A block of mass $4\, kg$ rests on an inclined plane. The inclination of the plane is gradually increased. it is found that when the inclination is $3$ in $5\left( {\sin \theta = \frac{3}{5}} \right)$, the block just begins to slide down the plane. The coefficient of friction between the block and the plane is
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$\sin \theta=\frac{3}{5}$
Clearly, base of the triangle is $4$ units
$\mu=\tan \theta=\frac{3}{4}=0.75$
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