2012/016) In a triangle ABC , 3sin A + 4 cos B = 6 and 3 cos A + 4 sin B = 1 , then find angle C

square both and add

(9 sin^2 A + 16 cos^2 B + 24 sin A cos B) + (9 cos^2 A + 16 cos^2 B + 24 cos A sin B) = 37

or 9+ 16 + 24( sin A cos B + cos A sin B) = 37

so sin (A+B) = 1/2

sin C = sin (A+B) as A+B+C = 180

so sin C = 1/2

C = 30 or 150

cos A < =1/3 so A > 60 degrees and C cannot be 120 so C = 30