jmc

The measure of an interior angle in a regular pentagon is $$\frac{180(5-2)}{5}=108^{\circ },$$ so $\angle QPT=108^{\circ }$. From isosceles triangle $PQT$, we have $\angle PQT=(180^{\circ }-\angle QPT)/2=36^{\circ }$. Similarly, $\angle RQS=36^{\circ }$. Finally, $△SQT$ is isosceles with $SQ=QT$, so median $\overline{QX}$ is also an angle bisector of $\angle SQT$. Since $\angle SQT=\angle PQR-36^{\circ }-36^{\circ }=108^{\circ }-36^{\circ }-36^{\circ }=36^{\circ }$, we have $\angle XQS=(\angle SQT)/2=\boxed{18^{\circ }}$.