TABLE II.—For the Solution of Oblique-Angled Spherical Triangles. An angle or a side being divided by a perpendicular, the first and second segments are denoted by 1 seg, and 2 seg. Values of the Ju ities required!. Sines of angles are as sines of oppos. sides. } Cot 1 seg. of this ang. = cos giv. side x tan adj. angle. sin 1 seg. x cos ang. opp. given side Sin 2 seg. = Given. Required. between the } cos ang. adj. given side Sines of sides are as sines of their opposite angles. Let fall a perpen-Cot 1 seg. ang. req dicular from the included angle. Cos 2 seg = = tan given ang. × cos adj. side. cos 1 seg. x tan giv, side adj. giv. angle tan side opp. given angle req. = cos given ang. × tan adj. side. cos 1 seg. x cos side opp. given angle cos si e aj. given angle Tan 1 seg. of div. side = cos giv. ang. x tan side opp. ang. sought. from the third angle. Cos side sought cos side not div. x cos 2 seg. cos 1 seg. of side divided IV. Third angle. Let fall a perpen.Tan 1 seg. of div. side = cos giv. ang. x tan other given side. on one of the giv. sides. dicular on the third side. = Cot 1 seg. of div. ang. = cos giv. side x tan ang. opp. side sought. Tan side sought = tan giv. side x cos 1 seg. div. ang. cos 2 seg. of divided angle Let fall a perpen.) Cot 1 seg. div. ang. = cos giv. side x tan other giv. angle. giv. angles. from one of the Cos angle sought Let a, b, c, be the sides; A, B, C, the angles, b and c including the angle sought, A side and the to one of the A side opposite given angles. Let fall a perpen sine or cosine three of its half. three of its half. angles. An angle by the and s = a+b+c. Then, Let s be the sum of the angles A, B, and C; and let B and C be adjacent to a the side required. Then, TABLE III. For the Solution of all the cases of Oblique-Angled Spherical Triangles, by the Analogies of Napier. Given. Required. Values of the Terms required. By the common analogy, sines of angles as sines of opp. sides. {By the common analogy. By the common analogy. giv. side x sin diff. giv. angles sinsum of those angles tangiv. side x cos diff. giv. angles cos sum of those angles {By the common analogy. { Let fall a perpen. on the side adjacent to the angle sought. Cos angle sought = tan adj. seg. x cot adja. side. Will be obtained by finding its correspondent angle, in a triangle which has all its parts supplemental to those of the triangle whose three angles are given. Questions for Exercise in Spherical Trigonometry. Er. 1. In the right-angled spherical triangle BAC, rightangled at A, the hypothenuse a = 78° 20′, and one leg = 76°52′, are given; to find the angles B, and C, and the other leg b. Here, by table 1 case 1, sin c = sin c sin a ; log sin clog sin a + 10. log tan c log cos B = Hence, 10+ log sin c = 10 + log sin 76°52′ : Remains, log sin a = log sin c = log tan a + 10. log cos c + 10. Here c is acute, because the given leg is less than 90°. log tan c = 10 + log tan 76°52′ B is here acute, because a and c are of like affection. Lastly, 10+ log cos a = 10+ log cos 78°20′ = 19.3058189 Remains, log cos c = log cos b = log cos 76°52′ = 9.3564426 log cos 27° 8′ = 9·9493763 where b is less than 90°, because a and c both are so. Ex. 2. In a right-angled spherical triangle, denoted as above, are given a = 78°20′, B = 27°45′; to find the other sides and angle. Ans. b = 27° 8', c = 76°52′, c = 83°56′. Ex. 3. In a spherical triangle, with A a right angle, given = 117°34′, c = 31°51′; to find the other parts. Ans. a 113°55′, c = 28°51′, B = 104° 8'. Ex. 4. Given b = 27°6′, c = 76°52′; to find the other parts. Ans. a 78° 20′, B = 27°45', c = 83°56'. Ex. 5. Given b= 42°12′, B=48°; to find the other parts. Ans. a 64°40'1, or its supplement, C= 54°44', or its supplement, c = 64°35', or its supplement. parts? Ex. 6. Given B = 48°, c = 64° 35'; required the other Ans. b 42°12', c 54°44', a 64°40'. |