In the diagram, $P R, P S, Q S, Q T$, and $R T$ are straight line segments. $Q T$ intersects $P R$ and $P S$ at $U$ and $V$, respectively. If $P U=P V, \angle U P V=24^{\circ}, \angle P S Q=x^{\circ}$, and $\angle T Q S=y^{\circ}$, what is the value of $x+y$ ?
(A) 48
(B) 66
(C) 72
(D) 78
(E) 156

![](https://cdn.mathpix.com/cropped/2024_04_20_6ed09463f225f8ba1f07g-076.jpg?height=518&width=494&top_left_y=1373&top_left_x=1268)

Part C: Each correct answer is worth 8.
Solution 1

Consider $\triangle U P V$.

Since $P U=P V$, then $\triangle U P V$ is isosceles, with

$$
\angle P U V=\angle P V U=\frac{1}{2}\left(180^{\circ}-\angle U P V\right)=\frac{1}{2}\left(180^{\circ}-24^{\circ}\right)=\frac{1}{2}\left(156^{\circ}\right)=78^{\circ}
$$

Since $P V S$ is a straight line, then $\angle Q V S=180^{\circ}-\angle P V U=180^{\circ}-78^{\circ}=102^{\circ}$.

Consider $\triangle Q V S$.

The sum of the angles in this triangle is $180^{\circ}$, and so $102^{\circ}+x^{\circ}+y^{\circ}=180^{\circ}$.

Therefore, $x+y=180-102=78$.

## Solution 2

Consider $\triangle U P V$.

Since $P U=P V$, then $\triangle U P V$ is isosceles, with

$$
\angle P U V=\angle P V U=\frac{1}{2}\left(180^{\circ}-\angle U P V\right)=\frac{1}{2}\left(180^{\circ}-24^{\circ}\right)=\frac{1}{2}\left(156^{\circ}\right)=78^{\circ}
$$

Since $\angle P V U$ is an exterior angle to $\triangle Q V S$, then $\angle P V U=\angle V Q S+\angle V S Q$. Therefore, $78^{\circ}=y^{\circ}+x^{\circ}$ or $x+y=78$.

ANSWER: (D)