Problem
1982 AHSME Problem 18
In the adjoining figure of a rectangular solid, \angle DHG=45^\circ and \angle FHB=60^\circ. Find the cosine of \angle BHD.
![[asy] import three;defaultpen(linewidth(0.7)+fontsize(10)); currentprojection=orthographic(1/3+1/10,1-1/10,1/3); real r=sqrt(3); triple A=(0,0,r), B=(0,r,r), C=(1,r,r), D=(1,0,r), E=O, F=(0,r,0), G=(1,0,0), H=(1,r,0); draw(D--G--H--D--A--B--C--D--B--F--H--B^^C--H); draw(A--E^^G--E^^F--E, linetype("4 4")); label("$A$", A, N); label("$B$", B, dir(0)); label("$C$", C, N); label("$D$", D, W); label("$E$", E, NW); label("$F$", F, S); label("$G$", G, W); label("$H$", H, S); triple H45=(1,r-0.15,0.1), H60=(1-0.05, r, 0.07); label("$45^\circ$", H45, dir(125), fontsize(8)); label("$60^\circ$", H60, dir(25), fontsize(8));[/asy]](/static/aops/248f6a269952debe23a90c89640844ea84749db2.png)
\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}
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