Problem: In triangle ABC, where A and B are acute angles, 4sinA+5cosB=5 and 5sinB+4cosA=6. Find the radian measure of C.
Work: (4sinA+5cosB=5)^2+(5sinB+4cosA=6)^2 => use c^2=b^2+a^2 to get c^2 => simplify to 41+40(cosBsinA+cosAsinB)=61 (equal to c^2) => 40(sinA+B)=20 ....
Not sure what to do next to solve for C.
Work: (4sinA+5cosB=5)^2+(5sinB+4cosA=6)^2 => use c^2=b^2+a^2 to get c^2 => simplify to 41+40(cosBsinA+cosAsinB)=61 (equal to c^2) => 40(sinA+B)=20 ....
Not sure what to do next to solve for C.