Let fall a perpen. Tan 1 seg. of div. side = cos giv. ang. x tan side opp. ang. sought. from the third angle. Tan ang. sought = tan giv. ang. x sin 1 seg. sin 2 seg. of div. side Let fall a perpen. Tan 1 seg. of div. side = cos giv. ang. x tan other given side. Third side. on one of the giv. sides. Cos side sought = cos 1 seg. of side divided A side opposite and the given angles. jacent Third angle. cos side not div. x cos 2 seg. Let fall a perpen-Cot 1 seg. of div. ang. = cos giv. side x tan ang. opp. side sought. dicular on the third side. Tan side sought = Let fall a perpen.) Cot 1 seg. div. ang. = cos giv. side x tan other giv. angle. Let a, b, c, be the sides; A, B, C, the angles, b and e including the angle sought, sin s.sin (s-a) from one of the IV. A side to one of the two ad angles. giv. angles. and s = a + b + c. cosine Then, three sides. of its half. sinA= sin (s-b).sin (sc) sin b. sin c • COSA = sin b. sin c. Let s be the sum of the angles A, B, and c; and let B and c be adjacent to a TABLE III. For the Solution of all the cases of Oblique-Angled Spherical Triangles, by the Analogies of Napier. Values of the Terms required. By the common analogy, sines of angles as sines of opp. sides. { By the common analogy. Third side. By the common analogy. tan diff. other two ang. x sin sum giv, sides sin diff. those sides opposite angle. Third angle. tan sum of other two ang. x cos sum giv. sides cos diff. of those sides Third side. { By the common analogy. Two angles, and the side IV. between them. Third angle. { By the common analogy. Either of the angles. VI. The three angles. Either of the sides. { Let fall a perpen. on the side adjacent to the angle sought. Cos angle sought = tan adj. seg. x cot adja. side. of the Will be obtained by finding its correspondent angle, in a triangle which has all its parts supplemental to those 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 lege = 76°52', are given; to find the angles B, and c, and the other leg b. Here, by table I case 1, sin c = sin e ; sin a Or, log sin c = log sin c - log sin a + 10. log cos B = log tan c - log tan a + 10. Hence, 10+ log sin c = 10 + log sin 76°52' = 19-9884894 log sin a = log sin 78°20′ = 9.9909338 Remains, log sin c = log sin 83°56′ = 9-9975556 Here c is acute, because the given leg is less than 90°. Again, 10 - log tan c = 10 + log tan 76°52′ = 20-6320468 Remains, log tan a = log tan 78°20′ = 10-6851149 log cos B = log cos 27°45′ = 9-9469319 B is here acute, because a and care of like affection. Lastly, 10 + log cos a = 10 + log cos 78°20′ = 19-3058189 log cos c = log cos 76°52′ = 9.3564426 Remains, log cos b = 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 b=117°34′, c = 31°51′; to find the other parts. Ans. a = 113°55′, с = 28°51′, в = 104° 8'. Ex. 4. Given b = 27°6′, c = 76°52′; to find the other parts. Ans. a = 78°20′, в = 27°45′, c = 83°56′. Ex. 5. Given b=42°12′, в=48°; to find the other parts. Ans. a = 64°40'1⁄2, or its supplement, c = 54°44′, or its supplement, c = 64°35′, or its supplement. Ex. 6. Given в = 48°, с = 64°35'; required the other Ans. b = 42°12', c = 54°44′, a = 64°40′. parts? Ex. 7. In the quadrantal triangle ABC, given the quadrantal side a = 90°, an adjacent angle c = 42°12′, and the opposite angle A = 64°40′; required the other parts of the triangle ? Ex. 8. In an oblique-angled spherical triangle are given the three sides, viz, a = 56°40′, b = 83°13′, c = 114°30'; to find the angles. Here, by the fifth case of table 2, we have sin A = sin (s-b).sin(sc). sin b, sinc : Or, log sin A = log sin (s-b)+log sin (s-c)+ar. comp. log sin b + ar. comp. log sin c: where s = a + b + c. log sin (s-b) = log sin 43°58′ = 9.8415749 log sin (s-c) = log sin 12°41′ = 9.3418385 A.c.log sin b = A.c.log sin 83°13′ = 0.0030508 A. c.log sin c = A. c.log sin 114°30′ = 0.0409771 Sum of the four logs .... 19-2274413 Half sum = log sin A = log sin 24°15′ = 9.6137206 Consequently the angle A is 48°31′. Then, by the common analogy, To, sin c... sin 125°19′ .. log = 9-9116507. So that the remaining angles are, в=62°56′, and c=125°19′. 2dly. By way of comparison of methods, let us find the angle A, by the analogies of Napier, according to case 5 table 3. In order to which, suppose a perpendicular demitted from the angle c on the opposite side c. Then shall we have tan diff. seg. of c = This in logarithms, is tan(b+a).tan(b-a) tan c log tan (b+a) = log tan 69°56′4 = 10.4375601 Subtract log tan c = log tan 57°15′ = 10-1916394 Therefore, |