Lesson 16 ~35 min Unit 3 · Geometry +85 XP

The Cosine Rule

When the sine rule is not enough, the cosine rule saves the day. Learn to find any side or angle in any triangle -- and discover that Pythagoras was just a special case all along.

Today's hook: In triangle ABC, side a = 7 cm, side b = 9 cm, and angle C = 60°. Can you find side c? Pythagoras will not work -- there is no right angle. What formula could connect these three pieces of information?

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Think First

warm-up

In triangle ABC, side a = 7 cm, side b = 9 cm, and angle C = 60°. Can you find side c using anything you already know? If not, what information would you need to use Pythagoras? How might the formula change if the angle were not 90°?

Record your answer in your workbook.

The Big Idea

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The cosine rule generalises Pythagoras. It works for any triangle -- acute, obtuse, or right-angled. When the included angle is 90°, the cosine rule collapses to $c^2 = a^2 + b^2$ because $\cos 90° = 0$.

The cosine rule has two forms. Finding a side (when you know SAS): $c^2 = a^2 + b^2 - 2ab \cos C$. Finding an angle (when you know SSS): $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$. The key insight is that the $-2ab \cos C$ term measures how much the triangle deviates from being right-angled.

$c^2 = a^2 + b^2 - 2ab \cos C$

C = 90° → Pythagoras

cos 90° = 0, so the extra term vanishes.

C < 90° → c < Pythagoras

The subtraction makes c smaller.

C > 90° → c > Pythagoras

cos is negative, so subtracting a negative adds.

What You'll Master

objectives

Know

The cosine rule formula in both forms (finding side and finding angle)
When to use the cosine rule instead of the sine rule

Understand

How the cosine rule generalises Pythagoras
Why the cosine rule is needed for SAS and SSS cases

Can Do

Use the cosine rule to find an unknown side given SAS
Use the cosine rule to find an unknown angle given SSS
Choose between sine and cosine rules appropriately

Words You Need

vocabulary

Cosine rule$c^2 = a^2 + b^2 - 2ab \cos C$. Relates three sides and one angle in any triangle.

Included angleThe angle between two known sides.

SASSide-Angle-Side: two sides and the included angle known.

SSSSide-Side-Side: all three sides known.

GeneralisationA broader rule that includes a simpler rule as a special case.

Inverse cosine$\cos^{-1}(x)$, used to find an angle from its cosine value.

Spot the Trap

heads-up

Wrong: Using the sine rule when you have two sides and the included angle. The sine rule cannot directly find the third side.

Right: For SAS, always use the cosine rule. For AAS or ASA, use the sine rule. For SSS, use the cosine rule to find one angle, then the sine rule for the rest.

Wrong: Forgetting that $\cos 90° = 0$, so the cosine rule becomes Pythagoras for right triangles.

Right: When $C = 90°$, $c^2 = a^2 + b^2 - 2ab \cos 90° = a^2 + b^2 - 0 = a^2 + b^2$. Pythagoras is a special case!

Finding a Side (SAS)

+5 XP

When you know two sides and the included angle (SAS), the cosine rule finds the third side directly. No need for right angles -- the formula works for any included angle.

Step 1: Identify the side opposite the known angle. This is the side you are finding.
Step 2: Write $c^2 = a^2 + b^2 - 2ab \cos C$ where $C$ is the included angle.
Step 3: Substitute and calculate $c^2$.
Step 4: Take the square root. Check reasonableness: if $C < 90°$, then $c < \sqrt{a^2 + b^2}$. If $C > 90°$, then $c > \sqrt{a^2 + b^2}$.

$c = \sqrt{a^2 + b^2 - 2ab \cos C}$

Opposite side

The side you want is opposite the known angle.

cos 60° = 0.5

Exact values save calculator errors.

Check with Pythagoras

For acute C, c should be less than the hypotenuse.

Finding an Angle (SSS)

+5 XP

When you know all three sides (SSS), rearrange the cosine rule to find any angle. The angle form is: $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$. Then use inverse cosine ($\cos^{-1}$) to find the angle.

Step 1: Identify which angle you want. It is opposite the side you put on top.
Step 2: Write $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$.
Step 3: Substitute all three side lengths.
Step 4: Calculate the fraction, then use $\cos^{-1}$. The sign of cos tells you the angle type: positive = acute, zero = right, negative = obtuse.

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

Sign tells the type

cos > 0: acute. cos = 0: right. cos < 0: obtuse.

Largest angle first

Find the angle opposite the longest side.

Verify the sum

All three angles should sum to 180°.

Choosing Between Sine and Cosine

+5 XP

Knowing which rule to use is half the battle. The decision depends entirely on what information you are given. Here is your decision guide.

AAS or ASA (two angles, one side) → Sine rule. You have an angle opposite a known side.
SSA (two sides, non-included angle) → Sine rule. Check ambiguous case.
SAS (two sides, included angle) → Cosine rule. Find the third side.
SSS (three sides) → Cosine rule. Find an angle, then optionally sine rule for others.

Given → Identify pattern → Choose rule

Look for the included angle

If the angle is between the two known sides, use cosine.

Opposite angle known?

If yes, sine rule is usually easier.

SSS always cosine first

Find one angle with cosine, then sine for others.

Common Pitfalls

heads-up

Using the wrong angle in the cosine rule

Putting a non-included angle into $c^2 = a^2 + b^2 - 2ab \cos C$. The angle $C$ must be between sides $a$ and $b$.

Fix: the angle in the cosine rule is always the included angle -- the one between the two sides you know. It is also opposite the side you are finding.

Sign errors with obtuse angles

When $C > 90°$, $\cos C$ is negative. The term $-2ab \cos C$ becomes positive, making $c$ larger than the Pythagorean prediction. Students sometimes forget this and get confused.

Fix: remember that for obtuse angles, cosine is negative. The subtraction of a negative is addition. The side opposite an obtuse angle is always the longest side.

Calculator in radian mode

Calculating $\cos 60°$ and getting $-0.952$ instead of $0.5$. This happens when the calculator is in radians instead of degrees.

Fix: always check your calculator is in DEGREE mode before starting. Do a quick sanity check: $\cos 60°$ should equal $0.5$.

Watch Me Solve It · 3 examples

Watch Me Solve It · Finding a side (SAS)

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PROBLEM

In $\triangle ABC$, $a = 7$ cm, $b = 9$ cm, and $\angle C = 60°$. Find side $c$.

1

Identify the setup

Two sides and the included angle (SAS) → cosine rule

We know $a$, $b$ and the angle between them ($C$). We want the side opposite ($c$).
2

Write the cosine rule

$c^2 = a^2 + b^2 - 2ab \cos C$
3

Substitute and calculate

$c^2 = 7^2 + 9^2 - 2(7)(9) \cos 60°$

$c^2 = 49 + 81 - 126 \times 0.5$

$c^2 = 130 - 63 = 67$
4

Find c and check

$c = \sqrt{67} \approx 8.2$ cm

Check: $C = 60° < 90°$, so $c < \sqrt{7^2 + 9^2} = \sqrt{130} \approx 11.4$. $8.2 < 11.4$ ✓

Answer$c = \sqrt{67} \approx 8.2$ cm

Watch Me Solve It · Finding an angle (SSS)

+15 XP per step

PROBLEM

In $\triangle ABC$, $a = 5$ cm, $b = 7$ cm, $c = 8$ cm. Find $\angle A$.

1

Identify the setup

Three sides known (SSS) → cosine rule for angles

We want angle $A$, which is opposite side $a$.
2

Write the angle form

$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$
3

Substitute and calculate

$\cos A = \frac{7^2 + 8^2 - 5^2}{2(7)(8)} = \frac{49 + 64 - 25}{112}$

$\cos A = \frac{88}{112} = 0.7857$
4

Find the angle

$A = \cos^{-1}(0.7857) \approx 38.2°$

Since $\cos A > 0$, angle $A$ is acute. This makes sense: side $a = 5$ is the shortest side, so it faces the smallest angle.

Answer$\angle A \approx 38.2°$

Watch Me Solve It · Cosine then sine

+15 XP per step

PROBLEM

In $\triangle ABC$, $a = 9$ cm, $b = 11$ cm, and $\angle C = 50°$. Find side $c$, then find $\angle A$ and $\angle B$.

1

Find side c using cosine rule

$c^2 = 9^2 + 11^2 - 2(9)(11) \cos 50°$

$c^2 = 81 + 121 - 198 \times 0.643 = 202 - 127.3 = 74.7$

$c = \sqrt{74.7} \approx 8.6$ cm
2

Find angle A using sine rule

$\frac{\sin A}{9} = \frac{\sin 50°}{8.6}$

$\sin A = \frac{9 \times 0.766}{8.6} = 0.802$

$A = \sin^{-1}(0.802) \approx 53.3°$
3

Find angle B

$B = 180° - 50° - 53.3° = 76.7°$

Angles in a triangle sum to 180°.
4

Verify with sine rule

$\frac{11}{\sin 76.7°} = \frac{11}{0.973} \approx 11.3$ and $\frac{8.6}{\sin 50°} = \frac{8.6}{0.766} \approx 11.2$

Close enough (rounding differences). The solution is consistent.

Answer$c \approx 8.6$ cm, $A \approx 53.3°$, $B \approx 76.7°$

Copy Into Your Books

Cosine Rule (side)

$c^2 = a^2 + b^2 - 2ab \cos C$
Use when you know SAS
$c$ is opposite angle $C$

Cosine Rule (angle)

$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$
Use when you know SSS
Then $\cos^{-1}$ to find angle

Choosing the Rule

AAS / ASA → Sine rule
SSA → Sine rule (ambiguous)
SAS → Cosine rule
SSS → Cosine rule

Special Case

$C = 90°$ → Pythagoras
cos 90° = 0
Cosine rule generalises Pythagoras

How are you completing this lesson?

Brain Trainer · 4 problems

Brain Trainer · Cosine rule drill

4 problems

Four quick problems on the cosine rule. Work each, then reveal the answer.

1 In $\triangle ABC$, $a = 8$ cm, $b = 10$ cm, $\angle C = 45°$. Find side $c$.

$c^2 = 8^2 + 10^2 - 2(8)(10) \cos 45° = 64 + 100 - 160 \times 0.707 = 164 - 113.1 = 50.9$. $c = \sqrt{50.9} \approx 7.1$ cm.$c \approx 7.1$ cm
2 In $\triangle ABC$, $a = 6$ cm, $b = 8$ cm, $c = 10$ cm. Find $\angle B$.

$\cos B = \frac{6^2 + 10^2 - 8^2}{2(6)(10)} = \frac{36 + 100 - 64}{120} = \frac{72}{120} = 0.6$. $B = \cos^{-1}(0.6) \approx 53.1°$.$B \approx 53.1°$
3 A triangle has sides 5 cm, 6 cm, 7 cm. Find the largest angle.

Largest angle is opposite longest side (7). $\cos C = \frac{5^2 + 6^2 - 7^2}{2(5)(6)} = \frac{25 + 36 - 49}{60} = \frac{12}{60} = 0.2$. $C = \cos^{-1}(0.2) \approx 78.5°$.$C \approx 78.5°$
4 A triangle has sides 8 cm, 10 cm, 14 cm. Is it acute, right, or obtuse? Justify.

Check largest angle (opposite 14): $\cos C = \frac{8^2 + 10^2 - 14^2}{2(8)(10)} = \frac{64 + 100 - 196}{160} = \frac{-32}{160} = -0.2$. Since $\cos C < 0$, $C > 90°$. The triangle is obtuse.Obtuse

Complete in your workbook.

Multiple Choice · 5 questions

MC1

When to use cosine rule

+10 XP

The cosine rule is used when you know:

MC2

Cosine rule calculation

+10 XP

In $\triangle ABC$, $a = 5$, $b = 7$, $\angle C = 120°$. Side $c$ is:

MC3

Largest angle from sides

+10 XP

In a triangle with sides 5, 6, 7, the largest angle is closest to:

MC4

Special case

+10 XP

When $\angle C = 90°$, the cosine rule becomes:

MC5

Classify the triangle

+10 XP

A triangle has sides 8 cm, 10 cm, 14 cm. It is:

Short Answer · 3 questions

Cosine then sine

+15 XP

SHORT ANSWER

In $\triangle ABC$, $a = 9$ cm, $b = 11$ cm, and $\angle C = 50°$.
(a) Find side $c$.
(b) Find $\angle A$ using the sine rule.
(c) Find $\angle B$.

Write your working in your book.

SSS analysis

+15 XP

SHORT ANSWER

A triangle has sides 7 cm, 9 cm, and 12 cm.
(a) Find all three angles.
(b) Classify the triangle as acute, right, or obtuse. Justify your answer.
(c) Verify that the three angles sum to 180°.

Write your working in your book.

Navigation problem

+15 XP

SHORT ANSWER

Two ships leave port at the same time. Ship A sails on a bearing of 060° at 15 km/h. Ship B sails on a bearing of 150° at 20 km/h.
(a) Find the angle between their paths.
(b) Find the distance between the ships after 2 hours.
(c) Find the bearing of ship B from ship A at this time.
(d) A third ship wants to sail directly from ship A to ship B. What course should it steer?

Write your working in your book.

Stretch Challenge · The quadrilateral

+20 XP

STRETCH

Quadrilateral $ABCD$ has $AB = 8$ cm, $BC = 10$ cm, $CD = 7$ cm, $DA = 9$ cm, and diagonal $AC = 12$ cm.
(a) Find $\angle ABC$ using $\triangle ABC$.
(b) Find $\angle ADC$ using $\triangle ADC$.
(c) Hence find the sum of angles $B$ and $D$ in the quadrilateral.
(d) A property of cyclic quadrilaterals is that opposite angles sum to 180°. Is $ABCD$ cyclic? Justify.

Record in your book -- full marks require clear working.

Reveal solution

(a) In $\triangle ABC$: $\cos B = \frac{8^2 + 10^2 - 12^2}{2(8)(10)} = \frac{64 + 100 - 144}{160} = \frac{20}{160} = 0.125$. $B = \cos^{-1}(0.125) \approx 82.8°$.

(b) In $\triangle ADC$: $\cos D = \frac{9^2 + 7^2 - 12^2}{2(9)(7)} = \frac{81 + 49 - 144}{126} = \frac{-14}{126} = -0.111$. $D = \cos^{-1}(-0.111) \approx 96.4°$.

(c) $\angle B + \angle D = 82.8° + 96.4° = 179.2°$ (approximately 180° due to rounding).

(d) Since $\angle B + \angle D \approx 180°$, the quadrilateral is cyclic (or very close to cyclic, with the small difference due to rounding in intermediate steps). The exact values would sum to exactly 180°.

Quick Review

Side form

$c^2 = a^2 + b^2 - 2ab \cos C$

Angle form

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

SAS

Cosine rule for side

SSS

Cosine rule for angle

Sign check

cos > 0 acute, < 0 obtuse

Pythagoras

Special case: C = 90°

Interactive: Cosine Rule Solver

Practice using the cosine rule to find missing sides and angles. Test your calculations with instant feedback.

Consolidation Game -- Doodle Jump Quiz

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