or, by logarithms, That is, log sin Blog blog sin A-log a. (130) The logarithmic sine of a required angle whose opposite side is given, is equal to the logarithm of that side, plus the logarithmic sine of the given angle, minus the logarithm of its opposite side. Read To find C. We have C=180° — (A + B). To find c. By (128), after Cis found, we have log clog a log sin C-log sin A. 127. Whenever the given angle is acute, and the side opposite to it is less than the side adjacent to it, there may be formed, as shown in Geometry (Prob. XI. Bk. V.), two triangles, each satisfying the given conditions, and, therefore, there will be two solutions. Thus (Fig. Art. 126) with two given sides a, b, equal respectively to B and A C, and the given acute angle A opposite the less side CB, there may always be formed two triangles, ABC, ABC, which have A, a, b in common, and the angles ABC, ABC supplements of each other. In one of them, therefore, the required angle is acute, and in the other it is obtuse. When the given angle is obtuse, the required angle must of necessity be acute, since a triangle can have but one obtuse angle. When the given angle is acute, and its opposite side is greater than the side opposite to the required angle, that must also be acute, since the greater angle must be opposite the greater side. When the side opposite the given angle is exactly a perpendicular let fall from Con AB, the required angle is a right angle. If the side opposite the given angle be less than the perpendicular, the solution is impossible, since there will be no triangle with the given parts. When two values are admissible for B, in case of ambiguity, two corresponding values will exist for C and c. EXAMPLES. 1. Given of any triangle AB C, the side b equal to 216 yards, the side a equal to 117 yards, and the angle opposite the side a equal to 22° 37'; to solve the triangle. Ans. Angle B, 45° 13′ 55′′, or 134° 46′ 5′′; angle C, 112° 9′ 5′′, or 22° 36′ 55′′; side c, 281.785 yd., or 116.99 yd. 2. Given two sides of a triangle equal to 9459.31 feet and 8032.28 feet, and the angle opposite the first side equal to 71° 3' 34"; to find the other parts. C=180°. log sin 9.975824 ar. co. log sin 0.024176 log sin 9.904804 124° 29′ 34′′ = 55° 30′ 26′′ log sin 9.916032 c=8242.64 log 3.916067 Ans. Side, 8242.64 feet; angles, 53° 26′, 55° 30′ 26′′. 3. Given two sides of a triangle equal to 80 rods and 142.6 rods, and the angle opposite the second side equal to 96' to solve the triangle. 4. Given in a triangle ABC, the side a equal to 32.1098 rods, the side b equal to 125.701 rods, and the angle ▲ equal to 14° 48′; to solve the triangle. Ans. Angle B, 90°; angle C, 75° 12′; side c, 121.531 rods. 5. Given two sides of a triangle equal to 1540.37 feet and 760.9 feet, and the angle opposite the second equal to 30° 22' 8"; to find the other side and angles. Ans. Impossible. CASE III. 128. Given two sides and the included angle. Let there be given in the triangle ABC the sides a and b and the included angle C, to solve the triangle. To find A and B, we have A + B = 180° - C, and A B b whence, (A+B) = 90°C complement of C; = Then, the half difference of A and B is found by means of (95), α b a+b tan 1⁄2 (A — B) = tan † (4+B) = or, by logarithms, log tan (A-B)=log (a-b)—log (a+b)+ log tan † (A+B) = log (a — b) — log (a+b) + log cot That is, C. (133) The logarithmic tangent of half the difference of the two required angles is equal to the logarithm of the difference of the given sides, minus the logarithm of their sum, plus the logarithmic tangent of half the sum of the required angles, or plus the logarithmic cotangent of half the given angle. Since · have That is, (A+B) is known, when (AB) is found, we A = ≥ (4 + B) + 1⁄2 (A — B), B = 1 (A + B) — † (A — B). The GREATER of the two required angles is equal to half their sum, plus half their difference; and the SMALLER angle is equal to half their sum, minus half their difference. To find c. By (128), we have log c = log a log sin C - log sin A. EXAMPLES. 1. Given of any triangle ABC, the side a equal to 9459.31 feet, the side b equal to 8032.28 feet, and the included angle C equal to 55° 30 26"; to find the side c and the angles A and B. Solution. A + B = 180° C 124° 29′ 34′′, and ≥ (A + B) = 62° 14′ 47′′. Then, by (133), Ans. Side c, 8242.64 ft.; angle A, 71° 3′34′′; angle B, 53° 26'. 2. Given two sides of a triangle equal to 142.6 feet and 110 feet, and the included angle equal to 33° 55'; to solve the triangle. 3. Given the two sides of a triangle equal to 153 rods and 137 rods, and the included angle equal to 40° 33′ 12′′; to find the other parts. Ans. Side, 101.615 feet; angles, 78° 13′ 1′′ and 61° 13′ 47′′. CASE IV. 129. Given the three sides. Let there be given (Fig. Art. 128) the three sides a, b, and c; to solve the triangle. To jina A, B, and C. By (102), (103), and (104), we have (135) log sin C= That is, (136) log sin Blog (s—a) + log (s—c) — log a— log c log (s — a) + log (s — b) — log a — log b The logarithmic sine of half of any angle of a triangle is equal to the logarithm of the difference between half the sum of the sides and one of the adjacent sides, plus the logarithm of the difference between half the sum and the other adjacent side, minus the logarithms of those two sides, divided by 2. 130. A, B, and C can also be determined by formulæ (106), (107), and (108) for the cosine of half an angle, and by formulæ (109), (110), and (111) for the tangent of half an angle. When the half angle is less than 45°, the table will determine it from its sine with greater precision than from the cosine, and rice versa when the half angle is greater than 45o. |