Question 18In the adjoining figure of a rectangular solid, $\angle DHG=45^\circ$ and $\angle FHB=60^\circ.$ Find the cosine of $\angle BHD.$ $\frac{\sqrt3}{6}$$\frac{\sqrt2}{6}$$\frac{\sqrt6}{3}$$\frac{\sqrt6}{4}$$\frac{\sqrt6-\sqrt2}{4}$
