163. The formulæ exemplified in Arts. 160, 161, and 162, are known as Napier's Analogies. In each case there may be other forms, according as other elements are used. SOLUTION OF SPHERICAL OBLIQUE TRIANGLES. 164. In the solution of spherical oblique triangles, we may distinguish six cases: 1. Given a side and two adjacent angles. 2. Given two sides and their included angle. 3. Given the three sides. 4. Given the three angles. 5. Given two sides and the angle opposite to one of them. 6. Given two angles and the side opposite to one of them. 165. By application of the principles of Art. 136, (ƒ), the solution of any example under Cases 2, 4, and 6 may be made to depend upon the solution of another example under Cases 1, 3, and 5, respectively; and vice versa. Thus it is not essential to consider more than three cases in the solution of spherical oblique triangles. 166. The student must carefully bear in mind the remarks made in Arts. 151 and 152. CASE I. 167. Given a side and two adjacent angles. 1. Given A= 70°, B = 131° 20', c = 116°; find a, b, and C. By Napier's Analogies (Arts. 160, 161), we have Whence, tan (ba) sin (BA) csc (B+4) tan c, = + + and tan (ba) = cos (B − A) sec 1⁄2 (B+ A) tan c. From the data, (B — A) = 30° 40', † (B+A) = 100° 40', c = 58°. Then, and 180° (ba) = 82° 20.5' ..(b+a) = 97° 39.5'. a = 1 (b + a) — † (b− a) = 57° 56.7', To find C, we have by Art. 162, sin (b+a) cot C= sin (b − a) tan (BA) =sin(b+a) csc (b-a) tan (B-A). log sin (ba) = 9.9961 log csc (b− a) = 0.1946 log tan (B-4)=9.7730 log cotC9.9637 ...C47° 23.6', and C 94° 47.2'. = Note 1. The value of C may also be determined by the formula cos (b+a) Note 2. The triangle is always possible for any values of the given elements. and cot A tan (B-C) cot A tan (B+C) = 70°; find B, C, and a. taĥ 1⁄2 (B — C) = sin † (b − c) csc † (b + c) cot 1⁄2 A, tan (B+C) = cos(b − c) sec 1⁄2 (b+c) cot A. From the data, (bc) 10° 40', (b+c) = 126° 40', A= 35°. From which we find, Then, and (B-C) 18° 14.5', (B+C)=113° 2.9'. = B=(B+C) + (B-C) = 131° 17.4', To find a, we have by Art. 160, tana = sin (B+ C) tan † (b − c). From which we obtain ɑ = 57° 56.6'. Note. The triangle is always possible for any values of the given elements. 1. Given a = 60°, b = 137° 20', c = 116°; find A, B, and C. 2s = a+b+c=313° 20′, or s = 156° 40'. Whence, s — a = 96° 40', s — b = 19° 20', s — c = 40° 40'. log sin (s—b) = 9.5199 log sin (sc) = 9.8140 log csc s = 0.4022 log csc (sa) = 0.0029 log tan A 2)9.7390 = 9.8695 • ', A = 36° 31.2', and A = 73° 2.4'. In like manner we find, B=131° 32.2', and C-96° 55.4'. The values of A, B, and C may also be obtained by aid of the sine or cosine formulæ of Art. 158. If all the angles are to be computed, the tangent formulæ are the most convenient, as only four different logarithms are required. If but one angle is required, the cosine formulæ will be found to involve the least work. Note. The triangle is always possible for any values of the given elements which satisfy the conditions of Art. 138, (a) and (c); that is, if a+b+c is <360°, and no side is greater than the sum of the other two. 1. Given A=70°, B = 131° 10', C= 94° 50'; find a, b, |