Bearings with any triangle · Sine & cosine rules

When the angle inside a bearing diagram isn’t $90^\circ$ , Pythagoras won’t do. The plan is always the same:

Draw the diagram, with a north line at every point (Drawing the bearing diagram).
Find the angle inside the triangle at the turning point, from the bearings.
Use the cosine rule for the unknown distance.
Use the sine rule for an angle, then turn it into the bearing asked for.

Finding the angle at the turning point

At the point where the journey turns, draw a north line. The direction back along the first leg is the back bearing: add or take away $180^\circ$ . The angle inside the triangle is the gap between the back bearing and the new bearing.

A past question, step by step

Worked example · WAEC 2017 Paper 2, Q9(a)

WAEC 2017 · Paper 2 · Q9 (a)

An aeroplane flies 100 km from town $A$ on a bearing of $330^\circ$ to town $B$ . It then flies 300 km due west to town $C$ . (i) Illustrate this information in a diagram. (ii) Calculate the: (I) distance between $A$ and $C$ , correct to two decimal places; (II) bearing of $C$ from $A$ .

First leg
Draw north at $A$ . $330^\circ$ is $30^\circ$ west of north, so $AB$ goes up and to the left, 100 km long.
Think first. 330° is 30° short of north. Which way does AB point?
Second leg
Draw north at $B$ . Due west is $270^\circ$ : $BC$ goes straight left, 300 km long.
Think first. At B, what is the bearing back to A, and how far is it from due west?
The angle at B
- At $B$ , the direction back to $A$ is ${330^\circ - 180^\circ = 150^\circ}$ .
- The direction to $C$ is ${270^\circ}$ .
- So ${\angle ABC = 270^\circ - 150^\circ = 120^\circ}$ .
Cosine rule for AC
Two sides and the angle between them:
$\begin{aligned} &|AC|^2 \\ &= 100^2 + 300^2 \\ &\quad - 2(100)(300)\cos 120^\circ \\ &= 10\,000 + 90\,000 + 30\,000 \\ &= 130\,000 \end{aligned}$
so $|AC| = \sqrt{130\,000} \approx 360.56\text{ km}$ .
Think first. $\cos 120^\circ$ is negative. Will $AC$ be longer or shorter than Pythagoras would give?
Sine rule for the angle at A
$\begin{aligned} \frac{\sin\angle BAC}{300} &= \frac{\sin 120^\circ}{360.56} \\ \sin\angle BAC &= \frac{300 \times 0.8660}{360.56} \\ &\approx 0.7206 \end{aligned}$
so $\angle BAC \approx 46.1^\circ$ .
Turn it into a bearing
- At $A$ , $B$ is on ${330^\circ}$ .
- $C$ is further round anticlockwise (towards the west) by ${46.1^\circ}$ .
- The bearing of $C$ from $A$ is ${330^\circ - 46.1^\circ \approx 284^\circ}$ .

Your turn

WAEC 2019 · Paper 2 · Q12

A town $J$ is $20\text{ km}$ from a lorry station, $K$ , on a bearing $065^\circ$ . Another town, $T$ , is $8\text{ km}$ from $K$ on a bearing $155^\circ$ . Calculate:

(a)(i)
to the nearest kilometre, the distance of $T$ from $J$ ;
(a)(ii)
to the nearest degree, the bearing of $T$ from $J$ .

Try it on a graph

K is at the origin; north is up. The dashed line is JT.

Worked solution (try it first)

Draw north at $K$ .
$J$ is 20 km from $K$ on $065^\circ$ and $T$ is 8 km from $K$ on $155^\circ$ .
The angle between them is $\angle JKT = 155^\circ - 65^\circ$
$= 90^\circ$ , so triangle $JKT$ is right-angled at $K$ .

(a)(i)

$|TJ| = \sqrt{20^2 + 8^2}$
$= \sqrt{464}$
$\approx 21.54$ km, which is 22 km to the nearest kilometre.

(ii)

At $J$ : $\tan\angle KJT = \frac{8}{20} = 0.4$ , so $\angle KJT \approx 21.8^\circ$ .
At $J$ , the direction back to $K$ is $065^\circ + 180^\circ = 245^\circ$ , and $T$ is $21.8^\circ$ further round anticlockwise.
Bearing of $T$ from $J$ $= 245^\circ - 21.8^\circ$
$= 223.2^\circ$
$\approx 223^\circ$ .

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