In the diagram, $P R$ and $Q S$ meet at $X$. Also, $\triangle P Q X$ is right-angled at $Q$ with $\angle Q P X=62^{\circ}$ and $\triangle R X S$ is isosceles with $R X=S X$ and $\angle X S R=y^{\circ}$. The value of $y$ is
(A) 54
(B) 71
(C) 76
(D) 59
(E) 60

![](https://cdn.mathpix.com/cropped/2024_04_20_6ed09463f225f8ba1f07g-027.jpg?height=534&width=479&top_left_y=1451&top_left_x=1170)
Since $\triangle P Q X$ is right-angled at $Q$, then

$$
\angle P X Q=90^{\circ}-\angle Q P X=90^{\circ}-62^{\circ}=28^{\circ}
$$

Since $\angle P X Q$ and $\angle S X R$ are opposite, then $\angle S X R=\angle P X Q=28^{\circ}$.

Since $\triangle R X S$ is isosceles with $R X=S X$, then $\angle X R S=\angle X S R=y^{\circ}$. Since the angles in any triangle have a sum of $180^{\circ}$,

$$
\begin{aligned}
\angle X R S+\angle X S R+\angle S X R & =180^{\circ} \\
y^{\circ}+y^{\circ}+28^{\circ} & =180^{\circ} \\
2 y+28 & =180 \\
2 y & =152
\end{aligned}
$$

and so $y=76$.

ANSWER: (C)

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