In the diagram, $P Q R$ is a straight line segment and $Q S=Q T$. Also, $\angle P Q S=x^{\circ}$ and $\angle T Q R=3 x^{\circ}$. If $\angle Q T S=76^{\circ}$, the value of $x$ is
(A) 28
(B) 38
(C) 26
(D) 152
(E) 45

![](https://cdn.mathpix.com/cropped/2024_04_20_6ed09463f225f8ba1f07g-063.jpg?height=347&width=309&top_left_y=1946&top_left_x=1296)
Since $\angle Q T S=76^{\circ}$ and $\triangle Q T S$ has $Q S=Q T$, then $\angle Q S T=\angle Q T S=76^{\circ}$.

Since the angles in $\triangle Q T S$ add to $180^{\circ}$, then

$$
\angle S Q T=180^{\circ}-\angle Q T S-\angle Q S T=180^{\circ}-76^{\circ}-76^{\circ}=28^{\circ}
$$

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

Thus, $x^{\circ}+28^{\circ}+3 x^{\circ}=180^{\circ}$.

This gives $4 x+28=180$ or $4 x=152$ and so $x=38$.

ANsWER: (B)