2. In $\triangle A B C$, $\angle A$ and $\angle B$ satisfy $3 \sin A+4 \cos B=6, 4 \sin B+3 \cos A=1$. Then, the size of $\angle C$ is $\qquad$
2. $\frac{\pi}{6}$.

From the given, we have
$$
\begin{array}{l}
9 \sin ^{2} A+16 \cos ^{2} B+24 \sin A \cdot \cos B=36, \\
16 \sin ^{2} B+9 \cos ^{2} A+24 \sin B \cdot \cos A=1 .
\end{array}
$$

Adding the two equations, we get
$$
25+24(\sin A \cdot \cos B+\sin B \cdot \cos A)=37,
$$

which simplifies to $24 \sin (A+B)=12$,
$$
\begin{array}{l}
\therefore \sin C=\sin (A+B)=\frac{1}{2}, \\
\therefore \angle C=\frac{\pi}{6} \text { or } \frac{5 \pi}{6} . \\
\because 4 \sin B=1-3 \cos A>0, \\
\therefore \cos A<\frac{1}{3}, \text{ and since } \cos A>\cos \frac{\pi}{6}, \text{ it is clear that } \angle C \neq \frac{5 \pi}{6}.
\end{array}
$$
Thus, $\angle C=\frac{\pi}{6}$.