Finding a side | Year 9 | Ascent Library

01 / LEARN

Understand the idea.

Label, choose, substitute

Start by naming the reference angle and labelling opposite, adjacent and hypotenuse. Choose the ratio containing the known side and the unknown. Write that equation before calculating so that the order of the fraction remains clear.

If the unknown is in the numerator, multiply by the known denominator. If it is in the denominator, first multiply both sides by the unknown, then divide by the trigonometric value. These are ordinary equation-solving operations.

Check the size

Use side sizes to check the result. A leg found from a known hypotenuse must be shorter; a hypotenuse found from a leg must be longer. For a small acute angle, the opposite leg is small relative to the hypotenuse.

The examples use degree mode and round the final length to two decimal places. Keep the angle and trigonometric value unrounded during rearrangement, then attach the correct length unit to the answer.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Unknown opposite side

Hypotenuse = 12 cm and θ = 35°. Find opposite.

Worked solution

1
Choose sine for opposite and hypotenuse.
$sin 35^{\circ} = \frac{x}{12}$
The unknown opposite side and known hypotenuse form the sine ratio.
2
Multiply by the hypotenuse.
$x = 12 sin 35^{\circ}$
Multiplying both sides by twelve isolates the numerator.
3
Evaluate and round the length.
$\begin{aligned} x & = 6.882917 \dots cm \\ \approx 6.88 cm \end{aligned}$
Degree mode gives the stated angle, and the leg is shorter than the twelve-centimetre hypotenuse.

Answer

6.88 cm

EXAMPLE 02

Unknown hypotenuse

Adjacent = 9 m and θ = 42°. Find the hypotenuse.

Worked solution

1
Choose cosine for adjacent and hypotenuse.
$cos 42^{\circ} = \frac{9}{x}$
The unknown is the hypotenuse, so it belongs in the denominator.
2
Clear the unknown denominator.
$x cos 42^{\circ} = 9$
Multiply both sides by the positive length $x$ .
3
Divide by the cosine value.
$x = \frac{9}{cos 42^{\circ}}$
Division isolates $x$ because the cosine of this acute angle is nonzero.
4
Evaluate and check the result.
$\begin{aligned} x & = 12.110694 \dots m \\ \approx 12.11 m \end{aligned}$
The hypotenuse is longer than the nine-metre adjacent side, as required.

Answer

12.11 m

03 / YOUR TURN

Try it yourself.

Check your understanding

Hypotenuse = 10 and θ = 30°. Find opposite.

Show answer & reasoning

Opposite and hypotenuse select sine: $sin 30^{\circ} = \frac{o}{10}$ . Thus $o = 10 sin 30^{\circ} = 10 \times \frac{1}{2} = 5$ .

Keep practising

6 worksheet questions · 12 marks

Open worksheet Worked solutions

Read the questions online

Find opposite if $h = 14$ cm and $θ = 28^{\circ}$ . Give two decimal places.
2 marks
Find adjacent if $h = 18$ m and $θ = 57^{\circ}$ . Give two decimal places.
2 marks
Find opposite if $a = 11$ cm and $θ = 35^{\circ}$ . Give two decimal places.
2 marks
Find adjacent if $o = 8$ m and $θ = 41^{\circ}$ . Give two decimal places.
2 marks
Find hypotenuse if $a = 12$ cm and $θ = 27^{\circ}$ . Give two decimal places.
2 marks
Find hypotenuse if $o = 9$ m and $θ = 63^{\circ}$ . Give two decimal places.
2 marks

Read the worked solutions

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

Find opposite if $h = 14$ cm and $θ = 28^{\circ}$ . Give two decimal places.
2 marks
$o = 14 sin 28^{\circ}$ .
Evaluate in degrees and round the final length.
6.57 cm
Find adjacent if $h = 18$ m and $θ = 57^{\circ}$ . Give two decimal places.
2 marks
$a = 18 cos 57^{\circ}$ .
The result is shorter than 18 m.
9.80 m
Find opposite if $a = 11$ cm and $θ = 35^{\circ}$ . Give two decimal places.
2 marks
$o = 11 tan 35^{\circ}$ .
7.70 cm
Find adjacent if $o = 8$ m and $θ = 41^{\circ}$ . Give two decimal places.
2 marks
$tan 41^{\circ} = \frac{8}{a}$ .
$a = \frac{8}{tan 41^{\circ}}$ .
9.20 m
Find hypotenuse if $a = 12$ cm and $θ = 27^{\circ}$ . Give two decimal places.
2 marks
$cos 27^{\circ} = \frac{12}{h}$ .
$h = \frac{12}{cos 27^{\circ}}$ .
13.47 cm
Find hypotenuse if $o = 9$ m and $θ = 63^{\circ}$ . Give two decimal places.
2 marks
$sin 63^{\circ} = \frac{9}{h}$ .
$h = \frac{9}{sin 63^{\circ}}$ .
10.10 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 ↗

In this lesson

Understand the idea.

Label, choose, substitute

Check the size

See the method in action.

Unknown opposite side

Choose sine for opposite and hypotenuse.

Multiply by the hypotenuse.

Evaluate and round the length.

Unknown hypotenuse

Choose cosine for adjacent and hypotenuse.

Clear the unknown denominator.

Divide by the cosine value.

Evaluate and check the result.

Try it yourself.

Check your understanding

Keep practising