20. In $\triangle A B C$, $3 \sin A+4 \cos B=6, 4 \sin B+3 \cos A=1$, find the degree measure of angle $C$.

In $\triangle A B C$, $3 \sin A+4 \cos B=6, 4 \sin B+3 \cos A=1$, find the degree measure of angle $C$.
$$
\begin{array}{c}
20.6^{2}+1^{2}=(3 \sin A+4 \cos B)^{2}+(4 \sin B+3 \cos A)^{2}=3^{2}\left(\sin ^{2} A+\cos ^{2} A\right)+4^{2}\left(\cos ^{2} B+\sin ^{2} B\right)+ \\
24(\sin A \cos B+\cos A \sin B)=9+16+24 \sin (A+B)=9+16+24 \sin (\pi-C)=25+24 \sin C
\end{array}
$$

So $\sin C=\frac{1}{2}$, and $0<C<\pi$, hence $C=\frac{\pi}{6}$ or $\frac{5}{6} \pi$.
If $C=\frac{5}{6} \pi$, then $A<\frac{\pi}{6}$, at this time, $3 \sin A+4 \cos B<3 \times \frac{1}{2}+4<6$ which does not meet the requirement.
Therefore, $C=\frac{\pi}{6}$ is the solution.