More Pyramids

The desired angle is independent of $n$ and the work below verifies this.

For area of $n - gon$ :

Let $BC = x$ be a side of the $n - gon$ , $AC = AB= r$ , $AD = h^*$ , and $\angle{BAD} = \dfrac{180}{n}$ .

$\dfrac{x}{2} = r \sin(\dfrac{180}{n}) \implies r = \dfrac{x}{2 \sin(\dfrac{180}{n})} \implies h^* = \dfrac{x}{2} \cot(\dfrac{180}{n}) \implies A_{\triangle{ABC}} = \dfrac{1}{4} \cot(\dfrac{180}{n}) x^2 \implies$

$A_{n - gon} = \dfrac{n}{4} \cot(\dfrac{180}{n}) x^2 \implies$ the Volume of the pyramid $V_{p} = \dfrac{n}{12} \cot(\dfrac{180}{n}) x^2 H$

Let $AC = H$ be the height of the pyramid, $BC = h^*$ , and $AB = h'$ be the slant height of the pyramid and $m\angle{CBA} = \theta$

The lateral surface area $S = \dfrac{n}{2} x h'$ .

$h^* = \dfrac{x}{2} \cot(\dfrac{180}{n}) = h' \cos(\theta) \implies x = 2 \tan(\dfrac{180}{n}) \cos(\theta) h'$ and $H = h' \sin(\theta)$ and letting $u(n) = \tan(\dfrac{180}{n}) \implies$ $V_{p} = \dfrac{n}{3} u(n) \cos^2(\theta) \sin(\theta) h'^3 = K$ and $S = n * u(n) cos(\theta) h'^2.$

$V_{p} = \dfrac{n}{3} u(n) \cos^2(\theta) \sin(\theta) h'^3 = K \implies h' = (\dfrac{3K}{n * u(n_) cos^2(\theta) sin(\theta)})^{\dfrac{1}{3}} \implies S(\theta) = j(n) * (sec(\theta))^{\dfrac{1}{3}} (csc(\theta))^{\dfrac{2}{3}}$ , where $j(n) = (9 k^2 n * u(n))^{\dfrac{1}{3}}$

$\implies \dfrac{dS}{d\theta} = \dfrac{j(n)}{3} (sec\theta)^\dfrac{1}{3} (\csc\theta)^\dfrac{2}{3} * (\tan\theta - 2 \cot\theta) = 0 \implies$

$3 \sin^2\theta - 2 = 0 \implies \sin\theta = \pm \sqrt{\dfrac{2}{3}}$ . choosing $\sin\theta = \sqrt{\dfrac{2}{3}} \implies \theta = \boxed{54.7356^\circ}$

$\therefore$ The measure of the desired angle $\theta$ is independent of $n$ .