Sine rule | Year 10 | Ascent Library

01 / LEARN

Understand the idea.

Build opposite pairs

The sine rule connects each side with its opposite angle. In triangle $A B C$ , $a$ lies opposite $A$ , not beside it. Mark opposite pairs before selecting two equal ratios from the rule.

A complete known side-angle pair lets you compare with another pair containing one unknown. If two angles are known, the angle sum may supply a missing opposite angle before you use the rule.

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

An inverse sine result may not be the only angle

For angles inside a triangle, $sin θ = sin (180^{\circ} - θ)$ . An inverse-sine calculation can therefore hide a second possible triangle. Find the principal angle and its supplement, then test each with the known angle.

Every accepted candidate must leave a strictly positive third angle. If the calculated sine exceeds $1$ , no triangle satisfies the stated measurements. A calculator result alone does not establish how many triangles exist.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Find a side

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

Worked solution

1
Match each side with its opposite angle.
$\frac{b}{sin 65^{\circ}} = \frac{10}{sin 40^{\circ}}$
The known pair is $a = 10 cm$ opposite $A = 40^{\circ}$ .
2
Isolate the unknown side.
$b = \frac{10 sin 65^{\circ}}{sin 40^{\circ}}$
Multiply both sides by $sin 65^{\circ}$ .
3
Evaluate and round.
$b \approx 14.1 cm$
The larger angle faces a longer side, consistent with $b > 10$ .

Answer

14.1 cm

EXAMPLE 02

Two possible triangles

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

Worked solution

1
Use the known opposite pair.
$sin B = \frac{10 sin 35^{\circ}}{8} \approx 0.7169705$
Rearranging the sine rule puts the unknown sine alone.
2
Find both candidate angles.
$\begin{aligned} B_{1} & = {sin}^{- 1} (\frac{10 sin 35^{\circ}}{8}) \approx {45.8}^{\circ} \\ B_{2} & = 180^{\circ} - B_{1} \approx {134.2}^{\circ} \end{aligned}$
Supplementary angles have the same sine.
3
Test the angle sums.
$\begin{aligned} C_{1} & = 180^{\circ} - 35^{\circ} - B_{1} \approx {99.2}^{\circ} \\ C_{2} & = 180^{\circ} - 35^{\circ} - B_{2} \approx {10.8}^{\circ} \end{aligned}$
Both remaining angles are positive, so both triangles are possible.

Answer

45.8° or 134.2°

03 / YOUR TURN

Try it yourself.

Check your understanding

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

Show answer & reasoning

12 cm

The sine rule gives $b = \frac{6 sin 90^{\circ}}{sin 30^{\circ}} = \frac{6}{1 / 2} = 12 cm$ . The right angle faces the longest side.

Keep practising

6 worksheet questions · 16 marks

Open worksheet Worked solutions

Read the questions online

$A = 30^{\circ}$ , $a = 6$ cm and $B = 45^{\circ}$ . Find $b$ exactly.
3 marks
$A = 50^{\circ}$ , $a = 12$ cm and $B = 70^{\circ}$ . Find $b$ to 0.1 cm.
2 marks
$B = 40^{\circ}$ , $b = 9$ cm and $A = 55^{\circ}$ . Find $a$ to 0.1 cm.
2 marks
$A = 35^{\circ}$ , $a = 7$ cm and $b = 9$ cm. Find all possible $B$ to ${0.1}^{\circ}$ .
3 marks
Can $A = 40^{\circ}$ , $a = 5$ cm and $b = 12$ cm form a triangle? Explain.
3 marks
$A = 70^{\circ}$ , $B = 60^{\circ}$ 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.

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

In this lesson

Understand the idea.

Build opposite pairs

An inverse sine result may not be the only angle

See the method in action.

Find a side

Match each side with its opposite angle.

Isolate the unknown side.

Evaluate and round.

Two possible triangles

Use the known opposite pair.

Find both candidate angles.

Test the angle sums.

Try it yourself.

Check your understanding

Keep practising