In the diagram, points $R$ and $S$ lie on $P T$ and $P Q$, respectively. If $\angle P Q R=90^{\circ}, \angle Q R T=158^{\circ}$, and $\angle P R S=\angle Q R S$, what is the measure of $\angle Q S R$ ?
(A) $34^{\circ}$
(B) $22^{\circ}$
(C) $68^{\circ}$
(D) $11^{\circ}$
(E) $79^{\circ}$

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-032.jpg?height=233&width=504&top_left_y=1374&top_left_x=1228)
Since $\angle Q R T=158^{\circ}$, then $\angle Q R P=180^{\circ}-\angle Q R T=180^{\circ}-158^{\circ}=22^{\circ}$.

Since $\angle P R S=\angle Q R S$ and $\angle Q R P=\angle P R S+\angle Q R S$, then $\angle Q R S=\frac{1}{2} \angle Q R P=\frac{1}{2}\left(22^{\circ}\right)=11^{\circ}$. Since $\triangle Q S R$ is right-angled at $Q$, then $\angle Q S R=180^{\circ}-90^{\circ}-\angle Q R S=90^{\circ}-11^{\circ}=79^{\circ}$.

ANSWER: (E)