One of the adjacent angles with vertex $A$ is twice the other. Circles are inscribed in these angles with centers $O 1$ and $O 2$. Find the angles of triangle $O 1 A O 2$, if the ratio of the radii of the circles is $\sqrt{3}$.
One of the adjacent angles is $60^{\circ}$, and the second is $120^{\circ}$.

Let a circle with center $O 1$ and radius $r$ be inscribed in the angle equal to $60^{\circ}$, and a circle with center $O 2$ and radius $r \sqrt{3}$ be inscribed in the angle equal to $120^{\circ}$ (Fig.1), and the circles touch the line containing the additional sides of these angles at points $B$ and $C$ respectively.

The center of the circle inscribed in the angle lies on the bisector of the angle, therefore

$$
\angle O 1 A B=30^{\circ}, \angle O 2 A C=60^{\circ}, \angle O 1 A O 2=90^{\circ}
$$

From the right triangles $O 1 A B$ and $O 2 A C$ we find that

$$
A O 1=2 O 1 B=2 r, A O 2=\frac{O_{2} C}{\sin 60^{2}}=\frac{r \sqrt{3}}{\frac{\sqrt{3}}{2}}=2 r
$$

Triangle $O 1 A O 2$ is a right and isosceles triangle, hence its acute angles are each $45^{\circ}$. Now let the circle with center $O 1$ and radius $r$ be inscribed in the angle equal to $120^{\circ}$, and the circle with center $O 2$ and radius $r \sqrt{3}$ be inscribed in the angle equal to $60^{\circ}$ (Fig.2), and the circles touch the line containing the additional sides of these angles at points $B$ and $C$ respectively. From the right triangles $O 1 A B$ and $O 2 A C$ we find,

$$
\begin{gathered}
A O 1=\frac{O_{1} B}{\sin 60^{\circ}}=\frac{2 r}{\sqrt{3}}, A O 2=2 O 2 C=2 r \sqrt{3} . \\
\text { Therefore, } \\
\operatorname{tg} \angle A O 1 O 2=\frac{A O_{2}}{A O_{1}}=\frac{2 r \sqrt{3}}{\frac{2 r}{\sqrt{3}}}=3, \operatorname{tg} \angle A O 2 O 1=\frac{1}{3} .
\end{gathered}
$$

## Answer

$90^{\circ}, 45^{\circ}, 45^{\circ}$ or $90^{\circ}, \operatorname{arctg} 3, \operatorname{arcctg} 3$.

## Tangent Circles  Problem 115573 Topics: [ Trigonometric ratios in a right triangle ]

Circles with centers $O 1$ and $O 2$ touch each other externally at point $C$. A line touches these circles at different points $A$ and $B$ respectively. Find the angle $A O 2 B$, if it is known that $\operatorname{tg} \angle A B C=\frac{1}{2}$.

## Solution

Let $M$ be the point of intersection of the segment $A B$ with the common tangent to the given circles, drawn through their point of tangency $C$. Then $M A=M C=M B$, hence $\angle A C B=90^{\circ}$.

Drop a perpendicular $O 2 H$ from the center $O 2$ of the second circle to its chord $B C$. Then $H$ is the midpoint of $B C$. From the condition of the problem, it follows that $A C=\frac{1}{2} B C=B H$, and since $\angle B O 2 H=90^{\circ}-\angle O 2 B H=\angle A B C$, the right triangles $B O 2 H$ and $A B C$ are equal by the leg and the opposite acute angle. Therefore, $O 2 B=A B$. Hence, $\angle A O 2 B=\angle B A O 2=45^{\circ}$.

## Answer

$45^{\circ}$.