In a regular pentagon, the measure of each interior angle is $108^{\circ}$. If $P Q R S T$ is a regular pentagon, then the measure of $\angle P R S$ is
(A) $72^{\circ}$
(B) $54^{\circ}$
(C) $60^{\circ}$
(D) $45^{\circ}$
(E) $80^{\circ}$

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-027.jpg?height=414&width=422&top_left_y=796&top_left_x=1296)
Join $P$ to $R$.

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-054.jpg?height=431&width=442&top_left_y=1156&top_left_x=901)

Since $P Q R S T$ is a regular pentagon, then $\angle P Q R=\angle Q R S=108^{\circ}$.

Since $P Q=Q R$, then $\triangle P Q R$ is isosceles with $\angle Q P R=\angle Q R P$.

Since $\angle P Q R=108^{\circ}$, then

$$
\begin{aligned}
\angle P Q R+\angle Q P R+\angle Q R P & =180^{\circ} \\
108^{\circ}+2 \angle Q R P & =180^{\circ} \\
2 \angle Q R P & =72^{\circ} \\
\angle Q R P & =36^{\circ}
\end{aligned}
$$

Since $\angle Q R S=108^{\circ}$, then $\angle P R S=\angle Q R S-\angle Q R P=108^{\circ}-36^{\circ}=72^{\circ}$.