Elevation/depression and bearings | Year 11

01 / LEARN

Understand the idea.

Elevation and depression use a horizontal line

Angles of elevation and depression start at the observer's horizontal, measured upward and downward respectively. Parallel horizontal lines make a depression angle equal the corresponding elevation angle at the other end of the same sightline.

Trigonometry initially gives the vertical difference from eye or instrument level. Add eye height only when converting that rise to height above the same level ground. Distinguish a horizontal distance from the sloping line of sight.

A bearing is measured from north

A bearing is measured clockwise from north at the departure point and normally written with three digits. Draw the north line first: east is $090^{\circ}$ , south $180^{\circ}$ and west $270^{\circ}$ .

The reverse bearing differs by $180^{\circ}$ , adjusted into one full turn. For distance $d$ on bearing $β$ , components are $E = d sin β$ and $N = d cos β$ ; negative components indicate west or south.

\begin{aligned} E & = d sin β \\ N & = d cos β \end{aligned}

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Worked example 1

An observer’s eyes are 1.6 m above level ground. A tower is 20 m horizontally away and its top has elevation 32°. Find the tower height.

Worked solution

1
Define the rise above eye level.
$tan 32^{\circ} = \frac{h}{20}$
The rise is opposite the elevation angle and twenty metres is the horizontal adjacent leg.
2
Calculate that vertical rise.
$h = 20 tan 32^{\circ} \approx 12.4974 m$
Tangent connects the two perpendicular legs rather than the line-of-sight length.
3
Add the observer's eye height.
$H = 1.6 + 20 tan 32^{\circ} \approx 14.0974 m$
The triangle starts one point six metres above the level ground at the tower base.
4
Report the tower height.
$H \approx 14.1 m$
Round the total height, rather than an earlier intermediate measurement.

Answer

14.1 m

EXAMPLE 02

Worked example 2

A camp is 12 km east and 16 km north of a starting point. Find its distance and bearing from the start.

Worked solution

1
Find the direct distance.
$d = \sqrt{12^{2} + 16^{2}} = \sqrt{400} = 20 km$
East and north components are perpendicular legs of the displacement triangle.
2
Measure the direction from north.
$tan β = \frac{12}{16}$
East is opposite the angle from the north line; north is adjacent.
3
Calculate the angle and identify its quadrant.
$β = {tan}^{- 1} (\frac{12}{16}) \approx {36.8699}^{\circ}$
Both components are positive, so the camp is northeast and this acute angle is already the clockwise bearing.
4
Write the rounded three-digit bearing.
$037^{\circ}$
The nearest degree is thirty-seven, written with a leading zero for bearing notation.

Answer

20 km; bearing 037° to the nearest degree.

03 / YOUR TURN

Try it yourself.

Check your understanding

What is the reverse bearing of 248°?

Show answer & reasoning

068°

The reverse bearing is $248^{\circ} - 180^{\circ} = 68^{\circ}$ , written $068^{\circ}$ . Reversing a direction turns through half a full turn.

Keep practising

6 worksheet questions · 13 marks

Open worksheet Worked solutions

Read the questions online

A vertical tower is 16 m horizontally from an observer at ground level. Its top has elevation $35^{\circ}$ . Find its height to one decimal place.
2 marks
Repeat the tower calculation if the observing instrument is 1.5 m above the ground.
2 marks
From an 8 m high platform, a boat is seen at a depression of $25^{\circ}$ . Find its horizontal distance from the platform base.
2 marks
A point is 14 km east and 20 km north of a station. Find its bearing to the nearest degree.
2 marks
Find the reverse bearing of $032^{\circ}$ .
2 marks
A boat travels 24 km on bearing $210^{\circ}$ . Find its southward and westward distances, each to one decimal place.
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

A vertical tower is 16 m horizontally from an observer at ground level. Its top has elevation $35^{\circ}$ . Find its height to one decimal place.
2 marks
$h = 16 tan 35^{\circ}$ .
$h \approx 11.2 m$ .
11.2 m
Repeat the tower calculation if the observing instrument is 1.5 m above the ground.
2 marks
The trigonometric rise is still $16 tan 35^{\circ}$ .
Add the instrument height 1.5 m.
$H \approx 12.7 m$ .
12.7 m
From an 8 m high platform, a boat is seen at a depression of $25^{\circ}$ . Find its horizontal distance from the platform base.
2 marks
The boat’s elevation angle to the platform is $25^{\circ}$ .
$tan 25^{\circ} = \frac{8}{d}$ , so $d = \frac{8}{tan 25^{\circ}}$ .
$d \approx 17.2 m$ .
17.2 m
A point is 14 km east and 20 km north of a station. Find its bearing to the nearest degree.
2 marks
$β = {tan}^{- 1} (\frac{14}{20}) = 34.992^{\dots \circ}$ .
It lies in the north-east quadrant, so the bearing is $035^{\circ}$ .
$035^{\circ}$
Find the reverse bearing of $032^{\circ}$ .
2 marks
$032^{\circ} + 180^{\circ} = 212^{\circ}$ .
$212^{\circ}$
A boat travels 24 km on bearing $210^{\circ}$ . Find its southward and westward distances, each to one decimal place.
3 marks
$E = 24 sin 210^{\circ} = - 12 km$ .
$N = 24 cos 210^{\circ} \approx - 20.7846 km$ .
The negative signs mean west and south respectively.
20.8 km south; 12.0 km west

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 General Mathematics: practical trigonometry, including similarity, right triangles, triangle area, sine/cosine rules and contextual interpretation.

Official curriculum source ↗

In this lesson

Understand the idea.

Elevation and depression use a horizontal line

A bearing is measured from north

See the method in action.

Worked example 1

Define the rise above eye level.

Calculate that vertical rise.

Add the observer's eye height.

Report the tower height.

Worked example 2

Find the direct distance.

Measure the direction from north.

Calculate the angle and identify its quadrant.

Write the rounded three-digit bearing.

Try it yourself.

Check your understanding

Keep practising