It's about perspective!

Apply pythagorean theorem three times to get the lengths of the face diagonals.

$a=\sqrt{3^2+2^2}=\sqrt{13}~in.$

$b=\sqrt{4^2+3^2}=5~in.$

$c=\sqrt{4^2+2^2}=\sqrt{20}~in.$

Then apply cosine law to get the unknown angle.

$c^2=a^2+b^2-2ab \cos \theta$

$(\sqrt{20})^2=(\sqrt{13})^2+5^2-2(\sqrt{13})(5)(\cos \theta)$

$\theta \approx \boxed{60^\circ}$