d
$F _{1}= mg (\sin \theta+\mu \cos \theta)$
$F _{2}= mg (\sin \theta-\mu \cos \theta)$
$\frac{ F _{1}}{ F _{2}}=\frac{\sin \theta+\mu \cos \theta}{\sin \theta-\mu \cos \theta}$
$\frac{ F _{1}}{ F _{2}}=\frac{\tan \theta+\mu}{\tan \theta-\mu}$
$\frac{ F _{1}}{ F _{2}}=\frac{2 \mu+\mu}{2 \mu-\mu}$
$\frac{ F _{1}}{ F _{2}}=3$