Answer
$cos ^{-1}(\frac{2}{ \sqrt 6} )\approx 35.26 ^ \circ$
Work Step by Step
Consider a cube with side length $1$ , the angle between one diagonal of cube and a diagonal of one of its face is the angle between $ \lt 1,1,1 \gt $ to $\lt 1,1,0 \gt$ .
$ \theta = cos ^{-1}(\frac{\lt 1,1,1 \gt \cdot \lt 1,1,0 \gt}{\sqrt {1^2+1^2+1^2}\sqrt {1^2+1^2+0^2}})$
$\theta = cos ^{-1}(\frac{2}{ \sqrt 3 \sqrt 2} )$
$\theta = cos ^{-1}(\frac{2}{ \sqrt 6} )\approx 35.26 ^ \circ$
