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 $\mathrm{\angle }NOP$.

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.

2. Using the three-digit notation

POINTS	CARDINAL POINT NOTATION	3-DIGIT NOTATION
	$$N30{}^{\circ }E$$	$$030{}^{\circ }$$
	$$S50{}^{\circ }E$$	$$130{}^{\circ }$$
	$$S65{}^{\circ }W$$	$$245{}^{\circ }$$
	$$N25{}^{\circ }W$$	$$335{}^{\circ }$$

$$P_1=030{}^{\circ }$$

$$P_2=180{}^{\circ }-50{}^{\circ }=130{}^{\circ }$$

$$P_3=180{}^{\circ }+65{}^{\circ }=245{}^{\circ }$$

$$P_4=360{}^{\circ }-25{}^{\circ }=335{}^{\circ }$$

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, $\frac{a}{\sin A}=\frac{c}{\sin C}$

$$\frac{94.12}{\sin 38{}^{\circ }}=\frac{150}{\sin C}\to \sin C=\frac{\sin 38{}^{\circ }\times 150}{94.12}\to \sin C=\frac{0.6157\times 150}{94.12}\to \sin C=0.9812$$

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

$$Bearingofhisfinalpositionfromthestartingposition=360{}^{\circ }-79{}^{\circ }=281{}^{\circ }$$