jmc

We're given that $△ABQ\sim △QCP$ and thus $m\angle B=m\angle C.$ Therefore, $△ABC$ is isosceles. From the given $m\angle BAC=70^{\circ }$, we have that $m\angle ABC=m\angle BCA=55^{\circ }$. But we also know that $△ABC\sim △PAQ$, which means that $m\angle PAQ=55^{\circ }$ as well. Subtracting, $m\angle BAQ=15^{\circ }$. Finally, from similar triangles, we have $m\angle PQC=m\angle BAQ=\boxed{15^{\circ }}$.