Find the volume of a cone whose base radius is 2.2 metres and whose axial cross-section has an apex angle of 60 degrees.

Step-by-step solution

1. 1

   See what the apex angle gives you. The volume formula $V = \tfrac13\pi r^2 h$ needs a height, which is not stated. The axial cross-section is an isosceles triangle with apex angle $60^{\circ}$; the axis splits it into two right triangles with half-angle

   $\frac{60^{\circ}}{2} = 30^{\circ}$

2. 2

   Relate radius and height in that right triangle. The half-angle sits at the apex, with the radius opposite it and the height adjacent:

   $\tan 30^{\circ} = \frac{r}{h} \quad\Longrightarrow\quad h = \frac{r}{\tan 30^{\circ}} = r\sqrt{3}$

3. 3

   Compute the height.

   $h = 2.2\sqrt{3} = 3.810512\ \text{m}$

   Note the cone is taller than it is wide across the radius — a $60^{\circ}$ apex is fairly narrow.

4. 4

   Apply the volume formula.

   $V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (2.2)^2 (3.810512) = \frac{1}{3}\pi (4.84)(3.810512)$

   $V = 19.3133\ \text{m}^3$

5. 5

   Give the exact form and sanity-check. Exactly, $V = \tfrac13\pi r^3\sqrt3 = \tfrac{\sqrt3}{3}\pi (2.2)^3 = \tfrac{\sqrt3\,\pi}{3}(10.648) \approx 19.313\ \text{m}^3$. As a check, the enclosing cylinder of the same radius and height holds $\pi r^2 h = 57.94\ \text{m}^3$, and the cone is one third of that — $19.31\ \text{m}^3$.