$110^{\circ}$

As $\angle QPR=40^{\circ}$, $\angle PQR+\angle PRQ=180^{\circ}-40^{\circ} =140^{\circ}$.

So $\angle SQR+\angle SRQ=140^{\circ}/2=70^{\circ}$. Therefore $\angle QSR =180^{\circ}-70^{\circ}=110^{\circ}$.

*This problem is taken from the UKMT Mathematical Challenges.*