Ans: $220^\circ$
Note that $\angle PRQ=90^\circ$, then we have
Note that $\angle PRQ=90^\circ$, then we have
$\begin{array}{rcl}
\cos \angle QPR & = & \dfrac{9}{14} \\
\angle QPR & = & 49.994~799~12^\circ
\end{array}$
Therefore, the bearing of $Q$ from $P$
$\begin{array}{cl}
= & 270^\circ – \angle QPR \\
= & 270^\circ – 49.994~799~12^\circ \\
= & 220^\circ
\end{array}$
