Answer
The maximum angle for the hill where we can leave the car parked is $42^{\circ}$.
Work Step by Step
The component of the car's weight which is directed down the slope is $mg~sin(\theta)$.
As long as the force of static friction is enough to oppose this weight, the car will stay on the hill without sliding down. $F_f = F_N~\mu_s = mg~cos(\theta)~\mu_s$
$mg~sin(\theta) = mg~cos(\theta)~\mu_s$
$tan(\theta) = \mu_s$
$\theta = tan^{-1}(0.90)$
$\theta = 42^{\circ}$
The maximum angle for the hill where we can leave the car parked is $42^{\circ}$.
