Tetrahedron angles

In chemistry the angle 109.5 comes up a lot, as the angle from one corner of a regular tetrahedron to the center to another corner, as in a methane molecule. So, while I was waiting for the bus yesterday I tried to derive that angle by myself, and it turned out to be far more difficult than I expected. I used the projection of the tetrahedron on a plane parallel to one face and a cross section through two of the center-to-corner segments to get three equations and three variables, which I then solved for the angle I wanted. I got the right answer, but I’d like to know if there’s an easier way to do this. I googled a few ways of solving the problem and found slightly simpler methods to solve this problem, but nothing that was much simpler. One idea I had was to find two vectors parallel to the chemical bonds in CH[sub]4[/sub] and determine the angle between them, but thus far I’ve been unsuccessful. So, is there any easy way to find an exact solution to this problem?

The tetrahedron with vertices (1,1,1), (1,-1,-1), (-1,1,-1), (-1,-1,1) is a regular tetrahedron centered at the origin. So you can just figure out the angle between vectors [1,1,1] and [1,-1,-1].

To complete the thought, the dot product between these two cartesian vectors is 11+1-1+1*-1 = -1. This is equal to the product of the vector’s magnitudes (each magnitude = sqrt(3)) times the cosine of the angle between them. Thus cos(th) = -(1/3), so th = 109.47 degrees.

Thanks, that’s exactly what I was looking for.