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(D) Given: Height of the cone $h = 6.125 	ext{ cm}$ and semi-vertical angle $	heta = 30^{\circ}$.The radius $R$ of the base of the cone is related t

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$(25.5) \pi$

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(D) Given: Height of the cone $h = 6.125 	ext{ cm}$ and semi-vertical angle $	heta = 30^{\circ}$.The radius $R$ of the base of the cone is related to the height $h$ by the relation: $	an 	heta = \frac{R}{h}$.Substituting the values: $	an 30^{\circ} = \frac{R}{6.125} \Rightarrow \frac{1}{\sqrt{3}} = \frac{R}{6.125} \Rightarrow R = \frac{6.125}{\sqrt{3}} 	ext{ cm}$.The volume $V$ of a right circular cone is given by $V = \frac{1}{3} \pi R^2 h$.Substituting the values of $R$ and $h$: $V = \frac{1}{3} \pi \left( \frac{6.125}{\sqrt{3}} ight)^2 (6.125) = \frac{1}{3} \pi \left( \frac{6.125^2}{3} ight) (6.125) = \frac{\pi}{9} (6.125)^3$.Calculating $(6.125)^3$: $6.125 	imes 6.125 	imes 6.125 \approx 229.996$.Thus,$V \approx \frac{229.996}{9} \pi \approx 25.555 \pi 	ext{ cm}^3$.Rounding to the nearest given option,the approximate value is $(25.5) \pi 	ext{ cm}^3$.

The semi-vertical angle of a right circular cone is $30^{\circ}$. If the height of the cone is $6.125 \text{ cm}$,then the approximate value of the volume of the cone (in cubic $\text{cm}$) is

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