jmc

Join $PQ$, $PR$, $PS$, $RQ$, and $RS$. Since the circles with center $Q$, $R$ and $S$ are all tangent to $BC$, then $QR$ and $RS$ are each parallel to $BC$ (as the centres $Q$, $R$ and $S$ are each 1 unit above $BC$). This tells us that $QS$ passes through $R$. When the centers of tangent circles are joined, the line segments formed pass through the associated point of tangency, and so have lengths equal to the sum of the radii of those circles. Therefore, $QR=RS=PR=PS=1+1=2$.



Since $PR=PS=RS$, we know $△PRS$ is equilateral, so $\angle PSR=\angle PRS=60^{\circ }$. Since $\angle PRS=60^{\circ }$ and $QRS$ is a straight line, we have $\angle QRP=180^{\circ }-60^{\circ }=120^{\circ }$. Since $QR=RP$, we know $△QRP$ is isosceles, so $$\angle PQR=\frac{1}{2}(180^{\circ }-120^{\circ })=30^{\circ }.$$Since $\angle PQS=30^{\circ }$ and $\angle PSQ=60^{\circ }$, we have $\angle QPS=180^{\circ }-30^{\circ }-60^{\circ }=90^{\circ }$, so $△PQS$ is a $30^{\circ }$-$60^{\circ }$-$90^{\circ }$ triangle. Thus, the answer is $\boxed{30^{\circ }}$.