# The height of the cone is 4√3 cm, and the angle at the apex of the axial overcut is 120 degrees.

**The height of the cone is 4√3 cm, and the angle at the apex of the axial overcut is 120 degrees. find the area of the base of the cone.**

The diameter AB of the base of the cone and the generators AC and BC form an isosceles triangle ABC with an angle of 120 at its apex C.

The height of the OC is also the bisector and the median of the triangle ABC, then in the right-angled triangle AOC the angle ACO = ACB / 2 = 120/2 = 60.

In a right-angled triangle AOC, through the leg and the adjacent angle, we determine the length of the other leg.

tgACO = AO / OC.

AO = R = OC * tgACO = 4 * √3 * tg60 = 4 * √3 * √3 = 12 cm.

Determine the area of the circle at the base of the cone.

Sbn = n * R ^ 2 = 144 * n cm2.

Answer: The area of the base of the cone is 144 * n cm2.