To find a and b. (A+B)=81° 15' 44"-41, (A ~ B)=48° 49′ 38′′, and c = 25° 33′ 5′′-8. Sin (A+B) 9.9949302 Cos (A+B) Sin (AB) B) 9.8766379 Cos (A~ B) 9-8184449 Tanc 9-6795022 Tanc 9.1815880 9.6795022 19-5561401 19.4979471 Tan (ab) 9-5612099 Tan (a+b) 10-3163591 ~ And And since AB, therefore ab, whence (211, Part I.) a = 84° 14′ 29′′, b = 44° 13′ 45′′. To find angle C. Sin (ab) 20° 0′ 22′′ 9-5341789 Tan (A~ B) 48° 49′ 38′′ 10·0581929 20-0127184 10-4785395 Cot C 18° 22′ 44′′ 플 And C 36° 45′ 28′′. EXERCISES. 1. The angles A and B are 82° 27′ and 57° 30′, and side c is 126° 37′; what are the other parts? Ans. The angle C is 115° 21', and the sides b and c 48° 30' and 61° 41′. 2. Given A = 63° 39′ 57′′·8, B = 75° 0′ 51′′·3, and c = 42° 30′ 55′′, to find the other parts. Ans. a 66° 57′ 3′′·6, b = 97° 20′ 31′′-6, and C 41° 9′ 46′′. = 357. Case V.-When two sides and the angle opposite to one of them are given. The angle opposite to the other given side will be found by this theorem: 358. The sines of any two sides are proportional to the sines of their opposite angles." Let a, b, A, be the given parts, then And, by this analogy, B is found. When B is found, there are then two sides and their opposite angles known; and hence c and C can be found as in the third and fourth cases, thus: = sin (A~ B): sin (A + B) = tan (a ~ b): tan c, sin (ab) sin(a+b) tan (AB): cot C. There will, however, be sometimes two values of B, as in the analogous case (210) of plane trigonometry, and consequently two triangles can be formed from the same data; and hence this is an ambiguous case, as is also the next case, for a similar reason (see fig. to art. 314). When B has two values, so have c and C. The values of B are supplementary, and, by using first one of its values, the corresponding values of c and C will be found by the last two analogies, and then all the parts of one of the triangles are known. When the other value of B is taken, and the corresponding values of c and C are computed in the same manner, all the parts of the other triangle will then be known. Whenever a differs from 90° in excess or defect less than b does, there will be only one triangle, and therefore only one value of B, which will be of the same species as b; in other cases, B has two values that are supplementary.' When the difference of a from 90° is less than that of b, then it is evident that sin a sin b; that is, if († 7 ~ a) < (1⁄2T ~ b), then sin asin b. 1 EXAMPLE. The sides a, b, are 38° 30′ and 40°, and angle A 30° 28'; required the other parts. To find angle B. Sin a sin b sin A: sin B. Sin a 38° 30' Sin b 40° Sin A 30° 28' 9-794150 9.808067 9.705040 19-513107 9.718957 Sin B 31° 34' Or B 148° 26'. B has two values, for (~ a) 90°-38° 30′ 51° 30', and 90°-40-50°, or sin a sin b. (~b), since Taking the triangle that has B acute, then B = 31° 34′. There are now known a, b, A, and B, to find c and C, which are calculated exactly as in the third and fourth cases; and when this is done, all the parts of this triangle are known. Taking next the triangle that has B obtuse, then B = 148° 26'; and hence in this triangle are known a, b, A, and B; and hence c and C in it are found also as in the preceding triangle. It will be found, in the triangle in which B is acute, that C = 130° 3', and c = 70°. EXERCISES. 1. Given a 24° 4′, b = 30°, and A = 36° 8'; find the other parts. Ans. B 46° 18′, C 104° 0', and c = 42° 9′. Or, B = 133° 42', C= 11° 23′, and c = = 7° 51'. 2. Given a = 76° 35′ 36′′, b = 50° 10′ 30′′, and A = 121° 36′ 19′′-8, to find the other parts. Ans. B 42° 15′ 13′′-66, C = 34° 15′ 2′′-76, and c = 40° 0' 10". 359. Case VI.-When two angles and a side opposite to one of them are given. The side opposite to the other given angle will be found by the theorem in 358. Thus, if A, B, a, are the given parts, sin A: sin B sin a: sin b; by which analogy the side b is found. When b is found, there are then two sides and their opposite angles known; and hence c and C can be found as in the preceding case, thus: sin (A~ B): sin 1 (A + B) = tan sin (ab): sin (a + b) = tan (a ~ b): tanc, (AB): cot C. There will sometimes be two values of b admissible, as there were of B in the preceding case, and consequently also two triangles (see fig. in art. 315). When b has two values, so have c and C. The values of bare supplementary, and by taking one of them, there will then be known in one of the triangles the parts A, B, a, b ; and hence c and C can then be found, and all the parts of this triangle will be then known. Taking then the second value of b, the remaining parts c, C, of the other triangle can similarly be found. Whenever A differs from 90° in excess or defect by less than B does, there will be only one triangle, and therefore only one value of b, which will be of the same species as B; in other cases, b has two values that are supplementary." When (~A) ≤ (17~ B), then sin Asin B. EXAMPLE. The angles A and B are 31° 34' and 30° 28′, and the side a is 40°; required the other parts. To find the side b. Sin A: sin B = sin a: sin b. Sin A 31° 34' Sin B 30° 28' 9.718909 19-513107 Sin b 38° 30' 9-794198 The side b has only one value, for sin Asin B. In the triangle are now known the parts A, B, a, b, and the remaining parts c and C may be computed in the same manner as in the third and fourth cases. EXERCISES. 1. Given A = 51° 30', B = 59° 16′, and a = 63° 50′, to find the other parts. Ans. C131° 30′, b = 80° 19', and c = 120° 47'. Or, C=155° 22′, b = 99° 41′, and c = 151° 27'. 2. Given A = 97° 20′ 31′′·6, B = 66° 57′ 3′′-6, and a = 75° 0′ 51′′-3, to find the other parts. Ans. C 42° 30′ 55′′, b = 63° 39′ 57′′-8, and c = 41° 9' 46". Besides determining the species of the parts of oblique spherical triangles by means of the algebraical sines of the required parts, they can also be ascertained by certain theorems in spherical geometry (Geom. vol. II. Sph. Geom.) OTHER SOLUTIONS. The preceding methods of solution are generally the most convenient when all the parts of a spherical triangle are required; but when only one part is required, it will be more concise and simple to use some of the following methods:: The First Case. If the angle opposite to that side that is nearer to a quadrant than any of the other sides is first found by 349, then the other two angles may be found by the simple analogy in article 358. For the species of these angles are the same as of their opposite sides, and are therefore known. This rule depends on this principle:- Only that side of a spherical triangle can differ in species from its opposite angle, which is nearer to a quadrant than either of the other sides.' When only one angle is required, it must be found as in article 349. The Second Case. Find first the side opposite to that angle which is nearer to a quadrant than either of the other two by 351; then the other two sides can be found by the analogy in article 358. For the species of the other two sides will be the same as those of their opposite angles. The rule depends on the principle: Only that angle of a spherical triangle can differ in species from its opposite side, which is nearer to a quadrant than either of the other sides." When only one side is required, it must be found by 351. 360. The four following cases can be solved by the method for right-angled triangles, by dividing the given triangle into two right-angled triangles by means of a perpendicular from one of the angles upon the opposite side, so that one of the right-angled triangles shall contain two of the given parts. |