In the adjoining figure of a rectangular solid, $\angle DHG=45^\circ$ and $\angle FHB=60^\circ$. Find the cosine of $\angle BHD$.

$\text {(A)} \frac{\sqrt{3}}{6} \qquad  \text {(B)} \frac{\sqrt{2}}{6} \qquad  \text {(C)} \frac{\sqrt{6}}{3} \qquad  \text{(D)}\frac{\sqrt{6}}{4}\qquad \text{(E)}\frac{\sqrt{6}-\sqrt{2}}{4}$
WLOG, let $CD=1$. 
Looking at square GHDC, we see that $\angle DHC=45$, which implies that $DC=CH=1$ and $DH=\sqrt{2}$
Taking each cross-section one at a time, we look at square DHFB. We obviously know that CHB is a $30$ degree angle, giving $BH=\frac{2\sqrt{3}}{3}$, and $BC=\frac{\sqrt{3}}{3}$.
Looking at square ABCD, we see that $BD^2=1+\frac{1}{3}\implies BD^2=\frac{4}{3}=\frac{2\sqrt{3}}{3}$. 
We now look at triangle DBH, which has side lengths $BD=\frac{2\sqrt{3}}{3}$, $BH=\frac{2\sqrt{3}}{3}$ and $DH=\sqrt{2}$. Because $BD$ is opposite angle DHB, by the law of cosines, $\frac{4}{3}=\frac{4}{3}+2-2(\frac{2\sqrt{3}}{3})(\sqrt{2})\cos(\theta)\implies 2=\frac{4\sqrt{6}\cos(\theta)}{3} \implies \frac{2}{\frac{4\sqrt{6}}{3}}\implies \frac{6}{4\sqrt{6}} \implies \frac{\sqrt{6}}{4}$
Thus, the answer is $(\textbf{D})$