To find the side b. The side b will be the middle part, and the extremes adjacent: hence, As R Is to cot A So is tang a b R sin b=cot A tang a. ar.-comp. 64° 40' 42° 12' 25° 25′ 14′′ log. =64° 34′ 46" Hence, CA-90°-b-90°—25° 25′ 14′′ 48° 00′ 15′′. Ans. 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. Ans. 0.000000 9.675237 9.957485 9.632722 BAC, right angled 5. In the right angled spherical triangle at A, there are given c=116° 30′ 43′′, and b=29° 41′ 32′′: required the remaining parts. B-148° 56′ 45′′ C= 75° 30' 33" b=152° 13′ 50′′. 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. 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 thein. 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. Given two sides, and an angle opposite one of them, to find the remaining parts. CASE I. 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. a As sin So is sin A A To find the angle B. 44° 13′ 45′′ 84° 14′ 29′′ 49° 54′ 38′′ or sin E 130° 5′ 22′′ or the sign of the on that of cos b. ar.-comp. с log. B' D 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. 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 0.156437 9.997803 9.729445 9.883685 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 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. To find the angle C, in the triangle ACD. As cot A 32° 26′ 07′′ R ls to 10.000000 b 84° 14′ 29′′ To cot ACD 86° 21′ 06′′ 8.804570 +cos B: cos B To find the angle C in the triangie DCB. R As cot Hence 44° 13′ 45′′ 49° 35′ 38′′ ACB=135° 56′ 47′′. 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. A As sin So is sin a 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. Ec. 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 re maining parts. log. 50° 12' 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. To determine when there are two triangles, and also when there is but one, let us consider the second of equations (8.) R2 cos B-sin A sin C cos b-R cos A cos C, which gives Recos B+R cos A cos C cos b= sin A sin C Now, if cos B be greater than cos A we shall have and hence the sign of the second member of the equation will depend on that of cos B, and consequently cos b and cos B will have the same algebraic sign, or b and B will be of the same species. But when cos B>cos A the sin B<sin A: hence If the sine of the angle opposite the required side be less than the sine of the other given angle, there will be but one solution. 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 cos C, to render as R2 cos BR cos A cos C, or the sign of the second member of the equation 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 sin A sin C R2 cos B+R cos A cos C +cos b -cos b= and Hence, if the sine of the angle opposite the required side be greater than the sine of the other given angle there will be two solutions. Let us first suppose the side b to be less than 90°, or equal to 79° 12′ 10′′. If now, we let fall from the angle C a perpendicular on the base BA, the triangle will be divided into two right angled triangles, in each of which there will be two parts known besides the right angle. Calculating the parts by Napier's rules we find, C=130° 54′ 28 c=119° 03′ 26′′. If we take the side b=100° 47′ 50′′, we shall find C=156° 15' 06" c=152° 14′ 18′′. |