It's a Special Angle

Let $m\angle{BAC} = \omega$ .

$A_{\triangle{ABC}} = \dfrac{1}{2}r_{1}r_{2}\sin(\omega)a^2$ and $PD = \dfrac{r_{2}\sin(\omega)}{3}a$

$\overline{QD} = \dfrac{\sqrt{9h^2 + r_{2}^2\sin^2(\omega)a^2}}{3} \implies A = A_{\triangle{QDC}} =$ $\dfrac{r_{1}}{6}a\sqrt{9h^2 + r_{2}^2\sin^2(\omega)a^2}$

The volume $V = \dfrac{1}{6}(r_{2}r_{2}\sin(\omega))a^2h = K \implies h = \dfrac{6k}{(r_{1}r_{2}\sin(\omega))a^2} \implies$ $A(a) = \dfrac{\sqrt{324k^2 + r_{1}^2 r_{2}^4\sin^4(\omega)a^6}}{6r_{2}a\sin(\omega)} \implies$

$\dfrac{dA}{da} = \dfrac{r_{1}^2 r_{2}^4\sin^4(\omega)a^6 - 162k^2}{(3r_{2}\sin(\omega))\sqrt{324k^2 + r_{1}^2 r_{2}^4\sin^4(\omega)a^6}a^2} = 0$ $\implies a = (\dfrac{9\sqrt{2}k}{r_{1}r_{2}^2\sin^2(\omega)})^{\frac{1}{3}} \implies h = \dfrac{6k}{r_{1}r_{2}\sin(\omega)}(\dfrac{r_{1}r_{2}^2\sin^2(\omega)}{9\sqrt{2}k})^{\frac{2}{3}}$

$\dfrac{h}{a} = \dfrac{\sqrt{2}r_{2}\sin(\omega)}{3}$ $\implies \tan(\theta) = \dfrac{3}{r^2\sin(\omega)}(\dfrac{h}{a}) = \sqrt{2} \implies \theta \approx \boxed{54.7356103}$