SOLUTION OF RIGHT-ANGLED SPHERICAL TRIANGLES BY LOGARITHMS. 16. It is to be observed, that when any part of a triangle becomes known by means of its sine only, there may be two values for this part, and consequently two triangles that will satisfy the question; because, the same sine which corresponds to an angle or an arc, corresponds likewise to its supplement. This will not take place, when the unknown quantity is determined by means of its cosine, its tangent, or cotangent. In all these cases, the sign will enable us to decide whether the part in question is less or greater than 90°; the part is less than 90°, if its cosine, tangent, or cotangent, has the sign +; it is greater if one of these quantities has the sign —. In order to discover the species of the required part of the triangle, we shall annex the minus sign to the logarithms of all the elements whose cosines, tangents, or cotangents, are negative. Then, by recollecting that the product of the two extremes has the same sign as that of the means, we can at once determine the sign which is to be given to the required element, and then its species will be known. It has already been observed, that the tables are calcu lated to the radius R, whose logarithm is 10 (Plane Trig, Art. 100); hence, all expressions involving the circular functions, must be made homogeneous, to adapt them to the logarithmic formulas. EXAMPLES. 1. In the right-angled spherical triangle BAC, right-angled at A, there are given a = 64° 40′ and b 42° 12' required the remaining parts. = B First, to find the side c. The hypothenuse a corresponds to the middle part, and the extremes are opposite: hence, R cos a = cos b cos c, or, b 42° 12' Ꭱ COS a : COS :: 64° 40' : COS с 54° 43' 07" ar. comp. sin a 64° 40' sin b 42° 12' Ꭱ : sin B 48° 00′ 14′′. To find the angle B. The side b is the middle part and the extremes oppo site: hence, R sin b= cos (comp. a) x cos (comp. B) ar. comp. log. : :: : log. Ꭱ a 64° 40' : cot :: tang b 42° 12′ : COS C 64° 34′ 46′′ . To find the angle C. The angle C is the middle part and the extremes adjacent: hence, R cos C: = cot a tang b. ar. comp. log. 0.130296 10.000000 9.631326 9.761622 sin a sin B. 0.043911 9.827189 10.000000 9.871100 2. In a right-angled triangle BAC, there are given the hypothenuse a = 105° 34', and the angle B = 80° 40′: required the remaining parts. R cos a cot B cot C log. cot B 80° 40' COS a 105° 34' R C 148° 30′ 54′′. To find the angle C. The hypothenuse is the middle part and the extremes adjacent hence, 0.000000 9.675237 9.957485 9.632722 0.784220 + 9.428717 10.000000 + 10.212937 cot Since the cotangent of C is negative, the angle C is greater than 90°, and is the supplement of the arc which would correspond to the cotangent, if it were positive. To find the side c. The angle B corresponds to the middle part, and the extremes are adjacent: hence, R cos B cot a tang c. ar. comp. log. cot a : R :: Cos B 80° 40' : tang c 149° 47' 36" Ꭱ : sin α :: sin B : sin b 105° 34' To find the side 3. The side b is the middle part and the extremes are opposite: hence, a = R sin b = sin a sin B. log. 105° 34' 80° 40' 71° 54' 33" ar. comp. OF QUADRANTAL TRIANGLES. 17. A quadrantal spherical triangle is one which has one of its sides equal to 90°. Let BAC be a quadrantal triangle of which the side a 90°. If we pass to the corresponding polar triangle, we shall have 90°, B' = 180° - b, — A, 180° c, a' = 180° — B, c' = B from which we see, that the polar triangle will be right. angled at A', and hence, every case may be referred to a right-angled triangle. But we can solve the quadrantal triangle by means of the right-angled triangle in a manner still more simple. Let the side BC of the quadrantal triangle BAC, be equal to 90°; produce the side CA till CD is equal to 90°, and conceive the arc of a great circle to be drawn through B and D. Then C will be the pole of the arc BD, and the angle C will be measured by BD (B. IX., P. 4), and the angles CBD and D will be right angles. Now before the remaining parts of the quadrantal triangle can be found, at least two parts must be given in addition to the side BC= 90°; in which case two parts of the right-angled triangle BDA, together with the right angle, become known. Hence, the conditions which enable us to determine one of these triangles, will enable us also to determine the other. EXAMPLES. 1. In the quadrantal triangle BCA, there are given CB 90°, the angle = 42° 12', and the angle A = 115° 20'; required the remaining parts. = Having produced CA to D, making CD = 90°, and drawn the are BD, there will then be given in the rightangled triangle BAD, the side a C 42° 12', and the = = angle BAD = 180° — BAC = 180° — 115° 20′ = 64° 40', to find the remaining parts. :: sin : sin B To find the side d. The side a is the middle part, and the extremes opposite: hence, sin A 64° 40' R a d d R sin a sin A sin d. ar. comp. log. 42° 12' 48° 00' 14". 0.043911 10.000000 9.827189 9.871100 To find the angle B. The angle A corresponds to the middle part, and the extremes are opposite: hence, COS a Ꭱ :: cos A : sin B Ꭱ : cot A :: tang a : sin b To find the side b. The side b is the middle part, and the extremes are adjacent hence, R sin b = 64° 40' 42° 12' 25° 25′ 14′′. == ar. comp. cot A tang a. log. b = 90° 25° 25' 14" ABD = 90° - 35° 16' 53" log. Ans. Ans. B C b 2. In the right-angled triangle BAC, right-angled at A, there are given a 115° 25', and c = 60° 59' required. the remaining parts. = = 64° 34' 46" 54° 43′ 07′′ = = 48° 00′ 14′′ = 0.000000 9.675237 9.957485 9.632722 148° 56′ 45′′ = 75° 30′ 33′′ 3. In the right-angled spherical triangle BAC, rightangled at A, there are given c = 116° 30′ 43′′, and b = 29° 41′ 32": required the remaining parts. = 152° 13′ 50′′ 4. In a quadrantal triangle, there are given the quadrautal side = 90°, an adjacent side = 115° 09', and the included angle 115° 55': required the remaining parts. C=103° 52′ 46′′ Ans. B = 32° 30′ 22′′ a = 112° 48′ 58′′ side, 113° 18′ 19′′ angles,{ 40′ 07′′ |