A particle of mass 8 kg rests on a rough plane inclined at 30$^{\circ}$ to the horizontal and is on the point of slipping. Take $g = 9.81$ ms$^{-2}$.

The weight of the particle is $8g$ newtons, acting vertically downwards. Resolving it along the plane, the component down the slope is

$$8g\sin 30^{\circ}=8(9.81)\left(\frac{1}{2}\right)=4g=39.24 \text{ N},$$

which is 39 N correct to 2 significant figures.

The sine goes with the angle of inclination for the component along the plane; the cosine gives the component perpendicular to it, $8g\cos 30^{\circ}=67.96$ N, which the normal reaction balances.