To determine when there are two triangles, and also when there is but one, let us consider the second of equations (8.) R2 cos B=sin A sin C cos b-R cos A cos C, which gives R2cos B+R cos A cos C sin A sin C cos b Now, if cos B be greater than cos A we shall have and hence the sign of the second member of the equation will depend on that of cos B, and consequently cos b and cos B will have the same algebraic sign, or b and B will be of the same species. But when cos B>cos A the sin B<sin A: hence If the sine of the angle opposite the required side be less than the sine of the other given angle, there will be but one solution. If, however, sin B>sin A, the cos B will be less than cos A, and it is plain that such a value may then be given to cos C, as to render R2 cos BR cos A cos C, or the sign of the second member of the equation may be made to depend on cos C. We can therefore give such values to C as to satisfy the two equations +cos b -cos b: R2 cos B+R cos A cos C 9 sin A sin C R2 cos B+R cos A cos C and Hence, if the sine of the angle opposite the required side be greater than the sine of the other given angle there will be two solutions. Let us first suppose the side b to be less than 90°, or equa. to 79° 12′ 10′′. If now, we let fall from the angle C a perpendicular on the base BA, the triangle will be divided into two right angled triangles, in each of which there will be two parts known besides the right angle. Calculating the parts by Napier's rules we find, If we take the side b=100° 47′ 50′′, we shall find C=130° 54' 28" c=119° 03′ 26′′. C=156° 15' 06" c=152° 14′ 18′′. Ex. 2. In a spherical triangle ABC there are given A=103° 59′ 57′′, B=46° 18′ 7′′, and a=42° 8′ 48′′; required the remaining parts. There will but one triangle, since sin B<sin A. b=30° C=36° 7' 54" c=24° 3′ 56′′. CASE III. Ans. Having given the three sides of a spherical triangle to find the angles. For this case we use equations (3.). cos A=R√sins sin (†s—a) sin b sin c Ex. 1. In an oblique angled spherical triangle there are given a 56° 40', b=83° 13′ and c-114° 30'; required the angles. = =127° 11′ 30′′ (a+b+c)=s Log sin s 127° 11' 30" log sin (s-a) 70° 31′ 30′′ -log sin b 83° 13′ -log sin c 114° 30' Sum Half sum =log cos A 24° 15', 39′′ Hence, angle A=48° 31′ 18′′. The addition of twice the logarithm of radius, or 20, to the numerator of the quantity under the radical just cancels the 20 which is to be subtracted on account of the arithmetical complements, to that the 20, in both cases, may be omitted. Applying the same formulas to the angles B and C, we find, ar.-comp. ar.-comp. 9.901250 9.974413 0.003051 0.040977 19.919691 9.959845 Ex. 2. In a spherical triangle there are given a 40° 18′ 29′′, b-67° 14′ 28′′, and c=89° 47' 6": required the three angles. Ans. A= 34° 22′ 16′′ CASE IV. Having given the three angles of a spherical triangle, to find the three sides. For this case we employ equations (7.) cos a RV = cos(S-B)cos(S-C) sin B sin C (S-A) (S-C) Ex. 1. In a spherical triangle ABC there are given A=48° 30', B=125° 20', and C=62° 54'; required the sides. (A+B+C)=S= 118° 22′ = 69° 52' =― 6° 58' 55° 28' Log cos (S-B) —6° 58′ 125° 20' с 62° 54' Sum Half sum=log cos A=28° 19' 48" Hence, ar.-comp. ar.-comp. side a 56° 39′ 36′′. CASE V. Ans. 9.996782 9.753495 0.088415 0.050506 19.889198 9.944599 Ex. 2. In a spherical triangle ABC, there are given A=109° 55′ 42′′, B=116° 38′ 33′′, and C=120° 43′ 37′′; required the three sides. b=114° 29' 58" c= 83° 12′ 06′′. a= 98° 21' 40" b=109° 50′ 22′′ c=115° 13′ 26′′ Having given in a spherical triangle, two sides and their included angle, to find the remaining parts. For this case we employ the two first of Napier's Analogies. cos (a+b): cos (a-b): cot C: tang (A+B) sin (a+b) sin (a-b) :: cot C: tang (A-B). : Having found the half sum and the half difference of the angles A and B, the angles themselves become known; for, the greater angle is equal to the half sum plus the half difference, and the lesser is equal to the half sum minus the half diffe. rence. The greater angle is then to be placed opposite the greater side. The remaining side of the triangle can then be found by Case II. Ex. 1. In a spherical triangle ABC, there are given a=68° 46' 2", b=37° 10', and C-39° 23'; to find the remaining parts. (a+b)=52° 58′ 1′′, (a—b)=15° 48′ 1′′, C=19° 41′ 30′′. (a+b) 52° 58' 1" log. ar.-comp. 0.220210 (a-b) 15° 48′ 1′′ C 19° 41' 30" 9.983271 10.446254 (A+B) 77° 22′ 25′′ 10.649735 As cos To tang As sin (a+b) 52° 58′ 1" log. ar.-comp. 0.097840 9.435016 10.446254 Totang (A—B) 43° 37′ 21′′ 9.979110 Hence, A=77° 22′ 25′′ +43° 37′ 21′′-120° 59′ 46′′ Ex. 2. In a spherical triangle ABC, there are given b=83° 19′ 42′′, c=23° 27′ 46′′, the contained angle A=20° 39′ 48′′; to find the remaining parts. Ans. CASE VI. = B=156° 30' 16" a = 61° 32′ 12′′. In a spherical triangle, having given two angles and the included side to find the remaining parts. For this case we employ the second of Napier's Analogies. cos (A+B): cos (A—B) : : tang c: tang (a+b) sin (A+B) : sin (A—B) : : tang c: tang (a—b). From which a and b are found as in the last case. The remaining angle can then be found by Case I. Ex. 1. In a spherical triangle ABC, there are given A=81° 38′ 20′′, B=70° 9′ 38′′, c=59° 16′ 23′′; to find the remaining parts. (A+B)=75° 53′ 59′′, }(A—B)=5° 44′ 21′′, c=29° 38′ 11′′. 9.997818 c 29° 38′ 11′′ (a+b) 66° 42' 52′′ 10.366156 As cos To cos To tang As sin To sin So is tang To tang Hence (A+B) 75° 53′ 59′′ log. ar.-comp. 0.013286 (A-B) 5° 14′ 21′′ 9.000000 9.755051 29° 38′ 11′′ »c 8.768337 a=66° 42′ 52′′+3° 21′ 25′′ 70° 04′ 17′′ b=66° 42′ 52"-3° 21′ 25′′-63° 21′ 27′′ =64° 46′ 33′′. Ex. 2. In a spherical triangle ABC, there are given A=34° 15′ 3′′, B=42° 15′ 13′′, and c=' -76° 35′ 36′′; to find the remaining parts. |