To find the angle C. The hypothenuse will be 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 arc which would correspond to the cotangent, if it were positive. To find the side c. The angle B will correspond to the middle part, and the extremes will be adjacent: hence, The side b will be the middle part and the extremes oppo A quadrantal spherical triangle is one which has one of its sides equal to 90°. Let BAC be a quadrantal triangle in which the side a=90°. If we pass to the corresponding polar triangle, we shall have A'=180°-a=90°, B' 180°-b, C'=180°-c, a'=180°—A, b'=180°-B,c'=180°-C; from which B we see, that the polar triangle will be A 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. In the quadrantal triangle BAC, in which BC=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 B BD (Book IX. Prop. VI.), and the angles CBD and D will be right angles. Now before the remaining parts of the quadrantal triangle can α A 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. 3. In the quadrantal triangle BCA, there are given CB 90°, the angle C=42° 12', and the angle A=115° 20′ : required the remaining parts. Having produced CA to D, making CD=90° and drawn the arc BD, there will then be given in the right angled triangle BAD, the side a⇒C=42° 12′, and the angle BAD=180°— BAC=180°-115° 20' 64° 40', to find the remaining parts. To find the side d. The side a will be the middle part, and the extremes opposite: hence, The angle A will correspond to the middle part, and the ex tremes will be opposite : hence To find the side b. The side b will be the middle part, and the extremes adja Hence, BA=d ar.-comp. CA-90°-b-90°-25° 25′ 14′′ CBA-90°-ABD-90°-35° 16′ 53′′-54° 43′ 07′′. 48° 00′ 15′′. 4. In the right angled triangle BAC, right angled at A, there are given a=115° 25′, and c=60° 59′ : required the remaining parts. Ans. B-148° 56′ 45′′ 5. In the right angled spherical triangle BAC, right angled at A, there are given c=116° 30′ 43′′, and b=29° 41′ 32′′: required the remaining parts. Ans. C-103° 52′ 46′′ 6. 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. Ans. side, angles, { 113° 18′ 19′′ 117° 33′ 52" 101° 40′ 07′′. SOLUTION OF OBLIQUE ANGLED TRIANGLES BY LOGARITHMS There are six cases which occur in the solution of oblique angled spherical triangles. 1. Having given two sides, and an angle opposite one of them. 2. Having given two angles, and a side opposite one of them. 3. Having given the three sides of a triangle, to find the angles. 4. Having given the three angles of a triangle, to find the sides. 5. Having given two sides and the included angle. 6. Having given two angles and the included side. CASE I. Given two sides, and an angle opposite one of them, to find the remaining parts. For this case we employ equation (1.); As sin a sin b:: sin A: sin B. Ex. 1. Given the side a=44° 13′ 45′′, b 84° 14′ 29′′ and the angle A-32° 26′ 07′′: required the remaining parts. To find the angle B. A. B Since the sine of an arc is the same as the sine of its supplement, there will be two angles corresponding to the logarithmic sine 9.883685 and these angles will be supplements of each other. It does not follow however that both of them will satisfy all the other conditions of the question. If they do, there will be two triangles ACB', ACB; if not, there will be but one. To determine the circumstances under which this ambiguity arises, we will consider the 2d of equations (2.). R2 cos b=R cos a cos c+ sin a sin c cos B. from which we obtain Now if cos b be greater than cos a, we shall have or the sign of the on that of cos b. R2 cos b>R cos a cos c, second member of the equation will depend Hence cos B and cos b will have the same sign, or B and b will be of the same species, and there will be but one triangle. But when cos b>cos a, sin b<sin a: hence, If the sine of the side opposite the required angle be less than the sine of the other given side, there will be but one triangle. If however, sin b>sin a, the cos b will be less than cos a, and it is plain that such a value may then be given to c as to render R2 cos b<R cos a cos c, or the sign of the second member may be made to depend on COS C. We can therefore give such values to c as to satisfy the two equations Hence, if the sine of the side opposite the required angle be greater than the sine of the other given side, there will be two triangles which will fulfil the given conditions. Let us, however, consider the triangle ACB, in which we are yet to find the base AB and the angle C. We can find these parts most readily by dividing the triangle into two right angled triangles. Draw the arc CD perpendicular to the base AB: then in each of the triangles there will be given the hypothenuse and the angle at the base. And generally, when it is proposed to solve an oblique angled triangle by means of the right angled triangle, we must so draw the perpendicular that it shall pass through the extremity of a given side, and lie opposite to a given angle. |