A regular pentagonal pyramid's side edge is equal to its base edge. What angle does a side edge form with a base edge that does not intersect it and is not perpendicular to it?
We understand the angle between two skew lines as the angle between two lines passing through an arbitrary point in space and parallel to the given lines. Our task (according to the notation in the figure) is to determine the angle enclosed by the base edge $BC$ and the side edge $MA$ of the pyramid.

![](https://cdn.mathpix.com/cropped/2024_05_02_f435943bf90a326decdbg-1.jpg?height=391&width=637&top_left_y=260&top_left_x=730)

Since the diagonal $AD$ of the pentagon is parallel to $BC$, according to our previous definition, the sought angle is equal to the angle $MAD$. Given that the side edges of the pyramid are equal to the base edge, and due to the equality of the sides,

$$
\triangle ADM \cong \triangle ADE
$$

and thus the sought angle $\angle MAD = \angle EAD$.

However, the angle at vertex $E$ of the isosceles triangle $ADE$ is the angle of a regular pentagon, which is $108^\circ$, and thus the sought

$$
\angle MAD = \angle EAD = \frac{180^\circ - 108^\circ}{2} = 36^\circ.
$$

Rétey Piroska (Debrecen, Svetits III. o. t.)