N-gonal Truncated Pyramids.

$\triangle{ABC} \sim \triangle{DBE} \implies \dfrac{x}{H^{*}} = \dfrac{y}{H^{*} - H} \implies (x - y)H^{*} = xH \implies H^{*} = \dfrac{xH}{x - y}$

$\implies$

Volume of truncated pyramid $V_{T}(n) = \dfrac{n}{12}\cot(\dfrac{\pi}{n})(x^2(\dfrac{xH}{x - y}) - y^2(\dfrac{xH}{x - y} - H)) =$

$\dfrac{n}{12}\cot(\dfrac{\pi}{n})(\dfrac{x^3 - xy^2 + xy^2 - y^3}{x - y})H =$ $\dfrac{n}{12}\cot(\dfrac{\pi}{n})(x^2 + xy + y^2)H$ .

Using the fact that tangents to a circle from an outside point are congruent we obtain the diagram above.

The height $H$ of the truncated cone is $H = 2r$

Using right $\triangle{ABC}$ we have:

$(w + m)^2 = (w - m)^2 + 4r^2 \implies w^2 + 2wm + m^2 = w^2 - 2wm + m^2 + 4r^2 \implies 4wm = 4r^2 \implies r^2 = wm \implies$

$r = \sqrt{wm}$ .

The volume of the inscribed sphere $V_{s}(n) = \dfrac{4}{3}\pi(wm)^{\frac{3}{2}}$ .

The volume of the truncated $n$ - gonal pyramid is:

$V_{T}(n) = \dfrac{2n}{3}\tan(\dfrac{\pi}{n})(w^2 + wm + m^2)(wm)^{\frac{1}{2}}$

$V_{T}(n) = 2V_{s}(n) \implies w^2 + wm + m^2 = \dfrac{4\pi}{n}\cot(\dfrac{\pi}{n})wm \implies$

$w^2 - (\dfrac{4\pi}{n}\cot(\dfrac{\pi}{n}) - 1)wm + m^2 = 0$

For $n \geq 3 :$

Let $s(n) = \dfrac{4\pi}{n}\cot(\dfrac{\pi}{n}) - 1$

$(0 < s(n) < 3)$ and $s(n)$ is monotonic increasing for $n \geq 3$ and $\lim_{n \rightarrow \infty} s(n ) = 3$

To show: $\lim_{n \rightarrow \infty} s(n ) = 3$

To show: $\lim_{n \rightarrow \infty} n\sin(\dfrac{\pi}{n}) = \pi$

Using the inequality: $\cos(u) < \dfrac{\sin(u)}{u} < 1 \implies$ $\cos(\dfrac{\pi}{n} ) < \dfrac{\sin(\dfrac{\pi}{n})}{\dfrac{\pi}{n} } < 1 \implies$

$\pi\cos(\dfrac{\pi}{n}) < n\sin(\dfrac{\pi}{n}) < \pi$ and $\pi\lim_{n \rightarrow \infty} \cos(\dfrac{\pi}{n}) = \pi \implies \lim_{n \rightarrow \infty} n\sin(\dfrac{\pi}{n}) = \pi$

$\implies$

$4\pi\lim_{n \rightarrow \infty} \dfrac{\cot(\dfrac{\pi}{n})}{n} =$ $4\pi^2\lim_{n \rightarrow \infty} \dfrac{\csc^2(\dfrac{\pi}{n})}{n^2} =$

$4\pi^2\lim_{n \rightarrow \infty} \dfrac{1}{(n\sin(\dfrac{\pi}{n}))^2} = 4\pi^2(\dfrac{1}{\pi^2}) = 4$

$\implies \lim_{n \rightarrow \infty} s(n) = 3$ .

$w^2 - (\dfrac{4\pi}{n}\cot(\dfrac{\pi}{n}) - 1)wm + m^2 = 0$ $\implies$

$w(n) = (\dfrac{\dfrac{4\pi}{n}\cot(\dfrac{\pi}{n}) - 1 \pm \sqrt{(\dfrac{4\pi}{n}\cot(\dfrac{\pi}{n}) - 1)^2 - 4}}{2})m$ $= (\dfrac{s(n) \pm \sqrt{s(n)^2 - 4}}{2})m$

$\dfrac{w_{1}(n)}{m} = \dfrac{s(n) - \sqrt{s(n)^2 - 4}}{2} < \dfrac{3 - \sqrt{5}}{2} < 1$

and

$\dfrac{w_{2}(n)}{m} = \dfrac{s(n) + \sqrt{s(n)^2 - 4}}{2} > \dfrac{3 + \sqrt{5}}{2} > 1$

Since $\dfrac{w}{m} > 1$ we choose $w(n) = w_{2}(n) = (\dfrac{s(n) + \sqrt{s(n)^2 - 4}}{2})m$

$\implies \lim_{n \rightarrow \infty} w(n) = \dfrac{3 + \sqrt{5}}{2}m$

Let $\cos(\theta(n)) = \dfrac{w(n) - 1}{w(n) + 1}$

$\implies \lim_{n \rightarrow \infty} \dfrac{w(n) - m}{w(n) + m} = \dfrac{1 + \sqrt{5}}{5 + \sqrt{5}}$ $= \dfrac{1}{\sqrt{5}} = \cos(\theta)$

$\implies \theta = \boxed{63.43494882}$ .