\left.\begin{array}{l}{[\quad \text { Law of Sines }} \\ {\left[\begin{array}{l}\text { Angle between a tangent and a chord }]\end{array}\right]}\end{array}\right]

In triangle $ABC$, angle $\angle B$ is equal to $\frac{\pi}{6}$. A circle with a radius of 2 cm is drawn through points $A$ and $B$, tangent to line $AC$ at point $A$. A circle with a radius of 3 cm is drawn through points $B$ and $C$, tangent to line $AC$ at point $C$. Find the length of side $AC$.
Use the theorem about the angle between a tangent and a chord, then apply the Law of Sines.

## Solution

Let $O_{1}$ be the center of the circle with radius $R_{1}=2$, and $O_{2}$ be the center of the circle with radius $R_{2}=3$. Denote $\angle A C B=\boldsymbol{\alpha}$, $\angle B A C=\beta$. From the theorem about the angle between a tangent and a chord, it follows that $\angle C O_{2} B=2 \alpha$ (or $180^{\circ}-2 \alpha$) and $\angle A O_{1} B=2 \beta$ (or $180^{\circ}-2 \beta$). Then

$$
A B=2 R_{1} \cdot \sin \beta=4 \sin \beta, \quad B C=2 R_{2} \cdot \sin \alpha=6 \sin \alpha
$$

By the Law of Sines,

$$
\frac{A B}{\sin \alpha}=\frac{B C}{\sin \beta} \text{, or } \frac{4 \sin \beta}{\sin \alpha}=\frac{6 \sin \alpha}{\sin \beta} \text{. }
$$

From this, we find that $\frac{\sin ^{2} \beta}{\sin ^{2} \alpha}=\sqrt{\frac{3}{2}}$.

From the equality $\frac{A C}{\sin \angle A B C}=\frac{A B}{\sin \angle A C B}$, it follows that

$$
A C=\frac{A B}{\sin \angle A C B} \cdot \sin \angle A B C=\frac{4 \sin \beta}{\sin \alpha} \cdot \sin \frac{\pi}{6}=2 \sqrt{\frac{3}{2}}=\sqrt{6}
$$

## Answer

![](https://cdn.mathpix.com/cropped/2024_05_06_96a7587f2ba0a7b3aee2g-05.jpg?height=640&width=552&top_left_y=1855&top_left_x=753)

$\sqrt{6}$.