[

The Law of Sines

[Theorem of the length of a tangent and a secant; the product of the entire secant and its external part

A circle of radius $R$ with center at point $O$ passes through the vertices $A$ and $B$ of triangle $A B C$, intersects segment $B C$ at point $M$, and is tangent to line $A C$ at point $A$. Find $C M$, given that $\angle A C O=\alpha, \angle M A B=\beta$.

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Apply the tangent-secant theorem.

## Solution

By the tangent-secant theorem, $A C^{2}=B C \cdot C M=(B M+C M) C M$. Since $B M=2 R \sin \beta$ and $A C=R \operatorname{ctg} \alpha$, we have $R^{2} \operatorname{ctg}^{2} \alpha=C M^{2}+C M \cdot 2 R \sin \beta$. From this equation, we find that $C M=R\left(\sqrt{\sin ^{2} \beta+\operatorname{ctg}^{2} \alpha}-\sin \beta\right)$.

![](https://cdn.mathpix.com/cropped/2024_05_06_36c4956010bd91c4b4eeg-08.jpg?height=406&width=435&top_left_y=829&top_left_x=823)

## Answer

$R\left(\sqrt{\sin ^{2} \beta+\operatorname{ctg}^{2} \alpha}-\sin \beta\right)$