The force required just to move a body up an inclined plane is double the force required just to prevent the body sliding down. If the coefficient of friction is $\mu$ , the angle of inclination of the plane is

Retardation in upward motion $=g(\sin \theta +\mu \cos \theta )$
$\therefore$ Force required just to move up 
$F_{up}=mg(\sin \theta +\mu \cos \theta )$
Similarly for down ward motion $a$
$=g(\sin \theta -\mu \cos \theta )$
 $\therefore$Force required just to prevent the body sliding dow,Force required just to prevent the body sliding dow,
$F_{dn}=mg(\sin \theta -\mu \cos \theta )$
According to problem $F_{up}=2F_{dn}$
$mg(\sin \theta +\mu \cos \theta )=2mg(\sin \theta -\mu \cos \theta )$
$\Rightarrow \sin \theta +\mu \cos \theta =2\sin \theta -2\mu \cos \theta$
$\Rightarrow 3\mu \cos \theta =\sin \theta \Rightarrow \tan \theta =3\mu$
$\Rightarrow \theta ={\tan }^{-1}\left(3\mu \right)$