Star Pyramid

To find $A_{star}:$

$m\angle{AOP} = \dfrac{\pi}{5}$ and by Inscribed Angle Theorem $m\angle{EPF} = \dfrac{1}{2}m\angle{EOF} \implies m\angle{APO} = \dfrac{\pi}{10}$

$\dfrac{AC}{OC} = \tan(\dfrac{\pi}{5}) \implies OC = \dfrac{AC}{\tan(\dfrac{\pi}{5})}$ and $PC = \dfrac{AC}{\tan(\dfrac{\pi}{10})} = r - OC = r - \dfrac{AC}{\tan(\dfrac{\pi}{5})} \implies$

$AC = \dfrac{\tan(\dfrac{\pi}{10})\tan(\dfrac{\pi}{5})}{\tan(\dfrac{\pi}{5}) + \tan(\dfrac{\pi}{10})}r$

$\implies A_{\triangle{AOP}} = \dfrac{1}{2}(\dfrac{\tan(\dfrac{\pi}{10})\tan(\dfrac{\pi}{5})}{\tan(\dfrac{\pi}{5}) + \tan(\dfrac{\pi}{10})})r^2$

$\tan(\dfrac{\pi}{5}) = \dfrac{2\tan(\dfrac{\pi}{10})}{1 - \tan^2(\dfrac{\pi}{10})} \implies$ $A_{\triangle{AOP}} = \dfrac{\tan(\dfrac{\pi}{10})}{3 - \tan^2(\dfrac{\pi}{10})}r^2$

$\implies A_{star} = \dfrac{10\tan(\dfrac{\pi}{10})}{3 - \tan^2(\dfrac{\pi}{10})}r^2$

Let $f = \dfrac{10\tan(\dfrac{\pi}{10})}{3 - \tan^2(\dfrac{\pi}{10})}$ .

It will be shown that $\theta$ is independent $f$ .

Using the above diagram:

$EG^2 = 2r^2(1 - \cos(\dfrac{4\pi}{5}) = 4\sin^2(\dfrac{2\pi}{5})r^2 \implies EG = 2\sin(\dfrac{2\pi}{5})r$ $\implies OR = \sqrt{r^2 - r^2\sin^2(\dfrac{2\pi}{5})} = r\cos(\dfrac{2\pi}{5})$ and $RQ = \sqrt{h^2 + r^2\cos^2(\dfrac{2\pi}{5})}$

$\implies$ The lateral surface area $A = 5\sin(\dfrac{2\pi}{5})r\sqrt{h^2 + r^2\cos^2(\dfrac{2\pi}{5})}$

The volume $V = \dfrac{1}{3}fr^2h = k \implies k = \dfrac{3k}{fr^2} \implies$ $A(r) = 5\sin(\dfrac{2\pi}{5})r\sqrt{9k^2 + f^2\cos^2(\dfrac{2\pi}{5})r^6} \implies$ $\dfrac{dA}{dr} = (5\sin(\dfrac{2\pi}{5}))\dfrac{2f^2\cos^2(\dfrac{2\pi}{5})r^6 - 9k^2}{r^2\sqrt{9k^2 + f^2\cos^2(\dfrac{2\pi}{5})r^6}} \implies r = (\dfrac{3k}{\sqrt{2}f\cos(\dfrac{2\pi}{5})})^{\frac{1}{3}}$ $\implies h = \dfrac{3k}{f}(\dfrac{\sqrt{2} f\cos(\dfrac{2\pi}{5})}{3k})^{\frac{2}{3}}$

$\implies \tan(\theta) = \dfrac{h}{r} = \sqrt{2}\cos(\dfrac{2\pi}{5}) = \sqrt{2}\cos(72^\circ) \implies \theta \approx \boxed{23.606099^\circ}$ .