Inscribed Octagonal Pyramid.

$\vec{OP} = -\dfrac{\sqrt{2} - 1}{\sqrt{2}}r\vec{i} + \dfrac{r}{\sqrt{2}}\vec{j} + 0\vec{k}$

$\vec{OS} = -r\vec{i} + 0\vec{j} + H\vec{k}$

$\vec{OR} = -\dfrac{\sqrt{2} - 1}{\sqrt{2}}r\vec{i} - \dfrac{r}{\sqrt{2}}\vec{j} + 0\vec{k}$

$\implies$

$\vec{U} = \vec{OP} \:\ X \:\ \vec{OS} = \dfrac{rH}{\sqrt{2}}\vec{i} + \dfrac{\sqrt{2} - 1}{\sqrt{2}}rH\vec{j} + \dfrac{r^2}{\sqrt{2}}\vec{k}$

and

$\vec{V} = \vec{OR} \:\ X \:\ \vec{OS} = -\dfrac{rH}{\sqrt{2}}\vec{i} + \dfrac{\sqrt{2} - 1}{\sqrt{2}}rH\vec{j} - \dfrac{r^2}{\sqrt{2}}\vec{k}$

$\vec{U} \circ \vec{V} = |\vec{U}| |\vec{V}| \cos(\theta) \implies \cos(\theta) = \dfrac{\vec{U} \circ \vec{V}}{|\vec{U}| |\vec{V}|}$

$\implies \vec{U} \circ \vec{V} = -\dfrac{r^2}{2}(2(\sqrt{2} - 1)H^2 + r^2)$

$|\vec{U}| = \dfrac{r}{\sqrt{2}}\sqrt{2(2 - \sqrt{2})H^2 + r^2} = |\vec{V}|$

$\implies \cos(\theta) = -\dfrac{2(\sqrt{2} - 1)H^2 + r^2}{2(2 - \sqrt{2})H^2 + r^2}$

For octagonal base $x = 2r\sin(\dfrac{\pi}{8})$ and the height $h = r\cos(\dfrac{\pi}{8}) \implies$

The area of the octagonal base $A_{p} = 2\sqrt{2}r^2 \implies$ the volume of the octagonal pyramid

$V_{p} = \dfrac{2\sqrt{2}}{3}r^2 H$

$R^2 = H^2 - 2HR + R^2 + r^2 \implies H^2 - 2HR + r^2 = 0 \implies$

$r^2 = 2HR - H^2 \implies V_{p} = \dfrac{2\sqrt{2}}{3}(2H^2R - H^3) \implies$

$\dfrac{dV_{p}}{dH} = \dfrac{2\sqrt{2}}{3}H(4R - 3H) = 0 \:\ H \neq 0 \implies H = \dfrac{4}{3}R \implies r^2 = \dfrac{8}{9}R^2$

$\implies \cos(\theta) = -\dfrac{4\sqrt{2} - 3}{9 - 4\sqrt{2}} \implies \theta \approx \boxed{142.6285^{\circ}}$

$$

$$

Note: $\dfrac{d^2V_{p}}{dH^2}|_{H = \dfrac{4}{3}R} = -\dfrac{8\sqrt{2}}{3}R < 0 \implies$ a max occurs at $H = \dfrac{4}{3}R$ .