WAEC 2020 · Paper 1 · Q30

A force of 230 N acts in the direction $065^\circ$ . Find its horizontal component.

Worked solution (try it first)

A bearing of $065^\circ$ is measured from north, so the force makes $65^\circ$ with the vertical (north) direction.
The horizontal (east) component is $230\sin 65^\circ$ .
$230 \times 0.9063 = 208.45$ , which is 208.5 N, option A.