Slant Hexagonal Pyramid 2.

Let $r$ be the length of a side of the hexagon and $h = GA$ be the height of the hexagonal pyramid.

$C:(\dfrac{r}{2},\dfrac{\sqrt{3}}{2},0), \:\ B:(-\dfrac{r}{2},\dfrac{\sqrt{3}}{2},0)$ and $G(r,0,h)$ .

$\vec{u} = -r\vec{i} + 0\vec{j} + 0\vec{k}$ and $\vec{v} = \dfrac{r}{2}\vec{i} - \dfrac{\sqrt{3}}{2}\vec{j} + h\vec{k} \implies \vec{u} X \vec{v} = 0\vec{i} + rh\vec{j} + \dfrac{\sqrt{3}}{2}r^2\vec{k} \implies$

$d = \dfrac{\sqrt{4h^2 + 3r^2}}{2} \implies A_{\triangle{GCB}} = A_{\triangle{GEF}} = \dfrac{1}{4}\sqrt{4h^2 + 3r^2}r$

Let $A = \dfrac{1}{4}\sqrt{4h^2 + 3r^2}r$

The volume $V = \dfrac{1}{4\sqrt{3}}r^2h = k \implies h = \dfrac{4\sqrt{3}}{r^2} \implies$ $A(r) = \dfrac{1}{4}\dfrac{\sqrt{192k^2 + 3r^6}}{r} \implies$ $\dfrac{dA}{dr} = \dfrac{3(r^6 - 32k^2)}{2r^2\sqrt{192k^2 +3r^6}} = 0 \implies r = (4\sqrt{2}k)^{\frac{1}{3}} \implies h = 4\sqrt{3}k(\dfrac{1}{4\sqrt{2}k})^{\frac{2}{3}}$

$\vec{ED} = -\dfrac{r}{2}\vec{i} + \dfrac{\sqrt{3}}{2}r\vec{j} + 0\vec{k}$

$\vec{EF} = r\vec{i} + 0\vec{j} + 0\vec{k}$

$\vec{EG} = \dfrac{3}{2}r\vec{i} + \dfrac{\sqrt{3}}{2}\vec{j} + h\vec{k}$

$\implies \vec{p} = \vec{EG} \:\ X \:\ \vec{ED} = -\dfrac{\sqrt{3}}{2}rh\vec{i} - \dfrac{1}{2}rh\vec{j} + \sqrt{3}r^2\vec{k}$

and

$\vec{q} = \vec{EG} \:\ X \:\ \vec{EF} = 0\vec{i} + rh\vec{j} - \dfrac{\sqrt{3}}{2}r^2\vec{k}$

$\vec{p} \cdot \vec{q} = -\dfrac{1}{2}r^2(h^2 + 3r^2) < 0 \implies$

$\cos(\theta) = -\dfrac{|\vec{p} \cdot \vec{q}|}{|\vec{p}| |\vec{q}|} = -\dfrac{\sqrt{h^2 + 3r^2}}{\sqrt{4h^2 + 3r^2}} = -\dfrac{1}{\sqrt{2}} \implies \theta = \boxed{135^\circ}$ .