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 $24 \sin (A+B)=12$, hence $\sin C=\sin (A+B)=\frac{1}{2}, C=\frac{\pi}{6}$ or $\frac{5 \pi}{6}$.
Since $4 \sin B=1-3 \cos A>0$, it follows that $\cos A<\frac{1}{3}$, and thus $A>\frac{\pi}{6}$.
Clearly, $\angle C \neq \frac{5 \pi}{6}$, so $\angle C=\frac{\pi}{6}$.