10.287. Find the area of a triangle inscribed in a circle of radius 2 cm, if two angles of the triangle are $\pi / 3$ and $\pi / 4$.
## Solution.

Since $\angle A C B=\frac{\pi}{4}$ (Fig. 10.80), then $A B=2 \sqrt{2}$ cm (side of the inscribed square). Draw $B D \perp A C$. Considering that $\angle A B D=\frac{\pi}{6}$, we have $A D=\frac{1}{2} A B=\sqrt{2}$ (cm); further, since $\triangle B D C$ is isosceles, then
$D C=B D=\frac{\sqrt{3}}{2} A B=\frac{\sqrt{3}}{2} \cdot 2 \sqrt{2}=\sqrt{6}$ cm. Therefore, $A C=A D+D C=\sqrt{2}+\sqrt{6}$ (cm).

We obtain: $S=\frac{1}{2} A C \cdot B D=\frac{1}{2}(\sqrt{2}+\sqrt{6}) \sqrt{6}=(1+\sqrt{3}) \sqrt{3}=\sqrt{3}+3\left(\mathrm{~cm}^{2}\right)$.

Answer: $(\sqrt{3}+3) \mathrm{cm}^{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0770.jpg?height=452&width=715&top_left_y=422&top_left_x=319)

Fig. 10.81