# A ball of radius R is inscribed into the cone from the center of the ball; the generatrix

**A ball of radius R is inscribed into the cone from the center of the ball; the generatrix of the cone is seen at an angle alpha. Find the volume of the cone.**

In a right-angled triangle AOH, the angle AOH is adjacent to the angle ABO, then the angle AOH = 180 – AOB = (180 – α). Then the angle OAH = 180 – 90 – (180 – α) = (α – 90).

Then, in the triangle AOH tg (180 – α) = AH / OH = AH / R.

AH = R * tan (180 – α) = -R * tan α.

By the property of a tangent drawn from one point, the angle ОАH = ОАВ = (α – 90), then the angle НАВ = ОАН + ОАВ = (α – 90) + (α – 90) = 2 * α – 180.

In a right-angled triangle ABH tg (2 * α – 80) = BH / AH.

BH = AH * tan (2 * α – 180) = -R * tan α * tan (2 * α).

Determine the volume of the cone.

V = Sosn * h / 3 = n * AH2 * BH / 3 = n * (-R * tan α) ^ 2 * (-R * tan α * tan (2 * α)) / 3 = – (n * R3 * (tan α) ^ 3 * tan (2 * α)) / 3.

Answer: The volume of the cone is – (n * R ^ 3 * (tan α) ^ 3 * tan (2 * α)) / 3.