Problem:
In triangle ABC,3sinA+4cosB=6 and 4sinB+3cosA=1. Then ∠C in degrees is
Answer Choices:
A. 30
B. 60
C. 90
D. 120
E. 150
Solution:
Square both sides of the equations and add the results to obtain 9(sin2A+cos2A)+16(sin2B+cos2B)+24(sinAcosB+sinBcosA)=37.
Hence, 24sin(A+B)=12. Thus sinC=sin(180∘−A−B)=sin(A+B)=21, so ∠C=30∘ or ∠C=150∘. The latter is impossible because it would imply that A<30∘ and consequently that 3sinA+4cosB<3⋅21+4<6, a contradiction. Therefore ∠C=30∘.
Challenge. Prove that there is a unique such triangle (up to similarity), the one for which cosA=375−123 and cosB=7466−33.