How can I prove that the angles of the tetrahedral structure is $109.5^\circ$ with calculus. I could do it with geometry. How can I prove that the dihedral angles of the regular tetrahedral structure is $109.5^\circ$ with calculus or any other technique. I could do it with geometry by considering a cube.
 A: Four unit vectors pointing outward from the origin have equal angles between them.
By symmetry, their average, and hence their sum, is $0.$
Let a coordinate axis go out from the origin in the direction of one of those four vectors.
The component of that one vector on that axis is $1.$
Therefore sum of the components, on that axis, of the other three, must be $-1.$
Since there are three of them, the component of each on that axis is $-1/3.$
So the cosine of the angle is $-1/3.$
A: With dot products: Taking the axis-oriented cube of side length $2$ centered at the origin, you're looking for the angle $\theta$ between (say) the vectors $\mathbf{u} = (1, -1, 1)$ and $\mathbf{v} = (-1, 1, 1)$:
$$
\cos\theta = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\|\, \|\mathbf{v}\|}
  = -\frac{1}{3}.
$$
In response to Dave Renfro's request for a diagram (his triangles are bounded by a pair of vectors and a dashed line), here's a crossed-eyes stereogram:

A: There is an amazing geometric interpretation for figuring out the angle in that link ->
http://maze5.net/?page_id=367
