Any tetrahedral geometry theorems of methane bonding angles? For my 12 grade folio task I need to find alternate ways of finding the bonding angles in a methane molecule (regular tetrahedron). I have already done it through vector methods, co-ordinate geometry and triangle theorems.
I was wondering if anyone knows of, or can explain to me, any advanced maths that can find the bonding angle or theorems that can be used to deduce it.
If anyone could help it would be much appreciated.
 A: Consider a regular tetrahedron with each edge $a$ & having vertices A, B, C, D & center O.  $$AB=BC=AC=AD=BD=CD=a$$
Then the radius of the spherical surface passing through all 4 identical vertices of regular tetrahedron is given (see table or derivation) by the formula $$OA=OB=OC=OD=\frac{a}{2}\sqrt{\frac{3}{2}}$$
Now, consider any isosceles triangle say $\triangle OAB$ in which we have $$OA=OB=\frac{a}{2}\sqrt{\frac{3}{2}}, \ \ AB=a$$ Now, applying Cosine rule in $\triangle OAB$ to find $\angle AOB$ as follows $$cos \angle AOB=\frac{(OA)^2+(OB)^2-(AB)^2}{2(OA)(OB)}$$ Now, setting the corresponding values we get $$\cos \angle AOB=\frac{\left(\frac{a}{2}\sqrt{\frac{3}{2}}\right)^2+\left(\frac{a}{2}\sqrt{\frac{3}{2}}\right)^2-a^2}{2\left(\frac{a}{2}\sqrt{\frac{3}{2}}\right)\left(\frac{a}{2}\sqrt{\frac{3}{2}}\right)}$$ $$=\frac{\frac{3a^2}{4}-a^2}{\frac{3a^2}{4}}=\frac{-\frac{1}{4}}{\frac{3}{4}}$$ $$=-\frac{1}{3}$$
$$\angle AOB=\cos^{-1}\left(\frac{-1}{3}\right)=\pi-\cos^{-1}\left(\frac{1}{3}\right)$$
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{\text{Methane bonding angle }=\color{blue}{\pi-\cos^{-1}\left(\frac{1}{3}\right)\approx 109.47^\circ}}}$$
