In the diagram, $R$ is the point of intersection of $P T$ and $Q S, P Q=P R$, and $R S=R T$. If $\angle P Q R=2 x^{\circ}$, then the measure of $\angle R S T$, in degrees, is
(A) $45-x$
(D) $45+2 x$
(B) $90+\frac{1}{2} x$
(C) $90-\frac{1}{2} x$
(E) $90-x$

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-082.jpg?height=366&width=504&top_left_y=733&top_left_x=1209)
Since $\triangle P Q R$ is isosceles, then $\angle P R Q=\angle P Q R=2 x^{\circ}$.

Since $\angle P R Q$ and $\angle S R T$ are opposite angles, then $\angle S R T=\angle P R Q=2 x^{\circ}$.

Since $\triangle R S T$ is isosceles with $R S=R T$, then

$$
\angle R S T=\frac{1}{2}\left(180^{\circ}-\angle S R T\right)=\frac{1}{2}\left(180^{\circ}-2 x^{\circ}\right)=90^{\circ}-x^{\circ}=(90-x)^{\circ}
$$

ANSWER: (E)