=sin(b+c) cosa cos (b+c) sina [cos (BC) cos (B+C)] (Art. 198) cot r= 1 [cos S+cos(S-A)+cos (S-B)+cos (S-C)] (5) 2N 216. The Ex-Circles. To find the angular radii of the ex-circles of a triangle. A circle which touches one side of a triangle and the other two sides produced, is called an escribed circle, or ex-circle, of the triangle. It is clear that the three ex-circles of any triangle are the in-circles of its colunar triangles (Art. 191, Sch.). Since the circle escribed to the side a of the triangle ABC is the in-circle of the colunar triangle A'BC, the parts of which are α, π-b, π— C, A, B, C, the problem becomes identical with that A of Art. 215; and we obtain the value for the in-radius of B A the colunar triangle A'BC, by substituting for b, c, B, C, their supplements in the five equations of that article. Hence, denoting the radius by ra, we get 2 cos A sin B sin C 1 cotra= [-cos S―cos(S—A)+cos(S—B)+cos(S−C)] (5) 2N These formulæ may also be found independently by methods similar to those employed in Art. 215, for the in-circle, as the student may show. Sch. Similarly, another triangle may be formed by producing BC, BA to meet again, and another by producing CA, CB to meet again. The colunar triangles on the sides b and c have each two parts, b and B, c and C, equal to parts of the primitive triangle, while their remaining parts are the supplements in the former case of a, c, A, C, and in the latter, of a, b, A, B. The values for the radii r, and re are therefore found in the same way as the above values for ra; or they may be obtained from the values of r by advancing the letters. a The small circle passing through the vertices of a spherical triangle is called the circumscribing circle, or circumcircle, of the triangle. Let ABC be the triangle; bisect the sides CB, CA at D, E, and let O be the intersection of perpendiculars to CB, CA, at D, E; then O is the circumcentre. For, join OA, OB, OC; then (Art. 186) cos OB = cos BD cos OD, or = Now the angle ОАВ = OC. Similarly, OC=OA. OBA, OBC = OCB, OCA = OAC. ... OCB+A=}(A+B+C) = S. .. OCBS — A. Let OCR; then, in the triangle ODC, we have cos OCD = tan CD cot CO = tan a cot R. (Art. 186) Also cos (SA)= cos [(B+ C) − A] = cos(B+ C) cos 1⁄2 A + sin (B + C) sin ¦ A which may be reduced to the following: [sin(s− a)+sin(s — b) + sin(s — c) — sin s] (5) 218. Circumcircles of Colunar Triangles. To find the angular radii of the circumcircles of the three colunar triangles. Let R1, R2, R3 be the angular radii of the circumcircles of the colunar triangles on the sides a, b, c, respectively. Then, since R1 is the circumradius of the triangle A'BC whose parts are α, π — b, π - c, A, – В, π- С, we have, from Art. 217, π Π tan R1 = = Similarly, cos (SA) sina sin A sinb sinc 2 sina cosb cos c 1 2n n (1) (2) (3) (4) · [sins—sin(s—a)+sin(s—b)+sin(s—c)](5) 1. cos s + cos(s − a) + cos ( s − b) + cos (s — c) 2. cos (sb) + cos (s — c) — cos (s — a) — cos s = 4 cosa sinb sin c. = N 2 cos B sin C sin A N 2 cos C sinA sin B sins: sin (sa): sin (sb): sin(sc). 10. tan R1: tan R2: tan R2= cos(S-A): cos (S-B): cos(S-C). 1 11. cot R cot R1 cot R2 cot Rg = N2. 12. tan R cot R1 cot R2 cot R2 = cos2 S. AREAS OF TRIANGLES. 219. Problem. To find the area of a spherical triangle, having given the three angles. Let the radius of the sphere. E the spherical excess A+B+C-180°. K = = area of triangle ABC. = It is shown in Geometry (Art. 738) that the absolute area of a spherical triangle is to that of the surface of the sphere as its spherical excess, in degrees, is to 720°. |