136 8163 33 12.291 12.250 67 20' 50' 8426 8456 36 11.868 11.826 64 10' 5° .0 8716 .0 8749 1.00 38 11.474 11.430 .99 62 85° 10' 9005 9042 41 11.105 11.059 59 50' 20' 9295 9335 43 10.758 10.712 57 40' 30' 9585 9629 46 10.433 10.385 54 30' 40' 9874 9923 49 10.128 10.078 51 20' 50' .1 0164 .10216 52 9.8391 9.7882 48 10' 6° .1 0453 .10510 1.00 55 9.5668 9.5144 .99 45 84° 10' 0742 0805 58 9.3092 9.2553 42 50' 20' 1031 1099 61 9.0652 9.0098 39 40' 30' 1320 1394 65 8.8337 8.7769 36 30' 40' 1609 1688 68 8.6138 8.5555 32 20' 50' 1898 1983 72 8.4047 8.3450 29 10' 7° .1 2187 .1 2278 1.00 75 8.2055 8.1443 .99 25 83° cos/ cot/ cscy secy tan/ sin/ / NATURAL VALUES OF FUNCTIONS. X sinjif tanjr secjf csc;r cotr COSJf 70 .1 219 .1228 1.00 75 8. 2055 8. 1443 .99 25 83° 10' 248 257 79 0156 7. 9530 22 50' 20' 276 287 82 7. 8344 7704 18 40' 30' 305 317 86 6613 5958 14 30' 40' 334 346 90 4957 4287 11 20' 50' 363 376 94 3372 2687 07 10' 8° .1392 .1405 1.00 98 7. 1853 7. 1154 .99 03 82° 10' 421 435 1.01 02 0396 6. 9682 .98 99 50' 20' 449 465 07 6. 8998 8269 94 40' 30' 478 495 11 7655 6912 90 30' 40' 507 524 15 6363 5606 86 20' 50' 536 554 20 5121 4348 81 10' 9° .1564 .1584 1.01 25 6. 3925 6. 3138 .98 77 81° 10' 593 614 29 2772 1970 72 50' 20' 622 644 34 1661 0844 68 40' 30' 650 673 39 0589 5. 9758 63 30' 40' 679 703 44 5. 9554 8708 58 20' 50' 708 733 49 8554 7694 53 10' 10° .1736 .1763 1.01 54 5. 7588 5. 6713 .98 48 80° 10' 765 793 60 6653 5764 43 50' 20' 794 823 65 5749 4845 38 40' 30' 822 853 70 4874 3955 33 30' 40' 851 883 76 4026 3093 27 20' 50^ 880 914 81 3205 2257 22 10' 11° .1908 .1944 1.01 87 5. 2408 5. 1446 .98 16 79° 10' 937 974 93 1636 0658 11 50' 20' 965 .2 004 99 0886 4. 9894 05 40' 30' 994 035 1.02 05 0159 9152 .97 99 30' 40' .2 022 065 11 4. 9452 8430 93 20' 50' 051 095 17 8765 7729 87 10' 12° .2 079 .2 126 1.02 23 4. 8097 4. 7046 .97 81 78° 10' 108 156 30 7448 6382 75 50' 20' 136 186 36 6817 5736 69 40' 30' 164 217 43 6202 5107 63 30' 40' 193 247 49 5604 4494 57 20' 50' 221 278 56 5022 3897 50 10' 13° .2 250 .2 309 1.02 63 4. 4454 4. 3315 .97 44 77° 10' 278 339 70 3901 2747 37 50' 20' 306 370 77 3362 2193 30 40' 30' 334 401 84 2837 1653 24 30' 40' 363 432 91 2324 1126 17 20' 50' 391 462 99 1824 0611 10 10' 14° .2 419 .2 493 1.03 06 4. 1336 4. 0108 .97 03 76° cosy cot/ esc/ sec/ tan/ sin/ / Ill NATURAL VALUES OF FUNCTIONS. X sinx tanx secjr cscx cotjr COSJf 14° .2 419 .2 493 1.03 06 4. 1336 4. 0108 .97 03 76° 10' 447 524 14 0859 3. 9617 .96 96 50' 20' 476 555 21 0394 9136 89 40' 30' 504 586 29 3. 9939 8667 81 30' 40' 532 617 37 9495 8208 74 20' 50' 560 648 45 9061 7760 67 10' 15° .2 588 .2 679 1.03 53 3. 8637 3. 7321 .96 59 75° 10' 616 711 61 8222 6891 52 50' 20' 644 742 69 7817 6470 44 40' 30' 672 773 77 7420 6059 36 30' 40' 700 805 86 7032 5656 28 20' 50' 728 836 94 6652 5261 21 10' 16° .2 756 .2 867 1.04 03 3. 6280 3. 4874 .96 13 74° 10' 784 899 12 5915 4495 05 50' 20' 812 931 21 5559 4124 .95 96 40' 30' 840 962 29 5209 3759 88 30' 40' 868 994 39 4867 3402 80 20' 50' 896 .3 026 48 4532 3052 72 10' 17° .2 924 .3 057 1.04 57 3. 4203 3. 2709 .95 63 73° 10' 952 089 66 3881 2371 55 50' 20' 979 121 76 3565 2041 46 40' 30' .3 007 153 85 3255 1716 37 30' 40' 035 185 95 2951 1397 28 20' 50' 062 217 1.05 05 2653 1084 20 10' 18° .3 090 .3 249 1.05 15 3. 2361 3. 0777 .95 11 72° 10' 118 281 25 2074 0475 02 50' 20' 145 314 35 1792 0178 .94 92 40' 30' 173 346 45 1515 2. 9887 83 30' 40' 201 378 55 1244 9600 74 20' 50' 228 411 66 0977 9319 65 10' 19° .3 256 .3 443 1.05 76 3. 0716 2. 9042 .94 55 71° 10' 283 476 87 0458 8770 46 50' 20' 311 508 98 0206 8502 36 40' 30' 338 541 1.06 08 2. 9957 8239 26 30' 40' 365 574 19 9713 7980 17 20' 50' 393 607 31 9474 7725 07 10' 20° .3 420 .3 640 1.06 42 2. 9238 2. 7475 .93 97 70° 10' 448 673 53 9006 7228 87 50' 20' 475 706 65 8779 6985 77 40' 30' 502 739 76 8555 6746 67 30' 40' 529 772 88 8334 6511 56 20' 50' 557 805 1.07 00 8117 6279 46 10' 21° .3 584 .3 839 1.07 11 2. 7904 2. 6051 .93 36 69° cos/ cot/ CSC/ sec/ tan/ sin/ / NATURAL VALUES OF FUNCTIONS. IV X sinjr tanjr sec A" cscx COtAf cosjr 21° .3 584 .3 839 1.07 11 2.7904 2. 6051 .93 36 69° 10' 611 872 23 7695 5826 25 50' 20' 638 906 36 7488 5605 15 40' 30' 665 939 48 7285 5386 04 30' 40' 692 973 60 7085 5172 .92 93 20' 50' 719 .4 006 73 6888 4960 83 10' 22° .3 746 .4 040 1.07 85 2. 6695 2. 4751 .92 72 68° 10' 773 074 98 6504 4545 61 50' 20' 800 108 1.08 11 6316 4342 50 40' 30' 827 142 24 6131 4142 39 30' 40' 854 176 37 5949 3945 28 20' 50' 881 210 50 5770 3750 16 10' 23° .3 907 .4 245 1.08 64 2. 5593 2. 3559 .92 05 67° 10' 934 279 77 5419 3369 .9194 50' 20' 961 314 91 5247 3183 82 40' 30' 987 348 1.09 04 5078 2998 71 30' 40' .4 014 383 18 4912 2817 59 20' 50' 041 417 32 4748 2637 47 10' 24° .4 067 .4 452 1.09 46 2. 4586 2. 2460 .9135 66° 10' 094 487 61 4426 2286 24 50' 20' 120 522 75 4269 2113 12 40' 30' 147 557 89 4114 1943 00 30' 40' 173 592 1.10 04 3961 1775 .90 88 20' 50' 200 628 19 3811 1609 75 10' 25° .4 226 .4 663 1.10 34 2. 3662 2. 1445 .90 63 65° 10' 253 699 49 3515 1283 51 50' 20' 279 734 64 3371 1123 38 40' 30' 305 770 79 3228 0965 26 30' 40' 331 806 95 3087 0809 13 20' 50' 358 841 1.11 10 2949 0655 01 10' 26° .4 384 .4 877 1.11 26 2. 2812 2. 0503 .89 88 64° 10' 410 913 42 2677 0353 75 50' 20' 436 950 58 2543 0204 62 40' 30' 462 986 74 2412 0057 49 30' 40' 488 .5 022 90 2282 1. 9912 36 20' 50' 514 059 1.12 07 2153 9768 23 10' 27° .4 540 .5 095 1.12 23 2. 2027 1. 9626 .89 10 63° 10' 566 132 40 1902 9486 .88 97 50' 20' 592 169 57 1779 9347 84 40' 30' 617 206 74 1657 9210 70 30' 40' 643 243 91 1537 9074 57 20' 50' 669 280 1.13 08 1418 8940 43 10' 28° .4 695 .5 317 1.13 26 2. 1301 1. 8807 .88 29 62° cos/ cot/ CSC/ sec/ tan/ sin/ / NATURAL VALUES OF FUNCTIONS. X sinjif tanjf secx cscjr cotx cosx 28° .4 695 .5 317 1.1 326 2.1 301 1.8 807 .88 29 62° 10' 720 354 343 185 676 16 50' 20' 746 392 361 070 546 02 40' 30' 772 430 379 2.0 957 418 .87 88 30' +0' 797 467 397 846 291 74 20' 50' 823 505 415 736 165 60 10' 29° .4 848 .5 543 1.1 434 2.0 627 1.8 040 .87 46 61° 10' 874 581 452 519 1.7 917 32 50' 20' 899 619 471 413 796 18 40' 30' 924 658 490 308 675 04 30' 40' 950 696 509 204 556 .86 89 20' 50' 975 735 528 101 437 75 10' 30° .5 000 .5 774 1.1 547 2.0 000 1.7 321 .86 60 60° 10' 025 812 566 1.9 900 205 46 50' 20' 050 851 586 801 090 31 40' 30' 075 890 606 703 1.6 977 16 30' 40' 100 930 626 606 864 01 20' 50' 125 969 646 511 753 .85 87 10' 31° .5 150 .6 009 1.1 666 1.9 416 1.6 643 .85 72 59° 10' 175 048 687 323 534 57 50' 20' 200 088 707 230 426 42 40' 30' 225 128 728 139 319 26 30' 40' 250 168 749 048 212 U 20' 50' 275 208 770 1.8 959 107 .84 96 10' 32° .5 299 .6 249 1.1 792 1.8 871 1.6 003 .84 80 58° 10' 324 289 813 783 1.5 900 65 50' 20' 348 330 835 697 798 50 40' 30' 373 371 857 612 697 34 30' 40' 398 412 879 527 597 18 20' 50' 422 453 901 443 497 03 10' 33° .5 446 .6 494 1.1 924 1.8 361 1.5 399 .83 87 57° 10' 471 536 946 279 301 71 50' 20' 495 577 969 198 204 55 40' 30' 519 619 992 118 108 39 30' 40' 544 661 1.2 015 039 013 23 20' 50' 568 703 039 1.7 960 1.4 919 07 10' 34° .5 592 .6 745 1.2 062 1.7 883 1.4 826 .82 90 56° 10' 616 787 086 806 733 74 50' 20' 640 830 110 730 641 58 40' 30' 664 873 134 655 550 41 30' 40' 688 916 158 581 460 25 20' 50' 712 959 183 507 370 08 10' 35° .5 736 .7 002 1.2 208 1.7 434 1.4 281 .8192 55° cos/ cot/ CSC/ sec/ tan/ sin/ / NATURAL VALUES OF FUNCTIONS. VI X sin;r tanx secAf CSCA" cotjr COSJf "~~ 35° .5 736 .7 002 1.2 208 1.7 434 1.4 281 .8192 55° 10' 760 046 233 362 193 75 50' 20' 783 089 258 291 106 58 40' 30' 807 133 283 221 019 41 30' 40' 831 177 309 151 1.3 934 24 20' 50' 854 221 335 081 848 07 10' 36° .5 878 .7 265 1.2 361 1.7 013 1.3 764 .80 90 54° 10' 901 310 387 1.6 945 680 73 50' 20' 925 355 413 878 597 56 40' 30' 948 400 440 812 514 39 30' 40' 972 445 467 746 432 21 20' 50' 995 490 494 681 351 04 10' 37° .6 018 .7 536 1.2 521 1.6 616 1.3 270 .79 86 53° 10' 041 581 549 553 190 69 50' 20' 065 627 577 489 111 51 40' 30' 088 673 605 427 032 34 30' 40' 111 720 633 365 1.2 954 16 20' 50' 134 766 661 303 876 .78 98 10' 38° .6 157 .7 813 1.2 690 1.6 243 1.2 799 .78 80 52° 10' ISO 860 719 183 723 62 50' 20' 202 907 748 123 647 44 40' 30' 225 954 778 064 572 26 30' 40' 248 .8 002 807 005 497 08 20' 50' 271 050 837 1.5 948 423 .77 90 10' 39° .6 293 .8 098 1.2 868 1.5 890 1.2 349 .77 71 51° 10' 316 146 898 833 276 53 50' 20' 338 195 929 777 203 35 40' 30' 361 243 960 721 131 16 30' 40' 383 292 991 666 059 .76 98 20' 50' 406 342 1.3 022 611 1.1 988 79 10' 40° .6 428 .8 391 1.3 054 1.5 557 1.1 918 .76 60 50° 10' 450 441 086 504 847 42 50' 20' 472 491 118 450 778 23 40' 30' 494 541 151 398 708 04 30' 40' 517 591 184 345 640 .75 85 20' 50' 539 642 217 294 571 66 10' 41° .6 561 .8 693 1.3 250 1.5 243 1.1 504 .75 47 49° 10' 583 744 284 192 . 436 28 50' 20' 604 796 318 141 369 09 40' 30' 626 847 352 092 303 .74 90 30^ 40' 648 899 386 042 237 70 20' 50' 670 952 421 1.4 993 171 51 10' 42° .6 691 .9 004 1.3 456 1.4 945 1.1 106 .74 31 48° cos/ cot/ CSC/ sec/ tan/ sin/ / Vll NATURAL VALUES OF FUNCTIONS. X sin X tan jr sec x CSC X cot Jf cos jr 42° .6 691 .9 004 1.3 456 1.4 945 1.1 106 .74 31 48° 10' 713 057 492 897 041 12 50' 20' 734 110 527 849 1.0 977 .73 92 40' 30' 756 163 563 802 913 73 30' 40' 777 217 600 755 850 53 20' 50' 799 271 636 709 786 ZZ 10' 43° .6 820 .9 325 1.3 673 1.4 663 1.0 724 .73 14 47° 10' 841 380 711 617 661 .72 94 50' 20' 862 435 748 572 599 74 40' 30' 884 490 786 527 538 54 30' 40' 905 545 824 483 477 34 20' 50' 926 601 863 439 416 14 10' 44° .6 947 .9 657 1.3 902 1.4 396 1.0 355 .7193 46° 10' 967 713 941 352 295 73 50' 20' 988 770 980 310 235 53 40' 30' .7 009 827 1.4 020 267 176 33 30' 40' 030 884 061 225 117 12 20' 50' 050 942 101 183 058 .70 92 10' 45° .7 071 1.0 000 1.4 142 1.4 142 1.0 000 .70 71 45° cos/ cot/ CSC/ sec/ tan/ sin/ / Circular Measure. {»= X 180° Hyperbolic/ sinh u Functions. I cosh u tanx. sec jr. X e X e X u X u 1° .01745 V .00029 2° .0349 40° .7629 2° .03491 2' .00058 4° .0699 42° .8092 3° .05236 3' .00087 6° .1049 44° .8569 4° .06981 4' .00116 8° .1401 46° .9063 5° .08727 5' .00145 10° .1754 48° .9575 6° .10472 6' .00175 12° .2110 50° 1.0107 7° .12217 7' .00204 14° .2468 52° 1.0662 8° .13963 8' .00233 16° .2830 54° 1.1242 9° .15708 9' .00262 18° .3195 56° 1.1851 10° .17453 10' .00291 20° .3564 58° 1.2492 20° .34907 20' .00582 22° .3938 60° 1.3170 30° .52360 30' .00873 24° .4317 62° 1.3890 40° .69813 40' .01164 26° .4702 64° 1.4659 50° .87266 50' .01454 28° .5094 66° 1.5485 60° 1.04720 60' .01745 30° .5493 68° 1.6379 70° 1.22173 32° .5900 70° 1.7354 80° 1.39626 10" .00005 34° .6317 72° 1.8427 90° 1.57080 20" .00010 36° .6743 74° 1.9623 100° 1.74533 40" .00019 38° .7180 76° 2.0973 ELEMENTS OF TRIGONOMETRY. j;04c INTRODUCTION. Trigonometry investigates the numerical relations be- tween segments,* angles, and areas of figures. 1. The meaning of geometric operations will be retained in trigonometry. Thus, Fig. 1, if the segment CD, of length a, is added to the segment DE, of length h, the sum is the seg- ment CE, of length a -f 6. In Fig. 2, if DE, of length h, is subtracted from CD, of length a, the remainder CE is of length a — b. In Fig. 3, if CD is of length a, ^ ^ yig. 3. CE = a-}-a-fa-|-a = 4a. In Fig. 4, if CD is of length a, CE (two of the ten equal parts of CD) is of length .2 a. Hence if a segment . g of length a is mul- C E D tiplied by 2.4, the Fig. 4. product will be considered either a divided by 10 and the * In this work a straight line, or simply a line, is conceived to be infinitely long in two opposite directions from a point ; hence a half- line is conceived to be infinitely long in one direction from a point ; and a segment of a line, or simply a segment, is the name which will be applied to a limited portion of a line. 1 1 fl b C D Fig. 1. 1 E C E b Fig. 2. . a . a . a . n D C D E 2 INTRODUCTION. quotient multiplied by 24, or a multiplied by 24 and the product divided by 10. In Fig. 5, EF, of length h, may be conceived as being divided into CD, of length a. If CD contains EF five times, the ratio of CD to £ p EF is said to be 5. ' — ' If it is desired to , _ a _ _ , obtain the ratio of ^ G D Fir 5 EF to CD, CD is first divided into 10 equal parts and one of these parts CG applied to EF. If EF contains CG exactly two times, the ratio of EF to CD is said to be .2. The ratios - = 5 b ^ and - = .2 have a product of 1, and are said to be reciprocal ratios. In Fig. 6, EF is not contained exactly in CD, but is con- tained twice and a remainder GD more than three-tenths of EF, but less than three , ■■.■..... , and one-half tenths ; the ^ ^ ,_,_p-^ ratio of CD to EF is C G D said to be 2.3+. This F^^- ^■ value of the ratio contains an error not greater than .05. The ratio of the error to the assumed value is not greater than .05 in 2.3, or 5 in 230, or 1 in 46. If this error in 2.3, the assumed value of the ratio, is too great for use in a problem in which the ratio enters, EF is divided into 100, 1000, 10,000, etc., equal parts, and one of them applied to GD. Thus, if EF is divided into 10,000 parts and GD is found to contain one of these parts not less than 3470.5 times, but less than 3471 times, the ratio of CD to EF is said to be 2.3471 -. This value of the ratio contains an error not greater than .00005. The ratio of the error to the assumed value is not greater than .00005 in 2.3471, or 1 in 46,942. In the same way, if the ratio of CD to EF is desired, CD NEGATIVE SEGMENTS AND ANGLES. 3 is divided into 10, 100, 1000, 10,000, etc., equal parts, and so divided applied to EF. Suppose CD is divided into 10,000 parts, and the beginning of CD is placed at E. Then if the 4261st division of CD is the one nearest to F, the ratio of EF to CD is said to be approximately .4261 ; the ratio of the error to the assumed value of the ratio in this case is not greater than 1 in 8522. Operations upon angles. — All that has been said with reference to addition, subtraction, multi- plication, and division of segments will apply to corresponding operations per- formed upon angles. Thus, Fig. 7, the sum of angle BAC{=a) and angle CAD (= 6) is the angle BAD(= a + b). Fig. Similarly, the difference of BAC(=a) and CAD(—b), Fig. 8, is the angle BAD(=a-b). In a similar manner, the interpreta- tion of the multiplication of an angle by a number, the division of an angle by a number, and the division of an angle by ^^' ^' another angle will be conceived to be the same as in geometry. NEGATIVE SEGMENTS AND ANGLES. 2. Assume the direction of measurement of the segment CD, of length a, Fig. 9, to be from ^ left to right. From D, the ter- E C « 5 minal point of CD, lay off a seg- ^^^' ^' ment DE, of length b, backward on CD. Then if b is greater than a, E will be to the left of C, but a — b will still be said to be the length of the segment GE, measured from C to E, or in the direction opposite to that assumed to be positive. Now the length of a in the discussion is 4 INTRODUCTION. arbitrary, and may be conceived to be equal to the length of CD in Fig. 10, or zero ; in which case the segment CE (measured from right to left, or 5 p opposite the direction assumed to ^ C be positive) is equal to — h. ^^^' ^^' Hence, if any segment is multiplied by — 1, the effect of the operation is to measure the segment from the same point as at first but in an opposite direction ; thus, — 1 x a = —a, and — Ix — a= + a. The multiplication of a segment by a positive number will be considered to be repetition of the segment in the same direction the number of times indicated by the multi- plier (which may be either integral or fractional) ; thus, 4.3 times — a = — (4.3 a). The product of a segment by a negative number will be conceived to be : first, a reversal of the direction of the segment, and then the multiplication by a positive number ; thus, - 4.3 X a = 4.3 X - 1 X a = 4.3 X - a = -(4.3 a). Also -4.3 x-a = 4.3 x-1 x-a = 4.3 xa = + (4.3a). If division is defined as being the process of obtaining a quantity, which, multiplied with the divisor, will produce the dividend, each process of multiplication gives -(4-3 «) ■ 4.3. 3(i3a) = _„. — a 4.3 -(4-3 g) ^ ^3 -(4.3 g) ^ ^ a ' ' -4.3 + (4.3 g) ^ ^.3. ±M^=:-a. — a — 4.3 The same methods of reasoning will give the conception of negative angles. Thus, if turning a screw in one direction is positive turning, then turning the screw in the opposite direction is negative turning, because it cancels or undoes PRODUCTS AND QUOTIENTS OF SEGMENTS. 5 an equal amount of positive turning ; and negative angles so conceived may be multiplied by positive or negative num- bers, or divided by positive or negative numbers or angles. PRODUCTS AND QUOTIENTS OF SEGMENTS. 3. In geometry, the product of two segments is explained to be a term used to aid us in counting the number of units of area in a surface ; thus if a rectangle has sides of length 3 inches and 2 inches, respectively, the area is said to be 3 inches x 2 inches = 6 square inches. This statement is a short way of saying that in the rectangle there can be arranged 2 rows of inch-squares, 3 in a row, making a total of 6 inch-squares or 6 square inches. In the same way the operation 4 inches x 3 inches x 2 inches = 24 cubic inches, is a method of counting cubes in a rectangular par- allelopiped. In trigonometry, the same meaning is attached to these expressions as in geometry. Thus, the product of a segment and an abstract number, say 4.3 a, is said to be a length; the product (so called) of two segments and an abstract number, say 4.3 a^ or 4.3 ah, is said to be an area ; the product of three segments and an abstract number, say 4.3 ahc or 4.3 a^h, is said to be a volume. Since division is the inverse of multiplication, an area is indicated by the quotient of a volume by a segment, such 4.3 «6c C5- -1 1 4.3 a5c 4.3 a& / a o k ^/TWTTT^ represent segments. Expressions having no geometric meaning. — In geometry, expressions for lengths, areas, and volumes have meaning ; but many forms which occur in trigonometric reductions such as Va, — , -^abc, 4.3 abed, are said to have no geo- a metric meaning. 6 INTRODUCTION. Dimensions. — Any expression representing a length is said to be of one dimension ; an expression representing an area is said to be of two dimensions ; and any expression is of h dimensions if the algebraic sum of the exponents of the segments in each term is k. Thus a ratio has zero a 2 dimensions, since it is of the form - = a}h~'^ ; - has — 1 h a dimensions ; -yob = a^b^ is of |- dimensions ; abed is of 4 dimensions ; and so on. 4. Homogeneity of trigonometric equations. — No geometric meaning can be attached to the result obtained by adding an area and a length ; or by adding a volume and a length ; or by adding a volume and an area. And any equation involving terms of different dimensions is also without meaning. Hence in every trigonometric equation, each term must have the same dimensions. Equations in which this condition is fulfilled are called homogeneous equations. Therefore every trigonometric equation must be a homo- geneous equation. This property of trigonometric equations is frequently of value as a check on the correctness of reduction pro- cesses. For example, suppose an investigation of the rela- tions of the parts of any figure is undertaken. The first equation used must be homogeneous ; any other equation derived from it should also be homogeneous ; if it fails in homogeneity a mistake has been made in the process of reduction. Exercises. — 1. What are the dimensions of each of the following expressions : 3 7 a'bc 3 ab& a^U 1 l 2. Write an equation containing six terms of different form, each term representing a segment ; also an equation containing six terms, each representing an area. APPROXIMATE VALUES OF NUMBERS. 7 5. Significant figures. — Zeros are not significant figures, if they are supplied to a number merely for the purpose of putting the decimal point in the right position. Thus in the number .00908, the two zeros immediately following the decimal point are not significant figures, but the remaining figures, 9, 0, and 8, are significant figures. Also, in writing approximate values of large numbers, zeros are frequently supplied when necessary to get the first significant figure of the number in the proper position with reference to the unit's place. Thus in the number 92,500,000 miles, the 5 is understood to be the nearest fig- ure in the hundred-thousands place; the zeros following the 5 indicate a lack of knowledge as to the value of the figures in the ten-thousands place and in lower orders ; the 9, 2, and 5 are the only significant figures in this number. 6. Errors in approximate values of numbers. — Suppose an assumed value of a number, expressed with four significant figures, the fourth figure being the nearest figure in that place. The ratio of the error to the assumed value of the number is not greater than 1 to 2000. For the most un- favorable case is when 1000 is assumed as the value of 1000.5, in which case the ratio is 1 to 2000. In a similar manner the ratio of the error of an approxi- mate value of a number of five significant figures (the fifth figure being the nearest figure) cannot be greater than 1 to 20,000; and soon. 7. Errors arising in the processes of calculation. — Suppose that assumed values of numbers are introduced into a numerical calculation. From the principles of arithmetic all numbers arising subsequently in the process of calcula- tion will in general have about the same degree of precision as was contained in the assumed value of the least degree of precision which entered into the calculation. Thus, con- sider the following examples : 8 INTRODUCTION. Multiplication. — Suppose it is desired to multiply the approximate value .7862 by the number 43.59. The num- ber of which .7862 is the approximate value is between .78615 and .78625. Multiply each separately by 43.59. The products are 34.2682785 and 34.2726375. It will be seen that 34.27 will represent either product to four signifi- cant figures. Now multiply .7862 by 43.59 by .7862 first multiplying by 40, then by 3, then by .5, 43.59 then by .09. To reject all figures to the right of the vertical line drawn through the products, use all the figures of the multiplicand when multi- plying by 40, and put in the decimal point ; in multiplying by 3, strike out the 2 of the multi- plicand, but use it to obtain figures which may affect the column to the left of the vertical line ; thus, 3 times 2 are 6, and 6 being more than 5, 1 is carried ; 3 times 6 are 18, and 1 to carry is 19, and so on. In multiplying by 5, the 6 of the multiplicand is struck out, but is used to obtain 3 to carry ; thus, 5 times 6 are 30, 5 times 8 are 40, and 3 to carry are 43, and 34.27 so on. In multiplying by 9, the 8 is struck out, but the 86 now struck out being nearly 90, we proceed : 9 times 6 are 54, 9 times 8 are 72, and 5 to carry are 77, 9 times 7 are 63, and 8 to carry are 71 ; or simply considering the 86 thrown away to be 90, we say 9 times 9 are 81, 9 times 7 are 63, and 8 to carry are 71. In adding the partial products, the right-hand column adds to 21 ; reject the 1 and carry the 2 to the next column, as only four significant figures of the product can be correctly obtained. The method is expressed in the following Rule. — Use the figures of the multiplier in reverse order. Supply the decimal point in the first partial product. Eeject all unnecessary figures. 31.448 2.358 6 .393 10 .070 758 34.270 458 .7862 43.59 31.448 2 359 393 71 CONTKACTIONS IN COMPUTATIONS. 9 Division. — To reverse the process, divide the product 34.27 by 78.62. COMMON METHOD. CONTRACTED METHOD. 7862)34.27 3144 00000(43.589 8 .7862)34.27 3145 2 82 2 36 46 39 "7 2 82 2 35 46 39 20 S6 340 310 7 6 0300 2896 74040 70758 (43.59 Crosses are added to make the number of decimal places in the dividend equal to the number in the divisor. After the multiplication by the quotient figure 4, put as many crosses after the 4 as there are unused crosses in the divi- dend. Then supply the decimal point. The rest of the process is similar to the process of multiplication. The square root of .7862. COMMON METHOD. .78620000(.88667 64 CONTRACTED METHOD. 168 1462 1344 1766 118 105 00 96 17726 12 10 0400 6356 17732 1 4044 .7862(.8867 64 168 176 1462 1344 118 106 12 10 INTRODUCTION. A consideration of these examples will show that if data are correct to only three significant figures (say an error of 1 in 1000), the calculation may be kept in four significant figures ; for the result will be as precise as the data, and further figures are unnecessary and unreliable. Similarly, if the data are correct to four significant figures (say an error of 1 in 10,000), the calculation may be kept in five significant figures. And so on. Applications. — The summation of series is one of the many prac- tical applications of approximate values of numbers. Thus, sup- pose it is desired to compute to ten decimal places the value of the series : 1) 1. 2) 1. 3) .5 4) .166,666,666,67 5) 41,666,666,67 6) 8,333,333,33 7) 1,388,888,89 8) 198,412,70 9) 24,801,59 10) 2,755,73 11) 275,57 12) 25,05 13) 2,09 14) 16 1 2.718,281,828,46 e = 1 -h - + — — H = h etc., 1 1.2 1.2-3 ' in which each term is obtained from the one preceding by divi- sion, the divisors increasing by 1 each time. The work proceeds as in the margin. It will be seen that the nearest tenth decimal figure is 5. Problems. — Perform the following multiplications, keep- ing only the figures which affect the fourth significant figure of the product. After each multiplication, perform the re- verse operation of division or square root : 234.1 X 4.444 = 1040 1.414 x 1.414 = 2.000 .4567 X 9.009 = 4.114 .7454 x .7454 = .5000 7060 X .0302 = 213.2 57.33 x 57.33 = 3287 .7854 X 34.25 = 26.90 1.012 x 1.012 = 1.024 CHAPTER I. FUNCTIONS OF ANGLES. 8. Generation of an angle. — The angle BAC may be con- ceived to be generated by a turning of a half-line from either side of the angle q to the other side. Thus, the position AC may be reached by a turning of a half-line from AB through an angle x + any num- ber of revolutions; or the same position may be reached by a turning from AB in an opposite direction through an angle 360° — a; -f any number of revolutions. Also, if b, the length of AC, is regarded as negative, the turning from AB is either 180° -|- a; or a downward turning of 180°— .t. For example, if, in Fig. 11, x is 30° and b is 10 inches, the point C is determined by any of the following values of b and X : 6 = -1-10; a; = 30°, 390°, 750°, 1110°, etc. b = -10; x = 210°, 570°, 930°, 1290°, etc. 6 = + 10; a; =-330°, -690°, -1050°, etc. & = -10; a; = -150°, -510°, -870°, etc. The half-line AB from which angles will be conceived to be measured will be called the initial line ; the final position AC of the revolving half-line will be called the terminal 11 12 FUNCTIONS OF ANGLES. line of the angle ; the length, b, from the vertex of the angle to any point C on the terminal line will be called the distance of C. The direction of the initial line is entirely arbitrary. It will be drawn to the right in the discussions in this chapter. The angle will be regarded as positive when generated by a half-line turning from the initial line upward to the left, as indicated by the arrow, Fig. 11. It will be regarded as negative when generated by a half-line turning in the oppo- site direction. The distance will be regarded as positive when laid off from the vertex of the angle along the terminal line. 9. The four quadrants. — When a positive angle is under consideration, it will be conceived to be moved so that its vertex is brought at A, one side coincides with AB, and the angle extends in a positive direction. Then if the teitoiinal line falls in the angle BAD (where DDi is perpendicular to BABi), the angle is said to be of the first quadrant; if the terminal line falls in DAB^, the angle is of the second quadrant; if the terminal line is in BiAD^, the angle is of the third quadrant ; if the terminal line is in DiAB, the angle is of the fourth quadrant. Thus, 30°, 390°, -330°, -690° are angles of the first quadrant. 150°, 510°, -210°, -570° are angles of the second quadrant. 210°, 570°, -150°, -510° are angles of the third quadrant. 330°, 690°, - 30°, -390° are angles of the fourth quadrant. DEFINITION OF FUNCTION. 13 c Fig. 13. Abscissa and ordinate. — From any point C drop a perpen- dicular upon tlie initial line AB. The distance from A to the foot of this perpendicular is called the abscissa of C. The length of the perpendicular is called the ordinate of C. Abscissae measured to the right of A will be considered positive, and measured to the left, negative. Ordinates measured upward from the initial line (or ini- tial line drawn back- ward) will be considered positive, and measured downward, negative. ^ Thus, Fig. 14. C, In Quadrant I, a is + and c is -h In Quadrant II, a is + and c is — . In Quadrant III, a is — and c is — . In Quadrant IV, a is — and c is +. On the half-line AB, a is zero and c is -h. On the half-line AD, a is + and c is zero. On the half-line AB^^, a is zero and c is — . On the half-line AD^, a is — and c is zero. Definition of function. — Suppose a magnitude m, and a number of other magnitudes, p, q, r, so related to m, that when m changes in value, all of the others change in value ; 14 FUNCTIONS OF ANGLES. and when m remains fixed in value, all of the others remain fixed. In such a case, p, q, r are said to be functions of m. 10. The six ratios. — If we take in pairs the ordinate, distance, and abscissa of a point on the terminal line of an angle A, six different ratios can be obtained : a a h h c c It will now be shown that these six ratios are functions of the angle A. For suppose, first, that h remains the same length while the angle A changes in magnitude ; say A in- creases. Then if A is an angle of the first quadrant, the point C will move upward to the left, its or- dinate will increase, and its abscissa will decrease. Hence -, -, and - will in- h c c crease, and their reciprocal ratios will decrease. In a similar manner by assuming a or c to remain the same length, or by assuming the moving point C to be in any other quadrant, the same thing may be shown. Assume now that C moves so that the angle A is un- changed. If C moves either toward A or away from it to a point C, the similar tri- . angles ABC and AB'C ,.r give: >-'"■' ■■ BC:AC = B'C':AC' A^ y^ BC:AB = B'C':AB' ^ a AC:AB= AC : AB' ^ c the reciprocal ratios also A I J' B" B being equal in pairs. Fig. 16. NAMES OF THE RATIOS. 15 Names of the ratios. — For any point on the terminal line of an angle, The ratio of the ordinate to the distance is called the sine of the angle. The sine of x is written sin a;. The ratio of the ordinate to the abscissa is called the tangent of the angle. The tangent of x is written tan x. The ratio of the distance to the abscissa is called the secant of the angle. The secant of x is written sec x. These three ratios connect the ordinate, distance, and abscissa of a point ; but it is often convenient to use the reciprocal ratios. Imagine the triangle ABCi to be taken up from the plane and placed so that the vertex C'l falls at A, and the side CiB coincides with the initial line. B takes the position B\ and A the position A'. Then the ordinate of A' will equal the abscissa of (7i, and the abscissa of A' will equal the ordinate of C^. The three reciprocal ratios are : 16 FUNCTIONS OF ANGLES. First. The ratio of the distance of Ci (distance of A) to the ordinate of Oi (abscissa of J.') is the secant of B'AA' (=C, = y). Second. The ratio of the abscissa of Cy (ordinate of A') to the ordinate of (7i (abscissa of A') is the tangent of y. Tliird. The ratio of the abscissa of C] (ordinate of A^) to the distance of (7i (distance of A') is the sine of y. Now the angle Cj (=BAA' = y) is the complement of the angle BACi (=x); that is, y = 90°— x; on this account any function of y is called the function of the complement of X or the complement's function, or simply cof unction of x.* Although the prefix co was obtained by thinking of y, and therefore x as being acute, the functions of y or the cof unctions of x will be defined as applying to all angles. The co-named functions. — For any point on the terminal line of an angle : The ratio of the distance to the ordinate is called the cosecant of the angle. The cosecant of x is written esc x. The ratio of the abscissa to the ordinate is called the cotangent of the angle. The cotangent of x is written cot x. The ratio of the abscissa to the distance is called the cosine of the angle. The cosine of x is written cosic. When the group of functions, esc x, cot x, cos x, are referred to, they will be spoken of as the co-named func- tions of X. But it must be noticed that sin a;, tancc, secx are cof unc- tions ; that is, sin x is the cofunction of cos x, just as cos x is the cofunction of since. In the same way tanic and cotic are cofunctions of each other; secx and cscic are also cofunctions of each other. * Note that the complement of the complement of an angle is the angle itself [90° — (90° — x) = x] , and that any co-cof unction of the angle would be the function itself. RELATIONS BETWEEN FUNCTIONS. 17 RELATIONS BETWEEN FUNCTIONS AND SEGMENTS. 11. In each quadrant ordinate of C distance of _i distance of C ordinate of C That is, sin a; esc ic = 1 1 Similarly, tan x cot a; = 1 sec a? cos a; = 1 (1) Therefore sin x is the reciprocal of esc x, esc x is the reciprocal of sin x ; for this reason, sin x and esc x will be called reciprocal functions of a;; similar statements may be made of the other functions in group (1). In each quadrant, ordinate of C _ ordinate of C abscissa of C distance of C abscissa of C distance of G That is, Similarly, sin X = tan x cos x tan X = sec x sin x sec X = CSC X tan x csaa; = cot x sec x cot X = cos X CSC X cos a; = sin a; cot x (2) Solving the equations in group (2) for each function. sin a; tana; cos a; sec a; cot a; tana; sec a; CSC a; sin a; cos ay- esc a; tana; CSC a;: cota; = cot X sm X cos X = cot a; cos a;" cos a; sin a; sin a; sec a; tana; CSC a; sec a; cot a; tan X esc x (3) 18 FUNCTIONS OF ANGLES. Value of each function. If X is an angle of the first quadrant, the defini- tions give a g sm X = -'^ CSC X = tan x = ~: cot x = - c a sec a; cos a; = V (4) For the second quadrant, a a b b sm ic = - = + - ; CSC a; = - = 4- - 00 a a a ci ^ — c c tana;= = ; cota^ = = — — c c a a b b — c c sec X = = ; cos X = -^— = — - — c c' b b (5) And similarly for the other two quadrants. Values of the ordinate, distance, and abscissa of any point on the terminal line of an angle. — Clearing the equations in (4) of fractions gives for the first quadrant, a = b sin x = c tan x b = csecx = a(iSGx ■ (6) c =a cot x=b cos x The group in (5) gives for the second quadrant. a = b sin x = — c tan x b = — c secx = acsGx c = — a cot x = — b cos x And similarly for the other two quadrants. (7) MNEMONIC DEVICE. 19 sinjr coijr 12. Device for memorizing relations. — The diagram, Fig. 19, with the rules following, will be found useful in remem- bering the relations given in the groups of secjr equations (1) to (7). In " ^ tan^r the ligure, the angle x is drawn for the first quad- rant; but if the proper algebraic signs of a and c are supplied, the dia- gram and all the rules apply to an angle of any quadrant. It will be seen that three cross-lines are drawn through a and 6, a and c, h and c. Starting with the right angle and going in the positive direction of angles, the cross-lines are marked sin a;, tana;, seca;, esc a;, cot a;, cosaj, respectively. By going along the cross-lines, the relations in groups (4), (5), (6), and (7) are given by the two following rules : Rule I. Any segment is equal to the product of the two adja- cents ; thus along the horizontal cross-line, a = 5 sin a; and 5 = a CSC X. Rule II. Any segment or function is equal to the quotient of the adjacent by the next ; thus along the horizontal cross-line, sm a; = -, a CSC a; -, 6 = a CSC X sin a; By going around either way outside the triangle, consider- ing only the functions, equations (2) and (3) are given by two rules : Rule III. Any function is equal to the product of the two adjacents ; thus sin x = tan x cos x. 20 FUNCTIONS OF ANGLES. EuLE IV. Any function is equal to the quotient of the adja- cent by the next ; thus tancc . cos a; sm X = or sm x = sec a; cota^ Considering functions on the same cross-line, group (1) is given by KuLE V. Any function is equal to the reciprocal of its oppo- site function ; thus sin a; esc ic = 1. ABSOLUTE VALUES AND ALGEBRAIC SIGNS OF FUNCTIONS. 13. Definitions. The numerical value of a magnitude is the ratio of the magnitude to another magnitude of the same kind taken as a unit. The absolute value of a magni- tude is the numerical value of the magnitude considered without regard to sign. Thus the numerical value of — 5 feet is — 5, and the absolute value is simply 5. Absolute values of functions of angles in terms of functions of angles between 0° and 90°. — In Fig. 17, denote the positive acute value of angle BACi by x. Let the least positive value of angle BAC^ = 180° - x, of angle BAC^ = 180° + x, and of angle BAC^ = 360° — x. Also make the distances ACi, AC2, ACs, ACi all equal. Then the triangles BACiy B1AC2, BiACs, BAC^ are all equal. When the revolving half-line is at B or Bi, the angles generated from AB are all included in the expression n times 90°, where n is an even whole number. And when the revolving half-line is at D or Di, the angles generated from AB are all included in the expression n times 90°, where n is an odd number. Hence the angles generated when the terminal line is at ACi, AC2, ACs, or AC4 are included in the expressions 90n ± x where n is even, and 90n ± y where n is odd. REDUCTION- TO FIRST QUADRANT. 21 From the equality of the various triangles the ordinates and abscissse of Ci, Cg, Cg, and C^ are all equal in absolute value. Hence the functions of BACi, BAC23 -B-^Cs, BACi are equal in absolute value to the same functions of x ; and therefore also equal to the cof unctions of y. That is : I. The absolute value of any function of an angle dOn ± x where n is even, is the same as the absolute value of that func- tion of X. II. TJie absolute value of any function of an angle 90 /? ± / where n is odd is the same as the absolute value of that co- function of ^. Expressed in a formula, / standing for any function : /(90 n ± x)= ±fx when n is even = ± cofx when n is odd. For example : Any function of 150°, 210°, 330°, 390°, 510°, — 30°, — 150°, etc., is equal in absolute value to the same function of 30° or the co-named function of 60°. 22 FUNCTIONS OF "ANGLES. The algebraic signs of the functions may be remembered by this device : Write the functions in the order sin tan sec esc cot cos 2 3 4 2 3 4 The functions with positive values are : In Quadrant I. All of them. In Quadrant II. sin esc In Quadrant III. tan cot In Quadrant IV. sec cos In all other cases the functions are negative in value. The method of obtaining the value of a function of any angle in terms of an angle between 0° and 90° is given in these two rules : Rule I. Divide the angle by 90°. If the quotient is even, take the same function of the remainder. If the quotient is odd, take the cofunction of the remainder. Rule II. Prefix the proper algebraic sign. Thus sin 172°= cos 82°. esc 274° 9'=- sec 4° 9' tan 305° = - cot 35°. cot 329° 51 ' = - tan 59° 51 ' sec 595° = - sec 55°. cos 935° 22' = - cos 35° 22' Notice that the quadrant is indicated by 1 + the quotient. 14. Functions of 180° ± x, of 90° + x, and of - jr. — Con- sider any function, say the sine, of the two angles 180° — x and X. It has been shown that sin (180° — x) and sin x are equal in absolute value. Consider now their algebraic signs. If x is of the first quadrant, 180° — a; is of the second quadrant ; then sin x and sin (180° — x) are both +. REDUCTION TO FIRST QUADRANT. 23 If X is of the second quadrant, 180° — x is of the first quadrant ; then sin x and sin (180° — x) are both -|-. If x is of the third quadrant, 180° - a; is of I> the fourth quadrant ; then sinic and sin (180° -a;) are both -. If X is of the fourth quadrant, 180° — x is of the third quadrant; then sin a; and sin (180° -a;) are both -. Hence the equation sin (180° — a;) = sin x is true for all values of x. In a similar manner, it may be shown that the equation sin (90° + a;) = cos x is true for all values of x. Also that sin (180° + a;) = sin (— a;) = — sin a; are true for all values of X. In a similar manner, other functions of these angles may- be considered. The results are shown in the following table : sin (90° ■\-x)= cos x. tan (90° + a;) = — cot x. sec (90° -\-x) = — CSC x, esc (90° + a;) = sec x. cot (90° + a;) = - tan a;, cos (90° + a;) = — sin x. sin(— a;)= — sina;. tan (— a;) = — tan x. sec (—a;) = sec x. csc(— a;) = — CSC X. cot (—a;) = — cot x. cos(— a;)= cos a;. sin (180° -x)^ sin a?. tan (180° -x) = - - tanx. sec (180° - -x) = - - sec x. CSC (180° - -x) = CSC X. cot (180° -x) = - cot X. cos (180° — x) = — cos X. sin (180° -\-x) = — sin x. tan (180° + a;) = tana;, sec (180° + a;) = — sec x. esc (180° + a;) = — esc x. cot (180° + x)= cot X. cos (180° -\-x) = — cos X. 24 FUNCTIONS OF ANGLES. THE CURVE OF EACH FUNCTION. 15. If a point moves so that its distance remains the same, the ordinate of the moving point will be proportional to the sine of the varying angle; and the abscissa of the moving point will be proportional to the cosine of the angle ; for the sine of the angle equals the ordinate divided by the distance; and the cosine of the angle equals the abscissa divided by the distance ; and therefore if the distance re- mains the same, the sine varies as the ordinate ; and the cosine varies as the abscissa. Suppose, now, a second point to move in the plane so that its abscissa is proportional to the angle, and its ordi- nate is proportional to the sine of that angle. As the angle changes from 0° to 360°, this second point will trace out the Fig. 20. — Curve of sines. Curve of cosines. curve shown in Fig. 20, called a simple harmonic curve, or curve of sines. Or, if the ordinate of the second point was proportional to the cosine of the angle, the curve is called the curve of cosines. If the first point moves in a line at right angles to the initial line, the ordinate of this point will be proportional to the tangent of the angle, and the distance of this moving point will be proportional to the secant of the angle ; now if the abscissa of the second moving point was proportional to the angle and the ordinate proportional to the tangent of the angle, the curve traced out is called the curve of tangents (Fig. 21) ; or, if the ordinate of the second point CURVES OF FUNCTIONS. 26 \ / \ / V / \ / \ / \ / \ / \ / A / \ / \ / \ IW \ 1 \ / \ / \ / • \ / \ / \ / \ / /\ / \ / \ / \ / \ 36o' 90- \ / / \ / \ / \ / \ / \ 1 \ no' \ / \ / \ / /\ / \ / \ / \ / \ / \ / \ / \ Fig. 21. —Curve of tangents. Curve of cotangents. \ / / \ \ / / \ \ / \ / \ / / / \ \. / / \ 'y/ / / \ ^ ^^ ^ c" .^- iSo* »70* .,6o- /^ ^Vv^'' ^N / / \ / \ / \ \ \ \ / / \ \ / / \ / \ / \ \ \ \ Fig. 22. — Curve of secants. Curve of cosecants. 26 FUNCTIONS OF ANGLES. was proportional to the secant of the angle, the curve is called the curve of secants (Fig. 22). In a similar manner, the first point may be assumed to move parallel to the initial line, and the second point may be made to trace the curve of cosecants (Fig. 22), or the curve of cotangents (Fig. 21). LIMITING VALUES OF FUNCTIONS. 16. Since from geometry the hypotenuse of a right tri- angle can never be less than either side, the distance of a point can never be less than either the abscissa or ordinate of the point. Therefore the absolute value of the sine or of the cosine of any angle is never more than 1, and the absolute value of the secant or of the cosecant of any angle is never less than 1. The absolute value of the tangent or of the cotangent of angles is without limit. THE ANGLES EACH LESS THAN 360° CORRESPONDING TO A GIVEN VALUE OF A FUNCTION.* 17. From what has been said, it follows that there is but one value of any function of a given angle ; but for a given numerical value of any function of the angle, there will be 0, 1, or 2 values of the angle between 0° and 360°. 1. If sin X or cos x is more than 1 or less than — 1, or if sec X or esc x is between 1 and —l,x will be impossible. 2. If sin a? = H- 1 or esc x= -rl, x= 90°. If sin a; = — 1 or CSC a? = — 1, x = 270°. If sec a; =+ 1 or cos x = -|- 1, x = 0°. If sec a; = — 1 or cos a; = — 1, a; = 180°. 3. In all other cases, two values of x between 0° and 360° will be obtained. * This section can be more easily followed if the diagrams, Figs. 20, 21, and 22, are referred to. TO FIND AN ANGLE FROM A FUNCTION. 27 Suppose that sin jr = iV (or esc x = N), and that a^i is the corresponding acute value of x. Then x also equals 180° - X,. If the given equation is sin x= — N (or esc x= — N) and sin Xi = N (or esc x^ = N), where x^ is acute, then X = 180° + a^i or a; = 360° - Xi. If tan x= N (or cot x = N), and Xi is the acute value of x, another value of x is 180° + Xi. If the given equation is tan x = — N (or cot x = — N) and tan Xi = N (or cot Xi = N) where Xi is acute, then x = 180°-Xi or a; = 360°-aj. If sec jr = -^ (or cos x = N) and iCi is the acute value of x, another value of x is 360° — x^. If the given eqviation is sec a; = — ^ (or cos x = — N) and sec Xi= N (or cos x^ = N) where Xi is acute, then a; = 180°-a;i or a;=180° + a:i. Angle less than 180°. — If the problem of obtaining the values of an angle from a given function of the angle is still further limited by the condition that the angle shall be less than 180° (say an angle of a triangle), then : (1) If the given function is a sine or a cosecant with a negative value, no value of the angle could be obtained ; because all angles between 0° and 180° have a positive sine and a positive cosecant. (2) If the given function is a sine or a cosecant with a positive value, and the given value of the function was pos- sible, there will be two supplementary values of the angle ; because if Xi is the acute value, 180° — Xi will be the obtuse value. If x^ happens to be 90°, it may be said that the two values are equal, or there is but one value. (3) If the given function of the required angle is any other function than the sine or cosecant, there can be but 28 FUNCTIONS OF ANGLES. one value of the angle less than 180° ; acute if the value of the function is positive; obtuse if the value of the func- tion is negative. If it is known that a required angle was not greater than 90°, a negative value of any 'function of the angle would indicate that no value of the angle was possible. FUNCTIONS OF ^x AND 2 X. 18. Functions of i jr. — Bisect angle BAC, and draw BJE parallel to AD. Denote angle BAC by x. Then angle E = angle ABE = an- gle BAD = ix. Also AE = AB. BC^ = AC^-AB'. Divide by BC. 1 = CSC- X — cot^ X. (8) AB^EA^EC^AC AB BD BD BC BC BC' cot i cc = CSC X -f- cot X = smx EB EC , BD BC . EB ^ BD BC = — - and = ; whence — — x • = — -^• AD AC EA EC' AE AD AC 2 cos i a; sin i a; = sin x. (10) The quotient of (8) and (9) is tan 4- aj = esc a; — cot a; = -^ (11) ^ sma; ^ The product of (10) and (11) is 2 sin^ i a; = 1 — cos x. (12) The product of (9) and (10) is 2cos2|-a; = l-f cosa;. (13) The half difference of (13) and (12) is cos^ ^x — sin^ i a; = cos x. (14) FUNCTIONS OF PARTICULAR ANGLES. 29 The reciprocal of (10) is ^ sec ^ a; CSC ^ it- = esc x. (15) Functions of 2 x. — Denote now the angle BAD by x. Then from (10), (14), and (15) : sin 2 a; = 2 sin x cos x. (16) cos 2x = cos'^ X — sin^ x. (17) CSC 2 a; = "I" sec a; esc x. (18) The other functions can then be found in terms of these. FUNCTIONS OF PARTICULAR ANGLES. 19. Functions of 45°, 22° 30', etc. — Let angle ^ = 45°. Denote the length of AB by c ; then BC = c and AC = V^^T^^ = 1-4142 c. Hence tan 45°= cot 45° = 1. sec 45° = esc 45° = 1.4142. Also, sin 45° = cos 45° = .7071. AZ ^b Fig. 24. From these values of the functions of 45°, the values of the functions of 22° 30' can be found by means of formulae (9) to (13). Then from the values of the func- tions of 22° 30', the functions of 11° 15' can be found; and so on. Functions of 60°, 30°, 15°, etc. — The tri- C angle having angles of 90°, 60°, and 30° is one-half of an equilateral triangle. Take BC=a; then AC =2 a and AB =\^4.a'-a^ = 1.7321 a. ^^^,; Fig. 25. Hence the functions of 60° and 30° can be computed. And, therefore, the functions of 15°, 7° 30', etc., may be computed. 30 FUNCTIONS OF ANGLES. Functions of 72°, 36°, 18°, etc. — Let ^C= CD, and DE bisect angle D, CB bisect angle (7; take the value of angle A to be 72°. All the other angles on the fig- ure can then be found. From the equality of angles, AD = ED = EC ; also the tri- angle ACD is similar to the triangle ADE. Denote the length of ^C by h, and the length of ^5 =6 cos 72° by hx. Then AE=b-'2hx. From the similar triangles ACD and ADE, AG^^AD ^^ b ^ 2bx AD AE °^ 2bx b-2bx Solving : a? = cos 72° = sin 18° = i ( V5 - 1) = .3090. Denote the length of BC by by ; then BC = AC - AB' or by = 6^ _ 52^ That is, y'=l-x^ 01 sin^ 72° = cos^ 18° = 1 - .30901 Hence the other functions of 72° and 18° may be found. Also the functions of 36°, 9°, 4° 30', etc., may be found. Exercises. — For all values of x, prove that sin^ X -\- cos^ ic = 1, sec^aj — tannic = 1, csc^ X — cot^ X = 1, and explain how all the functions of an angle may be found when the value of only one function is known. CHAPTER II. EXPLANATION OF THE TABLES. I. NATURAL VALUES OF FUNCTIONS. 20. The values given for the functions of angles in The Table of Natural Values of Functions, page i to page vii, are approximate values to four significant figures at least. The last figure given in each value is the nearest figure in that place. Thus the value of sin 32°, which is not less than .52985 and not greater than .52995, is given on page v, line 25, as .5299. Any value printed without a decimal point will have one or more figures missing. The decimal point and the missing figures will be found to the left and slightly above the fig- ures printed for the value. Thus page v, line 29, sin 32° 40' is .5398, the 398 being printed in the proper place and .5 being found higher and to the left. tan 32° 40' = .6412, sec 32° 40' = 1.1879, etc. Given an angle to find any function of the angle — Angles between 0° and 45° will be found in the column marked x on the left of the page, and angles between 45° and 90° in the column marked^ on the right. When the angle is found on the left, the names of the functions are given at the top of the columns, x being added to the name of each function. When the angle is found on the right, the names of the functions (marked/) are given at the bottom of the page; thus : 31 32 EXPLANATION OF THE TABLES. Page V, line 39 sin 34° 20' = .5640 Pagevi, line 9 tan 53° 40' = 1.3597 Page iv, line 2 sec 21° 10' = 1.0723 Page vii, line 16 esc 45° 30' = 1.4020 Pageiv, line 26 cot 64° 50'= .4699 Page iv, line 7 cos 22° = .9272 To find function of angles which are not multiples of 10'. — When we desire the function of an angle, whose value is between two angles of the table, a process called interpolation is used. By reference to any part of the table, it will be seen that small changes in the angle produce nearly proportional changes in the values of any function of the angle. A four-place table usually gives the values of the functions of angles for every 10'. By assuming that the change in any function is pro- portional to small changes in the angle, the value of the function of an angle to a single minute can in general be found to four significant figures without appreciable error. Conversely, if the value of the function is correct to four significant figures, the value of the angle can in general be found correct to a single minute. A five-place table giving the values of the functions for each minute can be used to interpolate to 10"; a six-place table for every 10" can be interpolated to 1" ; a seven-place table for every second can be interpolated to .1", and so on. With every increase in the number of significant figures, the degree of precision is increased and also the labor of using the tables. If the data of a computation contain angles measured to a single minute, four-place tables can be used in the computation; if the angles are measured to one-tenth second, seven-place tables can be used; and so on. INTERPOLATION. 33 Let /stand for any function. Then if h is a small addition to angle x, and z is the change in any function of x due to a change k (10') in the angle, it is assumed f(x + h)-fx ^h z k or f{x^-h)=fx + h, K from which the value of a function of an angle x + h can be obtained by substitution. Thus : sin 34° 23' = sin 34° 20' + .3 (sin 34° 30' - sin 34° 20') = .5640 -f .3 X .0024 = .5647. tan 53° 44' = tan 53° 40' + .4(tan53° 50' - tan 53° 40') = 1.3597 + .4 X .0083 = 1.3630. sec 21° 11' = sec 21° 10' + .1 (sec 21° 20' - sec 21° 10') = 1.0723 + .1 X .0013 = 1.0724. esc 45° 38' = CSC 45° 30' + .8 (esc 45° 40' - esc 45° 30') = 1.4020 + .8 X (- .0040) = 1.3988. cot 64° 55' = cot 64° 50' + .5 (cot 65° - cot 64° 50') = .4699 + .5 (- .0036) = .4681. cos 22° 7' = cos 22° + .7 (cos 22° 10' - cos 22°) = .9272 + .7 (- .0011) = .9264. It should be noted that all of the co-named functions of acute angles decrease as the angle increases. Hence when the tabulated value of a co-named function of an acute angle is corrected for an addition to the angle, the cor- rection will be negative. If functions of angles greater that 90° are desired, they are obtained by applying the rules on page 22 ; thus (Quad- rant II), 34 EXPLANATION OF THE TABLES. sin 155° 12 ' = cos 65° 12 ' = .4195. tan 113° 47' = - cot 23° 47' = - 2.2691. sec 167° 9' = - CSC 77° 9' = -1.0257. CSC 123° 42' = sec 33° 42' = 1.2020. cot 147° 56' = - tan 57° 56' = - 1.5962. cos 134° 35' = - sin 44° 35' = - .7019. And in a similar manner for angles of Quadrants III and IV. Given the value of a function of the angle, to find the value of that angle. — The table is arranged to give the value of the function when the angle is known. The reverse operation of obtaining the angle when the function is known is some- what more inconvenient. Assume that we know the sine of the required angle ; the given value may be sought either in the column marked sin x at the top of the page or in the column marked sin/ at the bottom of the page. If the number is found in the column marked sinjr, and it is known that the angle is acute, the value of the angle will be found in the column marked jr on the left of the page. If the given value of the sine is found in the column marked sin/, the value of the acute angle will be found in the col- umn marked /. Any other function will be found in a similar manner; thus: Page V, line 27. If sin^ = .8450, A (acute) = 57° 40'. Page V, line 4. If tan A = 1.8418, A (acute) = 61° 30'. Page V, line 41. If sec A = 1.2158, A (acute) = 34° 40'. Page iv, line 33. If esc A = 2.2543, A (acute) = 26° 20'. Page V, line 34. If cot ^ = .6619, A (acute) = 56° 30'. Page vi, line 27. If cos A = .7735, A (acute) = 39° 20'. If the values of the functions of 45° are memorized (sin 45° = .7, tan 45° = 1, sec 45° = 1.4), INTERPOLATION TO FIND ANGLE. 35 it will be known whether the given value of the functions should be sought in the columns on the right of the page or on the left. Thus if sin A = .9112, the value of A must be more than 45°, because the sine of an acute angle increases as the angle increases, and the sine of 45° is .7. Since the angle is more than 45°, the given value is sought in the column marked sin y. As before, let / stand for any function, h a small addition to angle x, and z a change in any function due to a change Jc (10') in the angle. Then, since small changes in the angle are nearly proportional to the changes in any function, k z , f(x-\- Ji) — fx - x-\-h = x-{-' '^ ' — —k. z If sin A = .6479, the acute value of A = 40° 20' + •^^'^9 - •^^'^'^ 10' = 40° 23'. .6494 -.6472 If tan A = 3.7539, the acute value of A — 7K° _!_ 3.7539 — 3.7321 -tm _ 7^0 Kf If sec A = 1.2190, the acute value of A ^ 3^0 50, ^ 1.2190-1.2183 ^0, ^ 3^0 53, ^1.2208-1.2183 If esc A = 3.9984, the acute value of A = 14° 20' + ^-^^^^ - ^-^^^^ 10' = 14° 29'. 3.9939 - 4.0394 If cot A = 2.2062, the acute value of A = 24° 20' + ^-^Q^^ - 2.2113 ^Q, ^ 24° 23'. 2.1943-2.2113 If cos A = .5654, the acute value of A = 55° 30' -f '^^^^ ~ ^^^^ 10' = 55° 34'. ^ .5640 - 5664 36 EXPLANATION OF THE TABLES. If both values of the angle which are less than 360° are desired, any value, not acute, can be obtained by a reversal of the rules on page 22 ; thus. If sin A = .6479, A = 40° 23' or 139° 37'. If tan^ = 3.7539, A = 75° 5' or 255° 5'. If sec A = 1.2190, A = 34° 53' or 325° 7'. If CSC A = 3.9984, A = 14° 29' or 165° 31'. If cot A = 2.2062, A = 24° 23' or 204° 23'. If cos A = .5654, A = 55° 34' or 304° 26'. . Also when the given function is negative : If sin ^ = - .6479, A = 220° 23' or 319° 37'. If tan ^ = - 3.7539, A = 104° 55' or 284° 55'. If sec ^ = - 1.2190, A = 145° 7' or 214° 53'. If CSC ^ = - 3.9984, A = 194° 29' or 345° 31'. If cot ^ = - 2.2062, A = 155° 37' or 335° 37'. If COS ^ = - .5654, A = 124° 26' or 235° 34'. It should be noted that if the given value of the sine or cosine of an angle is greater than + 1 or less than — 1, no value of the angle can be obtained ; and if the given value of the secant or cosecant of an angle is between +1 and — 1, no value of the angle can be obtained. Exercises. — Copy one value in any row of the following table. Then find each of the remaining values in the row. sin 0°43' sin 179° 17' .0125 cos 89° 17' cos 270° 43' sin 10°18' sin 169° 42' .1788 cos 79° 42' cos280°18' sin 200° 41' sin 339° 19' -.3532 cos 110° 41' cos 249° 19' sin 210° 6' sin 329° 54' -.5015 cos 120° 6' cos 239° 54' sin 40° 23' sin 139° 37' .6479 cos 49° 37' cos 310° 23' LOGARITHMS. 3T tan 84° 15' tan 264° 15' tan 74° 18' tan 254° 18' tan 157° 13' tan 337° 13' tan 141° 33' tan 321° 33' tan 135° 19' tan 315° 19' sec 22° 34' sec 337° 26' sec 63° 12' sec 296° 48' sec 99° 11' sec 260° 49' sec 128° 43' sec 231° 17' 9.9310 cot 5° 45' cot 185° 45' 3.5576 cot 15° 42' cot 195° 42' -.4200 cot 112° 47' cot 292° 47' -.7940 cot 128° 27' cot 308° 27' -.9890 cot 134° 41' cot 314° 41' 1.0829 CSC 67° 26' esc 112° 34' 2.2179 CSC 26° 48' esc 153° 12' -6.2659 CSC 189° 11' esc 350° 49' -1.5988 csc218°43' csc321°17' sec 107° 47' sec 252° 13' -3.2742 esc 197° 47' esc 342° 13' II. LOGARITHMS OF NUMBERS. 21. Definition. — Logarithms are a series of numbers sever- ally assigned to the ordinary numbers, and so arranged that the multiplication and division of ordinary numbers is accom- plished by means of the addition or subtraction of their loga- rithms. Many systems of logarithms are possible ; but in the simplest systems the ordinary numbers are powers of a number called a base, and the logarithms are the corresponding exponents. Thus, if the base is 2, the num- bers are powers of 2, and the cor- responding exponents are called logarithms to the base 2. To multiply 64 by 16, we add the logarithms of 64 and 16 (6 and 4), producing 10 ; and 10 is the logarithm of 1024, the product. Numbers Logarithms OR OB Powers. Exponents. 1 2 1 4 2 8 3 16 4 32 5 64 6 128 7 256 8 512 9 1024 10 38 EXPLANATION OF THE TABLES. To divide 512 by 32, subtract their logarithms (9 and 5), giving 4 ; and 4 is the logarithm of 16, the desired quotient. To raise 4 to the fifth power, multiply the logarithm of 4 (2) by the index of the power (5), producing 10, the loga- rithm of 1024, the required power. To obtain the cube root of 512, divide the logarithm of 512 (9) by the index of the root (3), giving 3, the logarithm of 8, the desired root. Apply to logarithms the rules of algebra relating to exponents. Then we have Log of product = sum of logs of factors. Log of quotient = log of dividend - log of divisor. Log of power of a number = log of number x index of power. Log of root of a number = log of number ^ index of root. CHARACTERISTIC AND MANTISSA. If we think of integers as being arranged with zeros expressed decimally (thus, 7 = 7.0000), all logarithms may be said to consist of two parts, — an integral part and a fractional part. The integral part of a logarithm is called its characteristic. The fractional part is called its mantissa. Thus, if 2.1347 is a logarithm, 2 is its characteristic, and .1347 is its mantissa. The characteristic may be either positive or negative, but the logarithm is so arranged that the mantissa is always positive. Thus, if the logarithm was — 2.4178, the number 3 would be added and subtracted thus, 3 — 2.4178 — 3 = .5822 — 3. In this logarithm, — 3 is the characteristic, and .5822 is the mantissa. Exercises. — Find the characteristic and mantissa of the following logarithms : 4|-, — If, -|, — i^-, — f . CHARACTERISTIC AND MANTISSA. 39 DENARY LOGARITHMS. Although a practical table of logarithms could be con- structed by taking any positive number as the base of the system, the common system of logarithms has 10 for its The system whose base is 10 has two principal advan- tages : (1) The mantissa of a logarithm depends only on the figures of the number and their order; or, to express it another way, the mantissa of the logarithm of a number is independent of the position of the decimal point in the number. For example, the numbers 4.57 and 4570 have the same mantissa in their logarithms. (2) The characteristic of a logarithm depends only on the position from the units place of the first significant figure of the number ; or, to express it another way, the charac- teristic of the logarithm of a number is independent of the figures in the number and of the order of the figures. For example, the numbers 45.7 and 23.8 have the same charac- teristic in their logarithms. To prove the first of these principles, say the logarithm of 4.57 is 0.6599, and that the logarithm of A is a. Then 10«-^^= 4.57. 10- =A. But W =1000 103 =1000. Hence lO^-^^^ = 4570. 10" + ^ = looo A. It will be noticed that the decimal point of a number is moved to the right, when the number is multiplied by an integral power of 10. Therefore the logarithm of the prod- uct will be increased by this integer (the exponent of the multiplier). But adding an integer to a logarithm does not affect the fractional part of the logarithm. Hence the man- tissa of the logarithm is unchanged. 40 EXPLANATION OF THE TABLES. In a similar manner, it may be shown that a motion of the decimal point to the left will leave the mantissa of its logarithm unchanged. To prove the second of these principles consider the follow- ing table of logarithms, the base of the system being 10: From an inspection of the table it will be seen that all numbers between 1000 and 10,000 (numbers with the first significant figure three places to the left of the units place) will have logarithms between 3 and 4, and hence 3 + a proper fraction; and in general that numbers with the first significant figure n places to the left of the units place will have their logarithms between n and n + 1, and hence ?i + a proper fraction. Also that all numbers between .0001 and .001 (numbers with the first significant figure four places to the right of the units place) will have their logarithms between — 4 and — 3, and hence — 4 + a proper fraction ; and in general that numbers with the first significant figure n places to the right of the units place will have their loga- rithms between —n and — (?i — 1), and hence — n + a proper fraction. For example : The number 4570 is between 10^ and 10* ; hence its logarithm is 3 -f a proper fraction. The num- ber .00457 is between 10~^ and 10~^ ; hence its logarithm is — 3 + a proper fraction. It follows, from these two principles, that in the con- struction of a table of logarithms to the base 10 : Numbers. Logarithms. .0001 -4 .001 -3 .01 -2 .1 -1 1 10 1 100 2 1000 3 10000 4 TO OBTAIN A LOGARITHM. 41 (1) The mantissae need be given but once for numbers having the same figures in the same order. (2) The characteristics need not be given, but can be sup- plied by the following rule : Make the characteristic equal to the number of places by which the first significant figure of the number is removed from the units place : positive if the first significant figure is to the left of the decimal point, negative if to the right. THE USE OF THE TABLE OF LOGARITHMS. 22. To find the logarithm of a number. — The mantissae of the logarithms of numbers from to 1000 are printed on pages viii and ix of the tables at the back of the book. If any mantissa is printed without the decimal point, a figure of the mantissa has also been omitted ; the decimal point and missing figure will be found above and to the left of the figures in the table. The first two significant figures of the number will be found in the column marked N, and the third significant figure at the top of the columns. If a given number has but one significant figure, two zeros are added ; if but two significant figures, one zero is added. Thus : PAGE VIII. PAGE IX. Line. Number. Logarithm. Line. Number. Logarithm. 11 2.00 0.3010 11 .065 .8129 - 2 14 23.0 1.3617 14 .068 .8325 - 2 17 2.61 0.4166 17 .717 .8555 — 1 20 29.4 1.4683 20 .074 .8692 - 2 23 32.7 1.5145 23 .777 .8904 - 1 36 456 2.6590 36 .009 .9542 - 3 38 4.70 0.6721 38 .920 .9638 - 1 42 EXPLANATION OF THE TABLES. To find the logarithm of a number of four significant figures, the assumption is made that small changes in the numbers produce proportional changes in the logarithms. This assumption is only approximately true; that it is nearly true may be seen from an inspection of the table. The error in the assumption will not, in general, affect the fourth figure of the mantissa of the logarithm. Thus, let m stand for mantissa and log for logarithm. Then m log 200.2 = m log 200 + .2 (m log 201 - m log 200) ; = .3010 + .2 (.3032 - .3010). Therefore log of 200.2 = 2.3014. m log .02307 = m log 230 + .7 (m log 231 - m log 230) ; = .3617 + .7 (.3636 - .3617). Therefore log of .02307 = .3630 - 2. Given a logarithm, to find the corresponding number. — Let log iV= 1.4997. Then (page viii, line 22) the significant figures of N are 316. The characteristic indicates that the figure 3 is one figure to the left of the units place ; hence the figure 1 is in the units place and N= 31.6. Let log N = .5453 — 2. The significant figures of J^ are 351, and since the 3 is two figures to the right of the units place, N= 0.0351. Let log N= 2.3014. Then jsr= 200 + -^Qi^ - -^QiQ ^ 200.2. .3032 - .3010 Let log N = .3630 - 2. Then 3630 - .; .3636 - .3617 Q23Q -3630 -.3617 ^ Q2307. INTERPOLATION. 43 By assuming either column in this table find the other. Number. Logarithm. Number. Logarithm. 10.62 1.384 1886 273.1 1.0261 0.1411 3.2755 2.4363 4.426 56.81 6.282 733.5 0.6460 1.7544 0.7981 2.8654 Other methods of writing negative characteristics. — To avoid printing negative characteristics, it is usual to arbitrarily increase the characteristic by 10. Thus, log of .01657 = .2193 — 2 will sometiriies be written 8.2193, which is the logarithm of 165,700,000, or just 10,000,000,000 times as great as the real value. In such a case, L will be used instead of log. Thus log .01657 = 8.2193 -10 and X. 01657 = 8.2193. Suppose the cube root of .01657 is desired. Since log .01657 = .2193 - 2 or 8.2193 - 10, log V .01657 = .0731 - .6667 or 2.7398 - 3.3333. To avoid this negative mantissa, the logarithm of .01657 is arranged before division, thus : log .01657 = 1.2193 - 3 or 28.2193 - 30. Therefore log v .01657 = .4064 - 1 or 9.4064 - 10. That is : In taking the root of a proper fraction by loga- rithms, see that the negative part of the logarithm is divisi- ble by the index of the root. Negative numbers. — The suffix n placed after a logarithm indicates that the number corresponding to the logarithm is negative. If, in obtaining the product of several numbers, there is an odd number of negative factors, the suffix n will 44 EXPLANATION OF THE TABLES. occur an odd number of times ; therefore, by the principles of algebra, the logarithm of the product should have the suf- fix n. But if there is an even number of negative factors, the suffix n will appear an even number of times ; therefore the logarithm of the product will not contain the suffix n. In a similar manner, the rules of algebra will serve as a guide in the use of the suffix n, in the operations of division, involution, and evolution. 1. Compute X = 46.12 X .7827. log 46.12= 1.6639 log .7827= 9.8936 -10 log X = 11.5575 — 10 a;=:36.1 2. Compute X = 525.1 -- 78530. log 525.1 = 12.7202 - 10 log 78530= 4.8950 loga;= 7.8252 - 10 X = 0.006687 3. Compute X = 5850 H- 0.04754. log 5850 = 13.7672 - 10 log .04754 = 8.6771 - 10 log X = 5.0901 X = 123100 4. Compute 4 fl62.4 X 238.9 x 115.5 516.8 log 162.4 = 2.2106 log 238.9 = 2.3782 log 115.5 = 2.0626 6.6514 log 516.8 = 2.7133 log X' = 3.9381 log X = 1.9690 X = 93.12 5. Compute 3/129.97 X .0499 =4 278.7 log 129.97= 2.1138 log .0499= 8.6981-10 10.8119 - 10 log 278.7= 2.4451 log 0^ = 28.3668 - 30 logx= 9.4556-10 X = 0.2855 Exercises. — Use problems on page 10. EXERCISES. 45 III. LOGARITHMS OF FUNCTIONS. 23. The table, pages x to xvi, Logarithms of Functions, contains the logarithms of the natural functions of angles from 0° to 90°. To avoid printing negative characteristics, the characteristics of logarithms of numbers less than 1 have been printed 10 too large (indicated by the letter L in the first, second, and sixth columns). Hence —10 will be understood after such logarithms. Just as in the other tables, any number printed without the decimal point will have also one or more figures missing, which figures will be found above and to the left. The method of interpolation is the same as in the use of the table of natural functions. Exercises. — Copy a value in any row in the following table; then find each of the remaining values in the row: sin 18° 5' sin 161° 55' 9.4919 cos 71° 55' cos 288° 5' sin 26° 6' sin 153° 54' 9.6434 cos 63° 54' cos 296° 6' sin 25° 52' sin 154° 8' 9.6398 cos 64° 8' cos 295° 52' sin 196° 14' sin 343° 46' 9.4465 n cos 106° 14' cos 253° 46' sin 251° 31' sin 288° 29' 9.977071 cos 161° 31' cos 198° 29' tan 112° 14' tan 292° 14 tan 101° 46' tan 281° 46 tan 103° 17' tan 283° 17 tan 177° 37' tan 357° 37 tan 34° 44' tan 214° 44 sec 22° 13' sec 337° 47 sec 142° 23' sec 217° 37 sec 76° 8' sec 283° 52 sec 96° 26' sec 263° 34 sec 69° 36' sec 290° 24 0.3885 n cot 157° 46' cot 337° 46' 0.6813 ?i cot 168° 14' cot 348° 14' 0.6269 n cot 166° 43' cot 346° 43' 8.6193 n cot 92° 23' cot 272° 23' 9.8409 cot 55° 16' cot 235° 16' 0.0335 esc 67° 47' esc 112° 13' 0.2144 n CSC 232° 23' esc 307° 37' 0.6204 CSC 13° 52' esc 166° 8' 0.9506 n CSC 186° 26' esc 353° 34' 0.4577 CSC 20° 24' esc 159° 36' CHAPTER III. RIGHT TRIANGLES. 24. The primary object of trigonometry is the investiga- tion of the methods of computing the numerical values of the sides and angles of plane triangles. When, from any data, such numerical values have been obtained, the triangle will be said to be solved. With reference to the number of possible answers, a problem is said to be indeterminate, ambiguous, determinate, or impossible: (1) When there is an infinite number of answers, the problem is indeterminate. An algebraic illustration is : find X and y from the equation x — y = ^. (2) When there is more than one answer, but not an infinite number of answers, the problem is ambiguous. For example : find x from the equation a^ + 6 = 5 ic. (3) When there is one answer and but one answer for each unknown, the problem is determinate. For example : find X and y from the equations x-{- y = 7 and x — y = 3. (4) When there is no answer, the problem is impossible. For example : find x from the equation 2 x—5 = negative square root of (ot^ — 7). In trigonometry, negative and imaginary answers are, in general, rejected, as not indicating true solutions. It is known from geometry that at least three indepen- dent conditions are necessary to make the solution of a 46 RELATIONS OF ANGLES AND SIDES. 47 triangle determinate.* In most cases, three such conditions are sufficient. Two classes of triangles will be considered, — right and oblique. Any oblique tri- angle can be divided into two right triangles by let,- ting fall from any vertex a perpendicular upon the opposite side. Hence the solution of all triangles depends upon the solution of right triangles. Fig. 27. 25. In a right triangle, let A denote the value of angle A^ B denote the right angle, and C the value of angle C. Also let a and c denote the lengths of the sides of the triangle opposite A and C, respectively, and b denote the length of the hypotenuse. Relations of angles. — In a right triangle, one of the three conditions necessary to determine its size and its shape is explicitly given. For the angle B = 90°. Therefore A+C=90°; A = 90°-C', C = 90°-A. Hence, when either acute angle of a right triangle is given, the other acute angle is readily found. Relations of sides. — The relation existing between the sides of a right angle is well known. It is 6= Vc^ + a'; c= V(?> + a)(6-a); a= V(6 + c) (6 - c). Hence, if any two of the sides of a right triangle are given, the other side can be computed. * This is expressed by saying that the triangle is of triple mani- foldness. 48 RIGHT TRIANGLES. Compute a, given 6= 145, c = 143. a = V(145 -f 143) (145 - 143) = 24. Compute c, given a = 399, b = 401. c = V(401 + 399) (401 - 399) = 40. Exercises. — Given b and c ; find a in each of the following ; b c a 919.2 918.94 21.90 865.1 860.74 86.90 888.9 888.84 10.34 150.0 149.50 12.29 525.1 523.84 36.35 Relations of sides and angles. — Assume AB to be an initial line, c the abscissa, a the ordinate, and b the distance of C. tan A sin A The rules on page 19 give Any required side = product of adjacent side and adjacent function. SOLUTIONS USING NUMBERS. 49 Any function oi A = quotient of the adjacent side by the next side. If it is desired to use the angle C in a computation, the letters a and c, A and (7, can be interchanged in Fig. 28, and the rules applied to the resulting figure. SOLUTIONS. 26. 1. Given 6 = 120, A = 11° 25'; find a. The horizontal cross-line in Fig. 28 connects the given side b and the required side a ; hence by the rule, a = 6 sin A. That is, a = 120 sin 11° 25' = 120 X .1979 = 23.75. 2. Given c = 50.7, ^ = 57° 7'; find 6. Here we use the vertical cross-line, since that connects the given side and the required side. Hence b = c sec A. That is, 6 = 50.7 sec 57° 7' = 50.7 X 1.8419 = 93.38. 3. Given a = 71, A = 65° 32' ; find c. The oblique cross-line gives c = a cot ^ = 71 cot 65° 32' = 71 X .4550 = 32.3. 4. Given 6 = 75, a = 43.8 ; find A. Here A can be found by means of its sine or its cosecant. As the division by 75 is slightly easier, we select the sine. sin ^ = — =.5840; ^ = 35° 44'. 75 60 RIGHT TRIANGLES. 5. Given a = 849, c = 920 ; find A. Here the oblique cross-line connects the two given sides. As 920 is the easier divisor, 849 920 tan^ .9228; ^ = 42° 42'. 6. Given b = 401, c = 399 ; find A. Here cos .4 = |^ = .9950. 401 In this solution, A cannot be determined with much pre- cision as the table gives 5° 43', 5° 44', or 5° 45'. Hence find a first ; a = 40. Then csc^ = ^ = 10.025; A 5° 43^'. It will be noticed that the cosines and secants of small angles change slowly ; and that sines and cosecants of angles near 90° change slowly. When precision is desired, such functions of these angles are avoided. The following table gives the three sides and one angle of eight different right triangles. Assume any two parts of any one of these triangles and compute another part. a b c A 4 5 3 53° 8' 12 13 5 67° 23' 24 25 7 73° 44' 20 29 21 43° 36' 12 37 35 18° 55' 28 53 45 31° 53' 60 109 91 33° 24' 84 205 187 24° 11' APPLICATIONS. 61 Fig. 29. Fig. 30. PROBLEMS. [Answers to these problems will be found on page 177.] 27. 1. A tower, BC, stands on flat horizontal ground ; a distance of 250 feet (c) was measured from the base of the tower to a point A. At this point the angle between the horizontal {AE) and the visual ray (^0) to the top of the tower was found to be ^ = 22° 20'.* From these data, com- pute the height of the tower. 2. A lighthouse, BC, is 87 (a) feet high. From the top C, the angle between a horizontal line CD and the visual ray to a boat A is found to be 12° 40'.t From these data compute the distance of the boat from the base of the lighthouse. 3. Two forces, AB of 240 lbs. and AD of 100 lbs., act at right angles upon the same point of a body. Find the angle which their resultant AC makes with the larger force. 4. The length of the projection of ^C on a line AB is 23 inches ; the angle which AC makes with this projection is 38°. Find the length of AC. 5. A cannon ball, fired in the air at an angle of 33° 3' with the horizon, passes over 880 feet in one second. How high is it at the end of the second ? t * This angle is called the angle of elevation of the top of the tower, as seen from A. t This angle is called the angle of depression of the boat as seen from C. t This increase of height per second is called the vertical component of the velocity. 52 RIGHT TRIANGLES. « 6. In order to measure the breadth of a river, a base line AB was measured along the bank, the point B being directly opi)osite a tree C on the other bank. The base line AB was 96 (c) feet, and the angle BAG was found to be 52° 30' (A). What is the breadth of the river ? * 7. When the rays of the sun make with the horizontal plane an angle of 53° 35' (^),t a vertical pole throws a shadow 11.8 (c) feet long. Find the length of the pole if its lower end is 1 foot (d) underground. 8. From a window of a house, the angle of ele- vation of the top of a church tower on the op- posite side of the street was 56°36'(^), and the angle of depression of a^^- the base of the tower was 10° 24' (A)- The street being 60 feet (c) wide, find the height of the steeple. 9. A house is 24 feet (a) high. From the top of the house, the angle of elevation of a church tower on the opposite side of the street is 36° 54' (A), and the angle of depression of the base of the tower is 18° 56' (Ai). Find the height of the steeple. 10. From the top of a hill, the angles of depression of the top and of the bottom of a house 44 feet (a) high are found to be 28° 4' {A) and 43° 36' (^i), respectively. Find the height of the hill above the ground on which the house stands. * A segment drawn across an angle is said to subtend the angle. Thus in a triangle ABC, BC subtends the angle A. t The angle which the sun's rays make with the horizon is equal in degrees to the arc which measures the altitude of the sun. Jc. Fig. 31. c /i APPLICATIONS. 53 11. From the top of a bluff overhanging a river, the angle of elevation of the top of a tree on the other side of the river is 5° 22' (^), and the angle of depression of the bottom of the tree is 3° 40' (yli). The tree is known to be 79 feet (a) high. What is the breadth of the river ? 12. A man sailing due y / north observes two objects y' / directly west. After sail- ,'^ /' ing 12 miles (a), the direc- ^^^ / tions of the objects make X_ /_ B angles of 22° 38' (C)* and ^ of 16° 16' ((7i) with the ^'''•^^• ship's course. How far apart are the objects? 13. At a horizontal distance of 71 feet (c) from the middle of the base of a church tower, the angle of elevation of the top of the tower = 24° 27' {A), and of the top of the steeple = 66° 10' (A)- Find the height of the steeple. 14. Two observers, 570 (c) feet apart and facing each other, observe a balloon in the vertical plane passing through them, at angles of elevation of 67° 22' {A) and 61° 56' (^i), respectively. What is the height of the bal- loon above the ground ? 15. The angle of elevation of a house on the bank of a creek, observed from the opposite bank = 48° 35' {A). The observer walks straight back from the river a distance of 43.15 feet (c), and the angle of elevation is then 33° 24' (^i). Find the breadth of the creek. * The angle which a horizontal line makes with the meridian is called the bearing of the line. Thus \i BC is a meridian, and a point A is 22° 38' west of south from a point C, the bearing of A from C is written S. 22°38' W. 54 RIGHT TRIANGLES. 16. The angles of elevation of the top of a steeple at distances of 50 feet (c) and 450 feet (cj) are found to be complementary. Find the height of the steeple. 17. A man at the top of a lighthouse observes a boat heading straight for the lighthouse ; the angle of depression is 34° 40' (A) ; ten minutes (t) later, it is 55° 35' (A). How soon will the boat reach the lighthouse ? 18. Kain falls at an angle of 75° 30' (A) with the ground. The wind blows at right angles to a wall, 17.5 feet (a) high. How far from the wall can a man stand without getting wet, his height being 5.5 feet (h) ? 19. From the end of the diameter of a circle, a chord is drawn, the length of which is n times the diameter. Find the angle between the chord and the diameter. Take n=.62. 20. Find the altitude of the sun if the height of a man is n times the length of his shadow. Take n = .74. 21. From the top of a lighthouse 221 feet (a) above the ocean, two boats are ob- served, one due west hav- ,^ ing an angle of depression of 67° 37' (A), the other due south having an angle of depression of 74° 49' (Z>). Find the distance between the boats. A^'^--- 22. To measure the height BC of Si tower across a river, the tower ^^^ /' ^^^ being directly opposite a ^^J"^^ point A, a distance AD ^ = 500 feet (d) was meas- ^'^- ^^• ured along the bank. At A the angle of elevation of the top of the tower was 13° 58' (^), and at D the elevation was 10° (i>). Find the height of the tower. AREA. 65 THE AREA OF A RIGHT TRIANGLE. 'A. 28. From the right angle of the triangle, drop a perpen- dicular upon the hypotenuse. Denote the length of this c perpendicular by jp, and the segments of the hypotenuse by d and e. Denote the area by K. By geometry, K^\ac = \ bp. Exercises. — Prove the following relations involving the area of a right triangle: (7) 4ir2 = a2(62_a3). (8) 4.IP=c'(b'-(^, (9) 4.K =h^-(a- cf. (10) 16K^-4:a:'(PK^=a''(P. (11) 2K=d\tsinA-\-t3i.n^A). (12) 8^=(a + c)2-(a-c)l (13) 4 Jr=(a 4- 6 + c) (a - 6 + c). (14) 8 K= [(a + cy + (a - c)^] sin 2 A If K is eliminated between any two of these relations, a relation between the parts of the triangle is obtained. Thus if the 2d and 9th relations are used, there is obtained b'sm2A = b^-(a-cy. (1) 2K =a2cotA (2) 4:K =b'sm2A. (3) 2K =cHanA (4) K :=:p^csc2A. (5) 4.K =(c2-a2) tan 2 A (6) 4:K^=bH-b^d\ b6 RIGHT TRIANGLES. 23. A man having traveled 32 miles in a straight line, finds that he has made 17.3 miles more toward the north than toward the east. What direction has he been traveling ? 24. A triangular field is at the corner of two roads which form a right angle. The area of the field is 32,130 sq. ft., and the total length of the fence enclosing it is 918 feet. Find the frontage on each road. 25. A triangular field is at the corner of two roads which form a right angle. The area of the field is six-elevenths of an acre, and the length of the fence along the two roads is 487 feet (the total front fence). Find the length of the fence on the back of the field. SOLUTIONS USING LOGARITHMS. 29. 1. Given 3. Given a = .129, c = .089. a = 152,' ^ = 18° 25'. Find A and b. Find b and c. La 9.1106 log CSC ^0.5004 Lc 8.9494 log a 2.1818 log cot ^0.4776 ^=55° 24' log tan ^0.1612 log sec ^ 0.2458 5=481.1 log 6 2.6822 Lc 8.9494 c=456.5 logc 2.6594 & = .1567 Lb 9.1952 2. Given c = .3917, ^ = 65° 14'. Find a and b. log tan ^0.3360 Lc 9.5930 log sec ^0.3779 4. Given & = 86.53, ^ = 56° 3'. Find a and c. LsinA 9.9188 log b 1.9372 LgosA 9.7470 a=.8492 La 9.9290 a=71.78 log a 1.8560 5=.9352 Lb 9.9709 c=48.33 logc 1.6842 SOLUTIONS USING LOGARITHMS. 57 5. Given a = 50.7, h = 93.4. Find the unknown parts and the area. Check the work. h = 93.4 log 6 1.9703 a = 50.7 log a 1.7050 ^ = 32° 53' log CSC A 0.2653 .-. C=57°7' log cot A 0.1894 log a 1.7050^ c = 78.42 logc 1.8944 X.5 9.6990 K= 1988 log^ 3.2984 6 + a = 144.1 log (b + a) 2.1587 6 - a = 42.7 log (b — a) 1.6304 Check : logc^ 3.7891 6. Given b = 31.6, c = 31.4 ; find a and A. Here the hypotenuse and side are given nearly equal. By (11), tan ^A= CSC A a ^6 + c L(b-c) 9.3010 log (6 + c) 1.7993 loga^ 1.1003 itan^ 1^7.5017 a = 3.549 log a 0.5501 i^ = 3n3f; A = 6° 27' itan^^ 8.7508 Exercises. — Assume any two parts and find the remain- ing parts in the following right triangles: 1. a = 250 6 = 627 c = 575 ^ = 23° 30' A' = 71880 2. a = 13.13 6 = 21.94 c = 17.58 ^ = 36° 45' /f= 115.4 3. a = 41.54 6 = 42 c = 6.208 ^1 = 81° 30' 7^=1289 4. a = .3864 6 = 4.65 c = 4.634 A= 4° 46' 7f = .8953 58 RIGHT TRIANGLES. PROBLEMS. 30. Solve the following problems, using logarithms : 26. A wall is surrounded by a ditch. A ladder placed with its foot at the edge of the ditch just reaches the top of the wall on the other side of the ditch. The length of the ladder was 33.7 feet, and the angle which it formed with the horizontal plane was 24°. Find the height of the wall and the width of the ditch. 27. Two forces at right angles, one of which is 7.25 pounds, are acting at the same point of a body. Their resultant divides the angle between them into parts pro- portional to 3 and 5, the smaller angle being next to the given force. Calculate the other component force and the resultant of the two forces. 28. The diagonal of a sheet of paper is 16.5 inches, and it makes, with the shorter side, an angle of 57° 57'. Calcu- late the sides of the sheet. 29. On flat horizontal ground 154 feet from the foot of a tower, the angle of elevation of the top was 27° 10'. The axis of the telescope was five feet from the ground. Ee- quired the height of the tower. 30. The latitude of Philadelphia is 40° IN". Taking the radius of the earth at Philadelphia to be 3960 miles, find the radius and the length of 1° of the parallel passing through Philadelphia. 31. The circumference of the meridian passing through Paris is 40,000 kilometers. The length of 1° of the parallel passing through Paris = 73.13 km. Find the latitude of Paris, considering the earth a sphere. 32. Two forces, P = 7.25 lbs., and Q = 10.3 lbs., act at right angles on a body. Calculate the resultant and the angle that it makes with P. APPLICATIONS. 69 33. Calculate the velocity per second of a point in 40° latitude as the earth rotates on its axis. Take the radius of the earth = 3960 miles. 34. A frigate is 4 miles west from a steamer which is sailing north at the rate of 22 miles an hour. How far in front of the steamer must the gunner on the frigate aim in order to strike the steamer, if the average velocity of the cannon ball for that distance is 800 feet per second ? 35. A square is inscribed in another square ; the parts of the sides of the outer square made by the corners of the inner square are 965 (m) and 603 (?i). Find the angle made by the diagonal of the inner square and the side of the outer square. 36. An observer sees a cloud in the southwest at an angle of elevation of 43° 35'. Another observer 2526 feet (b) south of the first sees the same cloud in the northwest. How high is the cloud ? 37. A pedestal 12 feet high (a) supports a column 13 feet high (h). An observer, whose eye is in the same horizontal plane with the base of the pedestal, sees the pedestal and column each subtending the same angle. What is the hori- zontal distance of the observer from the pedestal ? CHAPTER IV. THE ISOSCELES TRIANGLE AND THE REGULAR POLYGON. 31. The isosceles triangle can be divided into two equal right triangles by drawing an altitude upon the un- equal side. Hence the formulae for right triangles will apply to the half- triangle ABiC. That is, p = bsmA = ^c tan A. b = ^cseGA=pcsGA. c = 2p cot A = 2h cos A. Area = K= ^Gp = ^hG sin A = \h''^m2A = \cH2iXiA. In these relations, A may be changed to ^ (7 and the co- functions taken. Exercises. — Assume any two parts in any row of the fol- lowing table and compute the remaining parts in that row : p h c A K 16.80 17.00 5.200 8ri2' 43.68 1.960 2.275 2.310 59° 29' 2.264 70.00 101.5 147.0 43° 36' 5145 10.80 33.30 63.00 18° ob' 340.2 3.179 4.264 5.684 48° 12' 9.034 25.00 62.70 115.0 23° 30' 1438 .5878 1.000 .1618 36° 00' .4755 18.08 83.50 163.0 12° 30' 1474 60 APPLICATIONS. 61 PROBLEMS. 32." 38. Two trains start at the same station at the same time on tracks making an angle of 67° 30'. Each goes with a velocity of 27 miles per hour. In how many minutes will the distance between them be equal to 5 miles? 39. The chord of a circle is 27 inches, and the angle at the center subtended by the chord is 116° 12'. Find the radius. 40. The angle at the vertex of an isosceles triangle is 87° 13', and the sum of the two unequal altitudes is 60. Find the base. 41. The ratio of the altitude from the vertex of an isos- celes triangle to one of the equal altitudes is 1.2, and each of the equal sides = 5.3. Find the base angle and the area. 42. From a point without a circle 12 feet (d) from the center, two tangents are drawn, making with each other an angle of 25° 3' (A). Calculate the area of an equilateral triangle inscribed in this circle. 43. From a point without a circle 23.2 feet from the cen- ter, two tangents are drawn. The radius of the circle equals 9.6 feet. Calculate the side of an equilateral triangle whose area is equal to that of the figure bounded by the two tan- gents and by the smaller arc included between the points of contact. 44. The slant height of a right cone is 13 feet, and it makes with the plane of the base an angle of 67° 23'. Find the convex surface of the cone. 45. When a cone was cut through its axis, the area of the section was found to be 10 sq. inches and the angle at the vertex 40° 30'. Find the convex surface of the cone. 62 ISOSCELES TRIANGLE AND REGULAR POLYGON. 46. The convex surface of a cone = 80 sq. inches (S), and the angle at the vertex of the principal section = 39° 13' (C). Find the volume of the cone. 47. When the convex surface of a right cone was spread out in a plane, it was found to be a sector of a circle whose central angle was 108° 36', and the chord subtending the corresponding arc was found to be 12 inches. Calculate the volume of the cone. 48. The altitude to the base of an isosceles triangle is 5.4 feet, and one of the equal altitudes is 7.12 feet. Find the area. 49. The area of an isosceles triangle is equal to the area of the semicircle on the base as a diameter. Find the angle at the base. 50. In a circle whose radius is 9.4 inches, a chord of 14.4 inches is drawn. Find the central angle corresponding to this chord. 51. Calculate the area of a circle in which a chord 5 feet long subtends a central angle of 77° 36'. 52. The lengths of the sides of a rectangle are 78 ft. and 71 ft. The middle points of its sides are the corners of an inscribed rhombus. Find an acute angle of this rhombus. 53. A point moves in the arc of a semicircle of radius b with a uniform velocity of a feet per second. In how many- seconds is it at a distance of c feet (measured on the chord) from its starting point ? Take b = 10, a = 3, c = 16. 54. Two equal forces act on a body, the angle between them being 146° 48'. Their resultant is 12 pounds. Find the component forces. 55. What must be the angle between two equal forces, each 3.5 pounds, if their resultant is 1 pound ? REGULAR POLYGONS AND STARS. 63 THE REGULAR POLYGON. 33. Denote the number of sides by n. Then the half 180° anerle at the center is , ° n which denote by A. Also de- note the side by a, the radius of the circumcircle by E, and the radius of the incircle by r. Then a = 2 R sin A = 2 r tsin A. R = rseGA = iacscA. r = ^a cot A = E cos A. Area polygon = ^ nar = J na^ cot A. Area incircle = v)-^ = \ ircv^ cot^ A. \ ttci^ csc^ a. Fio. 36. Area circumcircle = irE^ Prove the following properties of regular polygons : The area of the ring between the incircle and circumcircle of a regular polygon is equal to the area of the circle on one side as a diameter. The perimeter of a regular polygon : the perimeter of its incircle = area of polygon : area of incircle. The area of a regular inscribed polygon of an even num- ber of sides is a mean proportional between the areas of the regular inscribed polygon and the regular circumscribed polygon of half the number of sides. Stars. — Divide the circumference of a circle into n equal parts. Let s denote a whole number less than ^n-, join every sth point of division of the circumference (or every n — sth point). There is obtained an w-pointed star of species s. 64 ISOSCELES TRIANGLE AND REGULAR POLYGON. 1. If s is 1, the perimeter is continuous and non-inter- secting ; the figure consists of a regular polygon of n sides. 2. If n is a multiple of s, say n = rs, the perimeter is discontinuous ; the figure consists of s regular polygons each of r sides. For example, the fifteen-pointed star of species 3 consists of 3 regular pentagons. 3. If n and s have a common divisor, say n = ru and 5 = rv, the perimeter is discontinuous ; the figure consists of r different stars each of them having u corners. For example, the fifteen-pointed star of species 6 consists of 3 five-pointed stars. 4. If n and s are prime to each other, the perimeter is continuous and intersecting. For example, the fifteen- pointed star of species 4. Properties. — 1. In a star, prove that the sum of the 71 angles = (n- 2 s) 180°. 2. Prove that a = 2 i? sin- 180° = 2 r tan- 180°, 71 n R being the radius of the circumcircle, r the radius of tlie incircle, and a the length between any two points joined in the construction. Exercises. — The following table gives elements of seven regular polygons. Assume the number of sides and any one element to compute the other elements. n a B r K 5 24.68 21.00 16.99 1049 6 20.78 20.78 18.00 1122 7 13.88 16.00 14.41 700 8 33.14 43.30 40.00 5300 12 14.00 27.00 26.12 2195 20 6.257 20.00 19.75 1236 25 10.90 43.50 43.16 5880 APPLICATIONS. 65 PROBLEMS. 34. 56. The diagonal of a regular pentagon is 4.3 feet (d). Find the area of the pentagon. 57. The area of a circle is 15 sq. feet. Find the area of a regular heptagon whose perimeter is equal to the circum- ference of the given circle. 58. Find the largest and smallest diagonals of a regular polygon of 7 (2 n -f 1) sides, each side being 4 yards (a). 59. In a park it is desired to make a flower garden to contain one acre, and to be of the form of a regular five- pointed star. Find each side of the bounding decagon. 60. Kegular polygons of 27 sides are inscribed in and circumscribed about a circle whose radius is 9.2, the sides of the two polygons being respectively parallel. What is the area of the figure included between the perimeters of the two polygons ? 61. The area of a regular inscribed polygon of n sides is K. Find the area of a regular polygon of m sides circumscribed about the same circle. Put n = 9,m = 7, K= 100. 62. In a circle whose radius is 2 feet two parallel chords are drawn on the same side of the center. One is equal to the side of the regular inscribed polygon of 9 sides, the other to the side of the regular circumscribed hexagon (about the same circle). Calculate the area of the part included between these two chords. 63. The radius of a circle is 7 yards. A chord is drawn equal to the side of the regular inscribed polygon of 15 sides. In the larger of the two segments thus formed, a circle is inscribed touching the arc in its middle point. Calculate the area of a regular polygon of 9 sides circum- scribed about the latter circle. CHAPTER V. OBLIQUE TRIANGLES. Fig. 37. Fig. 38. 35. Notation. — In an oblique triangle, let A denote the value of angle A, B the value of angle B, C the value of angle C ; a the length of the side opposite A, b the length of the side opposite B, c the length of the side opposite (7; p the perpendicular from C on c, Cj the distance from A to the foot of the perpendicular, and Cg the distance from B to the foot of the perpendicular. The length Ci will be regarded as positive when drawn from A toward B, and negative when drawn in the opposite direction ; Cg will be regarded as positive when drawn from B toward A, and negative when drawn in the opposite direction. Relation between the angles. — From geometry, A + B-\-C = 180°. Hence, when any two angles of a triangle are given, the third angle can readily be found. 36. Methods of Solution. — At least three independent conditions must be given to make the solution of an oblique triangle determinate. Suppose that these three conditions are the values of sides or angles. When possible, draw an altitude such that two of the given parts are in one triangle. 66 METHODS OF SOLUTION. 67 Now find a value of this altitude in terms of two known parts in the one right triangle. Also in the other right triangle find a value of this same altitude in terms of the other given part and a desired part. These values of the altitude give an equation in which the desired part is the only unknown. Thus, Given one side and two angles : as 6, ^ and B. To find a, equate values of p thus : p = asm B = h sin A. Multiply by esc B: a = b esc B sin A. If C and c are also desired : C = 180° -(A-\- B). By drawing the altitude from A on a, c sin 2^ = 6 sin C, whence c = b esc B sin C. These values of a and c give the following relation : acscA = bcscB = c esc C. (19) Given two sides and the angle opposite one of these sides: as a, b, and A. Equate values of p : a s'm B = b sin A; whence sin B = - sin A, a C and c being found as in the last case. Given two sides and the included angle : as b, c, and A. First find Ci= b cos A and Cj = c — q. To find B, equate values of p thus : p = C2 tan B= Ci tan A ; whence tan B = — tan A. To find a, equate values of p^ : a^ — c^ = 6^ — Ci. Put c — Ci for Cg. a2 = 52 + (c _ c,)2 _ ci^ = 62 _^ c2 - 2 (x^. Put b cos A for Ci ; then a* = 6^ + c^ - 2 6c cos A. (20) Given the three sides : as a, 6, and c. From (20), cos A = ^' '^/~ ''\ (21) with similar relations for the other angles. 68 OBLIQUE TRIANGLES. It is usually inconvenient to compute cos A from (21) ; another method will now be found for computing the values of the angles of a triangle when the three sides are given. Fig. 40. Prolong the sides AB and ^C of triangle ABC. Bisect angles A, B, and GBC^, these bisectors meeting at O and Ox. The point is the center of the incircle which touches the sides of the triangle in Aq, Bq, and Co ; the point Oi is the center of the excircle in angle A which touches BC at Aiy AC prolonged at B^, and AB prolonged at Ci. From geometry, the tangents from a point outside of a circle are equal to each other ; thus ABq = ACq, ABy = AC^ and so on. Denote each of the tan- gents from A to the incircle by X, each of the tangents from B by y, each of the tangents from C by z, and half of the sum of the sides, |- (a 4- 6 + c) by s ; then Denote each of the tan- gents from A to the excircle by Xi, each of the tangents from B by y^, each of the tangents from G by z^, and half of the sum of the sides |(a + 6 + c) by s ; then THE INCIRCLE AND EXCIRCLE. 69 2^ 4- 2 = a, which give X = ABq = ACq = s — \i, y = BCo = BAo=8-b, z = CAq = CBq = s — c. yi + Zi = a, Xi-Zi = b, «i - 2/1 = c, which give Xi = ABi = ACi = s, y, = BCi = BA^ = s-c, z, = CA, = CB,=s-b. Now the two triangles AOCq and AOiCi are similar be- cause their sides are respectively parallel; and the two triangles BOCq and BOiCi are similar because their sides are respectively perpendicular ; hence ACo or r s — a r, s or - r s — b . ACi OCq^BCq BCi r and rj being the radii of the incircle and excircle respec- tively. Multiply these equations and solve for r. In triangle A00„, cot ^ j« = ?^^ P A Changing letters, cot^B = cot^C r — c (22) The equations in (22) furnish the solution for the angles of a triangle when the sides are given. This method is more suitable for logarithmic computation than the use of (21). 70 OBLIQUE TRIANGLES. SOLUTIONS. One side and two angles. 37. 1. Given a = 200, A = 74° 36', B = 81° 12' ; find b. By (19), b = acsGAsmB^ 200 x 1.0372 x .9882 = 205. 2. With the same data, find C and c. C= 180° -(A + B) = 180° - 155° 48' = 24° 12'. By (19), c = acscA sin (7=200 x 1.0372 x .4099 = 85. 3. Given & = 37, 5 = 53° 8', ^ = 107° 57'; find a. By (19), a = 6 CSC ^ sin ^ = 37 x 1.2500 x .9513 = 44. Two sides and the angle opposite one of them. 1. Given A = 30°, a = 1, ft = 4 ; find 5 and C. Here sin5 = f sin30° = 4 x .5 = 2. Since sin B>1, the solution is impossible. 2. Given^ = 30°, a = 2, 6 = 4; find ^ and C. Here sin 5 = f sin 30° = 1. Whence B = 90° and C = 180° -(A-\-B) = 60°. 3. Given^ = 30°, a = 3, 6 = 4; find ^ and C. Here sin ^ = J sin 30° = .6667. B = 41° 49' or 138° 11', and C = 108° 11' or 11° 49'. Note that the values of C are obtainable by subtracting A from the values of B. Explain this geometrically. 4. Given A = 30°, a = 4, 6 = 4 ; find 5 and C. Here sin ^ = J sin 30° = .5. B = 30° or 150°, and C = 120° or 0°. 5. Givenyl = 30°, a = 5, 6 = 4; find 5 and a Here sin B = ^sm 30° = .4. B = 23° 35' or 156° 25', and C = 126° 25' or - 6° 25 SOLUTIONS USING NUMBERS. 71 These five examples are illustrated in the geometric con- struction, Fig. 41. The base is drawn indefinitely, angle A is made equal to 30°, AC =6 = 4. Successive radii, Oj, a^, ttg, a^, Qs are used to draw arcs with the center at C. In order that the triangles may have the angle A = 30°, the vertex B of each of them must be on the half -line ABa drawn through A to the right. Also in order that the tri- angles may have the side a of the length given in each problem, the vertex B must be in the arc appropriate to that problem. To prevent confusion in the diagram, the side a has not been drawn, except for the negative solution of question 5. It is recommended that the boundary of the triangles be drawn with crayons of different colors. C " /' /' '' «. bT^ 7S /'b, /b, Fig. 41. It will be noticed that the first arc does not intersect the half -line AB^. The second arc touches it at one point. The third, arc cuts it at two points, both to the right of A. The fourth arc cuts it once to the right of A and once at A. The fifth arc cuts the half-line AB^ to the right of A and also cuts the half-line AB^ to the left of A\ the negative value of 0=:^-^° 25', being the angle ACBi, 72 OBLIQUE TRIANGLES. Solve the following five examples, and compare the results with the geometric construction, Fig. 42. 1. Given ^ = 150°, a = l, 6 = 4; find 5 and C. 2. Given ^ = 150°, a = 2, 5 = 4; 3. Given ^ = 150°, a = 3, & = 4; 4. Given ^ = 150°, a = 4, 6 = 4; 5. Given A = 150°, a = 5, 6 = 4; C find 5 and C. find 5 and C. find B and (7. find B and a ■A"'bV\'""bJ\T " \. Fig. 42, Assume a value in each of three columns in any row in the following table, and compute the remaining value (or values) in that row : a h ^ B 14 15 59° 29' 67° 23' or 112° 37' 37 40 67° 23' 93° 42' or 86° 18' 75 77 61° 56' 64° 57' or 115° 3' 69 139 21° 14' 46° 52' or 133° 8' 80 401 11° 27' 84° 17 'or 95° 43' 101 120 43° 36' 55° 1' or 124° 59' 148 153 53° 8' 55° 48' or 124° 12' 240 409 34° 7' 72° 56' or 107° 4' 449 560 51° 25' 77° 10' or 102° 50' SOLUTIONS USING NUMBERS. 73 6. Given a =.= 7, 6 = 8, A = 60°) find c directly. From (20), d" -2 be cos A = a" - h\ Substituting the given numerical values, c^ - 8 c = - 15, which gives c = 3 or 5. 7. Given a = 7, 6 = 5, ^ = 120°; find c. Here the equation is c^ + 5 c = 24. Therefore c = 3 or — 8. Two sides and the included angle. 1. Given 6 = 85, c = 200, ^ = 8ri2'; find ^. Ci = 6 cos ^ = 85 cos 81° 12' = 13. C2 = 200-13 = 187. p = Cg tan B = Ci tan A. tan J5 = ^ tan ^ = .4491 Therefore B = 24° 11. 2. With the same data, find a. From (20), a- = 6^ + c^ - 2 6c cos ^ = 7225 + 40000 - 5200 = 42025. Therefore a = 205. 3. Given 6 = 37, c = 15, ^ = 107° 57'; find 5. Here Ci = 6 cos A= — 11.4. C2= 15 -(-11.4)= 26.4. tan 5 = ^ tan ^ = - 1^4 X (- ^-^868) C2 26.4 = 1.3329. Therefore, B = 53° 7'. 4. With the same data, find a. a' = b^-\-€^-2bccosA = 1369 + 225 + 342 Therefore, a = 44. 74 OBLIQUE TRIANGLES. The three sides. 1. Given a — 7, b = 8, c = 5; find A. 2 be 2x8x5 Or the angle may be found by using the first and second equations in (22) ; thus, 2. Given a = 7, 6 = 5, c = 3 ; find J.. cos A = ^'+/-^' = _ .5. Therefore A = 120°. 2 be cot -^ ^ =-J ; '" ^ :"^ = .5774 ; 1^ = 60° 7.5 X .5 2"" \2.5x4.5 Exercises. — Assume any three parts in each of the fol- lowing five triangles. Then compute the fourth part : ^ = 60° a = 13 & = 15 c = 7Gr8 A = 120° a = 13 6=7 c = 8or-15 ^ = 53°8' a = 13 6 = 15 c = 4orl4 ^ = 61° 56' a = 65 6 = 68 c = 7or57 ^ = 67° 23' a = 25 6=26 c = 3 or 17 38. Given two sides and the included angle, to find a func- tion of another angle directly in terms of the given parts. Since c = c^ -f Cg : c=p (cot A + cot B). Multiply by esc A, and put 6 for ^ esc ^ : c CSC ^ = 6 (cot A + cot B). (23) Whence cotB = c^-^-cotA. (24) CSC A. After B has been found, a = GscB-. — - — BROCARD'S POINTS. 75 The objection to the use of (24) is that in general both tables of functions are used; the table of logarithms of functions in the computation of the value of c ^^^ , and b natural values of functions in completing the work. In general, this computation is not troublesome; and where results are to be carried through a series of triangles, as in a survey, it is of advantage to be able to write down the function ^,^ of a desired angle directly in /* >^\ terms of the data. / X ^^"\"\ Fia. 45. As a simple application of (23), Jc- consider the following problem : When a point (X) in a triangle is joined to the corners, it is found that the angles XBA^ XCB, and XAO are all equal. Find the value of one of these equal angles (jr). In triangle AXC, by (23) ; ^Xcsca;=6(cota;+cot CXA). In triangle AXB, by (19) ; AX cscx = c esc AXB. Now angle CXA = 180° -(C-x)-x = 180° - C. Similarly, AXB = 180° - A. Hence by substituting : b (cot x — cot C) = c esc A. But by (23), c esc A = b (cot A -h cot B). Hence cot X = cot ^ + cot -B 4- cot C. If the angles are 65°, 30°, and 85°, respectively, x = 23° 38'. The point X and another point found by counting the angles x from the other sides of a, b, and c are called Brocard's Points. They possess many interesting properties; for example, they are the foci of the ellipse inscribed in the triangle. 76 OBLIQUE TRIANGLES. THE AREA OF A TRIANGLE. 39. Denote the area by K. Then By geometry, K=\cp. Since ^ = 6 sin ^, -K" = ^ he sin A. Changing letters, K=^ca sin B. K=^ab sin C. b = acsGAsmB, K= ^a^ csGAsinBsinC. ' Changing letters, ^= ^ 6^ sin ^ esc B sin C. ^= i c^ sin J. sin 5 CSC (7. From Fig. 40, K=^ar +ibr + ^ cr = rs. Substitute from (22), K = Vs{s -a)(s -b)(s- c). Prove the following : K= \-wa?bh^ sin A sin B sin G. K=i^Qo\,\Acoi\BGoi^C. 7ir= ri^ cot i ^ tan 1 ^ tan 1 O. K=rriCot^A. Kz= ^ tan \ A tan \ B tan \ C. /ir= (s - a)nan 1 ^ cot i J5 cot -J C. By equating any two values of the area, a relation between segments and angles of the triangle is obtained. Exercises. — Without using logarithms, find the area of each of the following triangles : a = 115 ^ = 67° 23' (7 =46° 24' a = 240 6 = 53 B= 8° 10' a = 200 6 = 205 C=24°ll' a = 116 6 = 105 c = 143 SOLUTIONS USING LOGARITHMS. 77 LOGARITHMIC SOLUTIONS. 40. Given one side and two angles. — a = 24.31, ^ = 45° 18', 5 = 22° 11'. Solve the triangle. "v- = a CSC ^ sin 5, c — a esc A sin C isinS 9.5770 log a 1.3858 + ^= 67° 29' log CSC ^ 0.1483 (7 =112° 31' L sin C 9.9656 h= 12.92 log 6 1.1111 c= 31.6 logc 1.4997 Given two sides and the angle opposite one of them. — a = 215.9, h = 307.7, A = 25° 10'. Solve the triangle, sin 5= - sin ^, Ci — Bi — Af C2 = Bi — A, c = a csc J. sin C. (X log 6 log a LsinA LsinB L sin d log a log CSC A L sin C2 logCi l0gC2 2.4881 2.3343 0.1538 9.6286 JBi= 37° 18' 9.7824 ^2 = 142° 42' A= 25° 10' Ci = 117° 32' C2= 12° 8' 9.9478 2.3343 L 0.3714 9.3226 ci = 450.3 C2 = 109.2 2.6535 2.0383 78 OBLIQUE TRIANGLES. Given two sides and the included angle. — h = 767, c = 947, A = 10° 50'. Find the remaining parts. Ci = & cos A, Co= c — Cj, p = G2 tan B = Ci tan A j whence tan B = — tan A, a = C2 sec B. C2 c = 947 Ci = 753.3 C2 = 193.7 A= 10° 50' ^= 36° 40' 47° 30' 0=132° 30' a = 241.5 log 6 LcosA logci logca LtSinA LtSinB log sec B log a 2.8848 9.9922 2.8770 2.2871 0.5899 9.2819 9.8718 0.0958 2.2871 2.3829 Given b = 767, c = 947, A = 10° 50' ; find a. By (20), a = V6^ + c^ — 2 6c cos ^. 2 6c cos JL : 588300 : 896600 1484900 1427000 a' = 57900 a = 240.6 log 2 log 6 logc icos^ log 62 logc^ 0.3010 2.8848 2.9763 9.9922 5.7696 5.9526 log 2 6c cos ^ 6.1543 loga^ 4.7627 log a 2.3813 Two significant figures were lost in the subtraction of 2 6c cos A from 6^ + c-. Hence the lack of precision in the value of a. SOLUTIONS USING LOGARITHMS. 79 Given h = .049, c = .045, A = 143° ; find the remaining parts. I = b cos A, C2— c — •Ci, \ X tan 5 = - tan A, C'2 a = Cg sec B. L .^' \a c \ Fig. 47. c = .045 Lb LcosA Lci 8.6902 9.9023 n Ci = - .03913 8.5925 w C2 = .08413 Xtan^ L tan 5 log sec B LC2 La 8.9250 A = 143° 9.6675 n 9.8771 n J5= 19° 19' 9.5446 162° 19' C= 17°41' 0.0252 8.9250 a =.08917 . 8.9502 Given b = .049, c = .045, A = 143° ; find a. a = V62H-c2 — 2 be cos A. log 2 0.3010 Lb 8.6902 Lc 8.6532 LeosA Lb' 9.9023 n .002401 7.3804 .002025 Lc' 7.3064 .003521 L2be cos La' iA 7.5467 n .007947 7.9002 .08915 La 8.9501 80 OBLIQUE TRIANGLES. Given the three sides. — a = 704, Find the angles and the area. 6 = 302. c = 670. 4 Ks-a){s- s b)(s- ■""^ coti^ = ^~ a —, etc. ; 704 134 logs — a 2.1271 302 536 log s — 6 2.7292 670 168 log s — c 2.2253 1676 7.0816 868 868 logs log 7^ 2.9232 4.1584 logr 2.0792 83° 42' log cot ^ A 0.0479 25° 14' log cot ^ B 0.6500 71° 4' log cot ^ C 0.1461 100,550 logK 5.0024 /r= rs. Exercises. — Assume three values in any row of the follow- ing table and compute the remaining values in that row. a h c A B K 6.82 5.20 3.16 106° 47' 46° 53' 7.864 .317 .533 .510 35° 18' 76° 19' .07854 28.9 60.1 71.2 23° 32' 56° 9' 854.3 1.98 2.02 3.16 37° 22' 38° 16' 1.937 312 109 229 131° 25' 15° 11' 9360 14.8 17.5 15.3 53° 8' 71° 5' 107.1 1.75 1.62 1.19 67° 23' 73° 44' 99.96 145 119 156 61° 56' 46° 24' 8190 11.6 40.4 48.0 11° 25' 43° 36' 192.0 SOLUTIONS OP PROBLEMS. 81 41. Whenever it is possible, by means of the ruler and compasses, to obtain the geometric construction of a prob- lem, the trigonometric solution is always obtainable by follow- ing the order of the construction. It occasionally happens, however, that the geometric construction is not possible. When this occurs, the trigonometric solution produces an equation of a higher degree than the second. In many such cases, the method of solution by trial involves the least labor. As an illustration, consider the following problem : A ship is sailing in the direc- tion N. 35° W. When at B, two lighthouses are sighted, one due north, the other due west. After sailing 3 miles, the ship was equally distant from the light- houses; sailing 1 mile further on the same course, the ship and the lighthouses are in the same straight line. Find AC^ the dis- tance between the lights. From D, the second position of the ship, drop a perpendicular to F, the middle of AC. Denote angle A by a* ; then the geometry of the figure gives: ?i\\^\e BFE = 2 x, BEF= 125° -a;, BFD = 2x-^0°, and BDF = 215° -x. Apply (19) to triangles BFD and BFE : BF= 3 esc (2 a; - 90°) sin (215° -x) = 4.csc2x sin (125° - x), or -* 3 sec 2 a; cos (125° - x) = 4 esc 2 a; sin (125° - x). Multiply by sin 2 a; sec (125°— ar) and transpose : — 3 tan2a;- 4 tan(125° - a;) = 0. Now solve this by trial. Substituting 0° for x makes the left-hand member of the equation positive, 90°—, 45°+, 60°-, 50°+, 52°-, and so on; a; = 51° 7'. Hence AC=2BF=S esc 102° 14' sin 73° 53' = 7.864 miles. Fig. 48. 82 OBLIQUE TRIANGLES. In order to measure the dis- tance between two inaccessible points, C and D, a base line AB was laid off equal to 583 feet (a). At A angle CAD = 34° and angle DAB = 41° were measured. Also at B, the angle ABC = 44° and angle ^ CBD = 32°. Find CD. ^' I ; Fig. 49. ^(7=acsc^(7i?sin^J5(7; AD=acsGADBsmABD CD' = AC + AD'- -2 AC X AD COS CAD L sin ABC 9.8418 ^B (25) ACB = = 55° log CSC ACB 0.0866 log a 2.7657 ADB = = 57° log CSC ADB 0.0764 ABD = = 70° L sin ABD 9.9869 log AC 2.6941 AC' = = 244500 log AD 2.8290 AD''-- = 455000 log 2 0.3010 699500 L cos CAD 9.9186 553000 5.7427 log AC' 5.3882 log AD' 5.6580 CD'-. = 146500 log CD' 5.1658 CD-- = 382.7 log CD 2.5829 In making an extended survey, it may happen that CD, Fig. 49, is a base of known length, the angles being meas- ure(f from A and B. In this case, to find AB, assume the figure drawn to scale, AB being represented by 1 foot. In this reduced figure, CD may be computed ; then the numeri- cal value oi AB equals the ratio of the known length of CD to the computed length of CD in the reduced figure. SOLUTIONS OF PKOBLKMS. The sides of a trian- gle are : a = 57, b = 51, c = 64. A. point is taken in the side c, such that the segments of c are : /i = 21, A; = 43. Find I, the distance of this point from C. Denote the angle be- ^ tween h and I by D. By (20), l^ + h^-2 hi cos D = b^', By (20), P-\-k--2M cos (180° - D) = a\ Multiply the first by k; the second by h, and add : cl' -h chk = a'h + b'k. (26) Substituting, I is found to be 43.71. The top of a lighthouse, known to be 100 feet (h) high, is just seen in the horizon. What is the distance (6) of the lighthouse from the observer ? It is known that near the earth's surface light does not move in a horizontal line ; on account of refraction, a ray, which seems horizontal, moves in a curve, which is nearly the arc of a circle whose radius is 7 times the earth's radius. In (26) put E for h, (y R for k, 7 E for a or c, E + h for I Then 6b^ = T{2Eh-\- 1i'). In the problem under consideration li^ is very small com- pared to 2 Ell, and may be neglected. Now h is usually given in feet, and b is desired in miles. Neglect 1i^ and multiply the left side of the equation by 5280 : 31680 b^ (miles) = 14 E x (Ji in feet). Put 3960 for E and solve for b. b (in miles) = 1.323 V/i (in feet). Substituting, b = 13.23 miles. (27) 84 OBLIQUE TRIANGLES. Fig. 51. Snell's Problem. — On a coast are three prominent objects, A, C, and B. CB =- 1100 feet (a), ^IC-: 1800 feet (b), and the angle ACB = 150° (0). A buoy is placed at a point Xy and from a boat at the buoy the horizontal angles AXC = 80° {D) and CXB = 60° {E) are measured. Find the distance from X to C* The following values are obtained by following the geo- metric construction : angle AOC = 2 D (same arc), angle BO^C =2E, angle OCO, = C -{- D + E - 180°. Hence OC = \a esc D, and OiC = \h q^g E. Now denote angle COO^ = XAC by A. From (24), COtyl= CSC b DsmEcsG(C-\-D-i-E)-cot(C+D+E) 1 CX= a CSC D sin A (28) * By many writers the credit has been given to Pothenot for the suggestion of this method of determining the position of a point by means of angles subtended by segments joining three known points. The problem, however, was proposed and solved in the year 1617 by Willebrord Snell, the discoverer of the law of the refraction of light. John Collins gave a solution of the problem in the Philosophical Transactions for 1671 : Pothenot's solution was in 1692. SNELL'S PROBLEM. 85 3.0125 3.2553 i'xtermsof thedata, CX= ^^^^^^^' ^ - ab CSC D CSC E /It^ [C-^jyi^s) iL/a^ csc2 D 4- 6^ csc^ ^ + 2 aft esc 1) esc ^ cos (O + Z> + jEJ) The computation by (28) proceeds thus : log (- a) 3.0414 71 .5719 log a esc D 3.0480 log esc Z> 0.0066 nat cot 290° - .3640 Xsin^ 9.8634 L^x^E 9.9375 natcot^ .9359 log CX 2.9114 log esc 290° 0.0270 n A = 46° 54' CX = 815.5 log 6 9.7572 In laying out a railroad along the seashore, the dis- tances GB and DE were measured (C, B, D, and E all in a straight line) ; also the angles CAB, BAD, and DAE were measured. It was found that CB = 2200 (a), DE = 2300 (b), CAB = 21° 30' (A), 5^Z> = 20°26' (^2), and DAE = 22°A0' (As) ; find BD. Fig. 52. Denote BD by x. Then In triangle ADC, x -\-a = AG esc D sin (Ai -f A2). In triangle AEB, x -\- b = AE esc B sin (^2 + -^3)- In triangle ABC, AC = a esc A^ sin B. In triangle AED, AE = 6 esc ^3 sin D. The product of these four equations is (x -{-a)(x-\-b)= ab esc A^ esc ^3 sin (A^^ + ^2) sin (A^ + ^3) . (29) This quadratic equation gives x = 1800. 86 OBLIQUE TRIANGLES. There are two observatories on the same meridian : A in Sweden, latitude 65° 30' N. (I) ; B at the Cape of Good Hope, Fig 53. latitude 34° 20' S. (— T). When the moon passed this merid- ian, the angles ZAP = 62° 20' (z) and Z,BP = 39° 20' (z,) were measured. The radius of the earth being 3960 miles, find the distance from the moon to the earth's center. The angles of the quadrilateral AEBP are readily found : at E, the angle is l — lx; at A, it is 180° — 2; ; at P, it is z + z^-l + l^. Also the angle ABP is 90° + i -^ - 2 z^). In triangle AEB, AB = 2AE sin i AEB. In triangle ABP, AP = AB esc APB sin ABP. Denote AE by r, AP by v, EP by x, angle xiEP by w. v=2 r sin ^{l—lj) esc (z-^z^—l-^-li) cos^ Q-h-2z,) cot w = r = - CSC z + cot z V (30) x = = v(iSQw sin z log 2 0.3010 L^ 8.2257 V log CSC 2 0.0122 8.2379 .0173 nat cot 2 .2401 nat cot ic .2574 logv 5.3720 logr 3.5977 isin 49° 55' 9.8837 log CSC 1°30' 1.6332 Lcos 25° 15' 9.9564 logv 5.3720 log CSC w L sin z 9 5 a: = 236,40 .0139 .9878 .3737 Omi. APPLICATIONS. 87 PROBLEMS. 42. 64. Solve Problem 10 by logarithms. 65. Solve Problem 11 by logarithms. 66. Solve Problem 12 by logarithms. 67. Solve Problem 13 hy logarithms. 68. Solve Problem 14 by logarithms. 69. Solve Problem 15 by logarithms. 70. To measure the distance between two points A and C (C inaccessible), a base line AB = 093 feet was measured from A. At A, the angle subtended by BC was 68° 29' ; and at B, the angle subtended by AC was 66° 7'. Find the distance AC. 71. To find the distance AB across a pond, a third point C was taken. AC =735 yards, jB(7=840 yards, and the angle ACB= 55° 40' were measured. Fiml AB. 72. From a window on a level with the bottom of a steeple, the .angle of elevation of the top is 40°. From another window 18 feet directly above the former the angle of elevation is 37° 30'. Kequired the height of the steeple. 73. A tower 50 feet high is situated on a cliff over the sea. From the bottom of the tower, the angle of depres- sion of a ship at anchor is found to be 25° 17' ; and from the top the angle is 31° 12'. Find the height of the cliff and the direct distance of the ship from the top of the tower. 74. A frigate 10 miles S.W. of a harbor sees a ship sail from it in the direction S. 70° E. at the rate of 9 miles an hour. In what direction and at what rate must the frigate sail in order to come up with the ship in two hours ? 88 OBLIQUE TRIANGLES. 75. A person is at the bank of a river directly opposite a tower on the other side. Going backward from the bank, up a slope making an angle of 42° to the horizontal, he measures an oblique distance of 528 feet. He there finds the angle of depression of the top of the tower is 90°, and of the base is 27°. Required the height of the tower and the breadth of the river. 76. To measure the breadth of a stream, a base AB was chosen = 490 yards long, parallel to the bank and 50 yards from it. A tree C stands on the other bank of the river. The angle CAB wsiS found to be 62° 37' and CBAA0°2S\ What was the breadth of the stream ? 77. At the foot of a mountain the elevation of the sum- mit = 32° 57'. A person measures 775.3 feet along the slope toward the summit, making an angle of 7° 15' with the hori- zontal plane ; at this point, the elevation of the summit was 60° 32'. Find the height of the mountain. 78. A tower stands on a hill whose slope is uniformly 26° 37'. An observer at the foot of the hill finds that the tower subtends an angle of 10° 22'. Measuring horizontally up the slope a distance of 200 feet, he observes the tower subtends an angle of 20° 30'. Find the height of the tower. 79. A tower is situated on a hill. At a point in the hori- zontal plane through the foot of the hill, the angle of elevar tion of the top of the hill is 40° and of the top of the tower is 51°. Measuring in a line directly away from the hill a distance of 180 feet, the angle to the top of the tower is found to be 33° 45'. Find the height of the tower. 80. From a hill 200 feet above the sea, the top of a mast, known to be 150 feet above water, was just seen in the horizon. How far distant is it ? 81. A body is thrown with a velocity of 25 feet per second horizontally from the window of a railway carriage moving at the rate of 30 miles an hour; the direction in APPLICATIONS. 89 which the body is thrown makes an angle of 30° with the rear of the train. Find the direction of the vertical plane in which the body moves. 82. A man traveling at the rate of 6 miles an hour on a road that went due east observed that the wind struck him from the northeast. But having occasion to stop, he found that it actually came from the direction N. 35° E. Find the velocity of the wind. 83. A ship's apparent course is N. 33° 45' E., 8 knots an hour ; but the tide sets her S. 67° 30' E. at the rate of 3 knots an hour. What is her true course and rate of progress ? 84. Two forces, P= 6.2 lbs. and Q = 10.9 lbs., making an angle of 112° 4', act at the same point on a body. Find the magnitude of the resultant and the angle which it makes with the force P. 85. Three forces in a plane acting on a body produce equilibrium. The greatest is 5 lbs., the smallest 3 lbs. The angle between the 3 lb. force and the unknown force is 79° 43'. Find the third force. 86. On the same meridian are two places whose difference of latitude is 68° 13'. From these places, the zenith dis- tances of the moon at its culmination were found to be 25° and 44° 18', respectively. The radius of the earth being 3960 miles, find the distance of the moon from the earth's center. 87. In measuring a distance AD, a portion of it (BC) between A and D is inaccessible. AB = 244, CD = 520. From a point E outside of AD, angles were measured : AEB = 82° 8', BEC = 21° 2', CED = 40° 14'. Find BC. 88. A man walking along a straight road observes the angle of elevation of a tower to be 28° 40'. At a point 57 feet further on, the angle of elevation of the same tower is 48° 42'. And 143 feet beyond that, the angle is 33° 10'. Find the height of the tower. 90 OBLIQUE TRIANGLES. 89. From a ship, a rock is seen in the direction N. 11° 48' E. The ship sails due southeast 5 miles when the direction of the rock is N. 1° 36' E. Find the distance from the rock to each position of the ship. 90. The sides of a triangle are three consecutive whole numbers; and the greatest angle is double the least (A). Find the average side (c). 91. The distance between the centers of two circles is 238 (c). The outer common tangents make with each other an angle of 36° 8' (A). The inner common tangents make with each other an angle of 104° 12' (B). Find the length of the radius of the larger circle. 92. The distances between three points A, B, C are as follows: AC =1200 feet, BC= 1320 feet, and ^jB = 1400 feet. An observer at X in the plane ABC finds angle ^X(7=:25°32'and BXC = AO°. Find the distances of X from the points A, B, and C, if C and X are on different sides of AB and angle AXB = AXC+ CXB. 93. From the top of a cliff 300 feet high, the altitude of the sun was observed to be 70°, the angle of elevation of a balloon was 52° 12', the angle of depression of its shadow on the sea was 68° 14' The sun is in front of the observer, and the cliff, sun, balloon, and shadow are all in the same vertical plane. Find the height of the balloon above the sea. 94. From a bluff 100 feet above the surface of a lake, the angle of elevation of a balloon is 48° 20', and the angle of depression of its image reflected from the surface of the water is 39° 50'. Find the height of the balloon above the lake. 95. From the deck of a ship sailing due east, two con- spicuous headlands are observed to bear 63° and 24°, respectively, to the northward of the ship's course. After APPLICATIONS. 91 sailing 8 miles, the corresponding angles were found to be 150° and 98°. Find the distance of the headlands from one another. 96. Calculate the distance of two inaccessible points A and B, knowing a base CD = 600 feet, the angle BCD = 40°, ACD = 69°, ADC = 38° 30', and the angle BDC = 70° 30'. 97. Three points A, C, and B being given on a chart of a coast, it is desired to determine the position of a fourth point D. ^(7=1000 feet, 5(7=850 feet, angle ^C5=114°40', angle ADC=W\1\ angle C/)i? = 30° 9'. Find CD. [CD is between AD and BD; C* and D are on opposite sides of AB.^ 98. From the top of a lighthouse 103.5 feet above the sea, the horizontal angle between two boats was found to be 41° 20', and their angles of depression were 13° 23' and 17° 41' respectively. Find the distance between the boats. 99. A balloon is observed from two stations 1 mile apart. At the first station, the horizontal angle of the ^^^ balloon and the other ^.'' station = 82°, and the ^-'' / elevation of the balloon ^-^^ / = 12°. At the second ^^-'' / station, the horizontal an- ^-'' / gle between the first sta- AV / ;^B tion and the balloon = 65°. ^^^ C Find the height of the . ^\^ ' balloon. \ / y^ 100. To determine the ^^J^^ height BC of 'd tower, two points A and D 90 feet apart were taken in the horizontal plane through its base. The angle BAD = 56° 33', BAC = 65'' 23', BDC =56° 26'. Find BC. 92 OBLIQUE TRIANGLES. 101. To measure the breadth of a river AD, a point B was taken in AD prolonged. From B a base line BC = 400 feet was laid off, making an angle of 95° 16' with AB. At C, the angles BCA = 52° 48' and BCD = 24° 39' were meas- ured. Find AD. 102. From three points A, D, B in Si horizontal straight line, AD = 500 feet, DB = 300 feet, the angles of elevation of a tower were measured, and found to be 9° 30', 51° 14', and 27° 51 ', respectively. Find the height of the tower. 103. Three points A, D, B lie in a straight line. AD =232, DB = 80. To locate a fourth point C in the horizontal plane of these three, the angles ACD = 85° 12' and DCB = 46° 13' were measured. Find CA, CD, CB. 104. The pedestal of a monument is 9 feet high, the column is 8 feet high, and the statue is 10 feet high. A person is in such a position that he sees the three parts subtending the same angle. AVhat angle with the monu- ment is made by the line joining his eye and the base of the monument ; and what is the height of his eye above the base ? 105. Within an angle A = 73° 44', a point is chosen whose distances from the sides of the angle are 6 inches and 8 inches. A line is drawn through this point so that the portion included between the sides of the angle is bisected at the point. Find the area of the triangle thus formed. 106. The two non-parallel sides of a trapezoid are 7.5 feet and 6.3 feet. The angle formed by their continuations = 41° 21'. The base of the trapezoid is 10 feet. Find its area. 107. From two vertices of an equilateral triangle, whose side is 234 feet, two bodies begin to move simultaneously toward the third vertex ; one with a velocity of 4 feet per second, the other with a velocity of 3 feet per second. When will the distance between the moving points be equal to the altitude of the triangle ? APPLICATIONS. 93 Fki. 5r 108. Calculate the radius of an inaccessible cylindrical tower from the following data : A base line AB is measured = 70 feet. The angles formed with the base line by the pair of tangents are : From A, 60° and 20°; and from B, 75° and 25°. 109. At some distance from a circular pond, a base line AB = 250 feet is measured. At A the angles between the base line and the tangents to the pond are 71° 12' and 54° 12'. At B, the angle between the base line and the tangent to the right-hand side of the pond is 57°. Find the radius of the pond. 110. At some distance from a circular pond, a base line AB = 75 yards is measured. At A the angles made by the base line and the tangents to the pond are 72° 58' and 44° 2'. At B the angle between the base line and the tangent to the left side of the pond is 38° 20'. Find the area of the pond. 111. To find tbe height of a hill, three points, A, J5, and C, not in a straight line, were found, such that at each of the three points the angle of elevation of the summit of the hill was the same ; viz. 52° 40'. CB was measured, and found to be 121.6 feet, and angle BAC = 15° 25'. Find the height of the hill. CHAPTER VI. PROPERTIES OF TRIANGLES. 43. The altitudes. — p — a sm B = h sin A. Similar ex- pressions hold for the other altitudes. If the three altitudes are known, the triangle can be solved. Denote the altitudes by pi, p^, and p^. Then Substitute in (21) ; P-2P3 with similar expressions for cos B and cos C. Then b = 2^3 CSC A; c=p2 esc A, etc. The bisectors of the angles. — Let CQ (= q) be the bisector of the angle ACB. Then area BCQ + area QCA = area BCA ; J aq sin ^ C + ^ 6g sin i (7 = ^ ah sin C = ab sin ^Ccos^ O. 94 ALTITUDES, BISECTORS, AND MEDIANS. 95 Whence 2a&cos-^C Now angle FOQ = \ {A— B)-, hence q = i^ sec -J- {A — B). Hence q = a sin B sec -| {A — B). Similar values can be obtained for the other bisectors. AQ = b CSC ylQCsin i 0; QB = a esc CQB sin J (7. Since esc ^QC = esc CQB, ^ = -. QjB a But J(^4-Q^=c; hence ^(3 = -^ and QB "^ rt + ^ a -f- 6 The medians. — Let M be the middle point of AB. By (20), a^ = :^c^-\- w? — cm cos BMC; 6^ = J c^ + wi^ — cm cos ^ J/O. Now cos BMC + cos ^4Af(7 = 0. Add and solve for m. m = i^2a' + 2b'-c\ Similar relations hold for the other medians. If the angles are known and the medians are desired, find the angle at M first. Denote the middle point of BC by M„ the middle point of CA by M^. Then by (24), cot AMC = ^^^^ _ cot A A = l(GotB-cotA). Similarly, cot BM^A = J- (cot C — cot B). cot CM,B = I (cot A - cot C). Check : cot AMC + cot B3IiA + cot CM,B = 0. If the three medians are known, m^, ?%, m^, the sides can be found. Their values can be obtained from these equa- tions : 9a' = Sm,.,'-\-8ms'-4:7n^% 9b-r=8 7tii + 8 m^' - 4 m/, 9c^==8;».,--h8m,--4m32. 96 PROPERTIES OF TRIANGLES. Fig. 57. 44. The incircle and the excircles. — Let be the center of the incircle touching the sides of the triangle ABC in Aq, Bq, and Co ; Oj, the center of the excircle in angle A and A^, Bi, Ci, its points of tangency, etc. Then AOi = AB, = BA, = BC2 = CBs = CA^ = s, ACo = ABo = BA^ = BCs = CB^ =CA, = s-a, ACs = AB^ = BAo = BCo = CB^ =CA, = s-b, AC2 = AB2 = BA^ = BC^ = CBo = OAo = s-c. Denote the radius of the incircle by r, the radii of the excircles by Vi, r^, and r^ respectively. Values of the various segments in the figure will now be obtained in terms of the sides. OCo= 05 sin i 5; OB = a sin ^ C esc BOC(= 90° -h^ A), r = a sec J A sin i B sin i C = b sin ^A sec 1 B sin ^ C = c sin ^ J. sin 1 i? sec i C. THE INCIRCLE AND EXCIHCLES. 97 Since OjOi = OiB cos | B, and OiB = acsc GO,B{= 90° -\A) sin 0,CB{= 90° - ^C). Also, OiiB = c CSC ^Oi5(= 1 C) sin 1 A ri = a sec ^ ^ cos | B cos ^ C = 6 sin J ^ esc J- J5 cos \ C = c sin ^ ^ cos I JB esc J- C. Find 02C2(= r^ and 03(73(= ?3) in a similar manner. Since ACi = Vi cot \ A. s = tt esc I ^ cos \ Bcos^C = b cos ^A CSC ^B cos ^ C = C COS 1^ ^ COS I 5 CSC ^ C. ACo = rcot\A; BCo = r cot ^B-, CA)=r cot ^C. s — a = a csc^Asm^ Bsin^C = b cos ^ ^ sec J J3 sin ^ C = c cos ^Asin^B sec | C. s — b = asec^A cos ^ ^ sin ^ C = 6 sin ^ ^ esc 1^ J5 sin i (7 = c sin 1^ ^ cos ^Bsec^ C. s — c=aseG^Asm^Bcos^C = 6 sin I ^ sec I ^ cos \ C — c sin ^Asin^B esc ^ C. ^0=(s-a)seci^;l?0==(«-?>)sec|5; 00= (s-c) sec 1 C. ^Oi=ssec^^; ^'Oi=(.s-c)csc JJS; OOi =(«-?/) esc } O. ^02=(s-c)csci^; jB02=sseci7^; 00,= (s- a) esc 1 0. ^03=(s-&)csc^^; ^03= (6*— a) esc IB; 003=sseeiO 00i=^0i-^0=aseei^; OO.^ft sec iB; 003=cseciO. Oa03=^02+^03=a.csc|-^; 030i=6csc|-5; Oi02=ccsciO. Angle O3OiO2=90°-i^; OiO2O3=90°-i5; 0^0301= 90° -iO propp:rties of triangles. Functions of the half-angles of a triangle. ' be ' 2 (^s-b)(s-c)' s(.s-a) ^ be AVith similar relations for i jB and 4 (7. 45. The circumcircle. — Let F be the center of the cir- cumcircle, and M the middle point of AB. Then since angle AFB at the center of the circle is twice the angle C on the circumference, angle AFM= C. Now AF=AMcsgAFM. Therefore, changing letters, i2 = i a CSC ^ = i 6 CSC 5 Now Also, the sum and product of which give 7^ (sin A + sin B -|- sin C)= s, Tl^ sin A sin i? sin (7 == J ahc = i RK. 46. Relations between the sum or difference of two sides and the sum or difference of two angles. Let ABC be the triangle under consideration. With C as a center and radius equal to CA (b), describe a semicircle, cutting BC in JJ and E. Then BD = a-b and BE = a-\-b. i c CSC C. Fig . 58. K = i be sin A = rs. Hence R = abc abe 4 rs R sin^ = ia, R sin B = i?>, R sin C = ic; DIFFERENCE OF TWO ANGLES. 99 Angle DAE = 90° (in semicircle), angle E (on circnmfer- ence)= | C, angle BDA (exterior) = 90° -f | 0, angle DAB = 180° ~B- (90° + ^C)=^(A-B). In triangle ^i5/>, by (19), AB sin Z>^5 = DB sin ^Z)2?. c sin i(^ -B) = (a-b) cos ^ (7. (31) In triangle AEB, by (19), AB sin ^^5 =^B sin E. c cos 1 (vl - JB) = (a -h b) sin | O. (32) The quotient of (31) and (32) reduces to (a + b) tani(yl - B) = (a - &)tan i(^ + B). The product and difference of the squares reduce to c^sin {A - B) = (a' - b") sin G. c^cos {A - B)=2 ab - (a" -f- b') cos 0. Eliminate a^ by (20), and a and b by (19). sin (^ - JS) = sin C - 2 sin B cos A, (33) cos (^ — 7?) = 2 sin ^ sin jB - cos C. (34) The other functions of ^ — J5 can then be found. Since A-\- B = 180° — C, any function of A -\- B can be expressed in terms of functions of C. Also, since B + ^C = 90° - ^(A - B), any function of B -\- ^C is equal to the cofunction of ^(A — B). 100 PROPERTIES OF TRIANGLES. 47. General method of solving a triangle, given three seg- ments in or about the triangle. — Suppose that to solve the triangle there is given any three segments (M, r, r^, s, s — a, etc.). Each of these segments may be expressed in terms of the sides and functions of the half angles ; these func- tions may be expressed in terms of the sides. Hence the sides may be found by solving the set of equations thus obtained. Relations between the segments. — By using values of any two of these segments a relation between the segments may be obtained. For example, find r and s in terms of M. r = a sec i J. sin i B sin a C. a = 2 jR sin J. = 4 i? sin i ^ cos ^ A. Hence r = 4 i2 sin i J. sin i JB sin i C. Similarly, s = iB cos i A cos ^ Bcos^ C. 48. Points of intersection of the circumcircle. — Let tri- angle ABC, Fig. GO, be the triangle under consideration, the bisectors of the interior and exterior angles being drawn as in Fig. 57. Also, let F be the center of the circumcircle of the tri- angle ABC. Since angle M"'CO is a right angle, it is in- scribed in a semicircle ; hence M"'F passes through K'", the intersection of the circumcircle and CO3. The lengths of OoM'" and of O^K'" will now be determined. Since FCB = 90° - ^ (see Fig. 58), angle FCK'^' = CK"'F= lC-(90° - A)=^{A - B). CM'" =2Rsm^(A- B), CK"' = 2 i2 cos i(^ - B). But i? = i c sec i- O CSC i C, c sin ^(A — B) = (a — b) cos J C, c cos i(^ -B) = {a + b) sin | C. CM'" =^(a-b) CSC i C; CK'" = i(a -f 6) sec 1 C. O2M'" =: i c CSC i (7 ; OsK"' = i c sec 1 (7. INTERSECTIONS OF THE CIRCUMCIRCLE. 101 Whence M" and A'"' are the middle points of Ofi, and 00 ^^ respectively. In the same manner it may be shown that M', K', M", K", are the middle points of O2O3, OOj, 0^0^, OO2, respectively. Fig. 60. If OCq(= r) is perpendicular to AB, the angle CqOK'" is also equal to ^{A — B), and, therefore, K"'F is parallel to OCq', it is, therefore, perpendicular to AB through its middle point M. Hence, K'" is the middle point of the 102 PROPERTIES OF TRIANGLES. arc AB, K' is the middle point of the arc BC, and K" is the middle point of the arc CA. The circiimcircle of ABC is called the nine-points circle of the triangle O1O2O3. 49. The nine-points circle. — Let ABC be the triangle under consideration. Draw the altitudes AP^, BP^, CPs, meeting in H. Join P1P2, AA> Ps^i- In the triangle AP^C, AP^ = b cos A, and in triangle AP2B, AP^ = c cos A. Therefore the triangles APiP^ and ABC having a com- mon angle and the including sides in each triangle propor- tional are similar ; hence P2P3 = « cos A. In a similar manner P^Px = h cos B and P^P^ = c cos C THE NINE-POINTS CIRCLE. 103 Also . angle B1\P^ = CP.Po = A, whence P,PJ\ = 180° -2 A, P.P^P;, = 180° -2B, P.P^P, = 180° - L> C, The intersection // of the altitudes is called tlie ortho- center of the triangle ABC\ the triangle PJ\P.^, joining the feet of the altitudes, is called the pedal triangle of the triangle ABC. It should be noticed that the orthocenter of a triangle is the center of the incircle of the pedal triangle. The circumcircle of the pedal triangle is the nine- points circle of the original triangle. All of the properties of any triangle ABC can be reduced to properties of its pedal triangle P1P2P3 by substituting in any formula referring to the triangle ABC, 180° — 2^1 wherever A occurs, 180° — 2B wherever B occurs, 180° — 2 C wherever C occurs. After completing these substitu- tions, put a cos A wherever a occurs, b cos B wherever b occurs, and c cos C wherever c occurs. For instance, make these substitutions in the value of U. The radius of the circumcircle of PiPJ\ or the nine-points circle of ABC = i a cos ^ sec i (180° -2 A) esc i (180° - 2 .1) = ^ a CSC ^ = ^ 7?. That is, the radius of the nine-points circle of any triangle is one-half the radius of the circumcircle of the triangle. All other properties of the nine-points circle of OiOM^ (circumcircle of ABC) can be reduced to properties of the nine-points circle of ABC (circumcircle of P^P^P:^ by the same substitutions. If F is the center of the circumcircle (see Fig. 58), K = area BFC + area CFA + area AFB, that is, rs = ^ aR cos A -{■ \bR qo& B -\- \ cR cos C. „ 2 s 2irR Hence = a cos A-\-b cos B -{- c cos C 2 irr That is, the perimeter of the triangle is to the perimeter of the pedal triangle as the circumference of the circum- circle is to the circumference of the incircle. 104 PROPERTIES OF TRIANGLES. 50. Abscissae and ordinates of various centers. — Count abscissae from A toward B. For F, the center of the drcumcircle. The abscissa is ^ c and the ordinate is ^ c cot G. For 0, the center of the incircle. The abscissa is s — a and the ordinate is (s — a) tan i A For Oi, the center of the excircle in A. The abscissa is s and the ordinate is s tan i A. For O2, the center of the excircle in B. The abscissa is — (s— c) and the ordinate is (s— c)cot^^. For O3, the center of the excircle in C. The abscissa is s—b and the ordinate is — (s— 6)cot^A For //, the iyitersection of the altitudes (prthocenter). The abscissa is h cos A and the ordinate is h cos A cot B. For 6r, the intersection of the medians. The abscissa is \{h cos A+c) and the ordinate is ^ & sin A. For N, the center of the nine-points circle. The abscissa is ^ 6 cos A-\-\c and the ordinate is |^ 5 sin A — \c cot (7. To obtain the abscissa and ordinate of N, note that the angle BP^N is 90° — (^ — B). Then these values can be deduced from (33) and (34). To obtain the distance between any two of these points, sub- tract their abscissa and square the difference ; subtract their ordinates and square the difference. The square root of the sum of the squares gives the required distance. These distances can also be obtained from the tri- angle. Thus to find the distance FH, find AH, AF, angle FAH, and apply (20) : AH =h(iosA sec (90° - B) (See Fig. 61) = h CSC B cos A= a esc A cos A= a cot A. Fia. 62. DISTANCES BETWEEN CENTERS. 106 Since a = 2 Esin A, AH = 2 R cos A. Angle BAH = 90° - jB ; BAF = 90° - C; hence angle FAH= C — B. Hence FH^ = i?^ [1 + 4 cos^ ^ - 4 cos (0 - B) cos A']. In (34) put 90° - C for A, 90° - B for ^, and 180° - A cos {C— B) = 2 cos O cos 5 + cos A. Hence i^//^ = R^{1-S cos ^ cos 5 cos C). To find (?F and (?/. vlO=r CSC L^ and angle F^O=J^-(90-C') = ^(C-fi). OF'=R- + r^csc^l^ - 2 72r csc ^^ cos ^ (0 - i^). Since r = 4 72 sin | ^ sin ^ ^ sin ^ C, r" csc2 1 ^ = 4 i^r CSC ^ .4 sin ^ jB sin 1 (7 = 2 i?r ^^^~^X a 6 + c By (32), CSC ^ ^ cos ^ (C - ^) = a Substituting OF'- = R^-2Rr. Similarly, O^F^ = R^ + 2 Rr^. Exercises: — Show that OH =2i'^ —AR^cosA cos B cos C. ON=iR-r. 0,N=iR-^n. PROBLEMS. 51. 112. The radius of the incircle of a triangle is 6; ^ = 38° 53', = 73° 44'. A line is drawn across angle A parallel to the side BC and tangent to the incircle. In the small triangle thus formed, a circle is inscribed. Find the radius of this latter circle. 113. In the triangle of the last problem (r=6, ^=67° 23% a small circle is drawn touching each side of angle A, and 106 PROPERTIES OF TRIANGLES. also tangent externally to the incircle. Find the radius of this small circle. 114. In the same triangle (r = 6,A = 67° 23', B = 38° 53'), the incircle touches the sides at A^, li^, and Cq. Find the radius of the incircle of the triangle A^, B^, C\. Malfatti's problem. — To inscribe in a triangle three circles, each touching the other two and also two sides of the tri- angle.* Let the sides of the triangle be a =148, b = 175, c = 153, and the angles A = 53° 8', i^=71°4', 0=55° 48'. Denote the radii of the Mal- fatti circles by x, y, and z. Hence ic cot \A -\- 2 -\/xy 4- y cot \B—c. Multiply by sin i B cos }jB = \ sin B. X cot^ A sm^ B CQs^ B + 2 sink B cosh B ^ xy -{-yzos^ -2 B =^csinB . * Steiner has given the following construction for tliis problem. Let be the center of the incircle of ABC. Through the point of con- tact D of the incircle of AOB, draw a tangent to the incircle of BOC, meeting BC m F. The incircle of the triangle BDF is a Malfatti cir- cle. Each of the other two Malfatti circles is constructed in a similar manner. MALFATTI'S PROBLEM. 107 o ri The coefficient of x equals sin^ ^B-\ ; a also c sin B = pi. Hence ( V^ sin ^ J3 + Vy cos i By = ^}h- ^'' ~ ''^^ - When y is changed to 2;, B is changed to C ( V^ sin i O -f- V^ cos i Cf = ^i), - ^-^ Equating and changing letters, ^/x sin ^B + Vy cos -^ J5 = Vx sin |^ 6^ + V^ cos ^ C. V// sin ^ O + V^ cos ^ C = Vy sin ^ .4 + Va; cos ^ ^. Adding these and arranging, V.V _ cos ^A — si n^ B + sin^ C _ ^..^ -y/^ ~ cos ^B — sin ^ yl -h sin 1^ C Substituting, a; = 30, 2/ = 26.28, 2 = 28.01. Properties of the Malfatti radii. — In the equation* sin A + sin i^ + sin C = 4 cos ^Acos^B cos ^ C, put 90° - i ^ instead of A, — ^ B instead of B, and 180° — ^ O instead of C; then change A and i?: cos 4^ — sin ^ 5 4- sin ^ C = 4 cos (45° — :J^ ^) cos :^ 2^ sin ^ C, cos ^ ^ — sin ^ ^ 4- sin ^ C = 4 cos J ^ cos (45° — -J- B) sin \ C. Hence ^ = ^""^ (— ~^^^ ^Q^?-^ V^ cos \ A cos (45° — \B) Now cos (45° — J ^) = cos 45° (cos \A-[- sin \ A),-\ Also cos (45° -\B) = cos 45° (cos i i5 -f sin 1 B). Hence V^(l + tani^) = V^(l + tan}5) = V2(l + tani C). * Proved on next page. t See equation (68), Appendix II, page 145. 108 PROPEKTIES OF TRIANGLES. 52. Relations between the angles of a triangle. — When two relations connecting the same two segments are obtained, their comparison gives a relation between the angles of the triangle. For example, 72 (sin ^ -f- sin JB + sin C)= s. s = 4 E cos 1 A cos ^ B cos i C. sin ^ + sin i5 + sin C = 4 cos ^^ A cos }^ B cos \ C. And since the only condition on A, B, and C is that their sum is 180°, this equation is true for any three angles having this sum. Thus, apply it to the pedal triangle, 180° - 2 ^ for A, 180° - 2 J3 for B, 180° - 2 C for C. sin 2 A-\- sin 2 J3 + sin 2 C = 4 sin A sin B sin C. By adding and multiplying equations in (22), coti^ + cot|J? + cotiC = ^^-^" + '' + ''> = g; TV cot i ^ cot i B cot 1 = (i^l^5fc*l(^^^ = ^ = ?. Hence cot i ^ + cot 1 JB -|- cot ^ C = cot i ^ cot ^ B cot 1 C. In this, put 180° -2 A for A, 180° -2B for B, 180° - 2 C for C. tan A + tan B + tan O = tan A tan _B tan C. Divide this equation by the right-hand member, cot B cot (7 + cot O cot J. 4- cot ^ cot J5 = 1. It will be seen that there is no limit to the number of re- lations which can thus be derived. Exercises. — For all triangles, prove sin^ A -f sin^ B + sin^ O = 2 -f- 2 cos A cos B cos C, sin^ A -f sin^ B + cos^ O = 1 -f 2 sin ^ sin B cos C, cos A -f- cos B -f cos C = 1 + 4 sin i ^ sin i B sin i (7, cos 2 J. + cos 2 ^ -f cos 2 C = — 1 — 4 cos y1 cos 5 cos C. ELEMENTS OF OBLIQUE TRIANGLES. 10^ Exercises. — The following table contains elements of six different triangles. Assume three independent elements in any one of the triangles and compute other elements. a 1 2 3 4 5 6 13 25 37 37 68 68 h 15 26 13 15 (j^y 75 <• 14 17 40 44 57 77 (^i 9 10 5 9 25 45 €■2 5 rr i 35 35 32 32 .S 21 34 45 48 95 110 s — a 8 9 8 11 27 42 s-b 6 8 32 33 30 35 s—c 7 17 5 4 38 33 A 53° 8' 67° 23' 67° 23' 53° 8' 67° 23' 53° 8' B 67° 23' 73° 44' 18° 55' 18° 55' 61° 56' 61° 56' G 59° 29' 38° 53' 93° 42' 107° 57' 50° 41' 64° 56' K 84.00 204.0 240.0 264.0 1710 2310 R 8.125 13.54 20.04 23.12 36.83 42.50 r 4.000 6.000 5.333 5.500 18.00 21.00 n 10.50 22.67 30.00 24.00 63.33 55.00 r2 14.00 25.50 7.500 8.000 57.00 66.00 n 12.00 12.00 48.00 66.00 45.00 *70.00 Pi 12.92 16.32 12.97 14.27 50.29 67.94 Pi 11.20 15.69 36.92 35.20 52.62 61.60 Ps 12.00 24.00 12.00 12.00 60.00 60.00 Qi 12.95 17.11 16.33 20.01 50.53 67.96 (J2 11.22 16.19 37.92 39.65 53.18 61.93 Qz 12.09 24.04 13.16 12.56 60.07 60.18 mi 12.97 22.49 23.29 27.17 50.80 68.09 m2 11.24 20.33 37.98 39.95 53.67 64.02 ms 12.17 24.19 19.21 17.69 60.10 61.39 CHAPTER Vir. SPHERICAL TRIANGLES. 53. If there are three points on the surface of a sphere, and each pair of points is joined by an arc of a great circle, the surface of the sphere is divided into two parts, each of which is called a Spheri- q cal Triangle. Join each ^--'''^^"Vx of the three points to 0, ^^^^"""^ \ ^ the center of the sphere. ^^^^-^"^^^ \ The arcs joining the C><^ 4b pairs of points on the ^^"-^^^^^ / / surface measure respec- ^"^-^-..^^^ / / tively the plane angles ^^^--.>^ formed at the center of the sphere. Thus the arc AB measures the plane angle AOB. Also the angles between the arcs are equal respectively to the dihedral angles made by the planes of the arcs. Thus the angle A of the spherical triangle is equal to the angle between the planes ACO and ABO. There is no limit to the number of different possible values of any side or angle of a spherical triangle. For the arc joining A and B may be any arc AB + any number of circumferences. And the angle A may be generated by turning AQ away from AB through any angle A + any number of revolutions. Since any function of n x 360° + x is equal to that function of x^ it follows that any single solution of a spherical triangle includes the solution of all spherical triangles which have any side or angle increased 110 THE GENERAL TRIANGLE. Ill by any number of times 360°. Therefore, without loss of generality, each side and each angle may be considered less than 360°. Taking each side less than 360°, there may be : Fi<>. G.5. ao = a Ao = 300'^ — A bo = b Bo = 360° - B co = c Co = SGir - C 1. None more than 180°. Fig. m. ai^nGO^-a y1i = 3r,0°-^ bi = b Bi= 180° - B ci.= c C'l = 180° - C 2. One more than 180°. Fig. G7.* ao = 3G0° -a Ao h.2 = 300° - h Bo C2 = C Ci 3. Two more than 180°. Fig. 68.* 180°- A as = 300° -a As = A 180°- B bs = 300° -b Bz = B 300° - - C C3 = 360° - c t\= C 80°. 4. Three more than 180°. * Out of perspective. It will be well if all the diagrams in this chapter are drawn with colored chalk on a spherical blackboard. 112 SPHERICAL TR'TANGLES. The four triangles in Eig. 69 are called Associated Triangles. The three triangles as- sociated with ABC are obtained by prolonging the sides of ABC. Spherical triangles, like plane triangles, are right, isosceles, equi- lateral, and oblique ; but a right spherical triangle may have more than one right angle. If one or more sides of the triangle is a quadrant, the triangle is said to be quadrantal. FORMULAE FOR RIGHT TRIANGLES. 54. Let ABC be a spherical right triangle, right angled at B. Let the angles A and C be acute, and each side of the triangle be less than 180°. Join A, B, and C to 0, the center of the sphere. Then the planes AOB and BOC are at right angles. Erom any point D in the edge 0(7, drop a perpendicular D^i to the plane AOB. Through DB^ pass a plane at right angles to the edge OA. Since DB^ is per- pendicular to the plane AOB, it is perpendicular to OBi and BiAi', and since AiBi and AiD are both perpendicular to OA, the angle B^AJJ measures the angle between the planes AOB and AOC. Fig. 70. FORMULAE FOR RIGHT TRIANGLES. 113 OD sin a = B^D = A^D sin A ; and A^D = OD sin 6. Therefore sin a = sin 6 sin >l. (35) OBi tan a = B^D = AiBi tan A ; and AiBi = OBi sin c. Therefore tan a = sin c tan A. (36) OAi tan b = AJ) = AiBi sec A ; and ^jSi = OAi tan c. Therefore tan 6 = tan c sec /!• (37) If the angle A-^B^Cs, Fig. 69, is a right angle, these for- mulae (35), (36), and (37) may be shown to apply to the tri- angle A^BqCs. Thus put for a the value 180° — ttg, put for b the value 180° -63, put for A the value 180°— ^3. There is obtained: sin (180° - a,) = sin (180° - 63) sin (180° - ^3) ; whence - sin a^ = sin 63 sin A^. tan (180° - as) = sin c tan (180° - ^3) ; whence tan a^ = sin C3 tan A^. tan (180° - 63) = tan c sec (180° - ^3) ; whence tan b^ — tan C3 sec A^. It therefore follows that the formulae (35), (36), (37) are true when one of the angles of the right spherical triangle is obtuse. In a similar manner, these formulae may be shown to apply to the triangle ^2-S2C2, Fig. 69, and are therefore true when two of the angles are obtuse. It therefore follows that formulae (35), (36), (37) are true for all right triangles in which each of the parts is less than 180°. In the same three formulae put successively the values of the parts of the various triangles in Figs. 66 to 68, in which assume B = 90°. It will be found that all three formulae apply ; they are therefore true for all right spheri- cal triangles. 114 SPHERICAL TRIANGLES. Collecting, Changing letters, Sq. root product (35) to (40), (41) -- product (36) and (39), (35) X (40) -f- (42) gives (37) X (38) ^ (42) gives sin a tana tan& sine tan c tan h sec h sec h sec c sec a sin h sin A, sin c tan A, tan c sec ^. : sin h sin C, sin rt tan C, tan a sec (7. : tan A tan (7. : sec a sec c. sin A sec C : sec A sin C. (35) (36) (37) (38) (39) (40) (41) (42) (43) (44) 55. The following diagrams with the accompanying rules will be found useful in rapidly picking out the relation which exists between any three parts of a spherical right triangle. sin a — sin A- ^sec G- a- = 12G°58', c = 56°52'; find i^T. APPENDIX I. THEORY OF LOGARITHMS. 67. Arithmetic and geometric series. — Logarithms are a series of numbers severally assigned to ordinary numbers, and so chosen that the multiplication and division of or- dinary numbers may be accomplished by the addition or subtraction of their logarithms. Thus, if the ordinary numbers are terms in a geometric progression (a multiplication series) and to each term we assign a term in an arithmetic progression (an addition series), it can be shown that the multiplication or division of the terms in the geometric progression can be accom- plished by the addition or subtraction of terms in the arithmetic progression. Thus suppose it is desired to multi- ply the numbers 9-^(32) and i^ (S). The logarithm of 1^ (32) isa + 5d (5), and the logarithm oi^^(8)isa+7d(7). Adding these loga- rithms gives 2 a + 12rf(12), which is larger than the logarithm of the 133 Numbers. Logs. ,,10 a r« a-\- d ,.8 «+ 2rf ,.7 a+ 3rf r« a-f 4rf 1^ a-f 5d ,A a-h 6rt 7-3 a+ Id 9-2 a+ 8c? r a-f 9(? 1 a + 10 d Numbers. Logs. 1024 512 1 256 2 128 3 64 4 32 5 16 6 8 7 4 8 2 9 1 10 134 THEORY OF LOGARITHMS. correct product [?•« (256)] by a + 10 d (10). But a + lOd (10) is the logarithm of 1. Hence log (r^ x t^) = log r^ + log r' — log 1, log (32 X 8) = log 32 + log 8 - log 1. And this relation is perfectly general in any two such series. Either by trial or reduction from this relation we may obtain log (pq) = \ogp-\- log q - log 1, log ^ = logp - log g + log 1, logp" = n logp — (n — 1) log 1, log -Vp = -(logp + log 1). n It will be seen that a logarithm is a function fulfilling the condition Function (ab) = function (a) H- function (6) — a constant. The theory of logarithms is principally devoted to ob- taining the logarithm of any number whatever, — in other words, interpolating in the two series. Logarithms may be considered geometrically, as they were by the inventor of the first system: NAPIERIAN LOGARITHMS.* 68. Suppose two points, Ao ^ A A and B, to start at the same time with the same initial velocity. A starts at Ao and moves along the 1-^^ — '^ 4 £ — k Fig. 79. half -line A^A^ with a uniform velocity. B starts at B„ and * This name has by some writers been incorrectly apphed to Speidell's Logarithms, which were published in 1619, five years after the publication of Napier's logarithmic tables. NAPIKUIAN LOGARITHMS. 135 moves along the segment -B^B^, with a changing velocity such that, when B arrives at any point U^, the velocity of B will be proportional to the remaining segment B^B^. Then, if at any time A is at A^ and at the same time B is at B^, the length of A^A^ was called by Napier the logarithm of The numerical value of the length of B^B, was taken by Napier to be 10', or 10,000,000. It will here be denoted by k ; hence from the definition, Napierian log k = zero. Suppose that during the first unit of time A moves a dis- tance 1, and B moves a distance which denote by (1 — r)fc.t The distance of the moving point B from B, will then be k —{1 —r)k = rk ; and hence Napierian log ?'A; = 1. Now during the second unit of time A will move another distance equal to 1, making its distance from A„ equal to 2. Since the law of the diminution of the velocity of B remains the same throughout its motion, we obtain the distance traversed by it in the second unit of time in the manner as it was obtained for the first unit of time. That is, multiply the distance which it is away from B^ at the beginning of the unit of time under consideration by the constant 1 —r. This gives for the space traveled over by B in the second unit of time (1 — r) rk ; so that B is then removed from B^ by the distance rk—{l— r) rk = r^k. Hence Napierian log i^k = 2. Similarly, Napierian log r^k = 3 ; and in general, Napierian log r"A: = n. * Mirifici Logarithmornm Canonis Constructio. English transla- tion by W. R. Macdonald, Edinburgh, 1889. t If the initial velocities of A and B are equal as assumed by Napier, and k is very large, the value of r will be nearly equal, but slightly less than 1. The distance gone by B in the first unit of time will be slightly less than 1 on account of the retardation of the velocity of ^ as it approaches Bg. 136 THEORY OF LOGARITHMS. It will be seen that the numbers form a geometric pro- gression and their logarithms form an arithmetic progression, and hence that Napierian logarithms are only a particular case of the more general logarithms. Napier solved the problem of finding corresponding intermediate terms in the two series ; his logarithms are no longer used, but the method by which we could interpolate in the Napierian series is quite similar to the method adopted for the same process applied to systems now in use. PRESENT SYSTEM OF LOGARITHMS. 69. Suppose two points A and B, moving in the same direc- tion, A with a uniform ve- locity along the line A„A,, 1" "' t" ■^-^— and B along the half-line . I ^„^,, with a velocity directly £ £ '^^ ET '+^~" proportional to its distance ^^" " from Bo- If at any time the distance of A from A„ is A^^, and at the same time the distance of B from B„ is B^B^, the length A„A^ is defined as the logarithm of B^B^. If it is assumed that A is at zero (yl„) when J5 is at 1 (Sj), the logarithm of 1 = 0. This assumption avoids the neces- sity of subtracting the logarithm of 1 when multiplying by means of logarithms. Denote the velocity which B has acquired when at B^ by V ; then at any other point the velocity of B will be v times its distance from B„ ; thus, the velocity of B at any point B^ = v times B„B^. In the first unit after B leaves B^, A will be supposed to move a distance 1, and B will move in the same time a distance, which may be denoted by cv times B„B^] for a particular system of logarithms, c and v are constants. Let us count from the instant that A was at A„ and B at ^1. At this instant B^ will coincide with Bi, and the dis- NATURAL LOGARITHMS. 137 tance of B from B„ = 1. In the first unit of time afterward, the distance moved over by B becomes cv, and its total dis- tance from B^ at the end of this first unit of time is 1 -\- cv\ therefore, log(l + cv)= 1. Under the same assumptions, in the second unit of time A moves another unit. This makes the distance A„A, = 2, and the motion of ^ = cv times its distance from B^ at the beginning of this unit of time = cv{\ + cv). Therefore, at the end of the unit of time, the distance of B from B^ = 1 -f cv + cv (1 + cv) = (1 + cvy. Therefore log (1 + cvf = 2. By a similar process of reasoning it may be shown that log (1 + cvy = 3, and in general that log (1 + cvy = n. 70. It will be seen that the values of the distances of A from A„ at the ends of successive units of time ; namely, 1, 2, 3 ••• n, form an arithmetic progression: and that the values of the distances of B from B„\ namely, (1 -\-cv), {l-\-cvy, (1 -\-cvy '" (l+cv)", form a geometric progres- sion. The value of c depends on the value assumed for y;, but V is perfectly arbitrary, and values for it may be assumed at pleasure ; now a system of logarithms may be constructed by assuming for v any value at all, since it has already been shown that the terms of any arithmetic progression may be used as logarithms of the terms of any geometric progression. But in systems of logarithms in actual use, special values have been given to v on account of special conveniences served thereby. Thus, in higher mathematics, the reciprocal of v arises as a factor ; this factor (denoted by M) is eliminated and processes simplified by making M= 1 (whence also v = 1). The system in which M and V are taken as unity is called the Natural System of Loga- rithms, and the logarithms in that system are called Natural Logarithms, or Speidell's logarithms. 138 THEORY OF LOGARITHMS. 71. The number corresponding to a given logarithm. Let us now imagine : (1) That B, having an initial velocity of v, moves for one unit of time divided into n parts. (2) That B, having an initial velocity of 1, moves for v units of time, each unit being divided into n parts. In each case denote the value of the constant, which is c for one unit of time, by —for each - of the unit of time, m n Then under the first assumption the motion of B is 1 + — ) ; and under the second assumption it is ( 1 4- — mj \ mj By the binomial formula, l + !iY^l + gH + »(»-l)f + "(»-l)(n-2)^V etc. mJ m 1 '2 ' m^ 1 • 2 • 3 • m^ -, , IV" 1 , ^v , n^v^ — nv , nV — 3nV-f-2nv . , 1 + - =1+ — + - — ^r -„H . „ „ ^„ hetc. mJ m 1 ' 2 ' m^ 1 • 2 • 3 • m^ Now when n is very great, the distance that B, with a velocity v (or 1 under the second assumption) will go from B^ in - of a unit of time will approach as a limit - f or - ]. But n . n\ nj this distance is also v (or 1) times — . Therefore, when ri is m infinite, m = n. Also, as m and n become very great, nin-^ and J^^ both approach ^ ""^ : n(n-l)(n-2)v^ ^^^ nh^-3n^v^+2nv ^ _nV 1.2.3.m3 1.2.3.m3 ^^ 1.2.3.m^ When these limiting values are put in the values of the motions, these values of the motions become identical. Denote the common result by b, which is either the motion of B from Bi for one unit of time with an initial velocity of /, or the BASE OF THE NATURAL SYSTEM. 139 motion of B from B^ for v units of time with an initial velocity of i. The value of h is In this value of &, make v = 1 and denote the result by e ; e is, therefore, the motion of B from B^ while A goes a dis- tance equal to 1, both ^ and 5 having the same initial velocity. The value of e is The numerical value of e, which is readily computed, is 2.71828 + . (See page 10.) After B was at By (or 1) its motion for a unit of time with an initial velocity of 1 made its distance from B„ equal to e; hence another unit of time will make its distance 1 x e x e=e^ ; and at the end of v units of time the distance of B from B„ will be e" ; but this = 6, the distance that B would be from jBo at the end of one unit of time, if its velocity when at 5, was V. Therefore at the end of the second unit of time, with an initial velocity of -y, the distance of B from Bq = 1 X 6 X 5 = 6^ = 1 X e" X e" = e^" ; and for n units of time with an initial velocity of v, the dis- tance of B from B^= 5" = e""; and hence from the defini- tion, 6" = e"" is the number whose logarithm is n. That is, if ^ is a number and n is its logarithm, 6" = N, where h = e*. Taking two numbers, Ni and N^, with logarithms Wj and n2, iVi X iVg = 6"! X 5"* = 6''^+"«. n JV' = 6"" and v^= b\ 140 THEORY OF LOGARITHMS. That is, log (N^ x N^) = log Ni + log iVg, log (N, - N,) = log N, - log N,, log N^ =p log iV, log -y/Nzzz log JV-7- r. It will be noticed that The numbers form a geometric progression ; Their respective logarithms form an arithmetic progression; These systems of logarithms are only particular cases of the more general form of logarithms ; Logarithms in these systems satisfy the condition : function {ah) = function {a) -f- function (h) ; The constant (log 1) which was introduced in the Napierian system will be zero in these systems. 72. Base. — In the relation 6" = N, we see that the loga- rithm n is that exponent of a number h corresponding to a number N. This is expressed by saying that in these systems, a logarithm is an exponent ; the number h which is affected by this exponent is called the base of the system of logarithms. It is evident that if N remains constant and h is arbitrarily made to change, then n will also change. Relations between logarithms in systems with different bases. — Let x be the logarithm of a in a system whose base is h (written x = log,, a), and z be the logarithm of & in a system whose base is a (written z = log, b). Then h'' = a and a" = b. Now in any of the present systems, taking the logarithms of each side of these equations, X log b = log a ; z log a = log b ; whence xz = 1. That is, log6 a x loga 6 = 1. (63) Hence if the logarithm of a number to any base is known, its logarithm to any other base can easily be found. THE LOGARITHMIC SERIES. 141 THE LOGARITHM CORRESPONDING TO A GIVEN NUMBER. 73. We have heretofore regarded the logarithms as in- creasing uniformly, and the rate of increase of the numbers continually changing. We will now find a relation giving the increase in the logarithm when the numbers increase uniformly. Now equation (62)* gives the value oi b = e", and hence putting nv for v, e-. = 6" = l + „.+^_ + ___ + etc., which is the number whose logarithm is n. Denote this number by 1 + s ; then s = nv -\- - — - + Y etc. 1.21.2.3 Assume now that n = As-\- Bs^ + Cs^ + etc., and substitute this assumed value of n in the value of s; and let us for the present, in squaring and cubing the assumed value of n, drop powers of s higher than the third. Equating the coefficients of the like powers of s, there is obtained Av = 1, whence A = -\ V Bv+^ = 0, whence JB = --^; Cv + ABv^+ — = 0, whence C= — ; 6 3 V and the law of the series being seen, it is unnecessary to repeat the operation retaining higher powers of s. Hence, „ = log(l + .)=l[.-|V|-lVf-etc.]. Changing the sign of s, 142 THEORY OF LOGARITHMS. Subtracting these two equations, and remembering that the subtraction of logarithms corresponds to a division of numbers, p — q 1. -\- s p Now let s = ; whence a _ = — -f =l?^^Kf^^Kfi|)■-«■} (". If g = 1 and 7? = 5, log 5 = 1. Substitute and solve for v : o[b-l , Ifb-iy , 1/b-iy , , "I whence v can be computed for any positive base. Note that if the base is 1, v is zero ; and that if the base is less than 1, v is negative. Modulus. — The reciprocal of v is called the modulus of the system of logarithms. Now v has been shown to be the ratio of the rate of increase in the numbers in a system whose base is b (A at A^, B at B^ to the rate of increase in the numbers in the natural system, the logarithms being conceived to increase uniformly. Hence the reciprocal of V, or the modulus of the system whose base is b, is the ratio of rate of increase in the logarithms in the system whose base is b {A at A^, B at ^1) to the rate of increase in the logarithms in the natural system, the numbers being conceived to increase uniformly. It will be seen from (64) that a logarithm in any system is inversely proportional to the value of v for that system. Now V is the reciprocal of M, and hence a logarithm in any system is directly proportional to the modulus of the sys- tem. The modulus of the natural system is 1 ; the logarithms of a number in any other system can therefore be obtained by multiplying the natural logarithm of the number by the modu- lus of the system. THE CURVE OF LOGAUITHMS. 143 If in (64) M is written instead of -, and logq is trans- posed, we obtain From which the logarithm of any positive number can be found when the logarithm of any other number and also the modulus of the system are known. 74. The logarithmic curve. — In the following diagram the abscissae of points in the curve represent numbers, and the corresponding ordinates represent the natural logarithms ot these numbers. Y Fig. 81. 75. Solve the following equations. — After obtaining the general solution, substitute 2 for a, 3 for, b 4 for c, 5 for d, 6 for e, 7 for/; obtain the numerical answer in each case. 130. a**+« = (Z«+^. ^^^ ^a'+f^b, 131. a*V' = e. 132. ab' = cd'\ i a' = lf>, ( a'+y = b, \ h^-y = a. 135. a^'b'^c = 1. 133. ( (f+ (Phi), etc. 150 GONIOMETRY. A table for converting degrees into radians, or for con- verting radians into degrees, will be found at the bottom of page vii of the tables. Exercises. — The following table gives the values of various angles, first, in degrees and minutes ; second, in radians ; third, in terms of an angle of half a revolution : Angle in Angle in Angle in Angle in Angle in Angle in Degrees. Radians. Terms of tt. Dk(;kees. Radians. Terms of tt. 172° 3.0020 .9556 144° 9' 2.5133 .8008 100° 1.7453 M56 169° 23' 2.9563 .9401 125° 2.1566 .6944 104° 51' 1.8300 • .5825 179° 3.1241 .9944 204° 12' 3.5639 1.1344 194° 3.3859 1.0778 195° 33' 3.4130 1.0864 300° 5.2360 1.6667 274° 10' 4.7851 1.5231 344° 6.0039 1.9111 289° 50' 5.0585 1.6102 354° 6.1784 1.9667 344° 19' 6.3355 2.9667 79. Arc functions. — The equation ^=sin^ gives the value of A^ as a function of 6; it is frequently convenient to express the value of ^ as a function of A^. The notation employed is : 6 = arc sin N* This will be read, 6 is the angle (or arc) whose sine is JSf. Similarly = arc tan N is read, 6 is the angle (or arc) whose tangent is N; and simi- larly for each of the other functions. It will be seen that if we omit the name of a trigono- metric function prefixed to one side of an equation, the name of the corresponding arc function is prefixed to the other side of the equation ; and reversely. It must be remembered, however, that in the equation A'= sin^, there is but one value for N for each value of 6; * By some authors this is written, d = sin-i iV. ARC FUNCTIONS. 151 but in the equation $ = arc sin N, there are an infinite num- ber of values of 6 for any value of the number N. Thus in the equation N= sinO, if = ^tt or 30*', N'= M; but in the equation d = arc sin N, if N= .5, 0=[n-{- ( — l)"^-]7r, n being any whole number. Exercises. — 1. Find 6 = arc tan 1. Here tan = 1, and = (n -{- ^)7r. 2. Find N= tan (arc cos ^). Let arc cos ^ = 0, then cos 6 = ^; 6 = (2n ± ^)7r. Hence N= tan (2n ± i)7r = ± 1.7321. 3. Prove that sin(arc sin N) = N, and that arc sin (sin 0) = 6. 2 N 4. Prove that 2 arc tan N= arc tan- -• 1 — -^^ 5. Arc sin N -\- arc sin -J- -^ = \ tt. Find N. 6. Arc sin 2 iV— arc sin iV V3 = arc sin N. Find N. PROBLEMS IN GONIOMETRY. 148. At a distance of 70 feet from the foot of a tower, the angle of elevation of the top is 21° greater than the angle of elevation seen at a distance of 160 feet from the foot of the tower. Find the height of the tower. 149. At a distance of 296 feet (a) from the base of a tower, the angle of elevation of the top is three times as great as the angle of elevation seen from a point 208 feet (6) nearer the base. Find the height of the tower. 150. The wind strikes a horseman at an angle of 50° with the road when he is traveling at the rate of 8 miles an hour ; and at an angle of 40° when his rate is 12 miles an hour. Find the direction and velocity of the wind, 152 COMPLEX QUANTITIES. 151. From the top of a hill 360 feet above the surround- ing plain, an officer sees a column of the enemy's infantry- advancing directly toward the hill. The column is known to be one mile long, and at a certain time the whole column is seen by the officer to subtend an angle of 2° 7'. At that time what is the distance of the front rank from the middle of the base of the hill ? 152. A circle has a radius of 20 inches (r). From the same point in the circumference two chords are drawn of lengths 10 inches (ki) and 15 inches (A^g), respectively. Find the area of the portion of the circle bounded by these chords and by the arc joining the ends of the chords. (Two solu- tions.) 153. A circular field, surrounded by a fence, contains 1 acre. A cow is fastened to the fence by a chain, which, when measured, was found to be 55.53 feet long. Over what area can the cow graze inside the circle? COMPLEX QUANTITIES. 80. A complex quantity is of the form a -\- hi, where i stands for V— 1 ; a is called the real part of the complex quantity, and hi is called the imaginary part. It has already been explained that the multiplication of a segment by — 1 reverses the direction of the segment. This reversal may be conceived as being made by means of two half-reversals. Since i'^ = — l, i will be taken as the multiplier which first turns OU (Fig. 84) into OUx, and then turns Ui into U^ ; multiplying again by i, U^. is turned into OlT'a; and another multiplication by i turns 0X1^ into OU. Real positive quantities have been conceived as laid hori- zontally to the right, and real negative quantities to the left. Positive imaginary quantities will be conceived as ARGUMENT AND MODULUS. 153 being laid vertically upward, and negative imaginaries ver- tically downward. Thus the complex quantity a-\-bi will be represented by laying off 0M= a to the right of the origin and MAi = h upward from M. If a is negative it is laid off to the left of 0, and if b is negative it is laid off downward. Fia. 84. Denote the angle MOA^ by and the length of the line OAi by m. Then a = h tan $ = m cos and 6 = m sin 0. Hence the complex quantity a -f bi may be written : m (cos 6 -\- i sin 0). The angle is called the argument of the complex quantity, and m is called its modulus. It will be noticed that the segment OU" (+ 1) is changed into the segment OAi by two distinct operations. First. A turning oi OU through the argument 6 into OUa' This turning is therefore equivalent to a multiplica- tion of + 1 by cos -\-i sin 6. Second. A multiplication of the length of OUa by the niodulus m, the positive square root of a^ + b^. 154 COMPLEX QUANTITIES. Fig. 85. Addition of complex quantities. — Suppose it is desired to add the complex quantities mi (cos ^ 4- ** sin B) and m^ (cos <^ -f- 1 sin <^). Make angle M0A=6, OA=m^ ; angle NAB=(fi, AB=m2', the required sum is OA-\-AB. The sum of the real parts is m^ cos 6 -\- m^, cos <^ {0M-\- AN)] and the sum of the imaginary parts is i (mi sin 6 -\- wig sin <^) ; that is, MA + NB. Had mg (cos <^ + i sin ) = OC been laid off first and mi (cos + i sin ) = CB been added to it, the result would have been precisely the same. Hence the segment OB rep- resents the sum both in mag- nitude and direction, A third complex quantity can be added to this sum by drawing from B a segment, say BD, representing in magni- tude and direction this third complex quantity. Then the segment OD will represent both in magnitude and direc- tion the sum of the three com- plex quantities, and so on. Subtraction. — To con- struct the difference of ma (cos ^ + ?* sin <;^) and mi (cos $ -\- i sin $), lay off the minuend (OA); let AB represent the sub- trahend; lay it off back- ward from AB to the point jBj ; then OB^ will represent the remainder both in magni- tude and direction. FUNDAMENTAL OPERATIONS. 156 Multiplication, — The product of the two complex quanti- ties m] (cos 6 4- 1 sin 6) and mg (cos -{- i sin <^) is 7iiim2 [(cos cos (^ — sin ^ sin ) -{-i (sin $ cos <^ + cos $ sin (UOB ov AOC) and then multiplied by m^ when operated upon by the multiplier m2{Gos). If (Fig. 87) angle UOA = e, UOB = , UOC=e + , then BOC=UOA. Also the length OC of the modulus = m^m^ is a fourth proportional to 1, ma, and mi; that is, OU: 0B= OA: OC. Hence the two triangles UOA and BOC are similar. To construct OC, therefore, it is sufficient to make angle BOC = e and angle B = angle OUA. If now OC is operated upon by a new multiplier, m2{cos\l/-\-ismif/)y the new argument will be 6 + cf)-{-if/, and the new modulus will be m-^tn^m^' The geometric construction can be obtained as before. Division. — Rationalizing the denominator of the fraction 7)iy (cos 6 -\-i sin 6) 'w?2(cos ^-f-i sin <^) Fig. 87. 156 COMPLEX QUANTITIES. gives ^ (cos ^ + I sin 6) (cos — i sin <^) = ^[cos ^ + ^ sin ^][cos (- <^) + i sin (- <^)] = ^ [cos {e~^) + i sin (0 - <^)], the arguments being subtracted and the moduli divided. A To construct the quotient OC, make angle AOC= and angle A = B. Then the argument UOC= UOA - UOB = 6-', and since OC : 0U= OA : OB, the length of 00 = the modulus of the quotient = — ^' Powers and roots. — The factor m (cos 6 -\-i sin B) repeated n times gives by (77) : [m (cos ^ -h i sin ^)]'* = m"(cos nO -h * sin nO). (78) n Write - instead of 6, m instead of m", and take the nth n root - / fi f)\ . - m" ( cos - 4- 1 sin - ) = [m (cos 0-\-i sin $)Y. \ n nj POWERS AND ROOTS. 157 Fia. 89. To construct the power, lay off, upward from the half-line OU,n angles each equal to UOAi = 6. Make OAi — m and join UA^. ""^^ Make angles OA^A^, OAiA^, etc., ^^^^ each equal to OCMi- Then 0C7= 1, OAi = m, OA2 = m^, OA3 = m*, etc. Now the arguments UOA^, UOA^f UOA^y etc., increase in arithmetical progression ; the segments OAi, OA2, OA^, etc., which represent the com- plex quantities m (cos $ + i sin 0), m^ (cos d-\- i sin Oy, m^ (cos 6-\-i sin fff, etc., increase in geometrical pro- gression. It follows that the arguments $, 2 6, 3 0, etc., are the logarithms of the corresponding complex quantities. Now denote m by k^ and take the ^h root of k^ (cos 6 + i sin 6). This root is k (cos 1 + I sin 1), which is the number whose logarithm is 1, and therefore the base of this system of logarithms. The curve UA1A2AS (Fig. j^ 89) is called the logarithmic spiral. It will be seen that it may be used as a graphical / / • \ ;^ table of logarithms ; the same thing is true of the 1 i^:^ lU curve in Fig. 81. When m = 1, the points Ai, A2, As, etc., will be ar- ranged at equal distances from each other on a circle of unit radius. Fig. 90. In this case k also equals 1 and the base of the system of logarithms is cos 1 + i sin 1 = .5403 -f .8415 i. 158 COMPLEX QUANTITIES. Construction of roots. — It has already been shown that the nth root of m (cos -{-i sin d) was mn( cos--f ism- ), \ n nj where m is the positive square r root of a^ + 6^ ; m» will here de- note the real positive nth root of m. Now the functions of 2 stt + are the same as those of 0, and hence m (cos + i sin 6) \n = ran cos 7i n and by taking values of s = 0, 1, 2, 3, etc., the n roots are 1 / ^ m"( COS- + I sin- m" ( cos f- 1 sm 2,7 + 71 Taking ^ = y5_^ ^ 50°^ m = 3.052, ti = 5, the roots are 1.25 (cos 10° + 1 sin 10°), 1.25 (cos 82° + ^ sin 82°), 1.25 (cos 154° + i sin 154°), 1.25 (cos 226° -f i sin 226°), 1.25 (cos 298° + ^ sin 298°), giving the points E^, R21 ^^3? -^45 ^s- By taking m=l and x=0, the 71 roots of +1 are obtained. By taking m=l and x=tt, the 71 roots of —1 are obtained. By taking 771= 1 and a;=|7r, the n roots of i are obtained. By taking m=l and a;= | tt, the n roots of —i are obtained. SERIES FOR 8IN0 AND COS 0. I59 81. Values of functions of multiple angles. — Put 1 for m in (78) : cos nO + i sin nO = (cos 6 -\- i sin Oy. (79) Expand by the binomial formula ; equate the imaginary parts and also the real parts : sin ne = n cos^-^^ sin - ^(^-^K^-^) cos"-^^ sin^ $ -f- etc. COS n6 = cos"^ - ^^r""-^) cos'*-^^ sin^ 9 + etc. 1 • ^ Thus to obtain the functions of 3 6, make w = 3, sin 3 (9 = 3 cos^ ^ sin ^ - sin^ = 3 sin 0-4: sin» 0, (80) cos 3 ^ = cos^ i9 - 3 cos ^ sin^ = 4: cos^ ^ - 3 cos 0. 82. Exponential value of cos 9 + / sin 6. — When is zero, cos i-i sin 0is 1. Now let us imagine a point generating a circle ; when the angle is zero, counted from a horizontal initial line, the horizontal component of the motion of the generating point is zero, and the vertical component of its motion is equal to the motion in the arc. Therefore when is at zero, the rate of increase of cos^ + i sin is i times the rate of the generating point. If we assume the gener- ating point to move a unit distance in unit time, the rate of increase (0 = zero) of cos ^ + 1 sin ^ is i. Now is the logarithm of cos -\-ism0; therefore, by the theory of logarithms, cos + i sin = e^, v being the rate of increase in the number cos -i-i sin 0, when is zero and increasing at unit rate. Therefore cos -\-ism0 = e^ ; Changing the sign of : cos ^ — ?' sin ^ = e~^; whence sin = — ^ (e^*" — e~^') _e 0^ , 0" 0' . .^.. ~1 1.2.3'^1.2.3.4.5 1.2.3.4.5.6.7^ ^ ^ cos(9 = i(e«* + e-^') 160 HYPERBOLIC FUNCTIONS. HYPERBOLIC FUNCTIONS. 83. Definition of an equilateral hyperbola. — Suppose a double cone (called a cone of two nappes) with vertex at V. If this cone is cut through the axis, two - ^ Singles, BVE Siud DVC, will be formed. If the cone is such that this angle is 90°, and the cone is again cut by a plane OA normal to the plane BVE, this new section is bounded by a curve of two branches QAR and Q^A^Ri; this curve is called an equi- lateral hyperbola. The points A and Ai are called the vertices of the hyperbola, the point O, midway between them, is called the center of the hyperbola, and AA^ is called the axis of the hyperbola. The two following properties of the equi- lateral hyperbola are proved in treatises deal- ing with this curve. First. If Q is a point on the curve, and O^is the abscissa ofQ, ON' - NC/ = 0A\ Fig. 92. Fig. 93. THE EQUILATERAL HYPERBOLA. 161 Second. Area segment AQN oi the hyperbola ^ONxNQ iOA' '-■Fin- Now the sector OAQ = triangle ONQ — segment AQN. Hence sector OAq = \ OAHogJ ^^^J^^ \ The quantity log ON ^ ] will be denoted by u: and OA J *^ ' the semi-axis OA will be denoted by a. Hence Sector OAQ = la^u. 84. Definitions of the hyperbolic functions. — Let be the center of an equilateral hyperbola, 0A(= a) its semi-axis, Q any point on the curve, ON the abscissa of Q, and NQ the ordinate. With a center and a radius a, describe a circumference; let P be a point on the cir- cumference ; let the length of arc AP be equal to aO ; then 6 represents the cir- cular measure of the angle AOP. The area of the cir- cular sector AOP equals its arc times ^ the radius = ^a^O. * 1. The equation of the two generators of the cone are x—y=0 and X + y = 0, their product being x'^—y'^=0. As these generators revolve around the axis of X, x remains constant and y^ becomes y^ + z'^. Hence the equation of the double cone is x'^ — y^ — z^ = 0. If now the cone is cut by the plane z= a, the projection of the section on the xy-plane is x^ — y^ = a^. 2. The integral of ydx or Vx"^ — a^ dx is Fia. »4. ^ xVx^ — a2 — 1 a'^ loge (x + Vx^ _ a^). Taking x for the upper limit and a for the lower, and putting y for Vx^ — a^, the area is I xy + y\ a I 162 HYPERBOLIC FUNCTIONS. From P, drop a perpendicular PM on OA. Then PM PM = a sin 6, 0M= a cos $, and — — = tan 6, and so on. From OM these properties of PM and OiHf, the trigonometric functions have been also called circular functions. From analogy, certain ratios of the semi-axis, the abscissa, and ordinate of a point on an equilateral hyperbola are called hyperbolic functions. They are defined as follows: Let u be such a quantity that Area sector OAQ = \d^u. Then 1. The ratio of NQ to OA is called the hyperbolic sine of u. Hyperbolic sine u will be written sinh u and will be read " shin ^^." It follows that QN= a sinh u. 2. The ratio of ON to OA is called the hyperbolic cosine of u ; written and pronounced cosh u. It follows that 3. The ratio of sinh u to cosh u {NQ : ON) is the hyperbolic tangent of u (tanh u). 4. The reciprocal of cosh u {OA : ON) is the hyperbolic secant of u (sech u). 5. The reciprocal of sinh ?* {NQ : OA) is the hyperbolic cosecant of u (csch u). 6. The reciprocal of tanh it {ONiNQ) is the hyperbolic cotangent of u (coth u). 85. Exponential values of hyperbolic functions. — By the property of the equilateral hyperbola, ON' - NQ" = OA'. Dividing by OA', cosh^ u — smlci' u = 1. But since u^log,(2^^y cosh u -f sinh u = e". Hence, by division, cosh u — sinh u = e~". RELATIONS OF THE FUNCTIONS. 163 Therefore, by subtraction and addition, sinh?^ = |(e«-e-«), (83) cosh u = ^ (e" + e-"). (84) GENERAL RELATIONS. 86. Relations of hyperbolic and circular functions. Now ^ (e" - e- ") = - i [~ (e"" - e" «")1 . Hence by (81) and (83) : sinh u = — i sin ui. (85) Also, |(e" + e-") = ^ (e"" + e-«'*). Hence by (82) and (84) : cosh u = cos ui. (86) Periodicity of hyperbolic functions. Now — I sin ui = — i sin (ui — 2 mr) = — i sin (u + 2 mri) L By (85), sinh u = sinh (w + 2 nTrt'). (87) Similarly, cosh u = cosh (u-{-2 mri). (88) The addition formulae. — From (S5), (S6), (65), and (66) : sinh (u -\- V) = — i sin (ui -f- vi) = — i (sin ui cos vi + cos ui sin vi) = sinh u cosh v -|- cosh u sinh v. (89) cosh (u-\-v) = cos (wi + vi) = cos wi cos vi — sin ?i i sin vi = cosh w cosh V + sinh m sinh v. (90) Functions of multiples of u. — From (89) and (90) there is obtained sinh 2u = 2 sinh u cosh u, cosh 2 u = cosh^ ^i + sinh^ ?/, sinh 3 w = 4 sinh^ 2t + 3 sinh w, (91) cosh 3 I* = 4 cosh^ u — 3 cosh u. (92) 164 HYPERBOLIC FUNCTIONS. 87. Longitude of u. — With, center N and radius equal to NQ, draw an arc intersect- ^ v > Q ing the circle in P. Join NP and OP. Since Fig. 95. (93) (94) ON'-NP' = OP\ Therefore, the angle OPN is a right angle. Denote the angle PON by x. sinh u = NQ:OA = NP: 0P= tan x, cosht^ = ON: OA = ON: 0P= sec x. In (11) put 90° 4- X for x ; then tan x-\-aecx = tan (45° -f- ^ x). Therefore tan (45° -\-^x)= sinh u + cosh u = e". Hence log tan (45° + 1 a;) = i^ log e = .43429 u. From which for any value of u, x can be found, and thence in (93) and (94), sinh u and cosh u can be found. The angle x was called by Gudermann the longitude of u. Some later writers, however, have referred to it as the Gudermannian function of u. A table at the bottom of page vii gives values of u corre- sponding to values of x, the longitude of u, within limits which admit of interpolation. 88. Solution of cubic equations. — Consider the cubic equa- tion 10^ ± aw = b. First. The coefficient of w is positive. Let w = 2 sinh u -i/ ^• The given equation reduces to sinh^ u + (95) X smh u = —- Sa Hence by (91), sinh 3 u b /3^ a ^a (96) CUBIC EQUATIONS. 165 Denote three values obtained from this equation by 3 u, 3 w + 2 TTi, Su — 2 iri. By (89), sinh(w ± I iri)= sinh u cosh f tti ± cosh u sinh | iri = sinh u cos 120° ± i cosh u sin 120°. Hence, to solve the equation under consideration, find u from (96) ; then by (95) the three roots of iv are 2 sinh u a/^, and ^^ ( — sinh u ± i V3 cosh u). Second. Consider the equation w^ — aw = b. Case I. 27 b^ is greater than 4 a^. Let m; = 2 cosh w a Z^- The given equation reduces to cosh^ w — f cosh i^ = TT— \/— 8a ^a Hence by (92), cosh 3u = ^ J?. (97) Find T^ from (97) ; then the three roots are 2 cosh u a/^, and a/- (— cosh u ± i V3 sinh it). Case II. 27 6^ is less than 4 a^. — This is the irreducible case of Tartaglia's solution, in which all the roots are real. Let w = 2 sin x a/-* The given equation reduces to sin' x — ^smx = — 8 a ^a Hence by (80), sin 3 a: = - 1-^ ^. (98) Find X from (98) ; then the three roots of iv are 2smxy[^, 2 sin (a; ± 120°) ^. 166 HYPERBOLIC FUNCTIONS. Exercises. — 1. Find the value of i\ Kaise to the i power the equation cos ^ + / sin ^ = e^\ (cos + i sin i$y = e-'. . + i)T. i' = e-l'^ X e^"'' = .2079e2n- where n is any whole number ; when n is zero, i* = .2079. 2. Solve by Horner's Method the equation and thus find sin 10°, sin 50°, and sin 130°. 3. By substituting ^ for in (81), find sin 10°. 18 4. Solve the following cubic equations : a? + 2Ax = 511 y?- Q>x= 9 t? - 9aj = 10 a^_,_54^^288 a:^- 7a;= 90 ar^- 7a;= 6 a;3 + 72a; = 400 . a^-48a; = 520 ar^-39a; = 70 PROBLEMS FOR EXAMINATION. [No answers are given for these problems.] RIGHT TRIANGLES. 154. A person standing at a distance of 143 feet from the middle of tlie base of a tower finds the altitude of the tower to be 43° 10'. Find the height of the tower, the eye of the observer being 5 feet from the ground. 155. From the top of a lighthouse 127 feet above the level of the sea, the angle of depression of a boat was seen to be 10° 20'. Find the distance of the boat from the foot of the lighthouse. 156. At what angle do we ascend a regular acclivity 6.5 miles long, attaining an altitude at the summit of 4268 feet ? 157. The horizontal distance between two columns is 124 feet, and the straight line joining the tops of the columns makes an angle of 12° 40' with the horizon. Find the dis- tance between the tops. 158. A canal boat is drawn along the middle of a straight canal by two men hauling upon ropes 60 feet and 70 feet long, respectively. Find the ratio of the strains upon them, the canal being 20 feet wide. 159. A row-boat, pulled at the rate of 5 miles an hour, is required to cross a river 1.3 miles wide to a poi-nt which is .78 mile lower down the stream than the point of starting. What angle must the direction of the boat's head make with the bank, supposing the stream to run at the rate of 4 miles an hour ? 167 168 PROBLEMS FOR EXAMINATION. 160. If d is the distance (in degrees) of the sun from the equator, show that the altitude of the sun at midday is 90° -f- d — Z, where I is the latitude of the place. Hence, find the length of the shadow at midday of a tower 120 feet high, the latitude of the tower being 40°, and the sun being vertically above a point in 10° north latitude. 161. At midday of the 21st of March, the length of the shadow of a 12-foot pole was 10.6 feet. What was the lati- tude of the place ? 162. The problem of Eratosthenes. At midday, when the sun was vertically over Syene, the shadow of a vertical rod at Alexandria was measured, the shadow being found to be .13 times the length of the rod. Taking Alexandria to be 5000 stadia due north of Syene, find the circumference of the earth. 163. The latitude of Berlin is 52° 30*. Taking the earth to be a sphere whose radius = 3960 miles, find the radius and length of 1° of the parallel passing through Berlin. 164. What will be the visual angle of a sphere 6 feet in diameter, if the eye of the observer is placed 7 feet from its center ? 165. On a hill over the sea is a 60 feet high tower. From a boat, the angle of elevation of the foot of the tower is 14°, and of the top of the tower is 20°. Find the height of the hill. 166. A flagpole, 20 feet high, is on the top of a tower. At a horizontal distance of 420 feet from the foot of the tower, the angle of elevation of the highest point of the flagpole is seen to be 15° 20'. Find the height of the tower. 167. The lu7ie of Hippocrates. A semi-circumference and a quadrant concave in the same direction are drawn to meet at their extremities so as to inclose space ; the radius of the semi-circumference is 10 feet ; find the area inclosed. RIGHT TRIANGLES. 169 168. On one side of an angle A = 50°, a distance AB = 10 feet is taken. From B, a perpendicular BC is let fall upon the other side of the angle ; from C, a perpendicular CD is let fall upon the first side ; from D a perpendicular DE is let fall upon the second side ; and so on, ad infinitum. Find the sum of all the perpendiculars. Put A instead of 50° and a instead of 10 feet ; then obtain the general solution of this problem.* 169. A circle has a radius of 22 inches. From the same point in the circumference, two chords are drawn of lengths 10 inches and 17 inches respectively. What angle is formed by these chords ? 170. Two towers stand 100 feet apart on the same hori- zontal plane. An observer in line with them, and 150 feet from the nearer one, sees the angle of elevation of both of them to be the same. He then walks directly toward them a distance of 130 feet, when he finds the angle of elevation of the nearer one to be twice as great as that of the farther one. Find the heights of the towers. Before performing the computation for this problem, obtain the general solution of it. 171. The top of a hill is 90 feet higher than the foot of a tower ; from the top of the hill the angle of depression of the top of the tower is 5°43|-'. The horizontal distance from the tower to the hill is 399 feet. Find the height of the tower. 172. I bought a lot having the shape of a right triangle ; the hypotenuse = 875 feet, and one of the acute angles = 37° 53'. How much was paid for the land if the price was 2.7 cents per square foot ? * If for the data of a problem, literal values are assumed instead of numerical values, the answer to the problem expressed in terms of these literal values is called the general solution of the problem. 170 PROBLEMS FOR EXAMINATION. OBLIQUE TRIANGLES. 173. To measure the distance between two points A and B separated by a pond, a third point O was chosen. AC was found to be 545 feet, CB = 305 feet, and the angle ACB = 67°. Find AB. 174. From a point on the bank of a river, an observer goes directly away from the river 165 yards up a slope inclined at an angle of 12° 12' to the horizontal; at this point the angle of depression of an object on the opposite bank and in the same plane with the first position was found to be 2° 10'. Find the breadth of the river. 175. The range finder. From a fort a torpedo-boat is observed to be exactly east; eighty seconds afterward it is exactly northeast, and eighty seconds after that it is N. 22° 30' E. The torpedo-boat is going at a uniform rate in a straight line. Find its bearing at the end of the next eighty seconds. 176. A vertical tower stands on a plane inclined 11° 20' to the horizontal. Measuring from the base of the tower down the slope an oblique distance of 59.18 feet, the tower was found to subtend an angle of 38°. Find the height of the tower. 177. Two inaccessible points X and T are observed from two stations A and B, 500 feet apart {A and X on the left). The angle XAY= 42°, angle YAB = 67°, angle ABX = 39^ angle X5 r= 47°. Find XY. 178. A house is seen from two ships A and B which are 300 feet apart. At A, the angle of elevation of the house is found to be 28° 20' and the horizontal angle from it to the ship B = 48° 40'. At B, the horizontal angle from the house to the ship A is found to be 20° 15'. Find the height of the house above the level of the water. OBLIQUE TRIANGLES. 171 179. To find the length BC of a lake running due north and south {C north of B), a point D was taken 530 feet north of C and a point A was found due west of B. It was found that AC = 1970 feet and AD = 2400 feet. Find the length of the lake. 180. To find the distance between two inaccessible points A and B, an observer chose a point C in the same plane and found angle ACB = 113°. Then he walked on the continua- tion of AC to a point 125 yards from C, where the angle subtended by AB was found to be one-half of angle ACB. Then he walked on the continuation of BC 173 yards from C and again found the angle subtended by AB to be one-half of angle ACB. Find AB. 181. Given the three sides of a triangle to find the length from the vertex to a point of trisection of the base. Take the base (c) = 200 feet, a = 205, b = 85, and find the length to the point of trisection near a. 182. How many square miles of the earth's surface can an observer see if stationed at the top of Mt. Everest, whose altitude is 29,000 feet ? The radius of the earth = 3960 miles. 183. A lighthouse 60 feet high is on a cliff over the sea. From the top of the lighthouse the angle of depression of a boat was seen to be 16° 20' ; and from the bottom of the lighthouse, the angle of depression of the boat was 12° 10'. Find the height of the cliff and the distance of the ship from the foot of the cliff. 184. Wishing to know the breadth of a river from a point X to a point A, an observer measures in the prolongation of XA a distance AB — 500 feet ; then from B, he measures at right angles to XAB a distance BC = 700 feet ; from the point C, the angle XCA was found to be 24° 20'. Find the breadth of the river. 172 PROBLEMS FOR EXAMINATION. 185. P, Q, and B are three known points in a straight line; PQ = 104, QP = 296 ; PQ and QR are observed to subtend equal angles = 46° 50' at a certain point S. Find SQ. 186. From the bottom of a wall 20 feet high, the angle of elevation of the top of a house = 40° 20' ; at the top of the wall the angle of elevation of the same is 19° 40'. Find the height of the house. 187. A person at a point A, wishing to know the distance of an inaccessible object C on the opposite side of a river, lays off a base AB = 898.8 ; he then measures the angle CAB = 63° 20' and angle CBA = 52° 40'. Find AC. 188. Of three towns the second is 11.6 miles from the first, the third is 10.5 miles from the second, and the first is 14.3 miles from the third. Find the angles of the triangle whose vertices are the three towns. 189. From the deck of a ship, 10 feet above the sea, the angle of elevation of the top of a cliff is 20° 20' ; from the top of the mast &o feet higher, the angle is 12° 30'. Find the height of the cliff from the sea and the distance of the ship from the foot of the cliff. 190. From the top of a tower 60 feet high, a column sub- tends an angle of 28° ; from the base of the tower, the col- umn subtends an angle of 40°. Find the height of the column. 191. A balloon is ascending vertically with a uniform velocity. When it is one mile high, its angle of elevation is 32° ; twenty minutes later, the elevation is 54°. How fast is the balloon rising ? 192. At two points, 70 and 50 feet from the foot of a tower, a flagstaff on the top of the tower subtends the same angle, viz., 3°. Find the length of the flagstaff. REDUCTION TO FIRST QUADRANT. 173 FUNCTIONS OF ANGLES. Find a function of an angle of the first quadrant equiva- lent to each of the following functions : 193. sin 283°. 201. sin 324°. 209. sec 247° 42'. 194. cot 196°. 202. CSC 562°. 210. esc 176° 28'. 195. cos 344°. 203. cos 897°. 211. cot 31 2° 53'. 196. sec 127°. 204. tan 631°. 212. sec 228° 12'. 197. tan 274°. 205. tan 189° 49'. 213. tan 142° 29'. 198. cos 245°. 206. cos 128° 11'. 214. cos 218° 47'. 199. CSC 144°. 207. sin 212° 14'. 215. sin 374° 55', 200. cot 133°. 208. cot 321° 37'. 216. esc 524° 19'. Reduce each of the following expressions to its simplest form: 217. a cos (180° -x)-\-b cos (90° + x) cot x. 218. k sin X cot x tan (180° + x) sec (90° - x). 219. tan (180° - x) cot (270° - x) -{- cot (90° + x) cot x. 220. cos (90° + X) sin (360° - x) - sin (270° - x) cos (- x). 221. (a + by sec (360° -x)-{a- bf esc (90° + x). 222. sec (90° + x) CSC (360° - x) + tan (90° -\- x) cot x. 223. tan a; cot a; + tan (— 2/) + tan (180° -f- y). 224. sin (180° -a;) COS (90° -t-ir) CSC (180° + a;) CSC «. 225. (a^ + b^) sec (270° -x)-{- (a' - b") esc (180° + x). 226. a sin (360° - x) tan (270° + x) sec (180° - x). 227. 2 cot 2 x COS a; + 2 sin x — esc x. 228. a? sin 90° + 2 a6 cos 180° + W sec 0°. 229. m sin 90° - n cos 360° + {m — n) cos 180°. 174 PROBLEMS FOR EXAMINATION. Prove the truth of the following equations : ___ sin X -\- tan x • , 230. — = smictana;. cot X + CSC X The proof consists in reducing the equation to an identity. Thus, by clearing of fractions, both sides of the equation are sin a; + tan x. By writing the work in reverse order, a direct proof of the given equation is obtained. 231. tana: = 2i^H^^^Hl+-^^^).. (1 + cos x) (1 H- CSC x) __- sec x cot x — CSC x tan x 232. CSC a; sec a; = — • 233. tan a; — cot a; = cos X — sm X 1 — 2 cos^ X sm X cos X 234. tan X + cot X = cot x sec^ x. 235. sin^ X + tan^ x = sec^ x — cos^ x. 236. sec X — cos X = sin x tan x. _-_ sin X + cos a^ _ sec a; + esc x ZOt, — ; • sin X — cos X sec x — esc x 238. (sin x + tan x) (cos x + cot a;) = (1 + sin x) (1 + cos x). 2 tan X 2 cot a; 239. sin 2 a; = 240. tan 2 a; = 1 + tan^ X 1 4- cot^ x 1 1 1 — tan X 1 + tan x .2, ^^, o sec'.'K sec a; CSC a; 241. sec 2 a; = 1 — tan- X cot X — tan x 242. CSC 2x = ^ cot a: cos^ x = ^ (tan a; + cot x). 243. cot 2 a; = 1^ (cot X — tan i») = ^ sec a; esc a; — tan x. cot^ a; — 1 cot a? — tan x 244. cos 2 a; cot^ a; + 1 cot a; + tan x GONIOMETRIC RELATIONS. I75 245. sin {x + y) = (tan x + tan y) cos x cosy. 246. tan (a: + 2/)= . ^^^'^-^^^'y sin X cos a; — sin y cos y 247. 5ec(x + y)= ^««^sec.v , 1 — tan X tan ?/ 248. sin (x — y) ^ tan a; - tan ?/ sin (a; + y) tan a; + tan y 249. tana;tan2/ = *Hi^+i5M. cot X + cot 2/ 250. tan(^ + y) + tan(x-y)= /tan^^f.'/^ ■ 1 — tan^ a.- tan^ ?/ 251. sin(a;+2/)-hcos(a;— ?/)=(sina;+cosa;)(sin!/4-cos2/). 252. tani(a;-2/)=^i2^^=-^iM = 22^r:i^2^. cos a; 4- cos y sin ?/ + sin x 253. tan (45° + a;) = ^^^^ + ^"^^ = L±taii^. cos X — sin x* 1 — tan x 254 tan a; ^ sin (a; + y) + sin (x — y) tan y sin (a; + ?/) — sin (x — y) 255. tan^ ^ x + 2 cot a; tan \x — l = 0. 256. tan- 1 a? — 2 CSC a; tan ^ a; + 1 = 0. Explain how the general quadratic equation may be solved by the use of 255 or 256. 257. tanla^ = l±4^?^^=-^^^. 1 + sin X + cos X 258. sin nx = 2 sin (^i — V)x cos x — sin (n — 2)x. 259. cos nx = 2 cos (n — l)a; cos x — cos (n — 2)a;. 260. tanna:= tan (n - l)a^ + tan a; ^ 1 — tan (11 — V)x tan x 176 PROBLEMS FOR EXAMINATION. Solve the following equations : 261. a tan^ a? + 6 = c sec a;. 262. cos^ X — sin^ x + tan^ ^ = f- 263. sin (20° + a;) cos (20° -a;) = .2. 264. tan X + cot aj = 4 cot 2 x. 265. cot X — tan x = 4 cos 2 a;. 266. tan 2 a? + tan 3 a? = 3 tan a;. 267. 1 — cos 2 aj = 3 sin 2 a;. 268. sin X + cos a; = V2. 269. 16 (cot a; — sec x) = tan x. 270. tan aj + cot a; = 2 sec 2 a;. 271. cot X tan 2 a; — tan a; cot 2 a; = 2. 272. sec^ X -f- tan a; = 3. 273. arctan(^+l) = 3arctan(i\r-l). — arc tan = ^-3- ■ N-1 N+1 '^ 274. arc tan^:= — r — arc tan^r — 7 = t-V tt. ANSWERS. 1. c tan ^ = 102.7. 21. aVcot^^ + cot^i) = 109. 2. a cot 12° 40' = 387.1. ^2 ___^L_^__ - 125 3. cot a; =2.4; 22° 37'. ' Vcot^ i) - cot^ ^ ~ 4. 23 sec 38° = 29.19. 23. b^ sm2 A=b' - (a - cY; 5. 880 sin 33° 3' = 480. ' N. 22° 32 ' E. 6. c tan ^ = 125.1. 24. 4ir=918(a- 6 -f c) ; 7. ctan^+d = 18. 340,189. 8. c (tan ^ + tan A) = 102. ^5. 6^ = (a + c)^ - 4 7^; 377. 9. a(l+tan^cot^i) = 76.5. ^^- 1^.7,30.8. ^Q atan^ ^^^^ 27. 4.844,8.72. tan ^1- tan ^ * 28. 14.25,8.75. 11. 9^ = 500. 29. 154tan27°10' + 5 = 84. tan^+tanA 30, 3034,52.94. 12. a (tan G — tan Cj) = 1.5. ^^ ^go ^q, 13. c(tanA-tan^)=128.4. ^^ 12.6, 54° 52'. c "• cot^ + cotA = ^^^' 33 27r^^os40_°^ 2206 ctany^ ' 86400 ' tan ^.- tan ^1 34. 2° 18.7', .1615m., 852.6 ft. 16. V^ = 150. 35. tana? = ^ + ^; 77°. aan^ _ *^^-^^ 18. (a - /i) cot ^ = 3.1. 37. aA/^i^ = 60. ^ h— a 19. cota; = ?2; 51° 41'. 2^ o^ ^ .^o.., 38- ^ CSC 33° 45' = 10. 20. tana; = n; 47° 44'. 27 177 178 39. 13.5 CSC 58° 6' = ANSWERS. : 15.9. cot 20° 15' H- 2 cos 20° 15' = 21.8. 41. sec^ 2.4; A 65° 23' ; K= 10.64. 42. 1.5fZ2sm60°sm2i^ 8.798. 43. Area = two triangles — sector = lic^ sin 60° ; 15. 44. 7r62 cos ^ = 204.2. 45. 7r/fseci(7 = 33.49. 46. 367rF2 = ^^sin2(7csci(75 73.43. 47. rad base = 1.81 CSC 54° 18' ; F= 36.65. 48. p^seGA = 2p', K=p^ Got A = 25.57. 49. tan^ = i7r; ^=57° 31'. 57. 2.1437rCot 25°43'= 14. 50. 100°. 51. 50. 52. 48° 55'. r, 90° 106.27 7r6 53. 6.185. 180 a 54. 21. 55. 163° 34'. 56. .3125 d^csc^ 36° = 12.15. 58. J a CSC a cos 2n + l 180° 2n + l 59. 143.1 feet. 60. 27 r" tan 6° 40' sin26° 40' = 3.6. = 8.988. 7.208. „ cyTYij^ 360°. 180° ^^^^ 61. 2— A^csc tan = 116.5. n n TYi 62. 1 7^ (16.96 TT + sin 70° 32 ' - sin 40°) = .4656. 63. 441 cos 46° tan 20° = 157. 67. c sin (A^-A) sec A sec A-,. 64. acsc(^i— J.)cos^sin^i. 68. ccsc (J.+J.1) sin^sin^i. 65. acsc(^-|-^i)cos^cos^i. 69. c esc (^— ^j) cos^ sin^i. 66. a sin (C- d) sec C sec Oj. 70. Apply (19) ; 890. 71. Apply (20) ; 741.2. 72. 18 sec 87° 30' cos 37° 30' sin 40° = 210. 73. 50 CSC 5° 55' cos 31° 12' sin 25° 17' = 177.1. 50 esc 5° 55' cos 25° 17' = 438.4. 74. Apply (24) ; N. 87° 49' E. Apply (19) ; 12 miles an hour. OBLIQUE TRIANGLES. I79 75. 528 CSC 37° sin 42° sec 29° sin 8° = 114.5. 528 CSC 37° sin 5° = 301. 76. c CSC C sin ^ sin J5 — 50 = 240. 77. 775.3 CSC 27° 35' sin 126° 43' sin 32° 57' = 730. 78. 200 sec2 26° 37' CSC 10° 8' sin 10° 22' sin 20° 30' = 89.63. 79. 180 CSC 17° 15' sin 33° 45' sec 40° sin 11° = 84. 80. Apply (27) ; 18.71 -|- 16.2 = 34.91. 81. Apply (24) ; cot a; = 1.76 esc 30° - cot 30°. 29° 15' with the front of the train. 82. 6 sin 45° CSC 10° = 24.44. 85. 3.5. 83. N. 70° 44' E. ; 9.983. 86. Apply (30) ; 229,000. 84. 10.32; 78° 14'. 87. Apply (29) ; 52. 88. Apply (26); h ^xGoiA,l = xcotD, a= XQotB\ 60. 89. Apply (19) ; 20.51 ; 23.63. 90. Apply (16), (19), and (20) ; cos^ = -^^+i- = -^±i-; c = 5. 2(c + l) 2(c-l)' 91. cco^\{B + A) ^\n\{B - A) = 51. 92. Snell's; 1034,1465,2047. 93. 300 CSC 68° 14' sin 120° 26' csc 17° 48' sin 70° = 856.2. 94. 100 sin 88° 10' csc 8° 20' = 676. 95. Apply (25) ; 5.715. 96. Apply (25) ; 331. 97. Apply (28) ; 1054. 98. Find AB and BD and apply (20) ; 287.2. 99. Find AB, then BC = .3537. 100. sin ^Z)5 = tan 56° 26' cot 65° 23' sin 56° 33'; 114. 101. Apply (19) ; 410. 102. Same as Problem 88 ; 100. 103. Snell's ; 229, 61, 109. 104. By(30)cosM=.85; ^=22° 47'. By (28) cot aj=1.125 csc 45° 34'-cot45° 34'; a;=59° 15'. By (19) ^=9 csc 22° 47' sin 82° 2' cos 59° 15'=11.77. 180 ANSWERS. 105. I (12 CSC A) (16 CSC A) sin ^ = 96 esc ^ = 100. 106. Divide into a triangle and a parallelogram ; 46.82. 107. i a ^ + «'±V3^ = 9 sec. or 1 min. 57 see. v^ + ^t7^ — i^it; 108. 70 sin 1 (^1 - A2) sin i (J5i + ^2) X esc 1 (A + ^2 + ^1 + -S2) = 18.34. 109. The pond is the incircle of the triangle formed by the three tangents ; 36.58. 110. The pond is the excircle of the triangle formed by the three tangents ; 768.3. 111. The hill is at the center of the circumcircle. 60.8 CSC 15° 25' tan 52° 40' = 300. 112. r tan ^5 tan 1(7=1.589. 113. r tan^ (45 -i^) =1.718. 114. 4r sin (45° - ^A) sin (45° -\B) sin (45°-J C)=2.925. 115. 3107. 116. 5057. 117. 2713. 118. 2410 geographical miles. 119. 7 hours. 120. lat. 37° 36' ; long. 68° 38'. 121. cosZ = -secZsind 238° 42'; 270°; 301° 18'; 180°. cos P = — tan I tan d. length = 2 P -r- 15. 14 hr. 50 min. 48 sec. ; 12 hr. ; 9 hr. 9 min. 12 sec. ; 24 hr. 122. sec P = tan I cot d. 3 hr. 55 min. 28 sec. ; 6 ; 12 p.m. 123. sin a = sin Z sin d. 14° 49 '. tan Z=secl cot d. 251° 37'. 124. tan Z = tan d sec 31° 30'; 14° 46'. 125. Let R be first position, S the secon'd. i?^ = 49°40', PP*S = 84°2', ZP^ = 50°31'; lat. = 47° 55'. 126. P = 32° 30', ZS = 45° 40', PS = 116° 6' ; lat. = 6° 56'. 127. 40° 6'. 128. .336 PI 129. 2.443 P2. 130. fl9K±Z^M±= -1.121. 131. -^ ^-^^ =.216. 6 log a— e log d 6 loga+dlog c 132. Eoots of x^ \ogd — x log 5 = log a — log c. Imaginary. GONIOMETRY. 181 133. ^ ^ log ?> (/log ^ - ^i log ^) ^ _ 21.93. log a log e — log h log c 134. „ = iri2g* + M = 1.108. 2 |_log a log 6J 135. Eoots of x^ log a + a; log 6 + log c = 0. Imaginary. 136. loglogc-loglogg^ g3o^ log 6 ,^„ , .6 sin 20° - sin 10° ^o KAf 137. tan a; = -— — -; 2° 56'. cos 10° -.6 cos 20°' 138. Reduce to sines and cosines and apply (69), sin {2x- 10°) = - 5 sin 10° ; 125° 7'. 139. Apply the value of tan (x ± y) and reduce cot2a; = l+2tan2 10°; 44° 8'. 140. Apply (65) and (75) ; cot a;=- (2 + cot 10°); 172° 34'. 141. cos (2x- 10°) = 1.2 - cos 10° ; 43° 47'. 142. By (13) ; 32 cos^ a; = 16 + 16 cos 2 x. 16cos22a; + 16cos2a; + 3 = 0; a; = 52°14'. 143. Apply (69); 29° 33'. 144. 4cosa;= V17-1; 38° 40'. 145. sin (2 a; -70°) = .6 CSC 70°; 105° 10'. 146. tan x= .4142 ; 22° 30'. 147. tan x = tan a tan h tan c. 148. 160 tan ^ = 70 tan {A + 21°) = 168. 149. a tan ^ = (a — b) tan 3 ^ = 59.2. 150. 97° 37' with the road; 11.44 miles. 151. 360 cot A = 360 cot (^ - 2° 7 ') = 4994. 152. ir'iO- sin 0) = 13.714 and 4.252 ; 9.462. 7rr2 - (13.714 + 4.252) = 1239. 153. Two segments,; 0, = 153° 44', 6^ = 54° 33'; .1 A. 182 FORMULAE FOR RIGHT TRIANGLES. tan A COS A PLANE RIGHT TRIANGLES. Any side = adjacent side times adjacent function. Any function of A = adjacent side divided by the next side. Any function of C = the corresponding cofunction of A. SPHERICAL RIGHT TRIANGLES. Any function is equal to the product of its adjacent functions. Any function is equal to the product of an adjacent function and the reciprocal of the next function. ^'"^ tan A~-^ tape —CSC A— sin fan a — cos G- A TABLE OF FOUR-PLACE MANTISSA OF LOGARITHMS OF NUMBERS FROM 1 TO 1000 A TABLE OF THE LOGARITHMS OF THE TRIGONOMETRIC FUNCTIONS OF ANGLES FROM 0° TO 90° FOR EVERY 10' A TABLE OF FORMULA A TABLE OF VALUES OF CONSTANTS viu LOGARITHMS OF NUMBERS. N 1 2 3 4 5 6 7 8 9 10 .0000 .0043 .0086 .0128 .0170 .0212 .0253 .0294 .0334 .0374 11 414 453 492 531 569 607 645 682 719 755 12 792 828 864 899 934 969 .1004 .1038 .1072 .1106 13 .1139 .1173 .1206 .1239 .1271 .1303 335 367 399 430 14 461 492 523 553 584 614 644 673 703 732 15 .1761 .1790 .1818 .1847 .1875 .1903 .1931 .1959 .1987 .2014 16 .2041 2068 .2095 .2122 .2148 .2175 .2201 .2227 .2253 279 17 304 330 355 380 405 430 455 480 504 529 18 553 577 601 625 648 672 695 718 742 765 19 788 810 833 856 878 900 923 945 967 989 20 .3010 .3032 .3054 .3075 .3096 .3118 .3139 .3160 .3181 .3201 21 222 243 263 284 304 324 345 365 385 404 22 424 444 464 483 502 522 541 560 579 598 23 617 636 655 674 692 711 729 747 766 784 24 802 820 838 856 874 892 909 927 945 962 25 .3979 .3997 .4014 .4031 .4048 .4065 .4082 .4099 .4116 .4133 26 .4150 .4166 183 200 216 232 249 265 281 298 27 314 330 346 362 378 393 409 425 440 456 28 472 487 502 518 533 548 564 579 594 609 29 624 639 654 669 683 698 713 728 742 757 30 .4771 .4786 .4800 .4814 .4829 .4843 .4857 .4871 .4886 .4900 31 914 928 942 955 969 983 997 .5011 .5024 .5038 32 .5051 .5065 .5079 .5092 .5105 .5119 .5132 145 159 172 33 185 198 211 224 237 250 263 276 289 302 34 315 328 340 353 366 378 391 403 416 428 35 .5441 .5453 .5465 .5478 .5490 .5502 .5514 .5527 .5539 .5551 36 563 575 587 599 611 623 635 647 658 670 37 682 694 705 717 729 740 752 763 775 786 38 798 809 821 832 843 855 866 877 888 899 39 911 922 933 944 955 966 977 988 999 .6010 40 .6021 .6031 .6042 .6053 .6064 .6075 .6085 .6096 .6107 .6117 41 128 138 149 160 170 180 191 201 212 222 42 232 243 253 263 274 284 294 304 314 325 43 335 345 355 365 375 385 395 405 415 425 44 435 444 454 464 474 484 493 503 513 522 45 .6532 .6542 .6551 .6561 .6571 .6580 .6590 .6599 .6609 .6618 46 628 637 646 656 665 675 684 693 702 712 47 721 730 739 749 758 767 776 785 794 803 48 812 821 830 839 848 857 866 875 884 893 49 902 911 920 928 937 946 955 964 972 981 50 .6990 .6998 .7007 .7016 .7024 .7033 .7042 .7050 .7059 .7067 51 .7076 .7084 093 101 110 118 126 135 143 152 52 160 168 177 185 193 202 210 218 226 235 53 243 251 259 267 275 284 2.2 300 308 316 54 324 332 340 348 356 364 372 380 388 396 LOGARITHMS OF NUMBERS. IX N 55 1 2 3 4 5 6 7 8 9 .7404 .7412 .7419 .7427 .7435 .7443 .7451 .7459 .7466 .7474 56 482 490 497 505 513 520 528 536 543 551 57 559 566 574 582 589 597 604 612 619 627 58 634 642 649 657 664 672 679 686 694 701 59 709 716 723 731 738 745 752 760 767 774 60 .7782 .7789 .7796 .7803 .7810 .7818 .7825 .7832 .7839 .7846 61 853 860 868 875 882 889 896 903 910 9i7 62 924 931 938 945 952 959 966 973 980 987 63 993 .8000 .8007 .8014 .8021 .8028 .8035 .8041 .8048 .8055 64 .8062 069 075 082 089 096 102 109 116 122 65 .8129 .8136 .8142 .8149 .8156 .8162 .8169 .8176 .8182 .8189 66 195 202 209 215 222 228 235 241 248 254 67 261 267 274 280 287 293 299 306 312 319 68 325 331 338 344 351 357 363 370 376 382 69 388 395 401 407 414 420 426 432 439 445 70 .8451 .8457 .8463 .8470 .8476 .8482 .8488 .8494 .8500 .8506 71 513 519 525 531 537 543 549 555 561 567 72 573 579 585 591 597 603 609 615 621 627 73 633 639 645 651 657 663 669 675 681 686 74 692 698 704 710 716 722 727 733 739 745 75 .8751 .8756 .8762 .8768 .8774 .8779 .8785 .8791 .8797 .8802 76 808 814 820 825 831 . 837 842 848 854 859 77 865 871 876 882 887 893 899 904 910 915 78 921 927 932 938 943 949 954 960 965 971 79 976 982 987 993 998 .9004 .9009 .9015 .9020 .9025 80 .9031 .9036 .9042 .9047 .9053 .9058 .9063 .9069 .9074 .9079 81 085 090 096 101 106 112 117 122 128 133 82 138 143 149 154 159 165 170 175 180 186 83 191 196 201 206 212 217 222 227 232 238 84 243 248 253 258 263 269 274 279 284 289 85 .9294 .9299 .9304 .9309 .9315 .9320 .9325 .9330 .9335 .9340 86 345 350 355 360 365 370 375 380 385 390 87 395 400 405 410 415 420 425 430 435 440 88 445 450 455 460 465 469 474 479 484 489 89 494 499 504 509 513 518 523 528 533 538 90 .9542 .9547 .9552 .9557 .9562 .9566 .9571 .9576 .9581 .9586 91 590 595 600 605 609 614 619 624 628 633 92 638 643 647 652 657 661 666 671 675 680 93 685 689 694 699 703 708 713 717 722 727 94 731 736 741 745 750 754 759 763 768 773 95 .9777 .9782 .9786 .9791 .9795 .9800 .9805 .9809 .9814 .9818 96 823 827 832 836 841 845 850 854 859 863 97 868 872 877,, 881 886 92 r 926 930 890 894 899 903 908 98 912 917 934 939 943 948 952 99 956 961 965 969 974 978 983 987 991 996 LOGAKITHMS OF FUNCTIONS. X L sivLX /.tanjr log sec X log CSC X log cot X L cosjr 0" Neg. Inf. Neg. Inf 000 Infinite Infinite 10.000 90= 10' 7. 4637 7. 4637 2. 5363 2. 5363 50' 20' 7648 7648 2352 2352 40' ^0' 9408 9409 0592 0591 30' 40' 8. 0658 8. 0658 1. 9342 1. 9342 20' 50' 1627 1627 8373 8373 10' 1= 8. 2419 8. 2419 0.000 1 1. 7581 1. 7581 9.999 9 89° 10' 3088 3089 1 6912 6911 9 50' 20' 3668 3669 1 6332 6331 9 40' 30' 4179 4181 1 5821 5819 9 30' 40' 4637 4638 2 5363 5362 8 20' 50' 5050 5053 2 4950 4947 8 10' 2° 8. 5428 8. 5431 0.000 3 1. 4572 1. 4569 9.999 7 88° 10' 5776 5779 3 4224 4221 7 50' 20' 6097 6101 4 3903 3899 6 40' 30' 6397 6401 4 3603 3599 6 30' 40' 6677 6682 5 3323 3318 5 20' 50' 6940 6945 5 3060 3055 5 10' 3° 8. 7188 8. 7194 0.000 6 1. 2812 1. 2806 9.999 4 87° 10' 7423 7429 7 2577 2571 3 50' 20' 7645 7652 7 2355 2348 3 40' 30' 7857 7865 8 2143 2135 2 30' 40' 8059 8067 9 1941 1933 1 20' 50' 8251 8261 0.001 1749 1739 10' 40 8. 8436 8. 8446 0.001 1 1. 1564 1. 1554 9.998 9 86° 10' 8613 8624 1 1387 1376 9 50' 20' 8783 8795 2 1217 1205 8 40' 30' 8946 8960 3 1054 1040 7 30' 40' 8. 9104 8. 9118 4 1. 0896 1. 0882 6 20' 50' 9256 9272 5 0744 0728 5 10' 5^-^ 8. 9403 8. 9420 0.001 7 1. 0597 1. 0580 9.998 3 85° 10' 9545 9563 8 0455 0437 2 50' 20' 9682 9701 9 0318 0299 1 40' 30' 9816 9836 0.002 0184 0164 30' 40' 9945 9966 1 0055 0034 9.997 9 20' 50' 9. 0070 9. 0093 3 0. 9930 0. 9907 7 10' 6^ 9. 0192 9. 0216 0.002 4 0. 9808 0. 9784 9.997 6 84° 10' 0311 0336 5 9689 9664 5 50' 20' 0426 0453 7 9574 9547 3 40' 30' 0539 0567 8 9461 9433 2 30' 40' 0648 0678 9 9352 9322 1 20' 50' 0755 0786 0.003 1 9245 9214 9996 9 10' T 0859 9. 0891 0.003 2 9141 9109 9.996 8 83° L cos/ Lcoty log CSC/ log sec/ log tan/ L sin/ / LOGARITHMS OF FUNCTIONS. XI X L sinjr L tan AT log sec X log CSC X Z.cot;r L cosjf T 9.0 859 9.0 891 0.00 32 0.9 141 0.9 109 9.99 68 83° 10' 961 995 34 039 005 66 50' 20' 9.1060 9.1096 36 0.8 940 0.8 904 64 40' 30' 157 194 37 843 806 63 30' 40' 252 291 39 748 709 61 20' 50' 345 385 0.00 41 665 615 59 10' 8^ 9.1 436 9.1 478 0.00 42 0.8 564 0.8 522 9.99 58 82° 10' 525 569 44 475 431 56 50' 20' 612 658 46 388 342 54 40' 30' 697 745 48 303 255 52 30' 40' 781 831 50 219 169 50 20' 50' 863 915 52 137 085 48 10' 9^ 9.1943 9.1997 0.00 54 0.8 057 0.8 003 9.99 46 sr 10' 9.2 022 9.2 078 56 0.7 978 0.7 922 44 50' 20' 100 158 58 900 842 42 40' 30' 176 236 60 824 764 40 30' 40' 251 313 62 749 687 38 20' 50' 324 389 64 676 611 36 10' 10^ 9.2 397 9.2 463 0.00 66 0.7 603 0.7 537 9.99 34 80° 10' 468 536 69 532 464 31 50' 20' 538 609 71 462 391 29 40' 30' 606 680 73 394 320 27 30' 40' 674 750 76 326 250 24 20' 50' 740 819 78 260 181 22 10' ir 9.2 806 9.2 887 0.00 81 0.7 194 0.7 113 9.99 19 79° 10' 870 953 83 130 047 17 50' 20' 934 9.3 020 86 066 0.6 980 14 40' 30' 997 085 88 003 915 12 30' 40' 9.3 058 149 91 0.6 942 851 09 20' 50' 119 212 93 881 788 07 10' 12^ 9.3 179 9.3 275 0.00 96 0.6 821 0.6 725 9.99 04 78^ ]0' 238 336 99 762 664 01 50' 20' 296 397 0.01 01 704 603 9.98 99 40' 30' 353 458 04 647 542 96 30' 40' 410 517 07 590 483 93 20' 50' 466 576 10 534 424 90 10' 13° 9.3 521 9.3 634 0.01 13 0.6 479 0.6 366 9.98 87 77° 10' 575 691 16 425 309 84 50' 20' 629 748 19 371 252 81 40' 30' 682 804 22 318 196 78 30' 40' 734 859 2"5 266 141 75 20' 50' 786 914 28 214 086 72 10' 140 9.3 837 9.3 968 0.01 31 0.6 163 0.6 032 9.98 69 76 L cos/ L cot/ log esc/ log sec/ log tan/ L sin/ / Xll LOGARITHMS OF FUNCTIONS. X L sinjr Ltanx log sec X log CSC X log cot X Z.cosjr 14° 9.3 837 9.3 968 0.01 31 0.6 163 0.6 032 9.98 69 76° 10' 887 9.4 021 34 113 0.5 979 66 50' 20' 937 074 37 063 926 63 40' 30' 986 127 41 014 873 59 30' 40' 9.4 035 178 44 0.5 965 822 56 20' 50' 083 230 47 917 770 53 10' 15" 9.4 130 9.4 281 0.01 51 0.5 870 0.5 719 9.98 49 75° 10' 177 331 54 823 669 46 50' 20' 223 381 57 777 619 43 40' 30' 269 430 61 731 570 39 30' 40' 314 479 64 686 521 36 20' 50' 359 527 68 641 473 32 10' 16° 9.4 403 9.4 575 0.01 72 0.5 597 0.5 425 9.98 28 74° 10' 447 622 75 553 378 25 50' 20' 491 669 79 509 331 21 40' 30' 533 716 83 467 284 17 30' 40' 576 762 86 424 238 14 20' 50' 618 808 90 382 192 10 10' 17° 9.4 659 9.4 853 0.01 94 0.5 341 0.5 147 9.98 06 73° 10' 700 898 98 300 102 02 50' 20' 741 943 0.02 02 259 057 9.97 98 40' 30' 781 987 06 219 013 94 30' 40' 821 9.5 031 10 179 0.4 969 90 20' 50' 861 075 14 139 925 86 10' 18° 9.4 900 9.5 118 0.02 18 0.5 100 0.4 882 9.97 82 72^ 10' 939 161 22 061 839 78 50' 20' 977 203 26 023 797 74 40' 30' 9.5 015 245 30 0.4 985 755 70 30' 40' 052 287 35 948 713 65 20' 50' 090 329 39 910 671 61 10' 19° 9.5 126 9.5 370 0.02 43 0.4 874 0.4 630 9.97 57 71° 10' 163 411 48 837 589 52 50' 20' 199 451 52 801 549 48 40' 30' 235 491 57 765 509 43 30' 40' 270 531 61 730 469 39 20' 50' 306 571 66 694 429 34 10' 20° 9.5 341 9.5 611 0.02 70 0.4 659 0.4 389 9.97 30 70° 10' 375 650 75 625 350 25 50' 20' 409 689 79 591 311 21 40' 30' 443 727 84 557 273 16 30' 40' 477 766 89 523 234 11 20' 50' 510 804 94 490 196 06 10' 21^ 9.5 543 9.5 842 0.02 98 0.4 457 0.4 158 9.97 02 69° L cos/ L cot/ log CSC/ log sec/ log tan/ L sin/ / LOGARITHMS OF FUNCTIONS. Xlll X Z-sinx Z.tanjr log sec X log CSC X log cot X Z.COSX 21° 9.5 543 9.5 842 0.02 98 0.4 457 0.4 158 9.97 02 69° 10' 576 879 0.03 03 424 121 9.96 97 50' 20' 609 917 08 391 083 92 40' 30' 641 954 13 359 046 87 30' 40' 673 991 18 327 009 82 20' 50' 704 9.6 028 23 296 0.3 972 77 10' 22° 9.5 736 9.6 064 0.03 28 0.4 264 0.3 936 9.96 72 68° 10' 767 100 33 233 900 67 50' 20' 798 136 39 202 864 61 40' 30' 828 172 44 172 828 56 30' 40' 859 208 49 141 792 51 20' 50' 889 243 54 111 757 46 10' 23° 9.5 919 9.6 279 0.03 60 0.4 081 0.3 721 9.96 40 67° 10' 948 314 65 052 686 35 50' 20' 978 348 71 022 652 29 40' 30' 9.6 007 383 76 0.3 993 617 24 30' 40' 036 417 82 964 583 18 20' 50' 065 452 87 935 548 13 10' 24° 9.6 093 9.6 486 003 93 0.3 907 0.3 514 9.96 07 66° 10' 121 520 98 879 480 02 50' 20' 149 553 0.04 04 851 447 9.95 96 40' 30' 177 587 10 823 413 90 30' 40' 205 620 16 795 380 84 20' 50' 232 654 21 768 346 79 10' 25° 9.6 259 9.6 687 0.04 27 0.3 741 0.3 313 9.95 73 65° 10' 286 720 2>-^ 714 280 67 50' 20' 313 752 39 687 248 61 40' 30' 340 785 45 660 215 55 30' 40' 366 817 51 634 183 49 20' 50' 392 850 57 608 150 43 10' 26° 9.6 418 9.6 882 0.04 63 0.3 582 0.3 118 9.95 37 64° 10' 444 914 70 556 086 30 50' 20' 470 946 76 530 054 24 40' 30' 495 977 82 505 023 18 30' 40' 521 9.7 009 88 479 0.2 991 12 20' 50' 546 040 95 454 960 05 10' 27° 9.6 570 9.7 072 0.05 01 0.3 430 0.2 928 9.94 99 63° 10' 595 103 08 405 897 92 50' 20' 620 134 14 380 866 86 40' 30' 644 165 21 356 835 79 30' 40' 668 196 27 332 804 73 20' 50' 692 226 34 308 774 66 10' 28° 9.6 716 9.7 257 0.05 41 0.3 284 0.2 743 9.94 59 62° L cos/ /.cot/ log CSC/ log sec/ log tan/ /.sin/ / XIV LOGARITHMS OF FUNCTIONS. X I sin A- L tan X log sec x log CSC x log cot X LCOBX 28° 9.6 716 9.1 2S1 0.05 41 0.3 284 0.2 743 9.94 59 62° 10' 740 287 47 260 713 53 50' 20' 763 317 54 237 683 46 40' 30' 787 348 61 213 652 39 30' 40' 810 378 68 190 622 32 20' 50' 833 408 75 167 592 25 10' 29° 9.6 856 9.7 438 0.05 82 0.3 144 0.2 562 9.94 18 61° 10' 878 467 89 122 533 11 50' 20' 901 497 96 099 503 04 40' 30' 923 526 0.06 03 077 474 9.93 97 30' 40' 946 556 10 054 444 90 20' 50' 968 585 17 032 415 83 10' 30° 9.6 990 9.7 614 0.06 25 0.3 010 0.2 386 9.93 75 60° 10' 9.7 012 644 32 0.2 988 356 68 50' 20' 033 673 39 967 327 61 40' 30' 055 701 47 945 399 53 30' 40' 076 730 54 924 270 46 20' 50' 097 759 62 903 241 38 10' 31° 9.7 118 9.7 788 0.06 69 0.2 882 0.2 212 9.93 31 59° 10' 139 816 77 861 184 23 50' 20' 160 845 85 840 155 15 40' 30' 181 873 92 819 127 08 30' 40' 201 902 0.07 00 799 098 00 20' 50' 222 930 08 778 070 9.92 92 10' 32° 9.7 242 9.7 958 0.07 16 0.2 758 0.2 042 9.92 84 58° 10' 262 986 24 738 014 76 50' 20' 282 9.8 014 32 718 0.1 986 68 40' 30' 302 042 40 698 958 60 30' 40' 322 070 48 678 930 52 20' 50' 342 097 56 658 903 44 10' 33° 9.7 361 9.8 125 0.07 64 0.2 639 0.1 875 9.92 36 57° 10' 380 153 72 620 847 28 50' 20' 400 180 81 600 820 19 40' 30' 419 208 89 581 792 11 30' 40' 438 235 97 562 765 03 20' 50' 457 263 0.08 06 543 737 9.91 94 10' 34° 9.7 476 9.8 290 0.08 14 0.2 524 0.1 710 9.91 86 56° 10' 494 317 23 506 683 77 50' 20' 513 344 31 487 656 69 40' 30' 531 371 40 469 629 60 30' 40' 550 398 49 450 602 51 20' 50' 568 425 58 432 575 42 10' 35° 9.7 586 9.8 452 0.08 66 0.2 414 0.1 548 9.91 34 55° L cos/ Lcoty log CSC/ log sec/ log tan/ L sin/ / LOGARITHMS OF FUNCTIONS. XV X Lsinjr /.tanr log sec X log CSC X log cot X /.cosjr 35° 9.7 586 9.8 452 0.08 66 0.2 414 0.1 548 9.91 34 55° 10' 604 479 75 396 521 25 50' 20' 622 506 84 378 494 16 40' 30' 640 533 93 360 467 07 30' 40' 657 559 0.09 02 343 441 9.90 98 20' 50' 675 586 11 325 414 89 10' 36° 9.7 692 9.8 613 0.09 20 0.2 308 0.1387 9.90 80 54° 10' 710 639 30 290 361 70 50' ZO' 727 666 39 273 334 61 40' 30' 744 692 48 256 308 52 30' 40' 761 718 58 239 282 42 20' 50' 778 745 67 222 255 33 10' 37° 9.7 795 9.8 771 0.09 77 0.2 205 0.1 229 9.90 23 53° 10' 811 797 86 189 203 14 50' 20' 828 824 96 172 176 04 40' 30' 844 850 0.10 05 156 150 9.89 95 30' 40' 861 876 15 139 124 85 20' 50' 877 902 25 123 098 75 10' 38^ 9.7 893 9.8 928 0.10 35 0.2 107 0.1 072 9.89 65 52° 10' 910 954 45 090 046 55 50' 20' 926 •980 55 074 020 45 40' 30' 941 9.9 006 65 059 0.0 994 35 30' 40' 957 032 75 043 968 25 20' 50' 973 058 85 027 942 15 10' 39° 9.7 989 9.9 084 0.10 95 0.2 011 0.0 916 9.89 05 51° 10' 9.8 004 110 0.11 05 0.1 996 890 9.88 95 50' 20' 020 135 16 980 865 84 40' 30' 035 161 26 965 839 74 30' 40' 050 187 36 950 813 64 20' 50' 066 212 47 934 788 53 10' 40° 9.8 081 9.9 238 0.11 57 0.1 919 0.0 762 9.88 43 50° 10' 096 264 68 904 736 32 50' 20' 111 289 79 889 711 21 40' 30' 125 315 90 875 685 10 30' 40' 140 341 0.12 00 860 659 00 20' 50' 155 366 11 845 634 9.87 89 10' 41° 9.8 169 9.9 392 0.12 22 0.1 831 0.0 608 9.87 78 49° 10' 184 417 ZZ 816 583 67 50' 20' 198 443 44 802 557 56 40' 30' 213 468 55 787 532 45 30' 40' 227 494 67 773 506 33 20' 50' 241 519 78 759 481 22 10' 42° 9.8 255 9.9 544 0.12 89 0.1 745 0.0 456 9.87 11 48° Z.COS/ /.cot/ log CSC/ log sec/ logtany L sin/ / XVI LOGARITHMS OF FUNCTIONS. X Lsinx Lt&nx log sec X log CSC X log cot ;f L cosx 48° 42° 9.8 255 9.9 544 0.12 89 0.1 745 0.0 456 9.87 11 10' 269 570 0.13 01 731 430 9.86 99 50' 20' 283 595 12 717 405 88 40' 30' 297 621 24 703 379 76 30' 40' 311 646 35 689 354 65 20' 50' 324 671 47 676 329 53 10' 43° 9.8 338 9.9 697 0.13 59 0.1 662 0.0 303 9.86 41 47° 10' 351 722 71 649 278 29 50' 20' 365 747 82 635 253 18 40' 30' 378 772 94 622 228 06 30' 40' 391 798 0.14 06 609 202 9.85 94 20' 50' 405 823 18 595 177 82 10' 44° 9.8 418 9.9 848 0.14 31 0.1 582 0.0 152 9.85 69 46° 10' 431 874 43 569 126 57 50' 20' 444 899 55 556 101 45 40' 30' 457 9.9 924 68 543 0.0 076 32 30' 40' 469 949 80 531 051 20 20' 50' 482 975 93 518 025 07 10' 45° 9.8 495 10.0 000 0.15 05 0.1 505 0.0 000 9.84 95 45° Z.COS/ L cot/ log CSC/ log sec/ log tan/ /.sin/ / FORMUL.^. a CSC J.= & CSC 5=0 CSC C. a^=b^-\-c'^-2 be cos A. CSC A cot B=c •cot A ^-■J (s-a)is-b)(s-c) cotlA= ; cotlB= 2 > 2 Circle: circuni = 2 7rr; area^Trr^. sector =:^rx arc; seg^ir^^— sin^). Sphere: sur=4 7r^2. vol=f wB^. zone=2 IT Bh; seg=l wJi^ (3 B-h). Natural log iV=2.30259 logio K Pendulum: ■nH=gt. Falling bodies: 2s=gt'^=vt=v^-T-g. Constants. NUMBER. LOG. radian 57.29578° 1.75812 TT 180° 2.25527 TT 10800' 4.03342 IT 648000" 5.81158 TT 3.14159 0.49715 l-TT 0.31831 9.50285 7r2 9.86960 0.99430 V^r 1.77245 0.24857 1 mile 5280 ft. 3.72263 1 acre 43560 sq. ft. 4.63909 e 2.71828 0.43429 l^M 2.30259 0.36221 vel. light 186330 5.27028 K40°) 39.0986 1.59216 ^(40°) 32.1573 ~ 1.50728 THIS BOOK IS DUE ON THE LAST DATE STAMPED BELOW AN INITIAL FINE OF 25 CENTS WILL BE ASSESSED FOR FAILURE TO RETURN THIS BOOK ON THE DATE DUE. THE PENALTY WILL INCREASE TO SO CENTS ON THE FOURTH DAY AND TO $1.00 ON THE SEVENTH DAY OVERDUE. DEC 4aE4m IML ^88-f X^W^'^^ w 2Nu'&UfiH ^i:tQl ^1 iOctsau — eJ^pi^52IllL 23Mar5 2L . 2lNov57P" REC'D >.d: NOV 15 gs 7 LD 21-100?n-12,'43 (8796s) ' D I / I Ou aA.53\ W5'4 THE UNIVERSITY OF CAUFORNIA UBRARY