That angle...

Let the required angle be $α$ , and the position vectors of the vertices of the tetrahedron be $\hat i-\frac{1}{\sqrt 3}\hat j, -\hat i-\frac{1}{\sqrt 3}\hat j, \frac{2}{\sqrt 3}\hat j, \frac{4}{\sqrt 6}\hat k$ .

Then the unit vectors along the normals to the two adjacent faces are

$\hat n_1=\hat k, \hat n_2=\dfrac {-\frac{2}{\sqrt 3}\hat k+\frac{8}{\sqrt 6}\hat j}{\sqrt {12}}$

Then the required angle is given by

$\cos α=\hat n_1.\hat n_2=\dfrac {2}{\sqrt {3\times 12}}=\dfrac {1}{3}$

$\implies α=\cos^{-1}(\frac{1}{3})\approx \boxed {70.528779\degree}$ .

$\theta = \cos^{-1}\frac {1}{3}≈70.529°$