It is to be observed, that the formulas employed are to be rendered homogeneous, in terms of R, as explained in Art. 30. This is done by simply multiplying the radius of the Tables, R, into the middle part. Formula (10), Art. 74, gives, for 90°C, middle part, log cos C = log cot a log tan b log cot a (105° 17′ 29′′) log tan b ( 38° 47′ 11′′) 9.436811 9.905055 10; B 9.341866 ... C 102° 41′ 33′′ = Formula (4), gives, for 90° - B, middle part, b log cos Blog sin C+ log cos - 10; log sin C (102° 41′ 33′′) 9.989256 9.891808 9.881064 ... B == 40° 29′ 50′′. Ans. c = 109° 46′ 32′′, B = 40° 29′ 50′′, C = 102° 41′ 33′′. 2. Given 51° 30′, and B = 58° 35', to find a, c, and C. Because b < B, there are two solutions OPERATION. Formula (7), gives for c, middle part, log sin c = log tan blog cot B-10; Formula (1), gives for 90° -a, middle part, log cos a = log cos blog cos c 10; Formula (10), gives for 90° - C, middle part, In a similar manner, all other cases may be solved. 3. Given a = 86° 51′, and B = 18° 03′ 32", b, c, and C. to find Ane. b 18° 01' 50", c = 80° 41′ 14′′, C 88° 58′ 25′′. find 4. Given b = 155° 27′ 54′′, Ans. and с =29° 46′ 08′′, to a = 142° 09′ 13", B = 137° 24′ 21′′, C 54° 01′ 16′′. 5. Given c = 73° 41′ 35′′, and B 99° 17' 33", to b, and C. find a, Ans. a = 92° 42′ 17′′, b = 99° 40′ 30′′, C 73° 54′ 47′′. 7. Given B = 47° 13′ 43′′, and C = 126° 40′ 24′′, to find a, b, and C. Ans. a = 133° 32′ 26', b = 32° 08′ 56′′, c 144° 27' 03". In certain cases, it may be necessary to find but a single part. This may be effected, either by one of the formulas given in Art. 74, or by a slight transformation of one of them. 90° Thus, let α and B be given, to find C. Regarding a, as a middle part, we have, log cos a + (a. c.) log cot B = log cot C; from which C may be found. In like manner, other cases may be treated. QUADRANTAL SPHERICAL TRIANGLES. 77. A QUADRANTAL SPHERICAL TRIANGLE is one in which one side is equal to 90°. To solve such a triangle, we pass to its polar triangle, by subtracting each side and each angle from 180° (B. IX., P. VI.). The resulting polar tri angle will be right-angled, and may be solved by the rules already given. The polar triangle of any quadrantal triangle being solved, the parts of the given triangle may be found by subtracting each part of the polar triangle from 180°. A = 90°, b = 104° 18', and C 161° 23'. Solving this triangle by previous rules, we find, B a = 76° 25' 11", c = 161° 55′ 20′′, B 94° 31′ 21′′; hence, the required parts of the given quadrantal triangle are, A' 103° 34' 49" C' 18° 04' 40", b′ = 85° 28′ 39′′. In a similar manner, other quadrantal triangles may be solved. FORMULAS USED IN SOLVING OBLIQUE-ANGLED SPHERICAL TRI ANGLES. C 78. Let ABC represent an oblique-angled spherical triangle. From either vertex, C, draw the arc of a great circle CB', perpendicular to the opposite side. The two triangles ACB' and BCB' will be rightangled at B'. From the triangle ACB', we have Formula (2), Art. 74, B A B' That is, in any spherical triangle, the sines of the sides are proportional to the sines of their opposite angles. Had the perpendicular fallen on the prolongation of AB, the same relation would have been found. |