A man moves 6.0m East and then 10.0m N30ºE. How far is he from his starting point?

This is a vector-addition problem, so the two journeys must be resolved into perpendicular components before they are combined. The bearing notation \(N30^\circ E\) means the direction is measured \(30^\circ\) away from north, turning towards the east. For a displacement of \(10.0\,\text{m}\) in that direction, north is the adjacent side and east the opposite side of the \(30^\circ\) angle:

• eastward part \(= 10.0\sin 30^\circ = 5.0\,\text{m}\)
• northward part \(= 10.0\cos 30^\circ = 8.66\,\text{m}\)

The first leg is entirely eastward, so the totals are \[x = 6.0 + 5.0 = 11.0\,\text{m (east)},\qquad y = 0 + 8.66 = 8.66\,\text{m (north)}.\] These two totals are at right angles, so Pythagoras gives the straight-line distance from the start: \[r = \sqrt{11.0^2 + 8.66^2} = \sqrt{121 + 75.0} = \sqrt{196} = 14.0\,\text{m}.\] The man is \(14.0\,\text{m}\) from his starting point.

The usual error is to add the magnitudes, \(6.0 + 10.0 = 16.0\,\text{m}\), or to interchange the sine and cosine because the angle was assumed to be measured from the east line. In bearings written as \(N\theta E\) the angle is measured from north, so north takes the cosine. Sketching the two arrows head-to-tail, as above, shows at once which component belongs to which trigonometric ratio.