A ball is inscribed into the cone, the angle at the apex of the axial section of which is 60 degrees.

A ball is inscribed into the cone, the angle at the apex of the axial section of which is 60 degrees. Find the volume of the cone if the volume of the ball is 2.

The volume of the ball is known by the problem statement. Let’s use the volume formula and find the radius of this ball:
V w. = 4 * π * R³ / 3 → R = ∛ (3 / 2π).
The axial section of the cone is a regular triangle (an apex angle of 60 ° – by condition). The height of this triangle is:
h = 3 * R = 3 * ∛ (3 / 2π).
Find the radius of the base of the cone:
r = h / tg 60 ° = 3 * ∛ (3 / 2π) / √3 = √3 * ∛ (3 / 2π).
We find the volume of the cone by the formula:
V к. = 1/3 * π * r² * h = 1/3 * π * (√3 * ∛ (3 / 2π)) ² * 3 * ∛ (3 / 2π) = 1/3 * π * 3 * ∛ (3 / 2π) ² * 3 * ∛ (3 / 2π) = π * 3/2 π * 3 = 9/2 = 4.5.
Answer: The volume of the cone is 4.5 cubic units.