Tetrahedrons

$\triangle{AEC}$ is an isosceles triangle $\implies EM$ is the perpendicular bisector of base $AC$ $\implies \triangle{MEC}$ is a right triangle and $AM = MC$ .

Since $\angle{ACB}$ is a common angle to both right triangle $ABC$ and $MEC \implies \triangle{ABC} \sim \triangle{MEC} \implies \dfrac{2m}{m} = \dfrac{x + a}{a} \implies x = a$ and the pythagorean theorem in $\triangle{ABC} \implies y = 2m = \sqrt{3}a$ .

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Let $h$ be the height of the tetrahedron.

$\vec{v} = -\sqrt{3}a\vec{i} + a\vec{j} + 0\vec{k}$

$\vec{u} = -\sqrt{3}a\vec{i} + 0\vec{j} + h\vec{k}$

$\implies \vec{u} X \vec{v} = -ah\vec{i} - \sqrt{3}ah\vec{j} - \sqrt{3}a^2\vec{k} \implies |\vec{u} X \vec{v}| = a\sqrt{4h^2 + 3a^2}$ and $|\vec{u}| = \sqrt{3a^2 + h^2}$

$\implies d = \dfrac{a\sqrt{4h^2 + 3a^2}}{\sqrt{3a^2 + h^2}} \implies A_{\triangle{BCD}} = \dfrac{1}{2}a\sqrt{4h^2 + 3a^2}$

$\triangle{BCD} \cong \triangle{BC'D}$

Let $A = A_{\triangle{BCD}} = A_{\triangle{BC'D}} = \dfrac{1}{2}a\sqrt{4h^2 + 3a^2}$ .

The volume $V = \dfrac{1}{\sqrt{3}} a^2 h = k \implies h = \dfrac{\sqrt{3}k}{a^2} \implies A(a) = \dfrac{1}{2}\dfrac{\sqrt{12k^2 + 3a^6}}{a} \implies$

$\dfrac{dA}{da} = \dfrac{3(a^6 - 2k^2)}{a^2\sqrt{12k^2 + 3a^6}} = 0 \:\ a \neq 0 \implies \boxed{a = (\sqrt{2}k)^{\frac{1}{3}}} \implies \boxed{h = \sqrt{3}(\dfrac{k}{2})^{\frac{1}{3}}}$

$\dfrac{dA}{da} < 0$ when $a < (\sqrt{2}k)^{\frac{1}{3}}$ and $\dfrac{dA}{da} > 0$ when $a > (\sqrt{2}k)^{\frac{1}{3}} \implies$ minimum at $a = (\sqrt{2}k)^{\frac{1}{3}}$ .

$|\vec{v}| = 2a$ and using $|\vec{u} X \vec{v}| = a\sqrt{4h^2 + 3a^2}$ and $|\vec{u}| = \sqrt{3a^2 + h^2}$ above $\implies$

$\sin(\theta) = \dfrac{|\vec{u} X \vec{v}|}{|\vec{u}||\vec{v}|} = \dfrac{1}{\sqrt{2}}$ $\implies \theta = \boxed{45^\circ}$ .

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Note: $h^2 = (\dfrac{3}{2^{\frac{2}{3}}})k^{\frac{2}{3}} = A_{\triangle{BCD}} = A_{\triangle{BC'D}}$ .