In the diagram, $Q R S$ is a straight line. What is the measure of $\angle R P S$ ?
(A) $27^{\circ}$
(B) $47^{\circ}$
(C) $48^{\circ}$
(D) $65^{\circ}$
(E) $67^{\circ}$

![](https://cdn.mathpix.com/cropped/2024_04_20_6ed09463f225f8ba1f07g-087.jpg?height=244&width=456&top_left_y=1268&top_left_x=1298)
Solution 1

Since the sum of the angles in a triangle is $180^{\circ}$, then

$$
\angle Q P S=180^{\circ}-\angle P Q S-\angle P S Q=180^{\circ}-48^{\circ}-38^{\circ}=94^{\circ}
$$

Therefore, $\angle R P S=\angle Q P S-\angle Q P R=94^{\circ}-67^{\circ}=27^{\circ}$.

## Solution 2

Since the sum of the angles in a triangle is $180^{\circ}$, then

$$
\angle Q R P=180^{\circ}-\angle P Q R-\angle Q P R=180^{\circ}-48^{\circ}-67^{\circ}=65^{\circ}
$$

Therefore, $\angle P R S=180^{\circ}-\angle P R Q=180^{\circ}-65^{\circ}=115^{\circ}$.

Using $\triangle P R S$,

$$
\angle R P S=180^{\circ}-\angle P R S-\angle P S R=180^{\circ}-115^{\circ}-38^{\circ}=27^{\circ}
$$

ANSWER: (A)