, 2019

Join $QU$ and $SU$.

Since $\triangle PUT$ is equilateral, then $PU=UT=TP$.

Since pentagon $PQRST$ is regular, then $QP = PT = TS$.

Thus, $PU = QP$ and $UT = TS$, which means that $\triangle QPU$ and $\triangle STU$ are isosceles.

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Each interior angle in a regular pentagon measures $108^\circ$.

Since $\angle UPT = 60^\circ$, then $\angle QPU = \angle QPT - \angle UPT = 108^\circ - 60^\circ = 48^\circ$.

Since $\triangle QPU$ is isosceles with $QP = PU$, then $\angle PQU = \angle PUQ$.

Thus, $\angle PUQ = \frac{1}{2}(180^\circ - \angle QPU) = \frac{1}{2}(180^\circ - 48^\circ) = 66^\circ$.

By symmetry, $\angle TUS = 66^\circ$

Finally, $\angle QUS = 360^\circ - \angle PUQ - \angle PUT - \angle TUS = 360^\circ - 66^\circ - 60^\circ - 66^\circ = 168^\circ$