It's All Angles

Using the law of sines $\implies \dfrac{\sin(2\beta)}{\sin(\beta)} = \dfrac{m}{m - 2} \implies \cos(\beta) = \dfrac{m}{2(m - 2)}$

Using the law of cosines with included $\angle{BAC} \implies m^2 - 4m + 4 = m^2 - 2m + 1 + m^2 - \dfrac{m^2(m - 1)}{m - 2} \implies m^2 - 7m + 6 = 0 \implies$ $(m - 6)(m - 1) = 0 \:\ m \neq 1 \implies m = 6 \implies \overline{AC} = 6a, \:\ \overline{AB} = 5a$ and $\overline{BC} = 4a$

$\cos(\beta) = \dfrac{3}{4} \implies \overline{BD} = \dfrac{5a\sqrt{7}}{4}a$ and $AD = \dfrac{15a}{4}$

Let $A:(0,0), \:\ C(6a,0)$ and Point $B:( \dfrac{15a}{4}, \dfrac{5a\sqrt{7}}{4}a)$

$\implies$ the midpoints of the sides are $M_{2}:(3a,0), M_{1}:(\dfrac{15}{8}a, \dfrac{5\sqrt{7}}{8}a), \:\ M_{3}(\dfrac{39}{8}a,\dfrac{5\sqrt{7}}{8}a)$

$m_{CM_{1}} = -\dfrac{5\sqrt{7}}{33} \implies y = -\dfrac{5\sqrt{7}}{33}(x - 6a)$

and,

$m_{AM_{3}} = \dfrac{5\sqrt{7}}{39} \implies y = \dfrac{5\sqrt{7}}{39}x$

Solving the two equations above $\implies x = \dfrac{13}{4}a, \:\ y = \dfrac{5\sqrt{7}}{12}a \implies$

The centroid $P(\dfrac{13}{4}a,\dfrac{5\sqrt{7}}{12}a)$ .and $\overline{PM_{2}} = \dfrac{\sqrt{46}}{6}a$

Note: You could also have found the centroid of the triangle by taking an average of the three vertices.

$A_{\triangle{ABC}} = \dfrac{15\sqrt{7}}{4}a^2$

Letting $\overline{QP} = h \implies \overline{QM_{2}} = \dfrac{\sqrt{36h^2 + 46a^2}}{6} \implies$ $A = A_{\triangle{AQC}} = \dfrac{a}{2}\sqrt{36h^2 + 46a^2}$

The volume $V = \dfrac{5\sqrt{7}}{4} a^2 h = k \implies h = \dfrac{4k}{5\sqrt{7} a^2} \implies$ $A(a) = \dfrac{\sqrt{576k^2 + 8050a^6}}{10\sqrt{7} a} \implies \dfrac{dA}{da} = \dfrac{8050a^6 - 288k^2}{5\sqrt{7}\sqrt{576k^2 + 8050a^6}} a^2 = 0 \implies a = (\dfrac{12k}{5\sqrt{161}})^{\frac{1}{3}}$ $\implies h = (\dfrac{92k}{45\sqrt{7}})^{\frac{1}{3}}$

Let $\theta = m\angle{QM_{2}P} \implies \tan(\theta) = \dfrac{6h}{\sqrt{46}a} = \sqrt{2} \implies \theta \approx \boxed{54.735610^\circ}$ .