12.133. The arc $AB$ of sector $AOB$ contains $\alpha$ radians. A line is drawn through point $B$ and the midpoint $C$ of radius $OA$. In what ratio does this line divide the area of the sector?
## Solution.

Let the radius of the given sector be $\mathrm{R}$.

The area of sector $A O B$ (Fig. 12.6) $S=\frac{R^{2} \alpha}{2}$.

$$
S_{\triangle B C O}=\frac{1}{2} O C \cdot O B \sin \angle B O C=\frac{R^{2} \sin \alpha}{4}
$$

The area of figure $A B C$

$$
S_{1}=S-S_{\triangle B C O}=\frac{R^{2} \alpha}{2}-\frac{R^{2} \sin \alpha}{4}=\frac{R^{2}}{4}(2 \alpha-\sin \alpha)
$$

Then $\frac{S_{\triangle B C O}}{S_{1}}=\frac{\sin \alpha}{2 \alpha-\sin \alpha}$.

Answer: $\frac{\sin \alpha}{2 \alpha-\sin \alpha}$.