A regular heptagram is given according to the image. What is the size of the marked angle?

(E. Patáková)

![](https://cdn.mathpix.com/cropped/2024_04_17_e8320636fb5c345b613fg-4.jpg?height=309&width=320&top_left_y=842&top_left_x=868)

Hint. Look for isosceles triangles, for which you can determine the sizes of the interior angles.
The angle we are looking for will be called $\alpha$. Further, let's denote the vertices and the center of the star as in the figure. If we connect all the vertices with the center $S$, we see many isosceles triangles, many of which are congruent. For example, triangles $A S C, B S D, \ldots, G S B$ are congruent: all these triangles have congruent legs and the angle at vertex $S$ (whose size is twice the size of angle $A S B$, i.e., $\gamma = 2 \cdot \frac{360^{\circ}}{7}$). The angle $\alpha$ is therefore equal to twice the angle at vertex $A$ in triangle $A S C$. Since this triangle is isosceles, the angle $\alpha$ is the same as the sum of the interior angles at vertices $A$ and $C$.

![](https://cdn.mathpix.com/cropped/2024_04_17_e8320636fb5c345b613fg-4.jpg?height=754&width=759&top_left_y=1733&top_left_x=657)

The sum of the sizes of the interior angles in any triangle is $180^{\circ}$, so $\alpha + \gamma = 180^{\circ}$, from which we can easily calculate the size of angle $\alpha$:

$$
\alpha = 180^{\circ} - \frac{2}{7} \cdot 360^{\circ} = \left(1 - \frac{4}{7}\right) \cdot 180^{\circ} = \left(77 \frac{1}{7}\right)^{\circ} \approx 77^{\circ} 8' 34''.
$$

Note. If you know or can justify that the size of the interior angle in a regular heptagon is $\frac{5}{7} \cdot 180^{\circ}$, then the problem can be solved as follows.

![](https://cdn.mathpix.com/cropped/2024_04_17_e8320636fb5c345b613fg-5.jpg?height=751&width=763&top_left_y=687&top_left_x=652)

The interior angles at vertices $K$ and $Q$ in the isosceles triangle $K A Q$ are exterior angles of the regular heptagon $K L M N O P Q$; their size is therefore $180^{\circ} - \frac{5}{7} \cdot 180^{\circ} = \frac{2}{7} \cdot 180^{\circ}$. The sum of the sizes of the interior angles in triangle $K A Q$ is $\alpha + \frac{4}{7} \cdot 180^{\circ} = 180^{\circ}$, from which we can express the unknown: $\alpha = \left(1 - \frac{4}{7}\right) \cdot 180^{\circ} = \ldots$