Chapter 12.5, Problem 19PPS

To determine

To find the volume of the given cone.

Answer to Problem 19PPS

cubic centimeters (approximately) $V\approx 190$

Explanation of Solution

Given:

  

  $\text{Height}(h)=11.5$ centimeters, $\theta ={20}^0$

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:

  $\tan \theta =\frac{perpendicular}{base}$

  $\tan {20}^0=0.36$

Calculation:

  

In $\Delta ABC$ ,

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

  $AC=h$

  $\Rightarrow \tan \theta =\frac{perpendicular}{base}$

  $\Rightarrow \tan \theta =\frac{AC}{BC}$

  $\Rightarrow \tan {20}^0=\frac{AC}{BC}$

  $\Rightarrow \tan {20}^0=\frac{h}{8}$

  $\Rightarrow 0.36=\frac{h}{8}$

On cross-multiplying,

  $\Rightarrow h=0.36\times 8$

  $\Rightarrow h=2.88$ centimeters

Now for the volume of the cone,

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

Put the values of $r=8$ and $h=2.88$ ,

  $\Rightarrow \frac{1}{3}\times 3.14\times {(8)}^2\times 2.88$

  $\left[\pi =3.14\right]$

  $\Rightarrow \frac{578.7648}{3}$

  $\Rightarrow V\approx 190$ cubic centimeters

Conclusion:

Therefore, the volume of the given cone is $190$ cubic centimeters (approximately).