Practical right-triangle problems and interpreting diagrams | Year 9

01 / LEARN

The idea, step by step

Make the model visible

A vertical wall or tree and level ground give perpendicular sides. A straight ladder, taut cable or sight line may be the hypotenuse. State these assumptions rather than treating every sloping sketch as a right triangle.

Label the requested quantity first. Height above eye level differs from height above ground. A drop is original height minus final height.

Return to the real situation

Include units and sensible precision. A ladder’s vertical reach is less than its length, and a ramp’s slope length exceeds its horizontal run.

When a situation changes, draw a second triangle. Calculate each relevant height from its own information before finding the difference.

A common trap

The new ladder height is the amount its top dropped.

Instead: The drop equals the original height minus the new height.

02 / WORKED EXAMPLES

See the thinking

Example 1

A ladder

A 10 m ladder has its foot 6 m from a vertical wall on level ground. How high does it reach?

$h^{2} + 6^{2} = 10^{2}$ .
$h = \sqrt{100 - 36} = 8$ m.

8 m

Example 2

Include eye height

An observer’s eyes are 1.5 m above level ground, 18 m horizontally from a vertical tree. Its top is 32° above horizontal. Find the tree height to two decimal places.

Height above eyes $= 18 tan 32^{\circ} \approx 11.2476$ m.
Total height $= 1.5 + 18 tan 32^{\circ} \approx 12.7476$ m.

12.75 m

Check your understanding

A ladder changes from reaching 12 m high to 5 m high. Find the drop.

Show answer

7 m

12 − 5 = 7 m.

03 / WORKSHEET

Your turn

6 questions · 20 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

A 13 m ladder has its foot 5 m from a wall. Find its vertical reach.
3 marks
The same 13 m ladder slides so its foot is now 12 m from the wall. How far does the top drop?
4 marks
A ramp makes 7° with level ground and has horizontal run 4.8 m. Find its sloping length to two decimal places.
3 marks
A vertical pole casts a 4.5 m shadow on level ground. The Sun’s elevation is 36°. Find the pole height to two decimal places.
3 marks
A taut 30 m kite string begins at ground level and makes 40° to horizontal. Find the kite height to two decimal places.
3 marks
Eyes are 1.6 m above level ground, 30 m horizontally from a building. Its top is at 25° above horizontal. Find total height to two decimal places.
4 marks

Read the worked solutions

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

A 13 m ladder has its foot 5 m from a wall. Find its vertical reach.
3 marks
$h = \sqrt{169 - 25} = 12$ m.
12 m
The same 13 m ladder slides so its foot is now 12 m from the wall. How far does the top drop?
4 marks
Original height $= \sqrt{169 - 25} = 12$ m.
New height $= \sqrt{169 - 144} = 5$ m.
Drop = 12 − 5 = 7 m.
7 m
A ramp makes 7° with level ground and has horizontal run 4.8 m. Find its sloping length to two decimal places.
3 marks
$cos 7^{\circ} = 4.8 / L$ .
$L = 4.8 / cos 7^{\circ}$ .
4.84 m
A vertical pole casts a 4.5 m shadow on level ground. The Sun’s elevation is 36°. Find the pole height to two decimal places.
3 marks
$h = 4.5 tan 36^{\circ}$ .
Assume a vertical pole and a straight ray of sunlight.
3.27 m
A taut 30 m kite string begins at ground level and makes 40° to horizontal. Find the kite height to two decimal places.
3 marks
$h = 30 sin 40^{\circ}$ .
The string is the hypotenuse.
19.28 m
Eyes are 1.6 m above level ground, 30 m horizontally from a building. Its top is at 25° above horizontal. Find total height to two decimal places.
4 marks
Height above eyes $= 30 tan 25^{\circ}$ .
Total height $= 1.6 + 30 tan 25^{\circ}$ .
15.59 m

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Curriculum reference & lesson scope

Australian Curriculum v9 and SA Mathematics

Year 9 similarity, Pythagoras and right-triangle trigonometric ratios, with prerequisite revision. Non-right-triangle sine/cosine rules are not required.

Official curriculum source ↗

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