A cone and a cylinder have the same height and volume. What is the opening angle of the cone if the surface area of the lateral surfaces of the two bodies is also equal?
Let the radius of the base circle of the cone be $r_{1}$, the radius of the base circle of the cylinder be $r_{2}$, their common height be $m$, their volume be $V$, and the slant height of the cone be $a$. Then,

$$
V_{\text {cone }}=\frac{r_{1}^{2} \pi m}{3}, \quad V_{\text {cylinder }}=r_{2}^{2} \pi m
$$

From the equality of the volumes, we get $r_{1}=\sqrt{3} r_{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_b6dc33ba85e96f90a1fcg-1.jpg?height=268&width=285&top_left_y=381&top_left_x=909)

The lateral surface area of the cone is $r_{1} \pi a$, and that of the cylinder is $2 r_{2} \pi m$. From the equality of the lateral surface areas and the ratio of the radii, it follows that $\frac{m}{a}=\frac{\sqrt{3}}{2}$.

A plane passing through the axis of the cone cuts an isosceles triangle from the cone. Let the angle at the base of this triangle be $\alpha$. Then $\sin \alpha=\frac{m}{a}=\frac{\sqrt{3}}{2}$, from which $\alpha=60^{\circ}$ (since $\alpha$ is an acute angle). Therefore, the plane section is an equilateral triangle, and thus the cone's opening angle is also $60^{\circ}$.