10.092. A triangle with angles of 15 and $60^{\circ}$ is inscribed in a circle of radius $R$ (Fig. 10.89). Find the area of the triangle.
Solution.

$$
\angle ABC = 105^{\circ}; \angle ABC = 180^{\circ} - \frac{1}{2} \angle AOC. \text{ Hence, } \angle AOC = 150^{\circ}. \text{ Then, }
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-600.jpg?height=428&width=470&top_left_y=96&top_left_x=137)

Fig. 10.89

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-600.jpg?height=290&width=596&top_left_y=207&top_left_x=654)

Fig. 10.90

we find that $\angle OAC = 15^{\circ}$. From $\triangle ADO$, we get $AD = R \cos 15^{\circ}, AC = 2AD = 2R \cos 15^{\circ}$. From $\triangle AOB$ (isosceles):

$AB^2 = R^2 + R^2 - 2R^2 \cos 120^{\circ} = 2R^2(1 + \cos 60^{\circ}) = 3R^2; AB = \sqrt{3}R$.

The area of triangle $ABC$:

$S = \frac{1}{2} AB \cdot AC \sin 15^{\circ} = \frac{1}{2} \sqrt{3}R \cdot 2R \cos 15^{\circ} \sin 15^{\circ} = \frac{\sqrt{3}}{4}R^2$.

Answer: $\frac{\sqrt{3}}{4}R^2$.