$A,B$ are adjacent vertices of a cube. $C$ lies on the edge diagonally opposite to $AB$ and divides that edge in the ratio $2:1.$

The angle $\mathrm{\angle }ACB$ is labelled $\theta .$ Find $\theta .$

Let the edge on which $C$ lies be called $DE.$

It is convenient to let $AB=3x$ so that $DC=2x$ and $CE=x.$

Use Pythagoras to find the diagonal $BE$ in terms of $x$ (and note that this is the same as $AD.$)

$\mathrm{△}BCE$ is a right-angled triangle so Pythagoras applies once again!