Oblique Triangular Prism 2

$V = \dfrac{\sqrt{3}}{4} r^2 H = b^2 \implies H = \dfrac{4b^2}{\sqrt{3}r^2}$ $A = 3rs = b \implies s = \dfrac{b}{3r} \implies Y(r) = y^2(r) = \dfrac{b^2}{3}(\dfrac{1}{3 r^2} - \dfrac{16b^2}{r^4}) \implies$ $\dfrac{dY}{dr} = \dfrac{2b^2}{3}(\dfrac{24 b^2}{r^5} - \dfrac{1}{3r^3}) = \dfrac{2b^2}{9r^5}(72 b^2 - r^2) = 0 \implies r = 6\sqrt{2} b$ for $r > 0 \implies H = \dfrac{1}{18\sqrt{3}}$ and $s = \dfrac{1}{18\sqrt{2}} \implies y = \sqrt{s^2 - H^2} = \dfrac{1}{18\sqrt{6}}$

Let $\lambda = m\angle{DFE}$ .

$\cos(\lambda) = \dfrac{y}{s} = \dfrac{1}{\sqrt{3}}$ and $\cos(\theta) = \cos(180 - \lambda) = -\cos(\lambda) = \dfrac{-1}{\sqrt{3}}$

$\boxed{\cos(\theta) = \dfrac{-1}{\sqrt{3}} \implies \theta \approx 125.26438968^\circ}$

$(0 < b < \sqrt{\dfrac{3}{5}}) \implies \dfrac{d^2Y}{dr^2}|_{r = 6\sqrt{2}b} = \dfrac{2b^2}{3}(\dfrac{72 - 120b^2}{72^3}) > 0$ $\implies \boxed{r = 6\sqrt{2}b}$ minimizes $y$ .

$d^2 = \dfrac{46656\sqrt{3}b^2 + 432b + \sqrt{3}}{648\sqrt{3}} \implies \boxed{d = \dfrac{\sqrt{46656\sqrt{3}b^2 + 432b + \sqrt{3}}}{18\sqrt{2}}}$

and $\:\ d^2 = \dfrac{b}{3\sqrt{3}} (216\sqrt{3} b + 4) \implies$

$46656\sqrt{3}b^2 + 432b + \sqrt{3} = 648\sqrt{3}(\dfrac{(216\sqrt{3}b + 4}{3\sqrt{3}})b =$ $216(216\sqrt{3}b^2 + 4b) = 46656\sqrt{3}b^2 + 864b \implies 432b = \sqrt{3} \implies \boxed{b = \dfrac{\sqrt{3}}{432}}$

$b = \dfrac{\sqrt{3}}{432} \implies\boxed{d^2 = \dfrac{\dfrac{3}{4}\sqrt{3} + 2\sqrt{3}}{2^3 9^2\sqrt{3}} = \dfrac{11}{2(36^2)}} \implies$ $\boxed{d = \dfrac{1}{36}\sqrt{\dfrac{11}{2}}}$ and $\boxed{r = 6\sqrt{2}(\dfrac{\sqrt{3}}{432}) = \dfrac{\sqrt{6}}{72} \implies r^2 = \dfrac{1}{6^3 * 2^2}}$

$r^2 = 2d^2 - 2d^2\cos(\alpha) \implies \cos(\alpha) = \dfrac{2d^2 - r^2}{2d^2} =$ $\dfrac{2(\dfrac{11}{2 * 6^4}) - \dfrac{1}{2^2 * 6^3}}{2(\dfrac{11}{6^4 *2})} =$ $(\dfrac{38}{2^2 * 6^4})(\dfrac{6^4}{11}) = \dfrac{38}{44} = \dfrac{19}{22} \implies$ $\boxed{\cos(\alpha) = \dfrac{19}{22} \implies \alpha \approx 30.27264^\circ}$

$\therefore \lfloor{\theta}\rfloor$ + $\lfloor{\alpha}\rfloor = \boxed{155}$ .