In the diagram, $\triangle Q U R$ and $\triangle S U R$ are equilateral triangles. Also, $\triangle Q U P, \triangle P U T$ and $\triangle T U S$ are isosceles triangles with $P U=Q U=S U=T U$ and $Q P=P T=T S$. The measure of $\angle U S T$, in degrees, is
(A) 50
(B) 54
(C) 60
(D) 70
(E) 80

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-069.jpg?height=390&width=380&top_left_y=976&top_left_x=1298)
Since $\triangle Q U R$ and $\triangle S U R$ are equilateral, then $\angle Q U R=\angle S U R=60^{\circ}$.

Since $Q U=P U=T U=S U$ and $Q P=P T=T S$, then $\triangle Q U P, \triangle P U T$ and $\triangle T U S$ are congruent.

Thus, $\angle Q U P=\angle P U T=\angle T U S$.

The angles around point $U$ add to $360^{\circ}$.

Thus, $\angle S U R+\angle Q U R+\angle Q U P+\angle P U T+\angle T U S=360^{\circ}$ and so $60^{\circ}+60^{\circ}+3 \angle T U S=360^{\circ}$ or $3 \angle T U S=240^{\circ}$ or $\angle T U S=80^{\circ}$.

Since $\triangle T U S$ is isosceles with $T U=S U$, then $\angle U S T=\angle U T S$.

Since the angles in $\triangle T U S$ add to $180^{\circ}$, then $\angle T U S+\angle U S T+\angle U T S=180^{\circ}$.

Therefore, $80^{\circ}+2 \angle U S T=180^{\circ}$ and so $2 \angle U S T=100^{\circ}$ or $\angle U S T=50^{\circ}$.

ANSWER: (A)