Vertices $A$ and $B$ of the prism $A B C A 1 B 1 C 1$ lie on the axis of the cylinder, while the other vertices lie on the lateral surface of the cylinder. Find the dihedral angle in this prism with edge $A B$.
When orthogonally projecting the data of a cylinder and a prism onto the plane of the cylinder's base, the lateral surface will transform into a circle equal to the base circle, vertices $A$ and $B$ will transform into the center $O$ of this circle, vertices $A 1$ and $B 1$ will transform into some point $M$ on this circle, and vertices $C$ and $C 1$ will transform into points $P$ and $Q$, respectively, lying on the circle (Fig. 2). Since $A A 1 \| C C 1$ and $A A 1 = C C 1$, then $O M = P Q$. Similarly, $O P = M Q$. In addition, $O M = O P$ as radii of the same circle. Therefore, quadrilateral $O M Q P$ is a rhombus, in which diagonal $O Q$ is equal to the side. Thus, $\angle M O Q = \angle P O Q = 60^{\circ}$, and $\angle M O P = 120^{\circ}$. The lines $O P$ and $O M$ lie in the faces of the dihedral angle with edge $A B$ and are perpendicular to the line $A B$. Therefore, $M O P$ is the linear angle of the sought dihedral angle.

## Answer

$120^{\circ}$.

Submit a comment