Chapter 12.5, Problem 20PPS

To determine

To find the volume of the given cone.

Answer to Problem 20PPS

  $V\approx 2.8$ cubic feet (approximately)

Explanation of Solution

Given:

  

Slant height, $(l)=2$ feet, $\theta ={47}^0$

Formula used:

Volume of cone, $V=\frac{1}{3}Bh=\frac{1}{3}\pi r^{2}h$

where $B=\pi r^2$ is the area of the base and $h$ is the height of the cone.

Trigonometric formulas:

  $\sin \theta =\frac{perpendicular}{hypotenuse}$

  $\cos \theta =\frac{base}{hypotenuse}$

Calculation:

  

In $\Delta ABC$ ,

  $AB=$ base, $BC=$ perpendicular, and $CA=$ hypotenuse.

  $AB=r$ , $BC=h$

  $\Rightarrow \sin \theta =\frac{perpendicular}{hypotenuse}$

  $\Rightarrow \sin \theta =\frac{BC}{CA}$

  $\Rightarrow \sin {47}^0=\frac{BC}{CA}$

  $\Rightarrow \sin {47}^0=\frac{h}{2}$

  $\Rightarrow 0.73=\frac{h}{2}$

  $[\because \sin {47}^0=0.73]$

On cross-multiplying,

  $\Rightarrow h=1.46$ feet

Now,

  $\Rightarrow \cos \theta =\frac{base}{hypotenuse}$

  $\Rightarrow \cos \theta =\frac{AB}{CA}$

  $\Rightarrow \cos {47}^0=\frac{AB}{CA}$

  $\Rightarrow \cos {47}^0=\frac{r}{2}$

  $\Rightarrow 0.68=\frac{r}{2}$

  $[\because \cos {47}^0=0.68]$

On cross-multiplying,

  $\Rightarrow r=1.36$ feet

Now for the volume ( $V$ ) of the cone,

  $V=\frac{1}{3}Bh=\frac{1}{3}\pi r^{2}h$

Put the values of $r=1.36$ and $h=1.46$ ,

  $\Rightarrow \frac{1}{3}\times 3.14\times {(1.36)}^2\times 1.46$

  $\left[\pi =3.14\right]$

  $\Rightarrow \frac{8.4793}{3}$

  $\Rightarrow 2.8$ cubic feet

  $\Rightarrow V\approx 2.8$ cubic feet

Conclusion:

Therefore, the volume of the given cone is $2.8$ cubic feet (approximately).