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In a circle of radius 12, chord $A B=6$, and chord $B C=4$. Find the chord connecting the ends of arc $A C$.
Apply the formula $a=2 R \sin \boldsymbol{\alpha}$. Consider two cases.

## Solution

Let $R=12$ be the radius of the circle. Then

$$
A B=2 R \sin \angle A C B, B C=2 R \sin \angle B A C
$$

From this, we find that $\sin \angle A C B=\frac{1}{4}$ and $\sin \angle B A C=\frac{1}{6}$. Since side $B C$ of triangle $A B C$ is not the largest, angle $A$ is acute and its cosine is positive. Therefore,

$$
\cos \angle B A C=\sqrt{1-\sin ^{2} \angle A}=\frac{\sqrt{35}}{6}
$$

$$
\cos \angle A C B= \pm \sqrt{1-\cos ^{2} \angle B}= \pm \frac{\sqrt{15}}{4}
$$

Then

$$
\begin{aligned}
& A C=2 R \sin \angle A B C=2 R \sin (\angle B A C+\angle A C B)= \\
& \quad=2 R(\sin \angle A \cdot \cos \angle C+\cos \angle A \cdot \sin \angle C)= \\
& =2 \cdot 12\left( \pm \frac{1}{6} \cdot \frac{\sqrt{15}}{4}+\frac{1}{4} \cdot \frac{\sqrt{35}}{6}\right)=\sqrt{35} \pm \sqrt{15}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_df2bbd76ac4dd2f0f20fg-14.jpg?height=444&width=489&top_left_y=807&top_left_x=790)

## Answer

$\sqrt{35} \pm \sqrt{15}$.