Answer:
Let \( h \) be the height of the cone, and \( r \) be the radius of the base. The semi-vertical angle \( \theta \) is given as 30°.
Using trigonometry, we can relate the height, radius, and the semi-vertical angle in the cone:
\[ \tan(\theta) = \frac{r}{h} \]
Given \( \theta = 30° \), we have:
\[ \tan(30°) = \frac{r}{h} \]
\[ \frac{1}{\sqrt{3}} = \frac{r}{h} \]
\[ h = r\sqrt{3} \]
Now, we'll show that the volume \( V \) of the cone is \( \frac{\pi r^3 \sqrt{3}}{3} \):
The volume of a cone is given by the formula:
\[ V = \frac{1}{3} \pi r^2 h \]
Substitute \( h = r\sqrt{3} \) into the formula:
\[ V = \frac{1}{3} \pi r^2 (r\sqrt{3}) \]
\[ V = \frac{\pi r^3 \sqrt{3}}{3} \]
So, the correct option is:
D) \( h = 2r\sqrt{3} \) , \( V = \frac{\pi r^3 \sqrt{3}}{3} \)