In the diagram, what is the measure of $\angle P Q R$ ?
(A) $45^{\circ}$
(B) $30^{\circ}$
(C) $60^{\circ}$
(D) $75^{\circ}$
(E) $15^{\circ}$

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-092.jpg?height=417&width=333&top_left_y=1407&top_left_x=1297)
Since $R Q=R S$, then $\angle R S Q=\angle R Q S$.

In $\triangle Q R S$, we have $\angle R Q S+\angle Q R S+\angle R S Q=180^{\circ}$, so $2(\angle R Q S)+60^{\circ}=180^{\circ}$.

Thus, $\angle R Q S=\frac{1}{2}\left(180^{\circ}-60^{\circ}\right)=60^{\circ}$.

Since $P Q=P S$, then $\angle P S Q=\angle P Q S$.

In $\triangle Q P S$, we have $\angle P Q S+\angle Q P S+\angle P S Q=180^{\circ}$, so $2(\angle P Q S)+30^{\circ}=180^{\circ}$.

Thus, $\angle P Q S=\frac{1}{2}\left(180^{\circ}-30^{\circ}\right)=75^{\circ}$.

Therefore, $\angle P Q R=\angle P Q S-\angle R Q S=75^{\circ}-60^{\circ}=15^{\circ}$.

ANSWER: (E)

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