A Special Angle 2

$\triangle{ABC} \cong \triangle{ADC} \implies A_{ABCD} = r_{1}r_{2}\sin(\omega)a^2$

The bi-medians of a quadrilateral bisect each other and the diagonals of a quadrilateral are perpendicular $\implies$ the bi-medians are congruent $\implies$ the centroid $P$ is the midpoint of $\overline{M_{1}}\overline{M_{2}} \implies P(\dfrac{2r_{2}\cos(\omega) + r_{1}}{4}a,0)$ .

$PM_{2} = \dfrac{a}{2}\sqrt{r_{1}^2 + 4r_{2}^2\sin^2(\omega)} \implies A = A_{\triangle{QDC}} =$ $\dfrac{ar_{3}}{4}\sqrt{4h^2 + (r_{1}^2 + 4r_{2}^2\sin^2(\omega))a^2}$

The volume of the pyramid $V = \dfrac{1}{3}(r_{1}r_{2}\sin(\omega))a^2h = K \implies h = \dfrac{3k}{r_{1}r_{2}\sin(\omega)a^2}$

Let $j = (r_{1}^2 r_{2}^2\sin^2(\omega))(r_{1}^2 + 4r_{2}^2\sin^2(\omega)) \implies \overline{PM_{2}} = \dfrac{\sqrt{j}a}{2r_{1}r_{2}\sin(\omega)}$ and $A(a) = r_{3}\dfrac{\sqrt{36k^2 + ja^6}}{4r_{1}r_{2}\sin(\omega)a} \implies$ $\dfrac{dA}{da} = r_{3}\dfrac{ja^6 - 18k^2}{(4r_{1}r_{2}\sin(\omega)a^2\sqrt{36k^2 + ja^6}} \implies a = (\dfrac{3\sqrt{2}k}{\sqrt{j}})^{\frac{1}{3}}$ $\implies h = \dfrac{3k}{r_{1}r_{2}\sin(\omega)}(\dfrac{\sqrt{j}}{3\sqrt{2}k})^{\frac{2}{3}}$

$\dfrac{h}{a} = \dfrac{\sqrt{j}}{\sqrt{2}r_{1}r_{2}\sin(\omega)} \implies \tan(\theta) = (\dfrac{2r_{1}r_{2}\sin(\omega)}{\sqrt{j}}(\dfrac{h}{a}) = \sqrt{2} \implies \theta \approx \boxed{54.7356103}$ .