3.21. From a point taken on a circle of radius $R$, two equal chords are drawn, forming an inscribed angle equal to $\alpha$ radians. Find the part of the area of the circle; enclosed by this inscribed angle.
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));

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

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-512.jpg?height=481&width=454&top_left_y=690&top_left_x=690)

Fig. 3.23

(by formula (1.2)). Therefore,

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