Cosine rule | Year 11 | General Mathematics

01 / LEARN

Understand the idea.

The cosine rule extends Pythagoras

The cosine rule finds a side when two sides and their included angle are known: $c^{2} = a^{2} + b^{2} - 2 a b cos C$ . The side $c$ must face the chosen angle $C$ ; the other two sides meet at that angle.

At $C = 90^{\circ}$ , the cosine term vanishes and the rule becomes Pythagoras. For an obtuse angle, cosine is negative, so subtracting the correction increases the opposite side's square. Keep the sign through the substitution.

c^{2} = a^{2} + b^{2} - 2 a b cos C

Rearrange to find an angle

With three side lengths, rearrange to $cos C = \frac{a^{2} + b^{2} - c^{2}}{2 a b}$ before using inverse cosine. The largest side faces the largest angle, so identify the required opposite side first.

Check triangle inequality before calculation: the two shorter sides must total more than the longest. Valid triangle angles lie between $0^{\circ}$ and $180^{\circ}$ , and their cosine ratios must lie between negative one and one.

cos C = \frac{a^{2} + b^{2} - c^{2}}{2 a b}

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Worked example 1

Two sides are 7 cm and 10 cm and their included angle is 60°. Find the opposite side.

Worked solution

1
Match the included angle with its opposite side.
$c^{2} = 7^{2} + 10^{2} - 2 (7) (10) cos 60^{\circ}$
The known angle lies between the seven- and ten-centimetre sides.
2
Evaluate the squared length.
$c^{2} = 49 + 100 - 140 (\frac{1}{2}) = 79$
Cosine of sixty degrees is one half, so the correction subtracts seventy.
3
Take the positive root and report it.
$c = \sqrt{79} cm \approx 8.9 cm$
A side length is positive; the exact square root is retained before final rounding.

Answer

√79 cm ≈ 8.9 cm

EXAMPLE 02

Worked example 2

A triangle has sides 3 cm, 5 cm and 7 cm. Find its largest angle.

Worked solution

1
Check that the lengths form a triangle.
$3 + 5 = 8 > 7$
The two shorter sides can reach beyond the longest, satisfying triangle inequality.
2
Place the largest angle opposite the longest side.
$c = 7, a = 3, b = 5$
The required largest angle C faces the seven-centimetre side.
3
Calculate its cosine.
$cos C = \frac{3^{2} + 5^{2} - 7^{2}}{2 (3) (5)} = \frac{- 15}{30} = - 0.5$
The rearranged cosine rule uses all three supplied side lengths.
4
Recover the angle.
$C = {cos}^{- 1} (- 0.5) = 120^{\circ}$
The negative cosine correctly gives an obtuse largest angle.

Answer

120°

03 / YOUR TURN

Try it yourself.

Check your understanding

Find the angle opposite 10 in a triangle with sides 6, 8 and 10.

Show answer & reasoning

90°

For the angle opposite ten, $cos C = \frac{6^{2} + 8^{2} - 10^{2}}{2 (6) (8)} = 0$ , so $C = 90^{\circ}$ . These side lengths also satisfy Pythagoras.

Keep practising

6 worksheet questions · 15 marks

Open worksheet Worked solutions

Read the questions online

Two sides are 8 m and 11 m and their included angle is $50^{\circ}$ . Find the third side to one decimal place.
2 marks
Find the angle opposite 13 in a triangle with sides 9, 10 and 13. Round to one decimal place.
2 marks
Sides of 12 cm and 17 cm enclose $40^{\circ}$ . Find the opposite side to one decimal place.
2 marks
Use cosine rule to show that sides 8, 15 and 17 form a right triangle.
3 marks
Could sides 4 cm, 5 cm and 12 cm form a triangle? Explain before calculating an angle.
3 marks
For $a = 6$ , $b = 9$ , $C = 110^{\circ}$ , a student adds $2 a b cos C$ instead of subtracting it. Correct the rule and find c.
3 marks

Read the worked solutions

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

Two sides are 8 m and 11 m and their included angle is $50^{\circ}$ . Find the third side to one decimal place.
2 marks
$c^{2} = 8^{2} + 11^{2} - 2 \times 8 \times 11 cos 50^{\circ}$ .
Take the positive square root.
8.5 m
Find the angle opposite 13 in a triangle with sides 9, 10 and 13. Round to one decimal place.
2 marks
$cos C = \frac{81 + 100 - 169}{2 \times 9 \times 10}$ .
$C = {cos}^{- 1} (\frac{12}{180})$ .
${86.2}^{\circ}$
Sides of 12 cm and 17 cm enclose $40^{\circ}$ . Find the opposite side to one decimal place.
2 marks
$c^{2} = 144 + 289 - 408 cos 40^{\circ}$ .
Take the positive square root.
11.0 cm
Use cosine rule to show that sides 8, 15 and 17 form a right triangle.
3 marks
$cos C = \frac{64 + 225 - 289}{2 \times 8 \times 15} = 0$ .
Thus $C = 90^{\circ}$ .
The angle opposite 17 is $90^{\circ}$ .
Could sides 4 cm, 5 cm and 12 cm form a triangle? Explain before calculating an angle.
3 marks
$4 + 5 = 9 < 12$ .
The two shorter sides cannot close a triangle around the longest side.
No.
For $a = 6$ , $b = 9$ , $C = 110^{\circ}$ , a student adds $2 a b cos C$ instead of subtracting it. Correct the rule and find c.
3 marks
Cosine rule subtracts $2 a b cos C$ .
$cos 110^{\circ}$ is negative, so subtracting this term increases $c^{2}$ .
$c^{2} = 117 - 108 cos 110^{\circ}$ ; $c \approx 12.4 units$ .

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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.

The cosine rule extends Pythagoras

Rearrange to find an angle

See the method in action.

Worked example 1

Match the included angle with its opposite side.

Evaluate the squared length.

Take the positive root and report it.

Worked example 2

Check that the lengths form a triangle.

Place the largest angle opposite the longest side.

Calculate its cosine.

Recover the angle.

Try it yourself.

Check your understanding

Keep practising