jmc

We have $\tan R=\frac{PQ}{PR}$ and $\cos Q=\frac{PQ}{QR}=\frac{PQ}{15}$, so $\tan R=5\cos Q$ gives us $\frac{PQ}{PR}=5\cdot \frac{PQ}{15}=\frac{PQ}{3}$. From $\frac{PQ}{PR}=\frac{PQ}{3}$, we have $PR=3$. Finally, the Pythagorean Theorem gives us $$\begin{array}{rl}PQ& =\sqrt{Q{R}^2-P{R}^2}\\ & =\sqrt{15^2-3^2}\\ & =\sqrt{(5\cdot 3{)}^2-3^2}\\ & =\sqrt{25\cdot 3^2-3^2}\\ & =\sqrt{24\cdot 3^2}\\ & =\sqrt{6\cdot 4\cdot 3^2}\\ & =\boxed{6\sqrt{6}}.\end{array}$$