[ Law of Sines]

In a circle, two chords $A B=a$ and $A C=b$ are drawn. The length of arc $A C$ is twice the length of arc $A B$. Find the radius of the circle.
Using the Law of Sines, form a trigonometric equation.

## Solution

Let $\angle ACB = \boldsymbol{\alpha}$. Then $\angle ABC = 2 \boldsymbol{\alpha}$. By the Law of Sines,

$$
\frac{a}{\sin \alpha} = \frac{b}{\sin 2 \alpha}
$$

Therefore, $\cos \alpha = \frac{b}{2a}$. Then

$$
\sin \alpha = \sqrt{1 - \left(\frac{b}{2a}\right)^2} = \frac{\sqrt{4a^2 - b^2}}{2a}
$$

If $R$ is the radius of the circle, then

$$
R = \frac{a}{2 \sin \alpha} = \frac{a^2}{\sqrt{4a^2 - b^2}}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_ae6ec2c916c776dd1759g-11.jpg?height=352&width=648&top_left_y=573&top_left_x=705)

Answer

$\frac{a^2}{\sqrt{4a^2 - b^2}}$.