Next, since (Art. 232. VIII.) cos. A" = tan S' tan S * It follows, from the equation marked (A), that the hypotenuse of a right-angled spherical triangle may be found, if the cosines of the sum, and of the difference, of the two oblique angles be known, even when those angles themselves are not known. P .. A” Also, (Art. 232. III.) cos S" = sin (S+S') = 1 cos S ; cos S' 1+ COS S' ()tan S" = √ [tan (S+S'). tan { (S— S')]. Lastly, (Art. 232. V.) sin S′ = sin S sin A. Let, therefore, sin S' = = 1 - sin S' 1 + sin S' 1 - tan x 1+tan x sin (90° - 2x) = cos 2% or (Introd. Art. 30.) 1- sin S sin A' 1 + sin Ssin Ä (Art. 23 2. V.) (x being such that tan x = tan (45°-x). (Introd. 29. 11.) If, therefore, a be first found, from the assumed equation, tan asin S sin A', and if z be next determined, from the equation, tan x=(45° - x), the side S', which is equal to 90° – 2 %, will be known. PROP. III. (262.) Problem. Two parts, besides the quadrant, of a quadrantal triangle, which has only one of its sides a quadrant, being given, to solve the triangle. It is evident, that the six theorems which have been applied (Art. 256.) to the solution of right-angled spherical triangles, may be readily translated, so as to express the relations existing between the parts of a quadrantal triangle, by means of Art. 78. For the polar triangle, in this case, is a quadrantal triangle. The four general Forms, also, which are marked (I.) (II.) (IX.) (X.) will give the same results, if a side in each of them be made equal to a quadrant. Thus, If S=90°; sin A' = sin A sin S' (IX.) S" = 90°; cos A′′ = - cot S cot S' (I.) Or, let CBD be a quadrantal triangle, having the side CB, and none other of its sides, a quadrant: from A D C as a pole, at the distance CB, let there be described (Art. 59.) the great circle BA, meeting CD produced, in A; wherefore, (Art. 50.) the angles CBA, CAB, are right angles, BD is common to the two triangles CBD, BAD; AD is the complement of DC; the angle ABD is the complement of DBC; and BA measures (Art. 54.) the angle C. It is manifest, then, that if two parts, besides the quadrant BC, of the triangle CBD, be given, the triangle BAD may be solved (Art. 256.): and if all the parts of the triangle ABD be known, all the parts of CBD are thereby known. PART II. THE ELEMENTS OF Spherical Trigonometry. SECTION III. ON THE SOLUTION OF OBLIQUE-ANGLED SPHERICAL TRIANGLES. (263.) Problem. THREE of the six parts of an oblique-angled spherical triangle being given, to solve the triangle. CASE 1. Let the three sides, S, S', S", be given, to find the three angles, A, A', A", of the triangle. |