when $\left.\theta=30^{\circ}, F-m g \sin 30^{\circ}-\mu m g \cos 30^{\circ}=0 \ldots \ldots \text { ( } 1\right)$

when $\left.\theta=60^{\circ}, F+\mu m g \cos 60^{\circ}-m g \sin 60^{\circ}=0 \ldots \ldots \text { ( } 2\right)$

using $(1),(2),$ becomes, $m g \sin 30^{\circ}+\mu m g \cos 30^{\circ}=-\mu m g \cos 60^{\circ}+m g \sin 60^{\circ}$

or

$\mu\left(\cos 60^{\circ}+\cos 30^{\circ}\right)=\sin 60^{\circ}-\sin 30^{\circ}$

Or

 $\mu\left(\frac{1}{2}+\frac{\sqrt{3}}{2}\right)=\frac{\sqrt{3}}{2}-\frac{1}{2}$

$\therefore \mu=\frac{\sqrt{3}-1}{\sqrt{3}+1}$