In the diagram, pentagon $T P S R Q$ is constructed from equilateral $\triangle P T Q$ and square $P Q R S$. The measure of $\angle S T R$ is equal to
(A) $10^{\circ}$
(B) $15^{\circ}$
(C) $20^{\circ}$
(D) $30^{\circ}$
(E) $45^{\circ}$

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-015.jpg?height=371&width=260&top_left_y=411&top_left_x=1301)

## Part B: Each correct answer is worth 6.
Join $S$ to $T$ and $R$ to $T$.

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-028.jpg?height=374&width=252&top_left_y=716&top_left_x=991)

Since $P Q R S$ is a square, $\angle S P Q=90^{\circ}$.

Since $\triangle P T Q$ is equilateral, $\angle T P Q=60^{\circ}$.

Therefore, $\angle S P T=\angle S P Q+\angle T P Q=90^{\circ}+60^{\circ}$.

Since $P Q R S$ is a square, $S P=P Q$.

Since $\triangle P T Q$ is equilateral, $T P=P Q$.

Since $S P=P Q$ and $T P=P Q$, then $S P=T P$ which means that $\triangle S P T$ is isosceles.

Thus, $\angle P T S=\frac{1}{2}\left(180^{\circ}-\angle S P T\right)=\frac{1}{2}\left(180^{\circ}-150^{\circ}\right)=15^{\circ}$.

Using a similar argument, we can show that $\angle Q T R=15^{\circ}$.

This means that $\angle S T R=\angle P T Q-\angle P T S-\angle Q T R=60^{\circ}-15^{\circ}-15^{\circ}=30^{\circ}$.

ANSWER: (D)