In the diagram, $\triangle P Q S$ is equilateral. Also, $\triangle P Q R$ and $\triangle P S R$ are isosceles with $P Q=P R=P S$. If $\angle R P Q=\angle R P S$, the measure of $\angle Q R S$ is
(A) $30^{\circ}$
(B) $60^{\circ}$
(C) $15^{\circ}$
(D) $20^{\circ}$
(E) $45^{\circ}$

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-033.jpg?height=518&width=307&top_left_y=1877&top_left_x=1386)
Since $\triangle P Q S$ is equilateral, then $\angle Q P S=60^{\circ}$.

Since $\angle R P Q, \angle R P S$ and $\angle Q P S$ completely surround point $P$, then the sum of their measures is $360^{\circ}$.

Since $\angle R P Q=\angle R P S$, this means that $2 \angle R P Q+\angle Q P S=360^{\circ}$ or $2 \angle R P Q=360^{\circ}-60^{\circ}$, which means that $\angle R P Q=\angle R P S=150^{\circ}$.

Since $P R=P Q$, then in isosceles $\triangle P Q R$, we have

$$
\angle P R Q=\angle P Q R=\frac{1}{2}\left(180^{\circ}-\angle R P Q\right)=\frac{1}{2}\left(180^{\circ}-150^{\circ}\right)=15^{\circ}
$$

Similarly, $\angle P R S=\angle P S R=15^{\circ}$.

This means that $\angle Q R S=\angle P R Q+\angle P R S=15^{\circ}+15^{\circ}=30^{\circ}$.