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, 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. that is, the sine of the angle opposite the given side, is to the sine of the angle opposite the required side, as the given side is to the required side. Applying logarithms, we have (Ex. 4, P. 15), = (a. c.) log sin A+ log sin Blog a 10 log b, and, (a. c.) log sin A+ log sin C+ log a 10 = log c. 2. Given A = 38° 25', B = 57° 42', and c = 400, to find C, α, and b. Ans. C 83° 53', a = 249.974, b = 340.04. 3. Given A = 15° 19′ 51′′, c = 250.4 yds, to find B, α, C = 72° 44' 05", and 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′′, a = 305.296 ft., to find A, b, and C. and 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). that is, the side opposite the given angle, is to the side op posite the required angle, as the sine of the given angle is to the sine of the required angle. Whence, by the application of logarithms, (a. c.) log a + log b + log sin A 10 log sin B; (a. c.) log a log b log sin A (22° 37') = = 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. First Case. Let ABC represent the triangle, in which the angle A, and the sides a and a longed if necessary, and denote its length by p. We shall have, from Formula (1), Art. 37, p = b sin A ; from which the value of p may be computed. If a is intermediate in value between p and b, there will be two solutions. For, if with C as a centre, and α as a radius, an arc be described, it will cut the line AB in two points, B and B', each of which being joined with C, will give a triangle which will conform to the conditions of the problem. In this case, the angles B' and B, of the two triangles AB'C and ABC, will be supplements of each other. - If a = P1 there will be but one solution. For, in this case, the arc will be tangent to AB, the two points B and B' will A unite, and there will be but a single triangle formed. If a is greater than both p and b, there will also be but one solution. For, although the arc cuts AB in two points, and consequently gives two triangles, only one of them conforms to the conditions of the problem. P a B In this case, the angle ABC will be less than A, and consequently acute. Second Case. When the given angle A is obtuse, the angle ABC will be acute; the side a will be greater than b, and there will be but one solu tion. In the example under considera tion, there are two solutions, the B A first corresponding to B = 45° 13′ 55", and the second to B' = 134° 46' 05". |