Theorem of the lengths of the tangent and secant; the product of the entire secant and its external part. The Law of Sines

In triangle $ABC$, a midline $MN$ is drawn connecting sides $AB$ and $BC$. A circle passing through points $M, N$, and $C$ is tangent to side $AB$, and its radius is $\sqrt{2}$. The length of side $AC$ is 2. Find the sine of angle $ACB$.
Apply the tangent-secant theorem and the sine theorem to triangle $B M N$.

## Solution

Let $B N=N C=x, \angle B A C=\alpha, \angle A C B=\beta$.

By the tangent-secant theorem,

$$
B M=\sqrt{B N \cdot B C}=\sqrt{x \cdot 2 x}=x \sqrt{2}.
$$

By the midline theorem of a triangle, $M N=\frac{1}{2} A C=\frac{1}{2} \cdot 2=1$ and $M N \| A C$. Therefore, $\angle B M N=\angle B A C=\alpha$ and $\angle B N M=\angle A C B=\beta$.

By the theorem of the angle between a tangent and a chord, $\angle N C M=\angle B M N=\alpha$. If $R$ is the radius of the given circle, then $M N=2 R \cdot \sin \angle M C N$, or $1=2 \sqrt{2} \sin \alpha$, from which we find that $\sin \alpha=\frac{1}{2 \sqrt{2}}$.

By the sine theorem,

$$
\frac{B M}{\sin \angle B N M}=\frac{B N}{\sin \angle B M N}, \text{ or } \frac{x \sqrt{2}}{\sin \beta}=\frac{x}{\sin \alpha}.
$$

Therefore,

$$
\sin \beta=\sqrt{2} \cdot \sin \alpha=\sqrt{2} \cdot \frac{1}{2 \sqrt{2}}=\frac{1}{2}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_96a7587f2ba0a7b3aee2g-06.jpg?height=260&width=820&top_left_y=2431&top_left_x=619)

Answer

## $\frac{1}{2}$. 

## Problem