In the diagram, if $\triangle A B C$ and $\triangle P Q R$ are equilateral, then $\angle C X Y$ equals
(A) $30^{\circ}$
(B) $35^{\circ}$
(C) $40^{\circ}$
(D) $45^{\circ}$
(E) $50^{\circ}$

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-099.jpg?height=393&width=515&top_left_y=1836&top_left_x=1203)
Since $\triangle A B C$ and $\triangle P Q R$ are equilateral, then $\angle A B C=\angle A C B=\angle R P Q=60^{\circ}$.

Therefore, $\angle Y B P=180^{\circ}-65^{\circ}-60^{\circ}=55^{\circ}$ and $\angle Y P B=180^{\circ}-75^{\circ}-60^{\circ}=45^{\circ}$.

In $\triangle B Y P$, we have $\angle B Y P=180^{\circ}-\angle Y B P-\angle Y P B=180^{\circ}-55^{\circ}-45^{\circ}=80^{\circ}$.

Since $\angle X Y C=\angle Y B P$, then $\angle X Y C=80^{\circ}$.

In $\triangle C X Y$, we have $\angle C X Y=180^{\circ}-60^{\circ}-80^{\circ}=40^{\circ}$.

ANSWER: (C)