How to know the volume and density of the material, as well as the angle $\alpha= 42^\circ,$ to find the radius and height of the semi-cone

Question

Is it possible, knowing the volume and density of the material, as well as the angle $\alpha= 42^\circ,$ to find the radius and height of the semi-cone

[pic of semi-cone] ?

Answer 1

The volume of this cone is

$ V = \dfrac{1}{3} \pi R^2 H $

Now you are given the angle $\alpha = 42^\circ$, and this angle satisfies

$ \tan(\alpha) = \tan(42^\circ) = \dfrac{H}{R}$

Therefore, $H = R \tan(42^\circ) $

Upon substituting this into the equation for the volume, one gets

$ V = \dfrac{1}{3} \pi R^3 \tan(42^\circ) $

Solving for $R$

$ R =\bigg( \dfrac{3 V}{\pi \tan(42^\circ) }\bigg)^{\dfrac{1}{3}} $

Once you have $R$ you can also calculate $H$, from the above equation relating $H$ and $R$.

Answer 2

By placing equally infinitesimally thin sliding discs between the base and a line parallel to a base we have a constant for volumes of oblique cones as $ \frac13$. So we need to solve together $$ \frac{H}{R}=\tan 42^{\circ}$$ $$ \frac13. \pi (R/2)^2 . H = 100. $$