3. In a circle with radius $R$, a triangle with angles $15^{\circ}$ and $60^{\circ}$ is inscribed. Find the area of the triangle.
Solution. From the fact that the central angle is twice the peripheral angle over the same arc, it follows that the angle $\gamma$ in the diagram is $150^{\circ}$. Then $\alpha=15^{\circ}$.

From the isosceles triangle $\triangle A O B$, it follows that $\overline{A B}=2 R \cos 15^{\circ}$. From the isosceles triangle $\triangle A O C$, it follows that

$$
\overline{A C}=2 R \cos 30^{\circ}=R \sqrt{3}
$$

Accordingly,

$$
P=\frac{\overline{A B} \cdot \overline{A C} \cdot \sin 15^{\circ}}{2}=\frac{2 R \cos 15^{\circ} \cdot R \sqrt{3} \cdot \sin 15^{\circ}}{2}=\frac{R^{2} \sqrt{3} \sin 30^{\circ}}{2}=\frac{R^{2} \sqrt{3}}{4}
$$