Elevation/depression | Year 10

01 / LEARN

Understand the idea.

Use the horizontal reference

An angle of elevation is measured upwards from the observer's horizontal; depression is measured downwards from it. Drawing that horizontal prevents accidentally using the complementary angle measured from a vertical wall.

Parallel horizontals make a depression angle equal to the elevation angle at the other end of the sight line. In a right triangle, tangent then compares the vertical difference with the horizontal distance.

Separate the height difference

The opposite side measures height above the observer's eyes, not necessarily height above the ground. On level ground, subtract eye height from total height when forming the triangle; add it back after finding the rise.

Label the horizontal distance separately from the sloping sight line. Use degree mode for these angles, and round the requested height or distance only after the entire calculation.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Include eye height

An observer stands 20 m from a tower. Eye height is 1.5 m and elevation is 35°. Find tower height to 0.1 m.

Worked solution

1
Identify the height above eye level.
$rise = h - 1.5$
The horizontal reference passes through the observer's eyes.
2
Set up the tangent ratio.
$tan 35^{\circ} = \frac{h - 1.5}{20}$
Rise is opposite the angle and horizontal distance is adjacent.
3
Rearrange and evaluate the whole height.
$h = 20 tan 35^{\circ} + 1.5 \approx 15.5042 m$
Multiply by the horizontal distance, then restore the eye height.
4
Round to one decimal place.
$h \approx 15.5 m$
The hundredths digit is zero.

Answer

15.5 m

EXAMPLE 02

Find a distance

A platform is 18 m high. An observer with eye height 1.6 m sees its top at elevation 32°. Find horizontal distance to 0.1 m.

Worked solution

1
Find the vertical difference.
$18 - 1.6 = 16.4 m$
The platform top is measured relative to eye level.
2
Set up the ratio for horizontal distance.
$tan 32^{\circ} = \frac{16.4}{d}$
Tangent is opposite divided by adjacent.
3
Solve and round the distance.
$d = \frac{16.4}{tan 32^{\circ}} \approx 26.2 m$
Dividing the rise by the tangent isolates the adjacent side.

Answer

26.2 m

03 / YOUR TURN

Try it yourself.

Check your understanding

A ground-level sight line has elevation 45° and horizontal distance 12 m. Find the height.

Show answer & reasoning

12 m

With a ground-level observer there is no eye-height adjustment: $h = 12 tan 45^{\circ} = 12 m$ because $tan 45^{\circ} = 1$ .

Keep practising

6 worksheet questions · 15 marks

Open worksheet Worked solutions

Read the questions online

From level ground 30 m away, roof elevation is $40^{\circ}$ . Find its height to 0.1 m.
2 marks
A tree top is 25 m above an observer sight line, at elevation $50^{\circ}$ . Find horizontal distance to 0.1 m.
2 marks
An observer is 16 m from a 12 m wall. Their eye is 1.6 m high. Find elevation to the top to ${0.1}^{\circ}$ .
3 marks
A cliff top is 40 m above sea level. Depression to a boat is $25^{\circ}$ . Find the horizontal distance to 0.1 m.
3 marks
A line of sight makes $38^{\circ}$ with a vertical wall. Find its angle with the horizontal.
2 marks
A camera 1.2 m high is 15 m from a flagpole. Top elevation is $55^{\circ}$ . Find total pole height to 0.1 m.
3 marks

Read the worked solutions

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

From level ground 30 m away, roof elevation is $40^{\circ}$ . Find its height to 0.1 m.
2 marks
$h = 30 tan 40^{\circ}$ .
Round the result, not the tangent value.
25.2 m
A tree top is 25 m above an observer sight line, at elevation $50^{\circ}$ . Find horizontal distance to 0.1 m.
2 marks
$tan 50^{\circ} = \frac{25}{d}$ .
$d = \frac{25}{tan 50^{\circ}}$ .
21.0 m
An observer is 16 m from a 12 m wall. Their eye is 1.6 m high. Find elevation to the top to ${0.1}^{\circ}$ .
3 marks
Vertical difference $= 12 - 1.6 = 10.4$ m.
$θ = {tan}^{- 1} (\frac{10.4}{16})$ .
${33.0}^{\circ}$
A cliff top is 40 m above sea level. Depression to a boat is $25^{\circ}$ . Find the horizontal distance to 0.1 m.
3 marks
Use the equal elevation angle at the boat.
$d = \frac{40}{tan 25^{\circ}}$ .
85.8 m
A line of sight makes $38^{\circ}$ with a vertical wall. Find its angle with the horizontal.
2 marks
$θ = 90^{\circ} - 38^{\circ} = 52^{\circ}$ .
$52^{\circ}$
A camera 1.2 m high is 15 m from a flagpole. Top elevation is $55^{\circ}$ . Find total pole height to 0.1 m.
3 marks
Height above camera $= 15 tan 55^{\circ}$ .
Add 1.2 m.
22.6 m

Checking saved progress…

Curriculum reference & lesson scope

Australian Curriculum v9 and South Australian school course scope

Right-triangle applications include AC9M10M03. The approved chapter also develops non-right-triangle methods, exact values and trigonometric functions within one normal Year 10 course.

Official curriculum source ↗

In this lesson

Understand the idea.

Use the horizontal reference

Separate the height difference

See the method in action.

Include eye height

Identify the height above eye level.

Set up the tangent ratio.

Rearrange and evaluate the whole height.

Round to one decimal place.

Find a distance

Find the vertical difference.

Set up the ratio for horizontal distance.

Solve and round the distance.

Try it yourself.

Check your understanding

Keep practising