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

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