Given the right circular cone of fixed height h and semi-vertical angle a. Let R be the radius of the base and H be the height of the right circular cylinder that can be inscribed in the right circular cone.
In the figure, ∠GAO = α, OG = r, OA = h, OE = R, CE = H.
We have, r/h = tan α
∴ r = h tan α ……(1)
Since ∆AOG and ∆CEG are similar.
Let V be the volume of the cylinder.
Hence, the height of the right circular cylinder is one-third of that of the cone.
