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. The solution of right-angled spherical triangles includes, also, the solution 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. a C B To effect the solution of either of the four quadrantal 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. = = To find the angle A, call it the R cos. CAB = R sin. PAC = middle part. = Then, cot. A C tan.AB. To find the angle C, call it the middle part. R cos. ACB = sin. CAB cos. AB. ACP = ACP = 117° 57' 15 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 quadrantal 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'. = b la A A' CIB 0 We here again operate on the triangle ABC. A'C, taken from 180°, gives 64° 51' = AC. To find BC, we call it the middle part. R sin. BC =sin.AC sin. BAC. To find AB, we call it the middle part. sin.AB = 62° 26′ 8′′ . 8.947674 180° A'B = 117° 33′ 52′′ the angle A'P'C. To find the angle C, we call it the middle part. R cos. C = cot. AC tan.BC. PRACTICAL PROBLEMS. 1. In a quadrantal triangle, given the quadrantal side, 90°, a side adjacent, 67° 3′, and the included angle, 49° 18', to find the other parts. Ans. The remaining side is 53° 5' 44"; the angle opposite the quadrantal side, 108° 32′ 29′′; and the remaining angle, 60° 48′ 54′′. 2. In a quadrantal triangle, given the quadrantal side, 90°, one angle adjacent, 118° 40′ 36′′, and the side opposite this last-mentioned angle, 113°2′ 28′′, to find the other parts. Ans. The remaining side is 54° 38′ 57′′; the angle opposite, 51° 2′ 35′′; and the angle opposite the quadrantal side 72° 26' 21". 8. In a quadrantal triangle, given the quadrantal side, 90°, and the two adjacent angles, one 69° 13′ 46′′, other 72° 12' 4", to find the other parts. Ans. the One of the remaining sides is 70° 8′ 39′′, the other is 73° 17' 29", and the angle opposite the quadrantal side is 96° 13′ 23′′. 4. In a quadrantal triangle, given the quadrantal side, 90°, one adjacent side, 86° 14' 40", and the angle opposite to that side, 37° 12′ 20′′, to find the other parts. Ans. The remaining side is 4° 43′ 2′′; the angle opposite, 2° 51′ 23′′; and the angle opposite the quadrantal side, 142° 42′ 3′′. 5. In a quadrantal triangle, given the quadrantal side, 90°, and the other two sides, one 118° 32′ 16′′, the other 67° 48′ 40′′, to find the other parts-the three angles. The angles are 64° 32′ 21′′, 121° 3′ 40′′, and Ans. 77° 11' 6"; the greater angle opposite the greater side, of course. 6. In a quadrantal triangle, given the quadrantal side, 90°, the angle opposite, 104° 41′ 17′′, and one adjacent side, 73° 21′ 6′′, to find the other parts. Ans. {Remaining Remaining side, 49° 42′ 16′′; remaining angles, 47° 32′ 38′′, and 67° 56′ 13′′. SOLUTION OF OBLIQUE-ANGLED SPHERICAL TRIANGLES. All cases of oblique-angled spherical trigonometry may be solved by right-angled Trigonometry, except two; because every oblique-angled spherical triangle is composed of the sum, or the difference, of two rightangled spherical triangles. When a side and two of the angles, or an angle and two of the sides are given, to find the other parts, conform to the following directions: Let a perpendicular be drawn from an extremity of a given side, and opposite a given angle or its supplement; this will form two right-angled spherical triangles; and |