15) The answer is (D)

Indeed, the full angle is equal to 5 times $P \hat{O Q Q}$, and thus $P \hat{O} Q = \frac{1}{5} 360^{\circ} = 72^{\circ}$. Since $O \hat{P} R$ and $O \hat{Q} R$ are right angles, we must have $P \hat{O} Q + P \hat{R} Q = 180^{\circ}$, from which $P \hat{R} Q = 108^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_df6134ed1e07189e6736g-2.jpg?height=252&width=371&top_left_y=2364&top_left_x=1482)
15) $P R$ and $Q R$ are tangents to the circle in the figure. Knowing that the arc $P S Q$ is four times the arc $P T Q$, then the angle $P \widehat{R} Q$ is

(A) $72^{\circ}$

(B) $90^{\circ}$

(C) $105^{\circ}$

(D) $108^{\circ}$

(E) $120^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_8047f9ac13dbecc510bdg-2.jpg?height=209&width=336&top_left_y=880&top_left_x=1126)