Solve for the angle $x$ in the right triangle without trigonometry.

Let $QM$ be an altitude of $\Delta PQB$.

Thus, $$MQ=QC=\frac{1}{2}PQ,$$ $$\measuredangle MPQ=30^{\circ},$$ $$\measuredangle PAQ=15^{\circ}.$$ Can you end it now?

Alternatively, from the Sine theorem: $$\frac{PQ}{\sin x}=\frac{BQ}{\sin \angle BPQ} \Rightarrow\\ \sin \angle BPQ=\frac{BQ\sin x}{PQ}=\frac{CQ}{PQ}=\frac12 \Rightarrow \angle BPQ=30^\circ \Rightarrow \angle PAQ=15^\circ \Rightarrow \\ 2x=90-\angle PAQ=75^\circ \Rightarrow x=37.5^\circ.$$