Answer
$\theta = 50.2^{\circ}$
Work Step by Step
We can let the friction force directed up the incline be equal in magnitude to the component of the person's weight directed down the incline.
$mg~sin(\theta) = mg~cos(\theta)~\mu_s$
$tan(\theta) = \mu_s$
$\theta = arctan(\mu_s) = arctan(1.20)$
$\theta = 50.2^{\circ}$