+585 Let A, B, and C be the angles of a triangle. Given that \tan \frac{A}{2} = \frac{1}{4} and \tan \frac{B}{2} = \frac{1}{3}, find \tan \frac{C}{2}.
+130060 tan A = 2 sin (A/2)cos (A/2) = 2 * 1/sqrt 17 * 4/sqrt 17 = 8/17 = 24 / 51
tan B = 2 sin (B/2) cos (B/2) = 2 * 1/sqrt 10 * 3/sqrt 10 = 6/10 = 3/5 = 24/ 40
AB = ( 51 + 40) = 91
AC = sqrt [ 24^2 + 51^2] = 3sqrt (353)
sin B = 24/ [8 sqrt (34)] = 3 /sqrt 34
Law of Sines
sin C / AB = sin B / AC
sin C / 91 = [3/sqrt 34] / [ 3sqrt (353)]
sin C = 91 / sqrt [ 34* 353] = 91 / sqrt [12002]
cos C = sqrt [ 12002 - 91^2 ] / sqrt [ 12002[ = 61/ sqrt [12002]
tan C = 91/61
arctan (91/61) = C = 56.16° or 180 - 56.16 = 123.84°
tan (C/2) = tan (123.84 / 2) ≈ 1.8744
