12.142. A triangle is inscribed in a circle of radius $R$, with its vertices dividing the circle into three parts in the ratio $2: 5: 17$. Find the area of the triangle.
## Solution.

Let $\cup A B=2 x$ (Fig. 12.15), $\cup B C=5 x$, $\cup A m C=17 x$ and $2 x+5 x+17 x=360^{\circ}, x=15^{\circ}$ $\Rightarrow \angle A O B=\cup A B=30^{\circ}, \angle B O C=\cup B C=75^{\circ}$, $\angle A O C=\angle A O B+\angle B O C=105^{\circ}$.

$S_{\triangle A B C}=S_{\triangle A O B}+S_{\triangle B O C}-S_{\triangle A O C}=$

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0902.jpg?height=469&width=474&top_left_y=706&top_left_x=756)

Fig. 12.15 $=\frac{1}{2} R^{2} \sin 30^{\circ}+\frac{1}{2} R^{2} \sin 75^{\circ}-\frac{1}{2} R^{2} \sin 105^{\circ}$. Since $\sin 75^{\circ}=\sin 105^{\circ}$, then $S_{\triangle A B C}=\frac{R^{2}}{4}$.

Answer: $\frac{R^{2}}{4}$.