2. Given A= 80° 20′; B=115° 30′; a=84° 20′, to find the Since the values of the half angles are given by these formulas, there will be no ambiguity in this case. EXAMPLES. 1. Given a = 100°; b=80°; c=75°, to find the angles. (a+b+c)=s=127° 30'; s-a=27° 30′; s-b= 47° 30'; 8-c=52° 30'. C = 0·100533 2)19.796775 tan. B = 9.898387 B=38° 21′ 27′′, and B=76° 42′ 54′′. 9.664406 9.867631 c) = 0·100533 = 0·100533 2)19-733103 tan. C 9.866551 36° 19' 57" and C=72° 39′ 54′′. 2. Given a = 49° 8'; b=57° 16'; c=96° 12', to find the angles. A 31° 32′ 42′′. = Ans. B 35° 35′ 15′′. C=136° 32' 48". CASE VI. 881. When the three angles are given. As in last case, we have three groups (D'), (D'), (D'''), either of which will give the sides. If we use (D"), we have B As in last case there can be no ambiguity in the values of a, b, c as here found. EXAMPLES. 1. Given A = 110°; B = 100°; C = 90°, to find the sides. A=110° C = 90° (A+B+C)=S=150°; S-A=40°; S-B-50°; S-C=60°. c=43° 9' 37" and c = = 86° 19' 14". Since the angle C is right, the solution might have been obtained by Case VI. of Right Triangles. (See Ex. 6, § 82.) 2. Given A = 85°; B=60°; C = 50°, to find the sides. § 82. The methods which we have given for the solution of the six cases of spherical oblique triangles, as well as for the six cases of spherical right triangles, are all direct and always applicable; but frequently other methods may be used for particular cases, which would lead to simpler methods of solution. We can, however, hardly expect any thing more simple or more easy to be retained by the memory, in the case of right triangles, than the Rules of Napier; but in oblique triangles we can frequently obtain simpler solutions than by the methods already given. We may obviously divide any oblique triangle into two right triangles, and thus make all the cases of oblique triangles depend, for their solution, upon the principles of Napier's Rules. As an example of this method, let us suppose in the triangle ABC, that we have the arcs AB, AC, and the angle A given, which corresponds with Case I. If we draw the perpendicular arc CD, we shall D have, by applying Napier's Rules to the right triangles ADC, BDC, as follows: cos. bcos. AD cos. CD. Dividing (1) by (2), we have Cos. AD = COS. b That is, if from an angle of a spheric oblique triangle an arc be drawn perpendicular to the opposite side, dividing it into two segments, we shall have the cosines of these segments to each other as the cosines of their adjacent sides. Returning to the triangle ADC, we have tan. AD tan. b cos. A. = (4.) Thus we know AD, which subtracted from AB= c, gives BD. The segments are then known, and condition (3) gives at once Cos. a. It is obvious that, by the aid of the great variety of formulas which we have given, the method of solving many of the cases may be varied to almost any extent we please. We will leave this kind of exercise wholly to the student, remarking that in our general solutions we have given those methods which, under all circumstances, seemed to be the safest and best. $83. EXAMPLES FOR PRACTICE. 1. Given, in a right spherical triangle, A = 91° 11'; B=111° 11', to find the remaining parts. 2. Given, in a right spheric triangle, a=35° 44′; A=37° 28', 3. Given, in a spheric oblique triangle, a=138°; A= 95°; C=104°, to find the other parts. 4. Given a = 81° 17'; b=114° 3′; c= 59° 12', to find the angles. A = 62° 39' 43". = Ans. B 124° 50′ 50′′. C= 50° 31' 43". 5. Given, in a right spheric triangle, a=118° 54'; B=12° 19′, 6. Given, in a right spheric triangle, A = 110°; B = 100°, to find the other part. (This is the same as first example under Case VI. of Oblique Spherical Trigonometry.) a = 110° 19' 20". Ans.b 100° 38′ 56′′. c = 86° 19' 14". |