The acute angle between any two faces of a regular tetrahedron is

$\cos^{-1}(\frac{1}{3})$

Let OABC be the regular tetrahedron. Taking O as the origin let the position vectors of A, B, C be $\vec a,\vec b$ and $\vec c$ respectively.

We have,

$OA OB=OC = AB = BC = CA$

$⇒|\vec a|=|\vec b|=|\vec c|$ and $\vec a.\vec b=\vec b.\vec c=\vec c.\vec a$

Let θ be the acute angle between focus OAB and OBC. Then, θ is also the angle between their normals

i.e. $\vec{n_1}\vec a×\vec b$ and $\vec{n_2}=\vec b×\vec c$

$∴\cos θ=\left|\frac{\vec{n_1}.\vec{n_2}}{|\vec{n_1}||\vec{n_2}|}\right|$

$⇒\cos θ=\left|\frac{\vec a.\{\vec b×(\vec b×\vec c)\}}{|\vec a×\vec b||\vec b×\vec c|}\right|$

$⇒\cos θ=\left|\frac{\vec a\{(\vec b.\vec c)\vec b-(\vec b.\vec b)\vec c\}}{|\vec a×\vec b||\vec b×\vec c|}\right|$

$⇒\cos θ=\left|\frac{\{(\vec a.\vec b)(\vec b.\vec c)-(\vec b.\vec b)(\vec a.\vec c)\}}{|\vec a×\vec b||\vec b×\vec c|}\right|$

Now,

$\vec a.\vec b=\vec b.\vec c=\vec c.\vec a=|\vec a||\vec a|\cos\frac{π}{3}=\frac{1}{2}|\vec a|^2$

and,

$|\vec a×\vec b|=|\vec a||\vec b|\sin\frac{π}{3}=\frac{\sqrt{3}}{2}|\vec a|^2=|\vec a×\vec c|=|\vec c×\vec a|$

$∴\cos θ=\frac{1}{3}⇒\cos^{-1}(\frac{1}{3})$