Calculating the tetrahedral bond angle

Every Chemistry textbook which covers molecular shapes will state with utmost authority that the bond angle in tetrahedral molecules is 109.5 degrees. Methane (CH4) is frequently quoted as the example, shown to be completely symmetrical and tetrahedral. And then the 109.5 degrees.  There's no proof given (after all, Chemistry textbooks aren't dealing with geometry, and there's no need to show something just for the sake of mathematical proof - rightly, the content is all about reactivity and structure).  However, the lack of proof has bugged me on-and-off for about 20 years, and recently I decided it was time to do something about it and prove it for myself.

There are various websites showing the geometry of a tetrahedron and how it relates to a cube, and those sites use the relationship between a cube and a tetrahedron in order to calculate the angle, but I'm going to demonstrate an alternative proof using solely the properties of a tetrahedron  - its symmetry and its equilateral triangular faces.

To start with, calculate the horizontal distance from one of the vertices to the centre of the opposite triangular face (the point directly below the central 'atom').  In this diagram, E is the top corner, D is the central "atom" (representing the centre of the tetrahedron) and C is the point directly below D, such that CDE is a straight line, and C is the centre of the shaded face (the base).

This gives a large right-angled triangle ACE, where the hypotenuse is one edge of the tetrahedron (length AE = l); one side is the line we'll be calculating (length AC, using the triangle ABC); and the third, CE, is the line extending from the top of the tetrahedron through the central atom down to the centre of the base.

In triangle ABC, length AB = l/2, angle A is 30 degrees, angle B is 90 degrees.  We need to calculate length AC:

cos 30 = l/2 / AC
AC = l /2 cos 30

Since we have two sides and an angle of a right-angled triangle, we can determine the other two angles; we're primarily interested in the angle at the top, labelled α.

sin α = AC / l

And as we know that AC = 1 / 2 cos 30 this simplifies to

sin α = 1 / (2 cos 30)

Evaluating:  1 / 2 cos 30 = 0.5773

sin α = 0.5773
α = 35.26 degrees.

Looking now at the triangle ADE which contains the tetrahedral bond angle at D:  the bond angle D can be calculating through symmetry, since ADE is an isosceles triangle.

D = 180 - (2*35.26) = 109.47 degrees, as we've been told all along.

QED