Example I. In the right angled spherical triangle ABC, right angled at A, given the hypothenuse a 64° 40', and one leg b= 42° 12′. Required the rest. This example falls under case 2, section 2, of the preceding chapter, where cos c, a req. angle = tan given side x cot hypoth. Hence the following logarithmic operation. From log sin 42° 12′..9-8271887 | From log cos 64° 40'..9-6313258 Take log sin 64° 40′:.9-9560886 Take log cos 42° 12′..9.8697037 Rem, sin B. 48° 0'..9.8711001 Rem. cos c..54° 43'..9.7616221 To log tan 42° 12′.... 9·9574850 Add log cot 64° 40′ ....9.6752372 The sum is log cos c 64° 35'....9.6327222 Here 10 are added to the index of the remainder, and taken from the index of the sum; conformably with note 2, art. 31 of the preceding chapter. 2dly. To compute the same by means of Napier's circular parts. Here, if the leg b be assumed as the middle term, (90° - a), and (90°-B) are the opposite parts. sin rect cos op. becomes, and sin mid. sin a sin B. this agrees with the foregoing process. Again, if (90°c) be the middle part, then (90°- α) and b are adjacent parts, and sin mid. rect tan adja. becomes cos ccot a tan b=cot hypoth. x tan given side; this also agrees with the preceding. 3dly. If (90° - a) be made the middle part, then b and c are the opposite parts, and sin mid. = rect cos oppos. becomes cos a = cos b cos c; whence cos c = cos hypoth. that is, cos c = cos given leg cos a cos b ; which also agrees with the preceding. Hence the logarithmic computation need not be repeated. Example II. Given in a spherical triangle, the quadrantal side CA = 90°, an adjacent angle c = 42° 12′, and the opposite angle в = 115° 20′; to find the other angle and sides. Suppose the side cв produced until CD = CA = 90°: then, it is evident from chap. vi. art. 33, that both the angles, CAD, and D, are = 90°, and consequently that AD is the measure of the angle c, and therefore = 42° 12′. It is also evident that ABD, as well as ACD, is a right angled triangle, and that ABD = 180° ABC 64° 40′. Hence, to find AB, &c. we make use of the triangle ABD, of which we determine the hypothenuse and oblique angles by case 4, of right angled triangles. Thus, sin given side sin req. an. = cos given angle cos given side From log sin 42° 12'..9.8271887 From log cos 64° 40'..9.6313258 Take log sin 64° 40'..9.9560886 Take log cos 42° 12′..9.8697037 Rem. sin AB .. 480..9.8711001 Rem. sin BAD 35°17'' Or COS CAB 54°43' 9-7616221 sin side req. tan given side x cot op, angle. In a right angled spherical triangle given the hypothenuse 64° 40', and an adjacent angle = 64° 35' to find the rest. Given one leg 48°; to find the rest. Given a leg to find the rest. 54° 43′, and its adjacent angle 45°; Example VI. = = Given the two legs 54° 13'; and 42° 12′, respectively; to find the rest. Example VII. Given the two oblique angles = 48° and 64° 35′ respectively; to find the rest. Example VIII. Given a quadrantal side, one of the other sides = 115° 9', and the angle comprehended between them = 115° 55'; to find the rest. Ans. Angles 101° 4' and 117° 34′, side 113° 18'. Example IX. Given in an oblique angled spherical triangle, the A side a 44° 13′ 45′′, b = 84° 14′ 29′′, and their included angle c = 36° 45′ 28′′; to find the rest. This example corresponds with case 2, prob. 2, of oblique angled spherical triangles; and may first be solved by means of the subsidiary arc, in the manner there explained. Thus, first find ø, so that tan cos c tan b. DB To log cos c = cos 36° 45′ 28′′.... 9·9037261 tan tan 82° 49′ 33′′....10.9000656 This arc exceeds a, therefore the perpendicular AÐ from the vertical angle falls on the base produced: hence the 2d expression becomes COS C cos b cos (a) cos b To find the remaining parts use the known propor tion of the sines of sides to the sines of their opposite angles; thus To sin A 51° 6' 11"....9-8911340 And so is sin b.. 84° 14′ 29′′....9.9978028 To sin B.... 130° 5′ 21′′....9.8836846 Here the logarithmic sine 9.8836846 answers either to 49° 54′ 39′′ or to its supplement 130° 5′ 21′′; the former of which is the exterior angle ABD, the latter the angle B of the triangle. 2d Method, by Napier's Analogies. Taking the 14th and 15th formulæ at the end of sect. 4, of the preceding chapter, we have tan (B — A) = cot c and tan (B+ a) = cot c cos (b + a) The log. computation will therefore stand thus: To log cot c....... 18° 22′ 44′′.... 10-4785395 Add log sin (b − a) 20° 0′ 22′′. 9.5341789 From the sum .... 20-0127184 Take log sin (b + a) 64° 14′ 7′′.... 9-9545255 Rem.logtan (BA) 48° 49′ 38′′....10-0581929 Also, to log cot c..18° 22′ 44′′....10-4785395 Add log cos(b — a) 20° 0′ 22′′.... From the sum 9-9729690 ..20-4515085 Take log cos(b + a) 64° 14′ 7′′.... 96381663 Rem.log tan (B+ A) 81° 15′ 44′′....10·8133422 Hence 81° 15′ 44′′ + 48° 49′ 38′′ and 81° 15′ 44′′ 48° 49′ 38′′ = 130° 5′ 22′′ = B, 32°26′ 6′′ = A ; agreeing nearly with the result of the former compu- . tation. Then to find c, use the proportion, as sin A: sin a :: sin c: sin c sin 51° 6′ 12′′. Here it would seem, from a comparison of the methods, that the first is rather quickest in operation, while the last is probably the easiest to remember, and provides best against the occasions of ambiguity. |