Now, if the two sides of a right-angled spherical triangle are of the same affection, the hypotenuse will be less than 90°; and if of different affection, the hypotenuse will be greater than 90°. If, in every instance, we make a natural construction of the figure, and use common judgment, it will be impossible to doubt whether an arc must be taken greater or less than 90°. We will now solve the triangle ACB. AC= 180° — 150° 33′ 20′′ 29° 26' 40". = To find BC, we use Eq. (3) or (13), Prop. 3, Sec. II., thus: To find AB, we use equation (1) or (11), thus: 1. In the right-angled spherical triangle ABC, given AB = 118° 21' A 4", and the angle A= 23° 40′ 12′′, to find the other parts. Ans. { B AC, 116° 17′ 55"; the angle C, 100° 59′ 26′′; and BC, 21° 5′ 42′′. 2. In the right-angled spherical triangle ABC, given AB 53° 14′ 20′′, and the angle A 91° 25′ 53′′, to find the other parts. 91° Ans. { AC, 91, 1' 9"; the angle C′, 53° 15′ 8′′; and BC, 91° 47′ 10′. 3. In the right-angled spherical triangle ABC, given AB 102° 50′ 25′′, and the angle A 113° 14′ 37′′, to find the other parts. Ans. { SAC, 84° 51′ 36′′; the angle C, 101° 46′ 56′′; and BC, 113° 46′ 27′′. 4. In the right-angled spherical triangle ABC, given AB 48° 24′ 16′′, and BC 59° 38′ 27′′, to find the other parts. 42′′; A, Ans. { AC, 700 23' 427; the angle 4, 66° 20′ 40′′; and the angle C, 52° 32′ 56′′. 5. In the right-angled spherical triangle ABC, given AB 151° 23' 9", and BC 16° 35′ 14′′, to find the other parts. Ans. {AC, 147° 16' 51'; the angle C, 117° 37′ 25′′; and the angle A, 31° 52′ 49′′. 6. In the right-angled spherical triangle ABC, given AB 73° 4′ 31′′, and AC 86° 12′ 15," to find the other parts. Ans. { BC, 76° 51' 20"; the angle A, 77° 24′ 23′′; and the angle C, 73° 29′ 40′′. 7. In the right-angled spherical triangle ABC, given AC 118° 32′ 12′′, and AB 47° 26' 35", to find the other parts. Ans. BC, 134° 56' 20"; the angle A, 126° 19' 2"; and the angle C, 56° 58′ 44′′. 8. In the right-angled spherical triangle ABC, given AB 40° 18' 23", and AC 100° 3' 7", to find the other parts. Ans. { The angle A, 98° 38′ 53′′; the angle 9. In the right-angled spherical triangle ABC, given AC 61° 3′ 22′′, and the angle A 49° 28' 12", to find the other parts. Ans. { AB, 490 364 6'; the angle C, 60° 29′ 20′′; and BC, 41° 41' 32". 10. In the right-angled spherical triangle ABC, given AB 29° 12′ 50′′, and the angle C 37° 26′ 21′′, to find the other parts. Ans. Ambiguous; the angle A, 65° 27′ 57′′, or its supplement; AC, 53° 24′ 13", or its supplement; BC, 46° 55′ 2′′, or its supplement. 11. In the right-angled spherical triangle ABC, given AB 100° 10' 3", and the angle C 90° 14′ 20′′, to find the other parts. Ans. AC, 100° 9′ 52", or its supplement; BC, 1° 19′ 55′′, or its supplement; and the angle A, 1° 21′ 12", or its supplement. 12. In the right-angled spherical triangle ABC, given AB 54° 21′ 35", and the angle C 61° 2' 15", to find the other parts. Ans. BC, 129° 28′ 28", or its supplement; AC, 111° 44' 34", or its supplement; and the angle A, 123° 47' 44", or its supplement. 13. In the right-angled spherical triangle ABC, given AB 121° 26' 25", and the angle C 111° 14′ 37′′, to find the other parts. Ans. The angle A, 136° 0′ 5′′, or its supplement; QUADRANTAL TRIANGLES. CD The solution of right-angled spherical triangles includes, also, the soluton of quadrantal triangles, as may be seen by inspecting the adjoining fig- A ure. When we have one quadrantal triangle, we have four, which with one right-angled triangle, fill up the whole hemisphere. P CB 0 A' To effect the solution of either of the four quadranta! triangles, APC, AP'C, A'PC, or A'P'C, it is sufficient to solve the small right-angled spherical triangle ABC. To the half lune AP'B, we add the triangle ABC, and we have the quadrantal triangle AP'C; and by subtracting the same from the equal half lune APB, we have the quadrantal triangle PAC. When we have the side, AC, of the same triangle, we have its supplement, A'C, which is a side of the triangles A'PC, and A'P'C. When we have the side, CB, of the small triangle, by adding it to 90°, we have P'C, a side of the triangle A'P'C; and subtracting it from 90°, we have PC, a side of the triangles APC, and A'PC. PROBLEM I. In a quadrantal triangle, there are given the quadrantal side, 90°, a side adjacent, 42° 21', and the angle opposite this last side, equal to 36° 31'. Required the other parts. = By this enunciation we cannot decide whether the triangle APC or AP'C, is the one required, for AC 42° 21' belongs equally to both triangles. The angle APC AP'C 36° 31′ AB We operate wholly on the triangle ABC. = == middle part. cot. A C tan.AB. 10.040231 Then, cot. A C 42° 21' To find the angle C, call it the middle part. R cos. ACB = sin. CAB cos. AB. To find the side BC, call it the middle part. R sin. BC tan.AB cot. A CB. We now have all the sides, and all the angles of the four triangles in question. PROBLEM II. In a quadrantal spherical triangle, having given the quad· rantal side, 90°, an adjacent side, 115° 09', and the included angle, 115° 55', to find the other parts. This enunciation clearly points out the particular triangle A'P'C. A'P' 90°; and conceive A'C= 115° 09'. Then the angle P'A'C 115° 55′ = P'D. = = From the angle P'A'C take 90°, or P'A'B, and the remainder is the angle OA'D = BAC 25° 55'. = A CD b CB ΤΟ A' P We here again operate on the triangle ABC. A'C, taken from 180°, gives To find BC, we call it the middle part. R sin. BC sin.AC sin. BAC. |