By Calculation. The angle C-90°-48° 17′ 41° 43'. Radius sine ZA :: AC: CB. Radius or sine 90 Radius Radius : sine 48 17′ :: 324 : CB. sine ZC:: AC: AB. sine 41 43′ :: 324: AB. Hence BC=AC × sine A÷R, and AB=AC x sine C÷R.. Solution by Natural Sines. 17464446 :: 324: BC=241.8480504. Solution by Logarithmic Sines. Log. BC=log. sine A+log. AC-log. Radius. Log. sine A 48° 17' 9.8729976 Log. AC 324 Log. Radius 10.0000000 Log. AB=log. sine C+log. AC-log. Radius. Log. sine 41° 43' 9.8231138 Log. 324 Log. Radius 10-0000000 Log. AB Whence BC=241.84805, and AB=215.605. The solution of the cases of triangles by logarithmic sines, &c. is generally more expeditious than that by natural sines, &c., and therefore is preferable in practice. Besides, some of the best tables do not contain natural sines, but all contain logarithmic sines, which are indispensable in trigonometrical calculations. Instrumentally. In the first proportion, extend the compasses from 90° to 48° 17′ on the line of sines. That extent will reach from 324 to 242 on the line of numbers, and will be the length of BC. In the second proportion, extend the compasses from 90° to 41° 43′ on the line of sines. That extent will reach from 324 to 215.6 on the line of numbers, and will be the length of AB. Note. The radius is 90° of sines (27), and 45' of tangents (28), and 0 of cosines (30) and secants (29). Ex. 2. Given the hypothenuse AC 121 yards, and the angle at the base A 55° 30′, to find the rest. Answer. BC=99-719 yards, AB=68.535, angle C= 34° 30'. Case 2. Given the side AB 125, and the angle A 51° 19′, to find the rest. By Construction. Draw an indefinite line AB, and from A to B set off 125, from a scale of equal parts. Make the angle A=51 19', as in case 1. At the point B erect a perpendicular, and from A through the extremity of the arc which measures the angle A draw AC meeting the perpendicular in C. Then AC, BC, measured on the same scale of equal parts, will be 199.5 and 156. The angle C-90-51 19-38 41. By Calculation. Angle C-90-51° 19' 38° 41'. R tan. 51° 19′ :: 125: BC. Cos. A or sine C: R:: AB: AC. Cos. 51° 19' or sine 38° 41': R:: 125: AC. or, R: sec. A :: AB : AC. R: sec. 51° 19′ :: 125: AC. Hence BC AB × tan. A÷R, and AC = AB × R÷ cos. A=ABX R÷sine C=AB x sec. A÷R. Solution by Natural Tangents and Secants.* 1: 1·2489484 :: 125 : BC = 156·11855. Solution by Logarithmic Sines, Tangents, and Secants. Log. tan. 51° 19′ Log. AB 125 Log. BC 156.1186 10.0965445 2.0969100 2.1934545 In this operation log. R, which is always 10·0000000, is subtracted, though not expressed. In other operations log. R will be seldom expressed, but will be added or subtracted tacitly. Log. AC = log. R+ log. AB-log. cos. A orlog. sine C. Log. R+log. AB 125 12.0969100 Log. cos. 51° 19′ or log. sine 38° 41′ 9.7958909 Log. AC 199.5603 2.3010191 Log. AC = log. sec. A+ log. AB-log. R. Log. sec. 51° 19' Log. AB 125 Log. AC 199-5603 10-2031641 2.0969100 2.3000741 Instrumentally. In the first proportion, extend the compasses from 45° to 51° on the line of tangents. That extent will * Some tables do not contain secants, but all contain sines, cosines, and tangents. E reach from 125 to 156 on the line of numbers, and will be the length of BC. In the second proportion, extend the compasses from 382° to 90° on the sines. That extent will reach from 125 to 199.5 on the line of numbers, and will be the length of AC. In the third analogy, extend the compasses from 0 to 511 on the secants. That extent will reach from 125 to 199.5 on the numbers, and will be the length of AC. Ex. 2. Given the base AB 50, and the angle at the base A 25° 17', to find the rest. Answer. Angle C = 64° 43', BC= 23.617, AC = 55.297. Case 3. Given the hypothenuse AC 415, and the base AB 249, to find the rest. By Construction. Draw AB 249. At the point B erect a perpendicular. From the centre A, with the radius AC = 415, describe an arc intersecting the perpendicular in C. Then BC measured on the line of numbers is 332, and the angles A and C measured on the scale of chords, or with a protractor, are 53° 8' and 36° 53′ nearly. By Calculation. AC: AB:: R: sine C or cos. A. R: tan. A: AB: BC. R: tan. A :: 249 : BC. Hence, sine C = cos. A = R >ABAC, and BC= AB tan. A÷R. 415 249 1: 0.6 sine C 36° 52′ 52′′. Log. sine Clog. R+ log. AB-log. AC. 1: 1.333: 249: BC 332. Log. BC= log. tan. A + log. AB-log. R. = Or BC may be found as follows. = Because AC AB + BC (47. 1), BC2=AC2-AB2 (AC + AB) × (AC-AB) (Cor. 5. 2), therefore BC =√(AC+ AB) × (AC — AB) = (415+249) x (415—249) 664 × 166= ✓ 110224 = 332. Or, log. 664 + log. 166= 1-411084+ 1.110054 2.521138, the number answering to which is 332. = Ex. 2. Given the hypothenuse 50, and the base 20, to find the rest. Answer. Angle A = 66° 25′, C = 23° 35′, and BC = 45.825. Case 4. Given the base AB 53, and the perpendicular BC 67, to find the other parts of the triangle. AB: BC:: R: tan. A. sine A: R:: BC: AC. sine A: R:: 67 : AC. Hence tan. A = R x BC÷AB, and AC = R x BC → sine A. 53: 67 :: 1 : 1·26415094 = tan. A 51° 39′ 15′′. = |