Geometry of Methane

Each C-H bond is a radius of circunsphere of regular tetrahedron with vertexes in each H.

Let $A, B, C, D$ be these 4 vertex. Let face $ABC$ be called "base" of our regular tetrahedron and $O$ be the center of insphere and circumsphere. The base is an equilateral triangle .

The circumradius lenght and inradius lenght are $3:1$ in a regular tetrahedron. Let the inradius be 1, and we can easy find by Pythagorean Theorem the projetion $AO´$ of circumradius $AO$ at base. Then, $AO´ = \sqrt{8}$ .

$AO´$ is the radius of circumcircle of base, so we can find the base side (edge of tetrahedron). It will be $AB = 2\sqrt{6}$ .

I would prefer find the asked angle ( $\angle AOB$ , known as the tetrahedral angle) applying cosine rule in triangle $AOB$ , which gives a prettier result: just ${ cos }^{ -1 }(-1/3)$ . But the question describe that angle as $2{ cos }^{ -1 }\sqrt{(a/b)}$ , so we need to consider the right triangle $OAM$ (M is the midpoint of AB).

By Pythagorean Theorem we get $OM = \sqrt{3}$ , so $cos( \angle AOM)=\frac { \sqrt { 3 } }{ 3 } = \frac { \sqrt { 3 } }{\sqrt{ 9 }}= \sqrt{1/3}$ .

Finally, the asked angle is $2{ cos }^{ -1 }\sqrt{(1/3)}$ and the answer is $a+b=\boxed{4}$ .