Question 44
Triangle $ABC$ obeys $AB = 2AC$ and $\angle BAC = 120^\circ$. Points $P$ and $Q$ lie on segment $BC$ such that
$$
\begin{aligned}
AB^2 + BC \cdot CP &= BC^2 \\
3AC^2 + 2BC \cdot CQ &= BC^2
\end{aligned}
$$
Find $\angle PAQ$ in degrees.
$$
\begin{aligned}
AB^2 + BC \cdot CP &= BC^2 \\
3AC^2 + 2BC \cdot CQ &= BC^2
\end{aligned}
$$
Find $\angle PAQ$ in degrees.