In the diagram, $\triangle P Q R$ is right-angled. Point $S$ lies on $P R$ so that $\triangle Q R S$ is equilateral and $\triangle P Q S$ is isosceles with $P S=Q S$. The measure of $\angle Q P R$ is
(A) $35^{\circ}$
(B) $37.5^{\circ}$
(C) $25^{\circ}$
(D) $32.5^{\circ}$
(E) $30^{\circ}$

![](https://cdn.mathpix.com/cropped/2024_04_20_6027bc27089ed4fc493cg-032.jpg?height=346&width=228&top_left_y=323&top_left_x=1339)
In equilateral triangle $Q R S$, each angle is equal in measure and so $\angle S Q R=60^{\circ}$.

Since $\angle P Q R=90^{\circ}$, then $\angle P Q S=\angle P Q R-\angle S Q R=90^{\circ}-60^{\circ}=30^{\circ}$.

In isosceles triangle $P Q S, \angle Q P S=\angle P Q S=30^{\circ}$.

Therefore, $\angle Q P R=\angle Q P S=30^{\circ}$.

ANSWER: (E)