HMMT 2008 U30

Triangle $A B C$ obeys $A B=2 A C$ and $\angle B A C=120^{\circ}$. Points $P$ and $Q$ lie on segment $B C$ such that

$$\begin{aligned} A B^{2}+B C \cdot C P & =B C^{2} \\ 3 A C^{2}+2 B C \cdot C Q & =B C^{2} \end{aligned}$$

Find $\angle P A Q$ in degrees.