In the diagram, $S$ lies on $R T, \angle Q T S=40^{\circ}, Q S=Q T$, and $\triangle P R S$ is equilateral. The value of $x$ is
(A) 50
(B) 60
(C) 80
(D) 90
(E) 100

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-074.jpg?height=220&width=570&top_left_y=1446&top_left_x=1100)
Since $\triangle P R S$ is equilateral, then all three of its angles equal $60^{\circ}$.

In particular, $\angle R S P=60^{\circ}$.

Since $Q S=Q T$, then $\triangle Q S T$ is isosceles and so $\angle T S Q=\angle S T Q=40^{\circ}$.

Since $R S T$ is a straight line segment, then $\angle R S P+\angle P S Q+\angle T S Q=180^{\circ}$.

Therefore, $60^{\circ}+x^{\circ}+40^{\circ}=180^{\circ}$ or $x=180-60-40=80$.

ANSWER: (C)