Explanation

(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 $F_{up} = 2F_{dn}$
$\Rightarrow mg(\sin\theta + \mu\cos{\theta) = 2mg(\sin\theta} - \mu\cos\theta)$
$\Rightarrow \sin\theta + \mu\cos\theta = 2\sin\theta - 2\mu\cos\theta$
$\Rightarrow 3\mu\cos\theta = \sin\theta \Rightarrow \tan\theta 3\mu$
$\Rightarrow \ \theta = \tan^{- 1}{(3\mu)} = \tan^{- 1}{(3 \times 0.25) = \tan^{- 1}{(0.75) = 36.8{^\circ}}}$