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 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. As sin A To find the side d. The side a will be the middle part, and the extremes opposite: hence, As cos a 42° 12' To sin B d 64° 40' 35° 16' 53" log. 0.043911 10.000000 9.827189 9.871100 To find the angle B. The angle A will correspond to the middle part, and the extremes will be opposite : hence R cos A sin B cos a. = ar.-comp. A log. 0.130296 10.000000 9.631326 9.761622 To find the side b. The side b will be the middle part, and the extremes adjacent: hence, Hence, CA-90°-b-90°-25° 25′ 14′′ 64° 34′ 46" CBA—90°—ABD=90°—35° 16′ 53′′=54° 43′ 07′′ 48° 00′ 15′′. Ans. log. 0.000000 9.675237 9.957485 9.632722 BA=d 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′′ C= 75° 30′ 33′′ b=152° 13′ 50′′. 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. C-103° 52′ 46′′ B 32° 30′ 22′′ a=112° 48′ 58′′. 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. Z 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. Given two sides, and an angle opposite one of them, to find the remaining parts. CASE I. or the sign of the on that of cos b. log. B' D a R2 cos b-R cos a cos c cos B sin a sin c Now if cos b be greater than cos a, we shall have R2 cos b>R cos a cos c, 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. 0.156437 9.997803 9.729445 9.883685 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 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 R2 cos b-R cos a cos c cos B R2 cos b-R cos a cos c 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. +cos B To find the angle C, in the triangle ACD. A 32° 26′ 07′′ ar.-comp. b 84° 14′ 29′′ As cot Is to As cot Is to So is cos Hence To find the angle C in the triangle DCB. B 49° 54′ 38′′ ar.-comp. R a 44° 13' 45" 49° 35′ 38′′ ACB=135° 56′ 47′′. log. 9.803105 10.000000 9.001465 8.804570 0.074810 10.000000 9.855250 9.930060 The arc 64° 43′ 31", which corresponds to sin c is not the value of the side AB: for the side AB must be greater than b, since it lies opposite to a greater angle. But b=84° 14′ 29′′ : hence the side AB must be the supplement of 64° 43′ 31′′, or 115° 16' 29". Ex. 2. Given b=91° 03′ 25′′, a=40° 36′ 37′′, and A=35° 57′ 15" required the remaining parts, when the obtuse angle B is taken. CASE II. log. Ans. As sin A To sin b Having given two angles and a side opposite one of them, to find the remaining parts. For this case, we employ the equation (1.) sin A sin B :: sin a: sin b. 0.270555 9.842191 9.843563 9.956309 B=115° 35′ 41′′ C= 58° 30′ 57′′ c = 70° 58′ 52′′ To find the side b. ar.-comp. Ex. 1. In a spherical triangle ABC, there are given the angle A=50° 12′, B=58° 8', and the side a=62° 42′ ; to find the remaining parts. log. 50° 12' 58° 08′ 62° 42' 79° 12′ 10′′, or 100° 47′ 50′′ 0.114478 9.929050 9.948715 9.992243 We see here, as in the last example, that there are two arcs corresponding to the 4th term of the proportion, and these arcs are supplements of each other, since they have the same sine. It does not follow, however, that both of them will satisfy all the conditions of the question. If they do, there will be two triangles; if not, there will be but one. |
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