3. If p of solutions. b indeterminate, B indeterminate, Infinite number of solutions. By a comparison of these results, we find, 1. If p differs more from 90' than P, there will be two solutions. 2. If p = P, and P < 90° or P> 90', there will be one solution. P 90', there will be an infinite number = 4. If p differs less from 90° than P, there will be no solution. 130. Remarks. 1. Napier's principles render it unnecessary to divide the subject of right-angled spherical triangles into cases. 2. Two parts will be given, and three required. 3. These parts or their complements will be circular parts. 4. Take the two given parts, if they are circular parts, otherwise their complements, and any one part required, if it is a circular part, otherwise its complement, and observe which is the middle part, and whether the other parts are adjacent to, or opposite, the middle part: if adjacent, the first of Napier's principles will give the formula; if opposite, the second. 5. Introduce R and apply logarithms. 6. Apply the principles which determine the species of the required part. cos b = b. B. R cos h cos p 10+ log cos h 90-P 1. To find b. From the second of Napier's principles, we have = cos b cos p, or cos h sin (90' — h): Finding cos b and introducing R, we have log cos b log cos h (110° 30′) 9.54433 log cos p ( 50° 45′) = 9.80120 + log cos b 9.74313 cos b cos h 90°- B B log cos p. cos b cos p. Since the hypotenuse is greater than 90°, the sides b and p are of different species; but p < 90°; .. b > 90°. But log cos b corresponds to 56° 23′ 29′′, and to its supplement 123° 36′ 31" which must be taken, since b> 90°. b=123° 36′ 31′′. The species of b can also be determined by the formula, Since h 90°, cos h is negative, and since p< 90°, cos p is positive; .. cos b is negative; .. b > 90°. The signs of the functions may be conveniently indicated by placing the signs after their logarithms. sin (90° sin ... log cos B log tan p ( 50° 45') log cot h (110° 30′) log cos B ... B=117° 14'. Since b and B are of the same species, and since b > 90', B > 90'. The species of B can also be determined from the sign of cos B. { h h p 3. To find P. p= = cos (90° — h) cos (90°—P), or sin p=sin h sin P. .*. sin P sin hi ... log sin P=10+ log sin plog sin h. log sin p ( 50° 45′) = 9.88896 + log sin h (110' 30') 9.97159 + log sin P 9.91737 + ... P=55° 45′ 57′′. B = S. N. 11. 2. To find B. P is of the same species as p, and since p< 90°, P < 90°. The species of P can not be determined by the sign of sin P, since the sign of sin P is plus from 0° to 180°. = tan p tan (90' — h), tan p cot h Ꭱ log tan p + log cot h—10. 10.08776 + 9.66050 = 94° 05'. 100° 45'. R sin p Req. 67° 33′ 27′′. 67° 54' 47". 99° 57′ 35′′. given triangle as poles. If one triangle is the polar of another, the second is the polar of the first. 70° 23′ 41′′. 48° 24′ 18′′. p= 59° 38' 27'. = 54° 01' 15". 142° 09' 12". 155° 27′ 55′′. 18° 07' 02" or 161° 52′ 58′′. 52° 34′ 31′′ or 127° 25′ 29′′. -23° 03′ 06′′ or 156° 56′ 54′′. 75° 13′ 01′′. 67° 27' 01". 58° 25′ 45′′. 8. If a line make an angle of 40° with a fixed plane, and a plane embracing this line be perpendicular to the fixed plane, how many degrees from its first position must the plane embracing the line revolve about it in order that it may make an angle of 45° with the fixed plane? Ans. 67° 22′ 44′′ or 112° 37′ 16′′. 132. Polar Triangles. The polar triangle of a given triangle is the triangle formed by the intersection of three arcs of great circles described about the vertices of the A B Thus, if A'B'C' is the polar of the triangle ABC, then ABC is the polar of A'B'C'. Each angle in one of two polar triangles is the supplement of the side lying opposite to it in the other; and each side is the supplement of the angle lying opposite to it in the other. Thus, 180' — b', - A = 180° — a', α = 180° — A', a, B' A' 180°. 1. Given Given h': 90°. = P: b B b = H= 90°. = 180' - B', 180° — b, 180° = 129° 15'. = Cor. If a' 90°, A 90°; hence, if one side of a triangle is 90', one angle of its polar triangle is 90°. 133. Quadrantal Triangles. A quadrantal triangle is a triangle one side of which is 90°. By the corollary of the last article, it follows that the polar of a quadrantal triangle is a right-angled triangle. B, A quadrantal triangle is solved by passing to its polar triangle, which is solved as a right-angled triangle, then by passing back to the quadrantal triangle, which is the polar of the right-angled triangle. 134. Examples. 62° 46′ 01′′. C 180°c. c = 180° - C'. C'= 180° C. 50° 45'. B=117° 13′ 59′′ = H': Req. B' p' Passing to the polar triangle, which is we have h b P = 69° 30'. |