In the diagram, point $T$ is on side $P R$ of $\triangle P Q R$ and $Q R S$ is a straight line segment. The value of $x$ is
(A) 55
(B) 70
(C) 75
(D) 60
(E) 50

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-056.jpg?height=537&width=642&top_left_y=1892&top_left_x=1056)
Solution 1

$\angle S R P$ is an exterior angle for $\triangle P Q R$.

Therefore, $\angle S R P=\angle R P Q+\angle R Q P$ or $(180-x)^{\circ}=30^{\circ}+2 x^{\circ}$.

Thus, $180-x=30+2 x$ or $3 x=150$ and so $x=50$.

## Solution 2

Since $Q R S$ is a straight line segment and $\angle S R P=(180-x)^{\circ}$, then $\angle P R Q$ is the supplement of $\angle S R P$ so $\angle P R Q=x^{\circ}$.

Since the angles in a triangle add to $180^{\circ}$, then $\angle P R Q+\angle P Q R+\angle R P Q=180^{\circ}$, and so $x^{\circ}+2 x^{\circ}+30^{\circ}=180^{\circ}$.

From this, we obtain $3 x=150$ and so $x=50$.

ANswer: (E)