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. The hypothenuse a corresponds to the middle part, and the extremes are opposite: hence, R cos a = cos b cos c, or, The side b is the middle part and the extremes oppo site: hence, R sin b = cos (comp. a) x cos (comp. B) sin a sin B. 0.043911 9.827189 10.000000 9.871100 The angle C is the middle part and the extremes adja 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. To find the angle C The hypothenuse is the middle part and the extremes adjacent: hence, Since the cotangent of C is negative, the angle C is greater than 90°, and is the supplement of the are 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, The side b is the middle part and the extremes are 17. A quadrantal spherical triangle is one which has one of its sides equal to 90°. Let BAC be a quadrantal tri angle of which the side a = 90°. If we pass to the corresponding polar triangle, we shall have A B d a from which we see, that the polar triangle will be rightangled 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. a A d 'D Then 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. C 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. making CD = 90°, and = be given in the rightC 42° 12', and the 115° 20′ = 64° 40′, to Having produced CA to D, drawn the arc BD, there will then angled triangle BAD, the side a angle BAD = 180° BAC 180° - 115° 20' find the remaining parts. To find the side d. The side a is the middle part, and the extremes oppo The angle A corresponds to the middle part, and the extremes are opposite: hence, The side b is the middle part, and the extremes are adjacent: hence, 2. In the right-angled triangle BAC, right-angled at A, there are given a = 115° 25', and c = 60° 59' required 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. 4. In a quadrantal triangle, there are given the quadrantal side = 90°, an adjacent side 115° 09', and the included angle = 115° 55': required the remaining parts. |