Review question

# If a cone sits on this sphere, what 's the cone's volume? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R9694

## Solution

A sphere of radius $\quantity{6}{cm}$ fits inside a cone of vertical angle $84^\circ$, so that $BA$ is a tangent to the sphere.

Calculate

1. the length $AB$;

Since $AB$ is tangent to the sphere, $\angle BAO = 90^\circ$. We are also given that $OA=6$, and that the vertical angle of the cone is $84^\circ$, so $\angle OBA = 42^\circ$, giving us

Now $\tan 42^\circ = \frac{\text{OA}}{\text{AB}} \implies \text{AB} = \frac{6}{\tan 42^\circ} = \quantity{6.66}{cm}.$

1. the radius of the base of the cone;
The length we require is $AC$. If we consider the triangle $CBA$, we have that \begin{align*} \sin 42^\circ = \frac{\text{AC}}{\text{AB}} \implies \text{AC} &= \frac{6 \sin 42^\circ}{\tan 42^\circ}\\ &= 6 \cos 42^\circ\\ &= \quantity{4.46}{cm}. \end{align*}
1. the shortest distance from $B$ to the sphere;

The point $D$ is the one on the sphere closest to B, so our goal is to find $BD$. We have $OD=6$, since it’s a radius of the sphere, and hence $\text{BD} = \text{BO} - 6.$ So we need to calculate the length of $BO$. Considering $\triangle AOB$, we have that $\sin 42^\circ = \frac{\text{AO}}{\text{BO}} \implies \text{BO} = \frac{6}{\sin 42^\circ}$ and therefore $\text{BD} = \frac{6}{\sin 42^\circ} - 6= \quantity{2.97}{cm}.$

1. the volume of the cone.
The volume of a cone is $\frac{1}{3} \times \text{area of the base} \times \text{height of the cone}.$ For us, the base is a disc with $AC$ as a radius, and the height of the cone is $BC$. Considering $\triangle ACB$, we see that $\tan 42^\circ = \frac{\text{AC}}{\text{BC}} \implies \text{BC} = \frac{\text{AC}}{\tan 42^\circ}.$ Thus the volume of the cone is \begin{align*} \frac{1}{3} \times \pi \times \text{AC}^2 \times \text{BC} &= \frac{\pi (6 \cos 42^\circ)^3}{3 \tan 42^\circ} \\ &= \frac{72 \pi (\cos 42^\circ)^4}{\sin 42^\circ}\\ &= \quantity{103.1}{cm^3}. \end{align*}