Find the volume of a cone whose diameter is 2.2 metres and whose apex angle is 60 degrees.

Step-by-step solution

1. 1

   Convert the diameter to a radius.

   $r=\frac{d}{2}=\frac{2.2}{2}=1.1\ \text{m}$

   The volume formula uses the radius, so this conversion has to come first.

2. 2

   Interpret "60-degree cone". The standard convention is that the quoted angle is the full apex angle of the axial cross-section — the triangle you see when slicing the cone through its axis. The axis bisects that angle, so the half-angle at the tip is

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

3. 3

   Get the height from the half-angle. In the right triangle formed by the axis, the base radius and the slant, the radius is opposite the $30^{\circ}$ half-angle and the height is adjacent to it:

   $\tan 30^{\circ}=\frac{r}{h}\quad\Rightarrow\quad h=\frac{r}{\tan 30^{\circ}}=r\sqrt3=1.1\sqrt3\approx 1.9052\ \text{m}$

   Using $\tan 30^{\circ}=1/\sqrt3$ keeps the answer exact.

4. 4

   Apply the cone volume formula.

   $V=\frac13\pi r^2h=\frac13\pi(1.1)^2\left(1.1\sqrt3\right)=\frac{1.331\sqrt3}{3}\pi\ \text{m}^3$

   since $(1.1)^2=1.21$ and $1.21\times 1.1=1.331$.

5. 5

   Evaluate numerically. With $\sqrt3=1.7320508$,

   $V=\frac{1.331(1.7320508)}{3}\pi=0.7684532\pi\approx 2.4142\ \text{m}^3$

6. 6

   Check against the enclosing cylinder, and note the ambiguity. A cylinder with the same radius and height holds $\pi r^2h=\pi(1.21)(1.9052)=7.2425$ m$^3$, and a cone is exactly one third of that: $7.2425/3=2.4142$ ✓. For completeness: if the $60^{\circ}$ had been meant as the half-angle, then $h=1.1/\tan 60^{\circ}=0.6351$ m and $V\approx 0.8047$ m$^3$ — a threefold difference, which is why stating the convention matters.