Pyramid with Rectangular Base 2.

$\triangle{AEC}$ 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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The lateral surface area $S = \dfrac{1}{2}(\sqrt{3}a\sqrt{4h^2 + a^2} + a\sqrt{4h^2 + 3a^2})$ and the volume $V = \dfrac{1}{2\sqrt{3}} a^2 h = k \implies h = \dfrac{2\sqrt{3}k}{a^2} \implies$

$S(a) = \dfrac{1}{2}(\dfrac{\sqrt{3}\sqrt{48k^2 + a^6}}{a} + \dfrac{\sqrt{48k^2 + 3a^6}}{a}) \implies$

$\dfrac{ds}{da} = \dfrac{(2\sqrt{3} + 6)a^6 - 48k^2(\sqrt{3} + 1)}{2a^2\sqrt{48k^2 + a^6}} = 0 \:\ a \neq 0 \implies a = (8\sqrt{3} k^2)^{\frac{1}{6}} \implies h = (3k)^{\frac{1}{3}}$

$\implies \tan(\theta) = \dfrac{2h}{a} =2^{\frac{1}{2}} * 3^{\frac{1}{4}} \implies \theta \approx \boxed{61.751530^{\circ}}$ and $\tan(\lambda) = \dfrac{2h}{\sqrt{3}a} =\dfrac{2^{\frac{1}{2}}}{3^{\frac{1}{4}}} \implies \lambda \approx \boxed{47.058597^{\circ}} \implies \dfrac{\theta}{\lambda} \approx \boxed{1.312226}$