In the diagram, $R$ lies on line segment $Q S$. What is the value of $x$ ?
(A) 50
(B) 55
(C) 75
(D) 100
(E) 105

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-074.jpg?height=290&width=488&top_left_y=806&top_left_x=1293)
Solution 1

Since $\angle P R S$ is an exterior angle to $\triangle P Q R$, then $\angle P Q R+\angle Q P R=\angle P R S$ or $x^{\circ}+75^{\circ}=125^{\circ}$. Thus, $x+75=125$ or $x=50$.

## Solution 2

Since $Q R S$ is a straight line, then $\angle P R Q=180^{\circ}-\angle P R S=180^{\circ}-125^{\circ}=55^{\circ}$.

Since the sum of the angles in a triangle is $180^{\circ}$, then $x^{\circ}+75^{\circ}+55^{\circ}=180^{\circ}$ or $x+130=180$ or $x=50$.

ANSWER: (A)