In a Triangle ABC, 3sin A + 4 cos B = 6 and 3 cos A + 4sin B = 1, then $\angle$C can be

Given,
3sinA + 4cosB =6 --------(i)
3cosA + 4sinB = 1 --------(ii)
On squaring and adding Eqs. (i) and (ii), we get
$\begin{array}{l}9+16+24\sin (A+B)=37\\ 24\sin (A+B)=12\\ \sin (A+B)=\frac{1}{2}\\ \Rightarrow \sin C=\frac{1}{2}\\ C={30}^{\circ }\text{ or }{150}^{\circ }\end{array}$
If C = 150°, then even of B = 0 and A = 30°. The quantity 3 sinA + 4cosB
$3\cdot \frac{1}{2}+4=5\frac{1}{2}<6$
Hence, C = 150° is not possible
$\Rightarrow \mathrm{\angle }\mathrm{C}={30}^{\circ }\text{ only }$