Sine rule and ambiguous cases | Year 11

01 / LEARN

Understand the idea.

Use opposite pairs

The sine rule links each side to the sine of its opposite angle. Mark the opposite pairs before substituting. A known side and its opposite angle supply the reference pair, whether the triangle is right angled or not.

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

Check the second angle

When solving for an angle from two sides and a non-included angle, inverse sine may hide a second triangle. An angle and its supplement have the same sine, so test both against the other given angle.

If the calculated sine exceeds 1, no triangle exists. Otherwise each candidate must leave a positive third angle. This is why the same SSA data can describe zero, one or two triangles.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Worked example 1

Given $A = 30^{\circ}$ , $B = 45^{\circ}$ , $a = 8$ cm, find b exactly.

Worked solution

1
Match the known opposite pairs.
$\frac{b}{sin 45^{\circ}} = \frac{8}{sin 30^{\circ}}$
The side $b$ faces $B = 45^{\circ}$ , and side 8 faces $A = 30^{\circ}$ .
2
Isolate the required side.
$b = \frac{8 sin 45^{\circ}}{sin 30^{\circ}}$
Multiplying by $sin 45^{\circ}$ preserves the opposite pairing.
3
Substitute exact trigonometric values.
$b = \frac{8 (\frac{\sqrt{2}}{2})}{\frac{1}{2}} = 8 \sqrt{2} cm$
The larger angle $45^{\circ}$ faces the longer side, consistent with $8 \sqrt{2} > 8$ .

Answer

$8 \sqrt{2} cm$

EXAMPLE 02

Worked example 2

Given $A = 30^{\circ}$ , $a = 8$ cm and $b = 8 \sqrt{3}$ cm, find possible B and C.

Worked solution

1
Rearrange the sine rule for the unknown angle.
$sin B = \frac{b sin A}{a} = \frac{8 \sqrt{3} (\frac{1}{2})}{8} = \frac{\sqrt{3}}{2}$
The known opposite pair fixes the sine of the second angle.
2
Find both angles with this sine.
$B = 60^{\circ} or B = 120^{\circ}$
Both the principal angle and its supplement lie between $0^{\circ}$ and $180^{\circ}$ .
3
Calculate the third angle in each case.
$\begin{aligned} B = 60^{\circ} & : C = 180^{\circ} - 30^{\circ} - 60^{\circ} = 90^{\circ} \\ B = 120^{\circ} & : C = 180^{\circ} - 30^{\circ} - 120^{\circ} = 30^{\circ} \end{aligned}$
Each third angle is positive, so both triangles are possible.

Answer

$(B, C) = (60^{\circ}, 90^{\circ})$ or $(120^{\circ}, 30^{\circ})$

03 / YOUR TURN

Try it yourself.

Check your understanding

If $sin B = 0.5$ and B is inside a triangle, what angles must be tested?

Show answer & reasoning

30° and 150°.

$sin 30^{\circ} = sin 150^{\circ} = \frac{1}{2}$ . Test both angles; another given angle may rule out one when the triangle angle sum is applied.

Keep practising

6 worksheet questions · 18 marks

Open worksheet Worked solutions

Read the questions online

$A = 30^{\circ}$ , $B = 90^{\circ}$ , $a = 6$ cm. Find $b$ .
3 marks
$A = 45^{\circ}$ , $B = 60^{\circ}$ , $a = 10$ cm. Find $b$ exactly.
3 marks
$A = 30^{\circ}$ , $a = 5$ cm, $b = 12$ cm. Decide how many triangles exist.
3 marks
Given $A = 30^{\circ}$ , $a = 5$ cm, $b = 5 \sqrt{2}$ cm, find all possible $B$ , $C$ pairs.
3 marks
$A = 120^{\circ}$ , $a = 10$ cm, $b = 5$ cm. Find $B$ and $C$ to one decimal place.
3 marks
A survey triangle has $A = 40^{\circ}$ , $B = 65^{\circ}$ and $a = 18$ m. Find $b$ to one decimal place and explain why $b > a$ .
3 marks

Read the worked solutions

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

$A = 30^{\circ}$ , $B = 90^{\circ}$ , $a = 6$ cm. Find $b$ .
3 marks
$b = \frac{6 sin 90^{\circ}}{sin 30^{\circ}} = 12$ .
12 cm
$A = 45^{\circ}$ , $B = 60^{\circ}$ , $a = 10$ cm. Find $b$ exactly.
3 marks
$b = \frac{10 (\frac{\sqrt{3}}{2})}{\frac{\sqrt{2}}{2}} = 5 \sqrt{6}$ .
$5 \sqrt{6} cm$
$A = 30^{\circ}$ , $a = 5$ cm, $b = 12$ cm. Decide how many triangles exist.
3 marks
$sin B = \frac{12 (\frac{1}{2})}{5} = 1.2 > 1$ .
None.
Given $A = 30^{\circ}$ , $a = 5$ cm, $b = 5 \sqrt{2}$ cm, find all possible $B$ , $C$ pairs.
3 marks
$sin B = \frac{\sqrt{2}}{2}$ .
Test $45^{\circ}$ and $135^{\circ}$ , then subtract each from $150^{\circ}$ .
$(45^{\circ}, 105^{\circ})$ or $(135^{\circ}, 15^{\circ})$
$A = 120^{\circ}$ , $a = 10$ cm, $b = 5$ cm. Find $B$ and $C$ to one decimal place.
3 marks
$sin B = \frac{\sqrt{3}}{4}$ .
$B \approx {25.6589}^{\circ}$ ; its supplement is incompatible with $A = 120^{\circ}$ .
$C \approx {34.3411}^{\circ}$ .
$B = {25.7}^{\circ}$ , $C = {34.3}^{\circ}$ .
A survey triangle has $A = 40^{\circ}$ , $B = 65^{\circ}$ and $a = 18$ m. Find $b$ to one decimal place and explain why $b > a$ .
3 marks
$b = \frac{18 sin 65^{\circ}}{sin 40^{\circ}} \approx 25.3793$ .
$65^{\circ}$ exceeds $40^{\circ}$ , so its opposite side is longer.
$b \approx 25.4$ m; $b$ faces the larger angle.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics: triangle rules, circular measure and trigonometric functions; school sequencing may vary.

Official curriculum source ↗

In this lesson

Understand the idea.

Use opposite pairs

Check the second angle

See the method in action.

Worked example 1

Match the known opposite pairs.

Isolate the required side.

Substitute exact trigonometric values.

Worked example 2

Rearrange the sine rule for the unknown angle.

Find both angles with this sine.

Calculate the third angle in each case.

Try it yourself.

Check your understanding

Keep practising