## 161. Math Puzzle 10/78

How large are the angles $\alpha$ at the tips of the five-pointed star?

If you take a closer look at the six sub-figures and consider the laws of the angles in them, you can easily calculate $\alpha$.
With the formula $\alpha=\frac{(n-2) \cdot 180^{\circ}}{n}$, one can calculate the angles of a regular $n$-gon. Therefore, the angles of the 5-gon in the center of the star are $108^{\circ}$, because

$$
\frac{(5-2) \cdot 180^{\circ}}{5}=108^{\circ}
$$

Their supplementary angles are $180^{\circ}-108^{\circ}=72^{\circ}$. They are simultaneously the base angles of the isosceles triangles in which the angle $\alpha$ lies.

Since the sum of the angles in each triangle is $180^{\circ}$, it must hold that: $72^{\circ}+72^{\circ}+\alpha=180^{\circ}$, thus $\alpha=36^{\circ}$.