2. In $\triangle A B C$, if $A C=\sqrt{2}, A B=2$, and $\frac{\sqrt{3} \sin A+\cos A}{\sqrt{3} \cos A-\sin A}=\tan \frac{5 \pi}{12}$, then $B C=$ $\qquad$
$$
\begin{array}{l}
\frac{\sqrt{3} \sin A+\cos A}{\sqrt{3} \cos A-\sin A} \\
\frac{2 \sin \left(A+\frac{\pi}{6}\right)}{2 \sin \left(A+\frac{\pi}{6}\right)}=\tan \left(A+\frac{\pi}{6}\right) \Rightarrow A=\frac{\pi}{4} .
\end{array}
$$

So $B C^{2}=2+4-2 \cdot \sqrt{2} \cdot 2 \cdot \frac{\sqrt{2}}{2}=2 \Rightarrow B C=\sqrt{2}$.