3 [Inscribed, circumscribed, and exscribed circles; their radii]

Triangle $A B C$ is isosceles. The radius $O A$ of the circumscribed circle forms an angle $O A C$ with the base $A C$, which is $20^{\circ}$. Find the angle $B A C$.
The inscribed angle is equal to half of the corresponding central angle. Consider two cases of the positions of points $O$ and $B$ relative to the line $A C$.

## Solution

If points $O$ and $B$ lie on opposite sides of the line $A C$, then

$$
\cup A B C=\angle A O C=180^{\circ}-2 \cdot 20^{\circ}=140^{\circ} \text {. }
$$

Then the angle $A B C$ subtends an arc whose angular measure is

$$
360^{\circ}-140^{\circ}=220^{\circ} .
$$

Therefore,

$$
\angle A B C=110^{\circ}, \angle B A C=\frac{180^{\circ}-110^{\circ}}{2}=35^{\circ}
$$

If points $O$ and $B$ lie on the same side of the line $A C$, then similarly we find that $\angle B A C=55^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_d1936174aeda27dcc10dg-13.jpg?height=395&width=420&top_left_y=54&top_left_x=390)

Fig. 1.

![](https://cdn.mathpix.com/cropped/2024_05_06_d1936174aeda27dcc10dg-13.jpg?height=394&width=554&top_left_y=60&top_left_x=1112)

## Answer

$35^{\circ}$ or $55^{\circ}$.