5. Two circles touch internally at point $A$. From the center of the larger circle $O$, a radius $O B$ of the larger circle is drawn, which is also tangent to the smaller circle at point $C$. Determine $\measuredangle B A C$.
Solution. Let $\theta=\measuredangle B A C$ be the desired angle and let $\Varangle S A C=\alpha$ and $\Varangle S B C=\beta$. By assumption, triangles $S A B$ and $A C O$ are isosceles ($S A=S B$ and $O A=O C$), and triangle $S O C$ is right-angled. Let $\measuredangle A S B=\varphi$. Then from triangle $A B S$ we have $2 \beta+\varphi=180^{\circ}$, i.e., $\beta=90^{\circ}-\frac{\varphi}{2}$. Furthermore,

![](https://cdn.mathpix.com/cropped/2024_06_05_c1452ffedcb4be8e059ag-2.jpg?height=394&width=407&top_left_y=1721&top_left_x=1075)
$\measuredangle S O C$ is an exterior angle for triangle $A C O$, so

$$
\measuredangle S O C=\measuredangle O A C=\measuredangle O C A=2 \alpha
$$

Then, from triangle SOC we have $\varphi+2 \alpha=90^{\circ}$, i.e., $\alpha=45^{\circ}-\frac{\varphi}{2}$. Finally,

$$
\theta=\measuredangle S A B-\alpha=\beta-\alpha=90^{\circ}-\frac{\varphi}{2}-\left(45^{\circ}-\frac{\varphi}{2}\right)=45^{\circ}
$$

## Eighth Section