Sine rule | Year 10 | Ascent Library

01 / LEARN

The idea, step by step

Build opposite pairs

In triangle ABC, side a is opposite angle A, b is opposite B and c is opposite C. The sine rule compares a side with the sine of its opposite angle. Mark pairs before substituting: a side beside an angle is not its opposite.

A known opposite pair and another side or angle allow a calculation. With two known angles, first find the third angle if that gives a useful pair.

\frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin C}

An inverse sine result may not be the only angle

For an angle between 0° and 180°, sine has the same value at θ and 180° − θ. A side-side-angle problem may therefore form two triangles. Test both candidates against the 180° angle sum.

If the calculated sine is greater than 1, the stated triangle cannot exist. Do not force a calculator error into an angle.

A common trap

Accepting only the inverse-sine calculator answer.

Instead: Test its supplement too, then reject candidates that make the triangle angle sum impossible.

02 / WORKED EXAMPLES

See the thinking

Example 1

Find a side

In triangle ABC, A = 40°, B = 65° and a = 10 cm. Find b to 0.1 cm.

Use the opposite pairs $(a, A)$ and $(b, B)$ .
$b = 10 sin 65^{\circ} / sin 40^{\circ}$ .

14.1 cm

Example 2

Two possible triangles

A triangle has A = 35°, a = 8 cm and b = 10 cm. Find all possible angles B to 0.1°.

$sin B = 10 sin 35^{\circ} / 8$ .
The calculator gives 45.8°.
The second candidate is 180° minus that value: 134.2°.
Both candidates leave a positive third angle with A = 35°.

45.8° or 134.2°

Check your understanding

A = 30°, a = 6 cm and B = 90°. Find b.

Show answer

12 cm

$b = 6 sin 90^{\circ} / sin 30^{\circ} = 12$ cm.

03 / WORKSHEET

Your turn

6 questions · 16 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

A = 30°, a = 6 cm and B = 45°. Find b exactly.
3 marks
A = 50°, a = 12 cm and B = 70°. Find b to 0.1 cm.
2 marks
B = 40°, b = 9 cm and A = 55°. Find a to 0.1 cm.
2 marks
A = 35°, a = 7 cm and b = 9 cm. Find all possible B to 0.1°.
3 marks
Can A = 40°, a = 5 cm and b = 12 cm form a triangle? Explain.
3 marks
A = 70°, B = 60° and a = 10 m. Find C and c to 0.1 m.
3 marks

Read the worked solutions

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

A = 30°, a = 6 cm and B = 45°. Find b exactly.
3 marks
$b = 6 sin 45^{\circ} / sin 30^{\circ}$ .
$b = 6 (\sqrt{2} / 2) / (1 / 2) = 6 \sqrt{2}$ cm.
$6 \sqrt{2}$ cm
A = 50°, a = 12 cm and B = 70°. Find b to 0.1 cm.
2 marks
$b = 12 sin 70^{\circ} / sin 50^{\circ}$ .
14.7 cm
B = 40°, b = 9 cm and A = 55°. Find a to 0.1 cm.
2 marks
$a = 9 sin 55^{\circ} / sin 40^{\circ}$ .
11.5 cm
A = 35°, a = 7 cm and b = 9 cm. Find all possible B to 0.1°.
3 marks
$sin B = 9 sin 35^{\circ} / 7$ .
Take inverse sine and its supplement.
Both have A + B less than 180°.
47.5° or 132.5°
Can A = 40°, a = 5 cm and b = 12 cm form a triangle? Explain.
3 marks
$sin B = 12 sin 40^{\circ} / 5 \approx 1.543$ .
A real angle cannot have sine greater than 1.
No triangle exists.
A = 70°, B = 60° and a = 10 m. Find C and c to 0.1 m.
3 marks
$C = 180^{\circ} - 70^{\circ} - 60^{\circ} = 50^{\circ}$ .
$c = 10 sin 50^{\circ} / sin 70^{\circ}$ .
C = 50°; c = 8.2 m

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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 ↗

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