a
(a) Angle of repose $\alpha = {\tan ^{ - 1}}(\mu ) = {\tan ^{ - 1}}(0.8) = 38.6^\circ $
Angle of inclined plane is given $\theta = 30^\circ $.
It means block is at rest therefore,
Static friction $=$ component of weight in downward direction $ = mg\sin \theta = 10\;N$
$\therefore m = \frac{{10}}{{9 \times \sin 30^\circ }} = 2\;kg$