opposite side AC, we must have employed the following propor tion: 9.991947 == 9.882101 9.874048 Therefore we conclude that the value of A-36° 48′, as we had before found it, 95. Let us now seek for the hypothenuse BC, supposing that we know the same quantities AB and B. Neither the first nor the second of the proportions announced above (92) can be immediately applied to the triangle ABC; but in the complemental triangle CDE, we know D and DE, we have therefore R: sin DE:: tang D: tang CE, or by substitution, neg. log radius log cos B (37° 19′) 9.189181 Therefore BC is 81° 13', since B and AB are of the same species. To obtain the value of the angle C upon the same suppositions, we must have recourse to the complemental triangle BFG, in which R: sin BF:; sin B: sin FG, which by substitution gives R: cos AB:: sin B: cos C. Hence the angle C is 83° 22′, or calculating as far as seconds, 83° 21' 33". This small difference arises from our supposing the side AB only 79°, whereas we have already found its value to be 79° 0' 20". 96. After having in two different cases determined the three unknown parts of the right-angled triangle ABC, it remains for us to examine the case in which this determination is impossible. And first, if we suppose the side AB and the opposite angle C to be known, it is evident that we cannot know of what species the hypothenuse is; for then the first proportion gives sin C sin AB:: R: sin BC But this last sine belongs indifferently to an hypothenuse less than 90°, or to its supplement, and to determine us in our choice we ought to know of what species the two oblique angles are. For want of this knowledge the case is doubtful; and therefore it is marked as such in the fourth column of the table. We fall into the same dilemma, if we desire to find the side AC, the same things being given. For, by the second proportion (92) we have tang C tang AB :: R : sin AC, but nothing in this case determines of what species the side AC is. The same difficulty respects the angle B, of which the value is deduced from the first 'proportion (92) applied first to the triangle BFG, and afterwards transported to the triangle ABC. For in the triangle BFG, we have sin BF: R:: sin FG : sin B, which in the triangle ABC gives cos AB: R:: cos C: sin B, from which we cannot ascertain of what species is the angle B. This is the first case of which the varieties offer three ambiguous solutions. Another case perfectly similar, is that in which we suppose the side AC, and the opposite angle B to be known. This case also presents three other ambiguous solutions. Generally, when in a right-angled spherical triangle we only know one of the oblique angles, and the side opposite to it, the value of the three other parts cannot be determined. 97. Hence there are six imperfect solutions among the thirty problems which any right-angled spherical triangle offers for solution: We might reduce these thirty to sixteen questions, by suppressing those which are absolutely similar. But it is desirable to find in a table, the proportion for which we have occasion, without being under the necessity of making any change. EXAMPLES OF RIGHT-ANGLED SPHERICAL TRIANGLES. 1. In the right-angled spherical triangle ABC, given the hypothenuse BC-64o 40', and the base AB=42° 10′, required the remaining side and the angles? Answer, C=47° 57′ 47′′, B-64° 36′40′′, 2. In the right-angled spherical triangle B C ABC, given the hypothenuse BC=50°, and the base AB=44° 20′; required the rest? Answer, C=65° 49′ 10′′, B-34° 56′ 8′′, AC=26° 1′ 10′′. 3. In the right-angled spherical triangle ABC, given the hypothenuse BC-63o 56', and the angle C=45o 41; required the other sides and angles. Answer, AC 55° 0′ 1′′, AB=39° 59′ 39′′, B=64° 46′ 15′′. 4. In the right-angled spherical triangle ABC, given the side AC-55°, and the opposite angle B-65° 46'; required the remaining parts? Answer, C=45° 41′ 33′′, AB=40° 0′ 12′′, BC=63° 56′ 12′′ 5. In the right-angled spherical triangle ABC, given the base AB=12° 30′, and the opposite angle C=24o 45'; required the other parts of the triangle? Answer, AB-28° 44′ 37′′, BC-31° 7′ 48′′, C=68° 27′ 53. 6. In the right-angled spherical triangle ABC, given AC=45o 15', and the angle C-63° 20′; required the rest? Answer, AB=54°44, B–51° 0 50”, BC=66 0 56, 7. In the right-angled spherical triangle ABC, given the leg AB-54o 30', and the adjacent angle B 44° 50'? required the rest? Answer, C=65° 49′ 53′′, AC=38° 59′ 11′′, BC=63° 10′ 4′′′. 8. In the right-angled spherical triangle ABC, given the sides about the right-angle respectively 55° 28′, and 63° 15'; required the hypothenuse and angles? Answer, BC 75° 13′ 2′′, C=67° 27′ 1′′, B=58° 25′ 46′′. 9. Given the two legs of a right-angled spherical triangle respectively 15o 20', and 31° 57; required the hypothenuse BC, and the angles B and C. Answer, C=67° 1′ 22′′, B=27° 23′ 27′′, BC=35o 5′ 3′′. 10. In the right-angled spherical triangle ABC, given the angle B=64° 40′, and the angle C-46° 15; required the sides? Answer, AC=40o 5′ 6′′, AB=53° 40′ 36′′, BC=63° 3′ 5′′. 11. In the spherical triangle ABC, right-angled at A, given the angles B and C respectively 72° 20′, and 24° 50′; required the sides? Answer, BC=46° 30′ 31′′, AC=17° 44′ 21′′, AB=13° 43′ 48′′. 12. In the spherical triangle ABC, right-angled at A, given AB=10° 39′ 40", and the adjacent angle B-23° 27′ 42′′; required the remaining angle C, and the sides AC and BC? Answer, BC=11° 35′ 40′′, AC=4° 35′ 26′′, C=66° 58′ 1′′. SOLUTION OF THE DIFFERENT CASES OF OBLIQUE-ANGLED SPHERICAL TRIANGLES. 98. Here there are as many varieties, as there are different combinations among the six parts of a triangle, taken four by four. But there are fifteen of these combinations (Page 111), and each admits of three different cases. Consequently there are 45 problems to solve relating to oblique-angled triangles. But as the resolution of one frequently includes that of several others, they may be reduced to the twelve following cases: I. 99. In an oblique-angled spherical triangle ABC, given the angles B and A, and the side opposite to the angle A, to find the side AC opposite to the other angle. Let A 61° 25′, B-82° 36′, BC 59° 40'. sin A sin B:: sin BC; sin AC. But arith. co. log sin A (61° 25') = 10.056445 log sin B (82° 36') = 9.996368 9.936062 B 9.988875 Whence AC 77° 5', or 102. 55'. Nor can we decide between these two results, unless we previously know of what species the side we seek for must be. gle. II. A B Let fall the perpendicular arc CD, and in the triangle BCD we shall have (90) and then (86) R: cos BC: tang B: cot BCD cos B: cos A:: sin BCD: sin ACD. Adding these two angles, or subtracting one from the other, according as the given angles A and B are of the same or of different species, we shall have the angle C required. Let A-61o 25', B=82° 36′, BC 59° 40', and we shall have neg. log radius log cos BC (59° 40') = 10. = 9.703317 log tang B (82° 36') = 10.886467 = 10.589784 Whence the angle BCD=14° 25′ (by 78). Consequently the angle ACD-67° 39', or 1120 21; which gives 82° 4′, or 126° 46', for the required angle ACB; and nothing in the question enables us to determine which value is to be preferred. III. 101. Given two angles A and B, with the opposite side BC, as in the preceding case, to find the side contained between these two angles (see preceding figure). The perpendicular arc CD forms the two right-angled triangles ACD and DCB, the last of which gives (87) R: cos B: tang BC: tang BD Besides we have (89) tang A tang B:: sin BD : sin AD. Consequently we shall have AB-AD+DB, according as the angles given are of the same or of different species. Preserving the same data as in the preceding case, neg. log radius Again, = 10. log cos B (82° 36') = 9.109901 log tang BC (59° 40′) = 10.232745 = 9.342646 Therefore BD=12° 25′. Hence the side AD=64o 25', or 115o 35'; and as in this example the angles A and B are of the same species, we must add the two segments BD and AD; consequently the side AB must be either 76° 50', or 128°. IV. 102. Given the two angles A and C, with the side on which these angles are formed, to find the third angle B (see preceding figure.) First, (90) R: cos AC:: tang A : cot ACD And then, (86) sin ACD: sin BCD:: cos A: cos B. The angle BCD is found by subtracting ACD from ACB, when the arc falls perpendicularly within the triangle; but when this arc falls without, we must subtract the angle ACB from the angle ACD, in order to find BCD. V. 103. Given as before, the two angles A and C, with the contained side AC, to find the two other sides, BC for example. The log taug. 61° 25' 10.263731, subtracting each figure from 9. except the first on the right, as before explained, we obtain 89.736269; and prefixing the negative unit as usual, the arithmetical equivalent becomes as above, 189.736269. The arithmetical equivalent for negative numbers is of considerable use in many arithmetical operations. See Nicholson's Treatise on Invo lution and Evolution. |