In the diagram, $P$ is on $R S$ so that $Q P$ bisects $\angle S Q R$. Also, $P Q=P R, \angle R S Q=2 y^{\circ}$, and $\angle R P Q=3 y^{\circ}$. The measure of $\angle R P Q$ is
(A) $90^{\circ}$
(B) $108^{\circ}$
(C) $120^{\circ}$
(D) $60^{\circ}$
(E) $72^{\circ}$

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-094.jpg?height=393&width=322&top_left_y=270&top_left_x=1292)
Solution 1

Since $R P S$ is a straight line, then $\angle S P Q=180^{\circ}-\angle R P Q=180^{\circ}-3 y^{\circ}$.

Using the angles in $\triangle P Q S$, we have $\angle P Q S+\angle Q S P+\angle S P Q=180^{\circ}$.

Thus, $x^{\circ}+2 y^{\circ}+\left(180^{\circ}-3 y^{\circ}\right)=180^{\circ}$ or $x-y+180=180$ or $x=y$.

(We could have instead looked at $\angle R P Q$ as being an external angle to $\triangle S P Q$.)

Since $x=y$, then $\angle R Q S=2 y^{\circ}$.

Since $R P=P Q$, then $\angle P R Q=\angle P Q R=x^{\circ}=y^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-179.jpg?height=304&width=412&top_left_y=1095&top_left_x=911)

Therefore, the angles of $\triangle R Q S$ are $y^{\circ}, 2 y^{\circ}$ and $2 y^{\circ}$.

Thus, $y^{\circ}+2 y^{\circ}+2 y^{\circ}=180^{\circ}$ or $5 y=180$ or $y=36$.

Therefore, $\angle R P Q=3 y^{\circ}=3(36)^{\circ}=108^{\circ}$.

## Solution 2

Since $R P=P Q$, then $\angle P R Q=\angle P Q R=x^{\circ}$.

Looking at the sum of the angles in $\triangle R P Q$ and $\triangle R S Q$, we have $x^{\circ}+3 y^{\circ}+x^{\circ}=180^{\circ}$ (or $2 x+3 y=180$ ) and $x^{\circ}+2 y^{\circ}+2 x^{\circ}=180$ (or $3 x+2 y=180$ ).

Adding these two equations gives $5 x+5 y=360$ or $x+y=\frac{1}{5}(360)=72$.

Thus, $2 x+2 y=2(72)=144$, so $2 x+3 y=180$, gives $y=180-(2 x+2 y)=180-144=36$.

Therfore, $\angle R P Q=3 y^{\circ}=3(36)^{\circ}=108^{\circ}$.

ANSWER: (B)