1.54. Find the area of a triangle inscribed in a circle of radius 2 cm, if two angles of the triangle are $\frac{\pi}{3}$ and $\frac{\pi}{4}$.
1.54. Since $\angle A C B=\frac{\pi}{4}$ (Fig.

1.42), then $A B=2 \sqrt{2}$ cm (side of the inscribed square). Draw $B D \perp A C$.

Then, considering that $\angle A B D=\frac{\pi}{6}$, we find $A D=\frac{1}{2} A B=\sqrt{2}$ (cm); further,

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-438.jpg?height=427&width=433&top_left_y=369&top_left_x=718)

Fig. 1.42 since $\triangle B D C$ is isosceles, $D C=B D=\frac{\sqrt{3}}{2} A B=\frac{\sqrt{3}}{2} \cdot 2 \sqrt{2}=\sqrt{6}$ (cm). Therefore, $A C=A D+D C=$ $=\sqrt{2}+\sqrt{6}$ (cm). Thus,

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

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