In the diagram, $R S P$ is a straight line and $\angle Q S P=80^{\circ}$. The measure of $\angle P Q R$ is
(A) $120^{\circ}$
(B) $90^{\circ}$
(C) $80^{\circ}$
(D) $60^{\circ}$
(E) $75^{\circ}$

![](https://cdn.mathpix.com/cropped/2024_04_20_6027bc27089ed4fc493cg-064.jpg?height=271&width=467&top_left_y=1106&top_left_x=1268)
Since $R S P$ is a straight line, then $\angle R S Q+\angle Q S P=180^{\circ}$, so $\angle R S Q=180^{\circ}-80^{\circ}=100^{\circ}$. Since $\triangle R S Q$ is isosceles with $R S=S Q$, then

$$
\angle R Q S=\frac{1}{2}\left(180^{\circ}-\angle R S Q\right)=\frac{1}{2}\left(180^{\circ}-100^{\circ}\right)=40^{\circ}
$$

Similarly, since $\triangle P S Q$ is isosceles with $P S=S Q$, then

$$
\angle P Q S=\frac{1}{2}\left(180^{\circ}-\angle P S Q\right)=\frac{1}{2}\left(180^{\circ}-80^{\circ}\right)=50^{\circ}
$$

Therefore, $\angle P Q R=\angle P Q S+\angle R Q S=50^{\circ}+40^{\circ}=90^{\circ}$.

ANSWER: (B)