In the diagram shown, $P Q R$ is a straight line segment. The measure of $\angle Q S R$ is
(A) $25^{\circ}$
(B) $30^{\circ}$
(C) $35^{\circ}$
(D) $40^{\circ}$
(E) $45^{\circ}$
![](https://cdn.mathpix.com/cropped/2024_04_20_46ead6524a8d61e21c51g-041.jpg?height=714&width=500&top_left_y=1474&top_left_x=1254)
Solution 1

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

Since $\angle S Q P+\angle S Q R=180^{\circ}$, then $\angle S Q R=180^{\circ}-\angle S Q P=180^{\circ}-75^{\circ}=105^{\circ}$.

The three angles in a triangle add to $180^{\circ}$, so $\angle Q S R+\angle S Q R+\angle Q R S=180^{\circ}$, or $\angle Q S R=180^{\circ}-\angle S Q R-\angle Q R S=180^{\circ}-105^{\circ}-30^{\circ}=45^{\circ}$.

## Solution 2

The exterior angle of a triangle is equal to the sum of the two non-adjacent interior angles of the triangle.

Since $\angle S Q P$ is an exterior angle of $\triangle S Q R$, and the two opposite interior angles are $\angle Q S R$ and $\angle Q R S$, then $\angle S Q P=\angle Q S R+\angle Q R S$.

Thus, $75^{\circ}=\angle Q S R+30^{\circ}$ or $\angle Q S R=75^{\circ}-30^{\circ}=45^{\circ}$.

ANswer: (E)