Case IX. Given two angles of an oblique-angled spherical triangle, and the side adjacent to both of them, to find a side opposite to one of the given angles. Here ZA, Z. c and the included side b are given; and formulae will be obtained by a change of letters, if Z A and ZH and the included side c are given; or if ZB and Z c and the included side a are given. CASE X. Given two angles of an oblique-angled spherical triangle, and the side adjacent to both of them, to find the other angle. SoLUTION. Find the other two sides by Case IX, and then find the other angle by Case VI. Or, the angles A, B, or c, respectively, may be found by the sixth set of equations, page 184. CASE XI. Given the three sides of an oblique-angled spherical triangle, to find the angles. Or, the angles A, B, or c, respectively, may be found by the formulae A. 182. CASE XII. Given the three angles of an oblique-angled spherical triangle, to find the sides. sine A. sine B II. Cos # c- nav/co-o) sine A. sine B (F) When two sides, and an angle opposite to one of them, are given, to find the rest (see Case Ist, IId, and IIId), the values of these required parts are sometimes ambiguous. (G) Practical Rules for determining whether the quantities sought are acute, obtuse, or ambiguous, are given in the solutions of the different cases. - The two following tables are the same as those given by Legendre at pages 400 and 401 of the 6th edition of his Geometry, and are deduced from Prop. xvi.1, page 149, and Prop. xvi.II, page 150 of this treatise. If A=90°, arb, or a +b= 180, the cases will not be ambiguous, but if b-90°, there will be two solutions. When two angles, and a side opposite to one of them, are given, to find the rest (see Case IVth, Vth, and VIth), the values of the required parts are subject to ambiguity; this triangle being supplemental to that wherein two sides and an angle opposite to one of them are given. TABLE II. Let A, B, and a be the given parts. Then, o AE-B one solution. 1. a-90°, B-290° } A-1B two solutions. o o s A+BE-180° one solution. 2. a-190°, BD-90 {. +B+, 180° two solutions. A + BE-180° two solutions. A +B+180° one solution. AE-B two solutions. A-1B one solution. * If a=90°, A=B, or A+B = 180°, there will be but one solution, but if B = 90°, there will be two solutions. (H) We may likewise remark, that since any side or angle of a spherical triangle is less than 180°, the half of any angle, or half the difference between any two sides, or half the difference between any two angles, must be acute. Hence in the equation, where cot # c. cos # (a-b)=tang ! (A+B), cos ? (a+b) (M. 188) it is plain that cot # c and cos; (a-b) are both positive (K. 100.), and therefore tang #(A+B) and cos 3 (a +b) must be both positive; consequently, half the sum of any two sides of a spherical triangle is of the same species as half the sum of their opposite angles. This rule is applied in the practical solutions of the different cases, and will frequently remove the ambiguity which would otherwise arise, where a quantity sought is to be determined by means of a sine. CHAP. VI. I. PRACTICAL RULES FOR THE SOLUTIONS OF ALL THE DIFFERENT CASES OF RIGHT-ANGLED SPHERICAL TRIANGLES, witH THEIR APPLICATION BY LOGARITHMs. Every spherical triangle consists of six parts, three sides, and three angles; any three of which, being given, the rest may be found. In a right-angled spherical triangle, two given parts, besides the right angle, are sufficient to determine the rest. The questions arising from a variation of the given and reguired parts are 16, but if distinguished by the data, the number of cases is 6. THE GIVEN QUANTITIES ARE, EITHER 1. The hypothenuse and an angle. |