#### 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:
$\phi = tan^{-1}(\frac{1-\mu_1\mu_2}{2\mu_1})$