BC is less than 90°, because AC and C are like (36). B is acute, because its op. side AC is less than 90°. Case 3. Given the side AB 42° 12', and its opposite angle C 48°, to find the rest. Note. In the remaining cases the analogies and the answers are given; and the logarithmic operations must be performed by the learner. Case 4. Given the hypothenuse BC 64° 40', and a side AB 42° 12′, to find the rest. Cos. AB 42° 12′ : cos. BC 64° 40′ :: R: cos. AC 54° 43′. AC is less than 90°, because AB and BC are like (37). S. BC 64° 40': s. AB 42° 12′ :: R: s. C 48°. C is acute, because its op. side is less than 90°. Tan. BC 64° 40′ tan. AB 42° 12′ :: R: cos. B 64° 35'. Case 5. Given the side AC 54° 43', and AB 42o 12', to find the rest. R: cos. AC 54° 43' :: cos. AB 42° 12': cos. BC 64° 40'. S. AB 42° 12′ : R :: tan. AC 54° 43′ : tan. B 64° 35′. S. AC 54° 43': R :: tan. AB 42° 12′ : tan. C 48°. C is acute, being like its op. side AB. Case 6. Given the two oblique angles B 64° 35', and C 48°, to find the sides. S. B 64° 35': cos. C 48° :: R: cos. AB 42o 12'. S. C 48° cos. B 64° 35' :: R: cos. AC 54° 43′. Tan. B 64° 35′ : cot. C 48° :: R : cos. BC 64° 40′. Miscellaneous Examples. 1. Given the hyp. BC 78° 20′, and the side AB 36° 31', to find the rest. Answer. AC=75° 25′, B = 81° 12′, C = 37° 25'. 2. Given the side AC 27° 54', and AB 11o 30', to find the rest. Answer. BC 30°, B = 69° 22′, C=23° 30'. 3. Given the two oblique angles, B 81° 12', and C 37° 25', to find the rest. Answer. AB = 36° 31', AC = 75° 25′, BC = 78° 20′. 4. Given the side AB 11° 30', and its opposite angle C 23° 30', to find the rest. Answer. AC 27° 54′ or 152° 6', B=69° 22′ or 110° 38", BC 30° or 150°. 5. Given the side AC 75° 25', and its adjacent angle C 37° 25', to find the rest. Answer. AB = 36° 31′, B = 81° 12′, BC = 78° 20′. 6. Given the hyp. BC 30°, and the angle C 23° 30', to find the rest. Answer. AB 11° 30′, AC = 27° 54′, B = 69° 22′. 7. In the right-angled spherical triangle ABC one of the oblique angles B is 60°, and the other C is 45°. Required the side AC opposite to the greater angle B. Answer. ÃC=45o. 8. In a right-angled isosceles spherical triangle each of the equal sides is 30°. Required the hypothenuse. Answer. 41° 24′ 35′′. 128. The following problems are solved by Napier's Rules, which may be applied to the preceding examples. Plate, Fig. 4 1. In the right-angled spherical triangle ABC, given one side AB 49° 17', and its adjacent angle A 23° 28', to find the other side BC, and the hyp. AC. Here AB is middle part, and BC and comp. A are adjacent parts; therefore R xs. AB=tan. BC and cot. A (100), therefore log. tan. BC=log. R+ log. s. AB-log, cot. A. Again, comp. A middle part, and AB and comp. AC adja-. cent parts; therefore Rx cos. A = tan. AB and cot. AC, therefore log. cot. AC = log. R+ log. cos. A-log. tan. AB. 10+ log. cos. 23° 28′ log. tan. 49° 17' 19.9625076 10.0651775 9.8973301 2. Given PS=68° 43′, and ZS = 72° 16', to find PZ and the angle Z. Fig. page 106. Here comp. ZS middle part, and ZP, PS opposite parts; therefore Rx cos. ZS cos. ZP x cos. PS; therefore log. cos. PZ=log. R+ log. cos. ZS-log. cos. PS. Again, PS middle part, and comp. ZS and comp. Z op. parts; therefore R x s. PS = s. ZS x s. Z; therefore log. s. Z= log. R+ log. s. PS-log. s. ZS. 10+ log. s. 68° 43' log. s. Z = 78° 2′ 12′′ 19.9693212 9.9788579 9.9904633 3. Given PS 66° 32′, and PZ 37° 48', to find ZS. Here comp. ZS middle part, and PZ and PS op. parts; therefore R x cos. ZS = cos. PZ x cos. PS; therefore log. cos. ZS=log. cos. PZ+ log. cos. PS-10. 4. In the quadrantal triangle ZPS, where ZS = 90°, given ZP 37° 48′, and SP 66° 32', to find the angles P and Z. and comp Here comp. P middle part, ZP and comp. SP adj. parts; therefore R x cos. P= cot. ZP x cot. SP; there fore log. cos. P= log. cot. ZP+log. cot. SP-10. log. cot. 37° 48′ log. cot. 66° 32′ log. cos. P=55° 58′ 2′′ 10.1103177 9.6376106 9.7479283 The angle in the tables answering to log. cos. P is 55° 58′ 2′′: but since PZ and PS are like, the angle P must be greater than 90° (107); therefore we must take the supplement of P, or 124° 1′ 58′′, for the angle sought. Again, comp. PS middle part, and Z and comp. PZ op. parts; therefore R x cos. PS=s. PZ x cos. Z; therefore log. cos. Z = 10+ log. cos. PS-log. s. PZ. 10+ log. cos. 66° 32′ log. s. 37° 48' log. cos. Z 49° 28′ 47′′ 19.6001181 9-7873946 - 9.8127235 From these examples the reader will perceive that Napier's Rules apply with facility to all the cases of right-angled spherical triangles. As they are concise and easy to remember, they are better adapted to the solution of numerical examples than the analogies in the preceding table. 129. Numerical Solution of the Cases of Oblique-angled Spherical Triangles. Plate, Fig. 6. Case 1. In the oblique-angled spherical triangle ABC, given the side AB 80o 19', the side AC 63° 50', and the opposite angle B 51° 30', to find the rest. If C had been taken 120° 44', A would have been 24° 36'. If C had been taken 120° 44', BC would have been 28° 34'. This example affords an opportunity of remarking, that the angle C opposite to the other given side, is always either an acute angle or its supplement. But it appears from the calculation that this may not be the case with respect to the remaining side or the remaining angle; for the two values of A are 131° 32′ and 24° 36′, and the two values of BC are 120° 46' and 28° 34', which two values, in both instances, are not supplements of each other. Case 2. Given the angle A 59° 16', the angle C 51° 30′, and its opposite side AB 63° 50', to find the rest. |