and log c (a. c.) log sin A+ log sin C + log a— 10: Ans. A 99° 16', b 351.024, and c = 158.976. 2. Given A = 38° 25', find C, a, and b. B = 57° 42', and c = 400, to Ans. C 83° 53', a = 249.974, b = 340.04. 3. Given A = 15° 19' 51", C = 72° 44′ 05", and c= 250.4 yds., to find B, a, and b. Ans. B 91° 56' 04", a 69.328 yds., b 262.066 yds. 4. Given B = 51° 15′ 35′′, C 37° 21′ 25′′, and a = 305.296 ft., to find A, b, and c. Ans. A = 91° 23', b = 238.1978 ft., c = 185.3 ft. CASE II. Given two sides and an angle opposite one of them, to find the remaining parts. 44. The solution, in this case, is commenced by finding a second angle by means of formula (13), after which we may proceed as in CASE I.; or, the solution may be completed by a continued application of formula (13). Examples. 1. Given A 22° 37', b = 216, and = B, C, and c. a = 117, to find From formula (13), we have α : b A: sin A sin B; that is, the side opposite the given angle, is to the side opposite the required angle, as the sine of the given angle is to the sine of the required angle. Whence, by the application of logarithms, log sin B = (a. c.) log a + log b + log sin A-10; Hence, we find two values of B, which are supplements of each other, because the sine of any angle is equal to the sine of its supplement. This would seem to indicate that the problem admits of two solutions. It now remains to determine under what conditions there will be two solutions, one solution, or no solution. There may be two cases: the given angle may be acute, or it may be obtuse. Represent the given parts of the triangle by A, a, b. The particular letters employed are of no consequence in the discussion, and, therefore, in the results, C or B may be substituted for A, provided that, at the same time, like changes are made in the corresponding small letters. 3. Given a = 100, and b = 60, to find B. C, and c. Ans. B = 36° 52′ 11′′, C = 53° 7′ 49′′, and c = 80. 4. Given a 19.209, and c = 15, to find B, C, and b Ans. B 38° 39′ 30′′, C 51° 20′ 30′′, b = 12. SOLUTION OF OBLIQUE-ANGLED TRIANGLES. 42. In the solution of oblique-angled triangles, four cases may arise. We shall discuss these cases in order. CASE I. Given one side and two angles, to determine the remaining parts. 43. Let ABC represent any oblique-angled triangle. From the vertex C, draw CD perpendicular to the base, forming two rightangled triangles ACD and BCD. Assume the notation of the figure. From formula (1), we have CD = b sin A, B Since a and b are any two sides, and A and B the angles lying opposite to them, we have the following principle: The sides of a plane triangle are proportional to the sines of their opposite angles. It is to be observed that formula (13) is true for any value of the radius. Hence, to solve a triangle, when a side and two angles are given: First find the third angle, by subtracting the sum of the given angles from 180°; then find each of the required sides by means of the principle just demonstrated. Examples. 1. Given B = 58° 07', C = 22° 37', and a = 408, to find A, b, and c. that is, the sine the sine of the sin A : sin B :: α : b; of the angle opposite the given side, is to angle opposite the required side, as the given side is to the required side. Applying logarithms, we have (Ex. 4, P. 15) log b = (a. c.) log sin A + log sin B + log a - 10; and log c = (a. c.) log sin A + log sin C + log a 10; Ans. C 83° 53', a = 249.974, b = 340.04. 3. Given A 15° 19' 51", C 72° 44' 05", and c= 250.4 yds., to find B, a, and b. Ans. B 91° 56' 04", a = 69.328 yds., b = 262.066 yds. = = 4. Given B = 51° 15′ 35′′, C 37° 21′ 25′′, 305.296 ft., to find A, b, and c. and a = Ans. A 91° 23', b = 238.1978 ft., c = 185.3 ft. = CASE II. Given two sides and an angle opposite one of them, to find the remaining parts. 44. The solution, in this case, is commenced by finding a second angle by means of formula (13), after which we may proceed as in CASE I.; or, the solution may be completed by a continued application of formula (13). Examples. 1. Given A 22° 37', b = 216, and a = 117, to find = B, C, and c. |