Chapter 12.5, Problem 8PSB

To determine

To calculate: The volume of a cone with a vertex angle of ${60}^{\circ }$ and slant height of 12.

Answer to Problem 8PSB

The volume of a cone is $391.8$ .

Explanation of Solution

Given information:

Vertex angle of cone= ${60}^{\circ }$ ,

Slant height of cone = 12.

Formula used:

Volume of cone $=\frac{1}{3}\times B\times h$

B = Base area of cone and h = height of cone

Area of a circle: $A=\pi r^2$

r = radius of circle

Calculation:

  

The triangle formed by the altitude of the cone, a radius, and a slant height is ${30}^{\circ }{60}^{\circ }{90}^{\circ }\Delta$

△ACB is ${30}^{\circ }{60}^{\circ }{90}^{\circ }\Delta$

In △ABC

  $\begin{array}{l}\sin {30}^{\circ }=\frac{BC}{AC}\\ \frac{1}{2}=\frac{BC}{12}\\ BC=\frac{12}{2}=6\\ r=BC=6\end{array}$

  $\begin{array}{l}\cos {30}^{\circ }=\frac{AB}{AC}\\ \frac{\sqrt{3}}{2}=\frac{AB}{12}\\ AB=\frac{12\sqrt{3}}{2}=6\sqrt{3}\\ h=AB=6\sqrt{3}\end{array}$

  $B=A_{Circle}=\pi r^2=\pi {\left(6\right)}^2=36\pi$

Volume of cone:

  $\begin{array}{l}{V}_{Cone}=\frac{1}{3}\times B\times h\\ V_{Cone}=\frac{1}{3}\times 36\pi \times 6\sqrt{3}\\ V_{Cone}=12\pi \times 6\sqrt{3}\\ V_{Cone}=72\pi \sqrt{3}\\ V_{Cone}\approx 391.8\end{array}$