Bearings - SS2 Mathematics Lesson Note

Consider the two points \(N\ \)and\(\ P\) away from a point \(O\) on a map, \(N\) is north of \(O\) and the bearing of \(P\) from \(O\) is \(\angle NOP\).

Bearing is always measured clockwise from \(ON\).

In measuring bearing angles, there are usually two ways to do this:

1. Using the cardinal point notation \(N,\ S,\ E,\ W\) for \(north,\ south,\ east\ \)and\(\ west\) respectively. Note, each quadrant between the cardinal points is \(90{^\circ}\) and measured from the north-south axis.

Using the three-digit notation

POINTS	CARDINAL POINT NOTATION	3-DIGIT NOTATION
	\[N\ 30{^\circ}\ E\]	\[030{^\circ}\]
	\[S50{^\circ}E\]	\[130{^\circ}\]
	\[S65{^\circ}W\]	\[245{^\circ}\]
	\[N25{^\circ}W\]	\[335{^\circ}\]

\[P_{1} = 030{^\circ}\]

\[P_{2} = 180{^\circ} - 50{^\circ} = 130{^\circ}\]

\[P_{3} = 180{^\circ} + 65{^\circ} = 245{^\circ}\]

\[P_{4} = 360{^\circ} - 25{^\circ} = 335{^\circ}\]

Example: A man walks \(100m\) due north and then \(150m\) on a bearing of \(S38{^\circ}W\). How far and on what bearing is he from his original position?

Solution

Using the cosine rule, \(a^{2} = b^{2} + c^{2} - 2bc\cos A\)

\[a^{2} = 100^{2} + 150^{2} - 2(100)(150)\cos{38{^\circ}}\]

\[a^{2} = 10,000 + 22,500 - (30,000)(0.7880)\]

\[a^{2} = 10,000 + 22,500 - 23,640\]

\[a^{2} = 8,860\]

\[a = \sqrt{8860} = \ 94.12m\]

Using the sine rule, \(\frac{a}{\sin A} = \frac{c}{\sin C}\)

\[\frac{94.12}{\sin{38{^\circ}}} = \frac{150}{\sin C}\ \ \ \rightarrow \ \ \sin C = \ \frac{\sin{38{^\circ}} \times 150}{94.12}\ \ \ \ \rightarrow \ \ \ \sin C = \frac{0.6157 \times 150}{94.12}\ \ \ \rightarrow \ \ \sin C = \ 0.9812\]

\[C = \ \sin^{- 1}{0.9812} \approx 79{^\circ}\]

\[Bearing\ of\ his\ final\ position\ from\ the\ starting\ position\ = \ 360{^\circ} - 79{^\circ}\ = 281{^\circ}\]