JAMB 1984 · UME · Q39

A right circular cone has base radius $r$ cm and vertical angle $2y^\circ$ . The height of the cone is

Worked solution (try it first)

The height drops from the vertex to the centre of the base and cuts the vertical angle in half, so the half-angle at the vertex is $y^\circ$ .
In that right-angled triangle, $r$ is opposite the angle $y^\circ$ and the height $h$ is adjacent: $\tan y^\circ = \dfrac rh$ .
Rearrange: $h = \dfrac{r}{\tan y^\circ}$
$= r\cot y^\circ$ cm, option C.