Answer: 12.

## Solution:

![](https://cdn.mathpix.com/cropped/2024_05_06_f5010c49868bbc23ccb1g-01.jpg?height=431&width=462&top_left_y=1949&top_left_x=794)

Consider point $B_{1}$, which is symmetric to point $B$ with respect to the line $O A$. It also lies on the circle and $\angle A M B=\alpha$. Note that points $B_{1}, M, C$ lie on the same straight line, and $\Delta B B_{1} M$ is isosceles. Therefore, the inscribed angle $\angle B B_{1} M=90^{\circ}-\alpha$, and the central angle $\angle B O C=180^{\circ}-2 \alpha$. $\triangle B O C$ is isosceles and $\angle O B C=\alpha \cdot \cos \alpha=\sqrt{1-\sin ^{2} \alpha}=\frac{2}{5}$. We find the base of the isosceles triangle using the formula $B C=2 * \cos \alpha * B O=2 * \frac{2}{5} * 15=12$
3. (7 points) On the radius $A O$ of a circle with center $O$, a point $M$ is chosen. On one side of $A O$ on the circle, points $B$ and $C$ are chosen such that $\angle A M B = \angle O M C = \alpha$. Find the length of $B C$ if the radius of the circle is $15$, and $\sin \alpha = \frac{\sqrt{21}}{5}$?