As the angle A is greater than C, the side BC will also be greater than AB (26). Now both the values of BC are greater than the value of AB, and therefore the answer is ambiguous. If BC had been taken 99° 41', AC would have been 151° 27'. From this example it appears that the side BC first found, must be either an arc less than 90° (80° 19′), or its sup. (99° 41'); and that the two values of the side AC (120° 47' and 151° 27′) are obtuse, and also the two values of the angle B (131° 30′ and 152° 22′). But these two values, in both instances, are not supplements of each other. Case 3. Given two sides AB 80° 19′ and BC 120° 47′, and the included angle B 51° 30', to find the rest. Cot. (AB + BC) 100° 33' · 9.2626729 9.9723380 10.3166443 11.0263094 In this example the latter number 95° 23′ must be taken to make the half sum of the angles A and C agree with the half sum of their opposite sides. Hence A = 36° 7′ + 95° 23′ = 131° 30′, and C = 95° 23′ Case 4. Given two angles A 131° 30′ and B 51° 30′, and the included side AB 80o 19', to find the rest. T. (AC+CB) 87°41′ or 92° 19′ 11.3925846 The greater number 92° 19′ must be taken, to make the half sum of the sides agree with the half sum of their opposite angles. 9.7344659 T. (BCAC) 28° 28′ Therefore BC = 92° 19′ + 28° 28′ = 120° 47′, and AC= Therefore the angle C = 59° 16'. Case 5. Given the three sides, AB 80° 19', BC 120° 47', AC 63° 50', to find the angles. 132° 28' 68° 38' sum of the sides = sum - AC Therefore the angle B= 51o 30'. In the same manner the angle C is found 59° 16', and the angle A 131° 30'. Or, the angles may be found by the rule for sine B, page 102. But when one angle is found by this rule, the other two angles may be more readily determined by the first case. In the triangle ZPS (fig. page 106) given ZP PS 74° 30', and ZS 108°, to find the angle P. 37° 48′, 9.9839105 Case 6. Given the three angles A 1310 30', B 51° 30', C 59° 16', to find the sides. 121° 8' = sum of the angles 61° 52' - sum C In the same manner AC is found 63° 50′, and BC 120° 47'. Or, AB may be found by the rule for cos. AB. See page 102. But when one side is found by this rule, the other two sides may be more readily determined by the second case. Miscellaneous Examples. 1. Given two sides and an opposite angle, namely, AC 58°, BC 110°, and the angle B 50°, to find the rest. Answer. A 121° 54′ 56′′, C 62° 34′ 6′′, AB 79° 17′ 14′′. 2. Given two sides and the included angle, namely, AC 58o, BC 110°, and the angle C 62° 34' 6", to find the rest. Answer. A 121° 54′ 56′′, B 50°, AB 79° 17′ 14′′. 3. Given two angles and an opposite side, namely, A 121o 54′ 56′′, B 50°, AC 58°, to find the rest. Answer. BC 110°, AB 79°17′ 14′′, C 62° 34′ 6′′. 4. Given two angles and the included side, namely, A 121° 54′ 56′′, C 62° 34′ 6′′, AC 58°, to find the rest. Answer. BC 110°, AB 79° 17 14", B 50°. 5. Given two sides 57° 30′ and 82° 26', and the included angle 126° 37', to find the other parts. Answer. The angles 61° 40′ and 48° 30′, and the third side 115° 20'. 6. Given two angles 61° 40′ and 126o 37', and the included side 57° 30', to find the other parts. Answer. The other sides 82° 26' and 115° 20′, and the third angle 48° 30'. 7. Given the three sides 65° 30', 96° 47' 50", 56° 40', to find the angles. Answer. 54° 40', 48° 30', 117° 5' 56". 8. Given the three sides 114° 30′, 83° 13', 50° 40′, to find the angles. Answer. 48° 31', 62° 56', 125o 19'. 9. Given the three angles 48° 31', 62° 52′, 125° 20' to find the sides. Answer. 114° 29′, 83° 9′, 56° 42'. 10. Given the three angles 61° 40', 126° 37', 48° 30', to find the sides. Answer. 57° 30', 82° 26', 115° 20′. 11. Each side of an equilateral triangle is 60°. Required the angles. Answer. Each angle is 70° 31′ 44′′. 12. In an isosceles triangle each of the equal sides is 95° 1, and their included angle is 100°. Required the base. Answer. 99° 22′ 24′′. 13. Given two sides of a triangle 96° 50' and 83° 10′, and their included angle 120°. Required the third side and the other two angles. Answer. Third side 120° 28' 10", and the angles 86° 4' 13", and 93° 55' 47". END OF SPHERICAL TRIGONOMETRY. |