Drawing the bearing diagram · Elevation, depression & bearings

In bearing questions, most lost marks come from the diagram. If the diagram is wrong, every calculation after it is wrong too. If it’s right, the rest is usually Pythagoras or one trig ratio (or, in harder questions, the cosine rule). This lesson is about getting it right.

A routine for any bearing diagram

Read one sentence at a time. Each sentence adds one point and one line.
Draw a north line at every point you leave from, before you draw the line from it.
Mark each bearing clockwise from that north line, and write the distance on the line.
Join the last point to the first if the question asks about them.
Find the angles inside the triangle using the parallel north lines: angles on a straight line, alternate angles and co-interior angles (which add up to $180^\circ$ ).

Build some journeys here. Change a bearing and watch the angle inside the triangle change with it.

Build the journeySet each leg's bearing and distance

13.2distance PR091°bearing of R from P271°bearing of P from R60°angle at Q

P → Qbearing = 050°distance = 15

Q → Rbearing = 170°distance = 10

Draw a north line at every point you leave from, and measure each bearing clockwise from it. The angle inside the triangle at Q comes from the two north lines, which are parallel: use angles on a straight line and co-interior or alternate angles. Then use Pythagoras (if there's a right angle) or the cosine rule to find PR.

The angle inside at QBack to P is 050° + 180° = 230°; on to R is 170°; the angle is 230° − 170° = 60°

A past question, step by step

Worked example · WAEC 2021 Paper 2, Q3

WAEC 2021 · Paper 2 · Q3

The points $X$ , $Y$ and $Z$ are located such that $Y$ is $15\text{ km}$ south of $X$ , and $Z$ is $20\text{ km}$ from $X$ on a bearing of $270^\circ$ . Calculate, correct to:

two significant figures, $|YZ|$ ;

the nearest degree, the bearing of $Y$ from $Z$ .

Place X and Z
$Z$ is 20 km from $X$ on a bearing of $270^\circ$ , which is due west. So $X$ is due east of $Z$ : draw $ZX$ across, 20 km long.
Think first. Z is on a bearing of 270° from X. Which direction is that, and so where is X seen from Z?
Place Y
$Y$ is 15 km due south of $X$ : draw $XY$ straight down from $X$ .
Think first. Y is due south of X. What angle does XY make with XZ?
Join Y and Z, and find the right angle
West and south are $90^\circ$ apart, so $\angle YXZ = 90^\circ$ . Pythagoras gives
$|YZ| = \sqrt{15^2 + 20^2} = \sqrt{625} = 25\text{ km}$
The angle at Z
In the right-angled triangle, at $Z$ the opposite side is $XY = 15$ and the adjacent side is $ZX = 20$ :
$\begin{aligned} \tan\angle XZY &= \frac{15}{20} \\ \angle XZY &\approx 36.87^\circ \end{aligned}$
Think first. At Z, which side is opposite the angle, and which is adjacent?
Turn it into a bearing
At $Z$ , draw north. $X$ is due east, on $090^\circ$ . $Y$ is further round clockwise by $36.87^\circ$ . So the bearing of $Y$ from $Z$ is $090^\circ + 36.87^\circ \approx 127^\circ$ .

More: right angles and isosceles triangles in bearing diagrams

When the angle inside the triangle isn’t $90^\circ$ , find the missing side with the cosine rule and the missing angle with the sine rule (see the sine and cosine rules). The diagram routine is exactly the same.

More: journeys that need the sine or cosine rule

Your turn

WAEC 2023 · Paper 2 · Q8 (a)

(a)
A tree is $8\text{ km}$ due south of a building. Musa is standing $8\text{ km}$ due west of the tree. (i) Illustrate the information on a diagram. (ii) How far is Musa from the building? (iii) Find the bearing of Musa from the building.
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)(i)

Draw the building $B$ , the tree $T$ 8 km due south of it, and Musa $M$ 8 km due west of the tree, so the angle at $T$ is a right angle.

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