4200.11 – Tetrahedron Cutoffs

Here is a rectangular solid with $\angle DGH$ = 45 degrees and $\angle FGB$ = 60 degrees. What is $\cos(\angle BGD)$ ?

Solution

Assume, for convenience, that $GH = 1$ . Then $DH = 1$ and $DG = \sqrt{2}$ . Also, $DC = 1$ , $BC = \sqrt{3}/3$ because $CG = 1 \text{ and } DB = 2 \sqrt{3}/3.$

Now, $\triangle BDG$ is iscosceles, so when we draw $BM$ to the midpoint of $DG$ , we have a right angle at $M$ .

So $\triangle BMG$ is a right triangle and now we can nail down the desired cosine.

\begin{aligned} \cos(\angle BGD) &=& \frac{\sqrt{2}/2}{2 \sqrt{3}/3} \\ &=& \frac{\sqrt{2}}{2} \cdot \frac{3}{2 \sqrt{2}} \\ &=& \frac{3 \sqrt{2} \cdot \sqrt{3}}{4 \sqrt{3} \cdot \sqrt{3}} \\ &=& \frac{3 \sqrt{6}}{12} = \frac{\sqrt{6}}{4}. \end{aligned}