Retardation in upward motion $=g(\sin \theta + \mu \cos \theta)$

Force required just to move up $F_{up} = mg(\sin \theta + \mu \cos \theta)$

Similarly, for down ward motion a $=g(\sin \theta - \mu \cos \theta)$

Force required just to prevent the body sliding down

$F_{dn} = mg(\sin \theta - \mu \cos \theta)$

According to problem $F_{\text{up}} = 2F_{dn}$

⇒ ⇒ $mg(\sin \theta + \mu \cos \theta) = 2mg(\sin \theta - \mu \cos \theta)$

⇒ ⇒ $\dot{s}\sin\theta + \mu \cos\theta = 2\dot{s}\sin\theta - 2\mu \cos\theta$

⇒ ⇒ $3 \mu \cos \theta = \sin \theta$ ⇒ ⇒ $\tan \theta = 3 \mu$

⇒ ⇒ $\theta = \tan^{-1}(3 \mu) = \tan^{-1}(3 \times 0.25) = \tan^{-1}(0.75)$ = 36.8^\circ