As a check, to test the accuracy of the above work As the test is satisfied, the work is probably correct. 3. Given a = 86° 51', and B = 18° 03' 32", to find b, c, and C. Ans. b 18° 01' 50", c = 86° 41′ 14′′, C = 88° 58' 25". = 4. Given b = 155° 27′ 54′′, and c = 29° 46′ 08′′, to find a, B, and C. Ans. a 142° 09′ 13", = 142° 09′ 13′′, B = 137° 24′ 21′′, = 5. Given c 73° 41′ 35′′, and B 99° 17′ 33′′, to find = = a, b, and C. Ans. a = 92° 42′ 17′′, b = 99° 40′ 30′′, C = 73° 54′ 47′′. 6. Given b = 115° 20', and B 91° 01' 47", to find a, c, and C. a = 64° 41' 11", с = 177° 49′ 27′′, C = 177° 35′ 36′′. a' 115° 18' 49", c' = 2° 10' 33", C' = 2° 24' 24". 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". 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 supplemental polar triangle, by subtracting each side and each angle from 180° (B. IX., P. VI.). The resulting polar triangle will be right-angled, and may be solved by the rules already given. The supplemental polar triangle of any quadrantal triangle being solved, the parts of the given triangle may be found by subtracting each part of the supplemental triangle from 180°. Passing to the supplemental polar triangle, we have A = 90°, b 104° 18', and C = = 161° 23'. Solving this triangle by previous rules, we find 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", Other quadrantal triangles may be solved in like manner. FORMULAS USED IN SOLVING OBLIQUE-ANGLED SPHERICAL TRIANGLES. 78. To show that, in a spherical triangle, the sines of the sides are proportional to the sines of their opposite angles. Let ABC represent an oblique-angled spherical triangle. From any vertex, as 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, sin CB' sin A sin b. From the triangle BCB', we have sin CB' = sin B sin a. Equating these values of sin CB', we have sin A sin b sin B sin ɑ; from which results the proportion, C B' B 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. 79. To find an expression for the cosine of any side of a spherical triangle. a B Let ABC represent any spherical triangle, and O the centre of the sphere on which it is situated. Draw the radii OA, OB, and OC; from C draw CP perpendicular to the plane AOB; from P, the foot of this perpendicular, draw PD and PE respectively perpendicular to OA and OB; join CD and CE, these lines will be respectively perpendicular to OA and OB C (B. VI., P. VI.), and the angles CDP and CEP will be equal to the angles A and B respectively. Draw DL and PQ, the one perpendicular, and the other parallel to OB. We then have In the right-angled triangle OLD, OL OD cos DOL = cos b cos c. The right-angled triangle PQD has its sides respectively perpendicular to those of OLD; it is, therefore, similar to it, and the angle QDP is equal to c, and we have QP = PD sin QDP = PD sin c. The right-angled triangle CPD gives PD = CD cos CDP = sin b cos A; substituting this value in (2), we have QP sin b sin c cos A; |