Whence, by the application of logarithms, log sin B = log b + log sin A + (a. c.) log 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. b are given. From C let fall a perpendicular upon AB, pro longed if necessary, and denote its length by p. We shall have, from Formula (1), Art. 37, pb 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 a 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. unite, and there will be but a single triangle formed. C In this case, the angle ABC will be equal to 90°. If α 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. A C α p 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 first corresponding to B = 45° 13′ 55′′, = 45° 13′ 55′′, and the second to B' 134° 46′ 05′′. log c = log b + log sin C+ (a. c.) log sin B – 10; and c = 281.785. Ans. B = 45° 13′ 55′′, C = 112° 09′ 05′′, and c Ans. B′ = 134° 46′ 05′′, С = 22° 36′ 55′′, and c = 116.993. 2. Given A = 32°, a a = 40, and b 50, B, C, and C. to find Ans. B = 41° 28′ 59′′, °C = 106° 31′ 01′′, c=72.368. Ans. B = 8° 56′ 05", C = 152° 11' 42", c = 39.611 yds. Given two sides and their included angle, to find the re maining parts. 45. Let ABC represent any E plane triangle, AB two sides, and A and AC any their included angle. With A as a centre, and AC, the shorter of the two sides, as a radius, describe a semi F H circle meeting AB in I, and the prolongation of AB in E. Draw CI and EC, and through I draw IH parallel to EC. Because ECI is an angle inscribed in a semicircle, it is a right angle (B. III. P. XVIII., C. 2); and consequently, both CE and I are perpendicular to CI. The angle EAC being external to the triangle ABC, is equal to the sum of the opposite interior angles, that is, equal to C plus B; the angle EAC being also external to the isosceles triangle AIC, it is equal to twice the angle AIC : hence, twice the angle AIC is equal to C plus B, AIC = (C + B). or, The angle ICB is equal to AIC diminished by the angle IBC (B. I., P. XXV., C. 6); that is, ICH = † ( C + B) − B : + (C – B). From the two right-angled triangles ICE and ICH, we have (Formula 3, Art. 37), EC = IC tan † (C + B), and IH IC tan (C — B) ; hence, from the preceding equations, we have, after omitting the equal factor IC (B. II., P. VII.), EC : IH :: tan (C + B) : tan (C – B). The triangles ECB and IHB being similar, their homologous sides are proportional; and because EB is equal to AB + AC, and IB to AB AC, we shall have the proportion, EC : IH :: AB + AC : AB – AC. Combining the preceding proportions, and substituting for AB and AC their representatives c and b, we have, c+b : c−b :: tan(C+B) : tan (C–B). Hence, we have the following principle: (14.) In any plane triangle, the sum of the sides including either angle, is to their difference, as the tangent of half the sum of the two other angles, is to the tangent of half their difference. The half sum of the angles may be found by subtracting the given angle from 180°, and dividing the remainder by 2; the half difference may be found by means of the principle just demonstrated. Knowing the half sum and the half |
