In the diagram, $\triangle P Q R$ has $\angle P Q R=120^{\circ}$. Also, $\angle Q P S=\angle R P S$ and $\angle Q R S=\angle P R S$. (In other words, $S P$ and $S R$ bisect $\angle Q P R$ and $\angle Q R P$, respectively.) What is the measure of $\angle P S R$ ?

![](https://cdn.mathpix.com/cropped/2024_04_20_6ed09463f225f8ba1f07g-034.jpg?height=317&width=805&top_left_y=595&top_left_x=622)
(A) $130^{\circ}$
(B) $120^{\circ}$
(C) $140^{\circ}$
(D) $160^{\circ}$
(E) $150^{\circ}$
Consider $\triangle P Q R$.

Since the sum of the angles in a triangle is $180^{\circ}$, then

$$
\angle Q P R+\angle Q R P=180^{\circ}-\angle P Q R=180^{\circ}-120^{\circ}=60^{\circ}
$$

Since $\angle Q P S=\angle R P S$, then $\angle R P S=\frac{1}{2} \angle Q P R$.

Since $\angle Q R S=\angle P R S$, then $\angle P R S=\frac{1}{2} \angle Q R P$.

Therefore,

$$
\begin{aligned}
\angle R P S+\angle P R S & =\frac{1}{2} \angle Q P R+\frac{1}{2} \angle Q R P \\
& =\frac{1}{2}(\angle Q P R+\angle Q R P) \\
& =\frac{1}{2} \times 60^{\circ} \\
& =30^{\circ}
\end{aligned}
$$

Finally, $\angle P S R=180^{\circ}-(\angle R P S+\angle P R S)=180^{\circ}-30^{\circ}=150^{\circ}$.