The length of segment $AB$ is $2r$. We draw a circle with center $A$ and radius $r$, and a circle with radius $r$ that is tangent to segment $AB$ at point $B$. What angle do the common tangents of these two circles form with segment $AB$?
The figure shows the construction.

![](https://cdn.mathpix.com/cropped/2024_05_02_d4fb9ef66c767703c57fg-1.jpg?height=591&width=621&top_left_y=195&top_left_x=752)

Since the radii of the two circles are equal, the common external tangents form the same angle $\alpha$ with the line $AB$ as the central line $AC$, because $\operatorname{tg} \alpha = \frac{r}{2r} = \frac{1}{2}$, hence $\alpha = 26^{\circ} 34'$.

At the midpoint of $AB$, the perpendicular $t_1$ to $AB$ is one of the common internal tangents of the two circles. Therefore, it forms a $90^{\circ}$ angle with $AB$. The internal tangents thus form an angle of $\left(90^{\circ} - \alpha\right)$ with the central line $AC$, and therefore an angle of $180^{\circ} - 2\left(90^{\circ} - \alpha\right) = 2\alpha$ with each other. Thus, the angle $\beta$ that $t_2$ forms with $AB$ is

$$
\beta = 90^{\circ} - 2\alpha = 90^{\circ} - 53^{\circ} 08' = 36^{\circ} 52'
$$

László Ványai (Sátoraljaújhely, Kossuth g. II. o. t.)