Let the force exerted by the door on the wedge be of magnitude $\displaystyle F_0$ . This force must act perpendicular to the surface of the wedge as it is smooth.

The wedge has a reaction force (from the ground) acting vertically upwards on it and a friction force $\displaystyle F_{\mu}$ acting horizontally on it.

In order for the wedge to remain where it is we must have (resolving vertically)

$\displaystyle R = F_0 \cos \alpha$

and (resolving horizontally)

$\displaystyle F_{\mu} \ge F_0 \sin \alpha$

but since

$\displaystyle F_{\mu} = \mu R$ then

$\displaystyle \mu F_0 \cos \alpha \ge F_0 \sin \alpha$,

$\displaystyle \color[rgb]{0,0,1} \boxed{\Rightarrow \alpha \le \arctan \mu}$.