Answer
The proof is below.
Work Step by Step
The ladder is at static equilibrium, so the net torques and the net forces are 0. Thus, we use the fact that the x and y forces are 0 to find:
$F_{n1}\mu_1=F_{n2}$
$mg = F_{n1}+F_{n2}\mu_2$
We know that the torques sum to zero, so using the bottom of the latter as the axis of rotation, we find:
$\frac{mgsin(90+\phi)}{2}=-\mu_2F_{n2}sin(90-\phi)-F{n2}sin(-\phi+180)$
Combining these equations, we obtain:
$\phi = tan^{-1}(\frac{1-\mu_1\mu_2}{2\mu_1})$
