I think the header summarises what I would like to do.
Regards
What is the apex angle measured against? Is it a vertical line which falls from the apex perpendicular to the base? (ie: bisecting the cone in the Z axis)
The volume of a cone is $\displaystyle V = \dfrac{1}{3}bh$ which for a circular base is $\displaystyle V = \dfrac{\pi r^2h}{3}$
We can also "draw" a triangle with the angle made with the apex being $\displaystyle \theta$. Since this is a right-angled triangle we can use trig: $\displaystyle \tan \theta = \dfrac{r}{h} \Leftrightarrow r = h\tan \theta$
Sub this expression for $\displaystyle r$ into $\displaystyle V$
$\displaystyle V = \dfrac{\pi (h \tan \theta)^2 h}{3} = \dfrac{\pi h^3 \tan^2 \theta}{3}$
Rearrange for h: $\displaystyle h = \sqrt[3]{\dfrac{3V}{\pi \tan^2 \theta}}$
As ever, you should always check I've not cocked up somewhere