In a triangle $ABC$, $3\sin A + 4\cos B = 6$ and $4\sin B + 3\cos A = 1$ hold. Then the angle $C$ equals (A) $30^\circ$ (B) $60^\circ$ (C) $120^\circ$ (D) $150^\circ$.
In a triangle $ABC$, $3\sin A + 4\cos B = 6$ and $4\sin B + 3\cos A = 1$ hold. Then the angle $C$ equals\\
(A) $30^\circ$\\
(B) $60^\circ$\\
(C) $120^\circ$\\
(D) $150^\circ$.