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

According to problem \({F_{up}} = 2{F_{dn}}\)

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

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

\(⇒\) \(3\mu \cos \theta = \sin \theta \)

\(⇒\)  \(\tan \theta = 3\mu \)

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