Answer
$\theta = 18^{\circ}$
Work Step by Step
We can write an equation when $\mu_s = 0$:
$F_1 = mg~sin~\theta$
We can write an equation when $\mu_s = 0.50$:
$F_2 = mg~sin~\theta+mg~\mu_s~cos~\theta$
We can divide the second equation by the first equation:
$\frac{F_2}{F_1} = \frac{sin~\theta+\mu_s~cos~\theta}{sin~\theta}$
$F_2~sin~\theta = F_1~(sin~\theta+\mu_s~cos~\theta)$
$(F_2-F_1)~sin~\theta = F_1~\mu_s~cos~\theta$
$tan~\theta = \frac{F_1~\mu_s}{F_2-F_1}$
$\theta = tan^{-1}~(\frac{F_1~\mu_s}{F_2-F_1})$
$\theta = tan^{-1}~[\frac{(2.0~N)(0.50)}{5.0~N-2.0~N}]$
$\theta = tan^{-1}~(\frac{1}{3})$
$\theta = 18^{\circ}$
