3.21. By the condition, $\angle B A C=\alpha, A B$ $=A C, A O=R$ (Fig. 3.23). Then $\cup B C=2 \alpha, \cup A B=\cup A C=$ $=\frac{2 \pi-2 \alpha}{2}=\pi-\alpha$. The area of the part of the circle $S_{B A C}=S_{\text {sect } B O C}+$ $+2 S_{\triangle O A C}$. But

$$S_{\text {sect } \cdot B O C}=\frac{1}{2} R^{2} \cdot 2 \alpha=R^{2} \alpha$$

(by formula (1.34));

$$S_{\triangle O A C}=\frac{1}{2} R^{2} \sin (\pi-\alpha)=\frac{1}{2} R^{2} \sin \alpha$$

!

Fig. 3.23

(by formula (1.2)). Therefore,

$$S_{B A C}=R^{2} \alpha+R^{2} \sin \alpha=R^{2}(\alpha+\sin \alpha)$$