2.28. Through the vertex $A$ of a right circular cone, a section of maximum area is drawn. Its area is twice the area of the section passing through the axis of the cone. Find the angle at the vertex of the axial section of the cone.
2.28. Consider an arbitrary section passing through the vertex $A$. This section is a triangle $A B C$, and its sides $A B$ and $A C$ are the generators of the cone, i.e., they have a constant length. Therefore, the area of the section is proportional to the sine of the angle $B A C$. The angle $B A C$ varies from $0^{\circ}$ to $\varphi$,
where $\varphi$ is the angle at the vertex of the axial section of the cone. If $\varphi \leqslant 90^{\circ}$, then the axial section has the maximum area, and if $\varphi > 90^{\circ}$, then the section with a right angle at the vertex $A$ has the maximum area. Thus, from the problem statement, it follows that $\boldsymbol{\operatorname { s i n }} \varphi=0.5$ and $\varphi > 90^{\circ}$, i.e., $\varphi=120^{\circ}$.