+585 In triangle ABC, \cos A=\sqrt{\frac{2}{5}} and \cos B= 1/2. Find \cos C.
+130060 cos A = 2/5 → tan A = sqrt 21 / 2 → sqrt (63) / sqrt (12)
cos B = 1/2 → tan B = sqrt 3 / 1 → sqrt (63) / sqrt (21)
AC = sqrt (63 +12) = sqrt (75)
Law of Sines
sin C / AB = sin B / AC
sin C / ( sqrt (12) +sqrt (21)) = (sqrt (3) / 2) / sqrt (75)
sin C / ( sqrt (12) + sqrt (21)) = 1 / [ 2 sqrt 25 ]
sin C / (sqrt (12) + sqrt (21)) = 1/10
sin C = [ sqrt (12) + sqrt (21) ] /10
cos C = sqrt [ 10^2 - ( sqrt 12 + sqrt 21)^2] /10 =
sqrt [ 100 - ( 12 + 21 + 2sqrt (12)sqrt (21) ] /10 =
sqrt [ 67 - 12sqrt 7] /10
