Sine rule | Year 11 | Ascent Library

01 / LEARN

Understand the idea.

Match each side to its opposite angle

Name opposite pairs before applying the sine rule: side $a$ faces angle $A$ , and similarly for $b, B$ and $c, C$ . One complete known pair provides the reference ratio for a second pair.

Use $\frac{a}{sin A} = \frac{b}{sin B}$ for a missing side, or invert both ratios when isolating an angle's sine. Keep each side matched to its own opposite angle during rearrangement and calculation.

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

SSA can be ambiguous

With two sides and a non-included angle, the data may form zero, one or two triangles. Inverse sine returns an acute or right candidate, but an obtuse supplement can have the same sine.

Test both candidates against the angle sum and any stated restriction. A third angle must be positive. If the calculated sine genuinely exceeds one, no matching angle exists; rounding it to one would create a different problem.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Worked example 1

A=40°, B=65° and b=12 m. Find a to one decimal place.

Worked solution

1
Identify the complete pair.
$b = 12 m, B = 65^{\circ}$
The side b is opposite the known angle B.
2
Match the required side with its angle.
$\frac{a}{sin 40^{\circ}} = \frac{12}{sin 65^{\circ}}$
Each side remains paired with its opposite angle.
3
Isolate and calculate the side.
$a = \frac{12 sin 40^{\circ}}{sin 65^{\circ}} \approx 8.5109 m$
Multiply through by the sine of the angle facing a.
4
Round and check relative sizes.
$a \approx 8.5 m$
The forty-degree angle is smaller than sixty-five degrees, so its opposite side should be shorter than twelve metres.

Answer

8.5 m

EXAMPLE 02

Worked example 2

A=30°, a=8 cm and b=10 cm. Find every possible B.

Worked solution

1
Isolate the unknown angle's sine.
$sin B = \frac{10 sin 30^{\circ}}{8} = 0.625$
The known reference pair is a opposite A.
2
Find the first candidate.
$B_{1} = {sin}^{- 1} (0.625) \approx {38.6822}^{\circ}$
Inverse sine returns the acute candidate.
3
Find the supplementary candidate.
$B_{2} = 180^{\circ} - B_{1} \approx {141.3178}^{\circ}$
These two angles have the same sine.
4
Check the remaining angles.
$\begin{aligned} C_{1} & = 180^{\circ} - 30^{\circ} - B_{1} \approx {111.3178}^{\circ} \\ C_{2} & = 180^{\circ} - 30^{\circ} - B_{2} \approx {8.6822}^{\circ} \end{aligned}$
Both third angles are positive, so both triangles fit.
5
Report every possible B.
$B \approx {38.7}^{\circ} or {141.3}^{\circ}$
Neither valid candidate can be discarded when the task asks for all possibilities.

Answer

38.7° or 141.3°

03 / YOUR TURN

Try it yourself.

Check your understanding

A=70° and B=45°. Find C.

Show answer & reasoning

65°

The remaining angle is $C = 180^{\circ} - 70^{\circ} - 45^{\circ} = 65^{\circ}$ . Every triangle's three interior angles sum to one hundred eighty degrees.

Keep practising

6 worksheet questions · 17 marks

Open worksheet Worked solutions

Read the questions online

$A = 50^{\circ}$ , $B = 70^{\circ}$ and $a = 9 cm$ . Find b to one decimal place.
2 marks
$A = 70^{\circ}$ , $a = 20 cm$ and $b = 15 cm$ . Find B to one decimal place and explain whether its supplement fits.
3 marks
$A = 45^{\circ}$ , $B = 60^{\circ}$ and $a = 10 m$ . Find C and c.
3 marks
$A = 30^{\circ}$ , $a = 10 cm$ and $b = 14 cm$ . Find both B angles to one decimal place.
3 marks
$A = 50^{\circ}$ , $a = 8 cm$ and $b = 13 cm$ . Does a triangle exist? Explain.
3 marks
$A = 35^{\circ}$ , $B = 85^{\circ}$ and $c = 18 m$ . Find a to one decimal place.
3 marks

Read the worked solutions

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

$A = 50^{\circ}$ , $B = 70^{\circ}$ and $a = 9 cm$ . Find b to one decimal place.
2 marks
$b = \frac{9 sin 70^{\circ}}{sin 50^{\circ}}$ .
$b \approx 11.0 cm$ .
11.0 cm
$A = 70^{\circ}$ , $a = 20 cm$ and $b = 15 cm$ . Find B to one decimal place and explain whether its supplement fits.
3 marks
$sin B = \frac{15 sin 70^{\circ}}{20}$ .
The supplementary candidate plus $A = 70^{\circ}$ exceeds $180^{\circ}$ .
${44.8}^{\circ}$ ; the supplementary angle does not fit.
$A = 45^{\circ}$ , $B = 60^{\circ}$ and $a = 10 m$ . Find C and c.
3 marks
$C = 180^{\circ} - 45^{\circ} - 60^{\circ} = 75^{\circ}$ .
$c = \frac{10 sin 75^{\circ}}{sin 45^{\circ}}$ .
$C = 75^{\circ}$ ; $c \approx 13.7 m$
$A = 30^{\circ}$ , $a = 10 cm$ and $b = 14 cm$ . Find both B angles to one decimal place.
3 marks
$sin B = \frac{14 sin 30^{\circ}}{10} = 0.7$ .
Both inverse-sine and supplementary candidates leave a positive third angle.
${44.4}^{\circ}$ or ${135.6}^{\circ}$
$A = 50^{\circ}$ , $a = 8 cm$ and $b = 13 cm$ . Does a triangle exist? Explain.
3 marks
$sin B = \frac{13 sin 50^{\circ}}{8} \approx 1.245 > 1$ .
No angle can have this sine.
No triangle exists.
$A = 35^{\circ}$ , $B = 85^{\circ}$ and $c = 18 m$ . Find a to one decimal place.
3 marks
$C = 60^{\circ}$ .
$a = \frac{18 sin 35^{\circ}}{sin 60^{\circ}}$ .
11.9 m

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 General Mathematics: practical trigonometry, including similarity, right triangles, triangle area, sine/cosine rules and contextual interpretation.

Official curriculum source ↗

In this lesson

Understand the idea.

Match each side to its opposite angle

SSA can be ambiguous

See the method in action.

Worked example 1

Identify the complete pair.

Match the required side with its angle.

Isolate and calculate the side.

Round and check relative sizes.

Worked example 2

Isolate the unknown angle's sine.

Find the first candidate.

Find the supplementary candidate.

Check the remaining angles.

Report every possible B.

Try it yourself.

Check your understanding

Keep practising