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hscscience Maths Std · Y12
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MST-12-S2-04 ~55 min ⚡ +95 XP available

The Cosine Rule and the Area of a Triangle

The sine rule needs a side paired with the angle opposite it. Plenty of triangles never give you that pair. The cosine rule works from two sides and the angle between them, or from all three sides, and the area rule finds the area from the same two sides and that same angle.

Today's hook, A farmer needs to fence a triangular paddock. Two fences already run from one corner, one of $240\text{ m}$ and one of $180\text{ m}$, and they meet at $110°$. How much wire does the third side need, and how much spray does the paddock take? Not one angle is opposite a side you know.
0/5QUESTS
1

Orient to the cosine rule

Meet the included angle, set the goal and settle the key terms.

Worksheets

Practise this lesson

Three printable worksheets that build from foundations to mastery, or build your own from any question in this focus area.

Build Foundations & guided practice Apply Application practice Master Mastery challenge Build custom Build your own from any question
01
Recall, your gut answer first
+5 XP warm-up

A triangle gives you two sides, $8\text{ cm}$ and $11\text{ cm}$, and the angle of $47°$ sitting between them. Last lesson's sine rule needs a side paired with the angle opposite it.

Without calculating write down whether the sine rule can start here, and if not, say exactly which piece of information it is missing.

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02
The cosine rule, and the included angle
+5 XP to read

The sine rule fails the moment you have no matched pair. Two sides with the angle squeezed between them is exactly that situation, and it is one of the two most common triangles in an exam. The cosine rule handles it, and it handles the other one too, where all three sides are known and no angle is.

The included angle. The angle $C$ is included between sides $a$ and $b$ when those two sides are the ones that form it. In the cosine rule the side you are finding, $c$, is always the one opposite the angle you are given.

It is Pythagoras with a correction. If $C = 90°$ then $\cos C = 0$, the last term vanishes, and what is left is $c^2 = a^2 + b^2$. The $-2ab\cos C$ term is the price of the angle not being a right angle.

$c^2 = a^2 + b^2 - 2ab\cos C$
Finding a SIDE (SAS)
Two sides and the angle between them. Substitute straight into $c^2 = a^2 + b^2 - 2ab\cos C$, then square root at the very end.
Finding an ANGLE (SSS)
All three sides. Rearranged, $\cos C = \dfrac{a^2 + b^2 - c^2}{2ab}$, then take the inverse cosine.
Finding the AREA
The same two sides and the same included angle give the area rule, $A = \tfrac{1}{2}ab\sin C$. No height needed.
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What you'll master
Know

Key facts

  • The cosine rule: $c^2 = a^2 + b^2 - 2ab\cos C$
  • Its angle form: $\cos C = \dfrac{a^2 + b^2 - c^2}{2ab}$
  • The area rule: $A = \tfrac{1}{2}ab\sin C$, with $C$ the included angle
  • Side $c$ is opposite angle $C$, exactly as in the sine rule
Understand

Concepts

  • Why the cosine rule is Pythagoras with a correction term
  • Why a negative cosine means the angle is obtuse, with no ambiguity to resolve
  • Why the area rule needs the angle between the two sides and no other
  • How to choose between the sine rule and the cosine rule from what you were given
Can do

Skills

  • Find the third side from two sides and the included angle
  • Find any angle from three sides, to a degree or to the nearest minute
  • Find the area of any triangle from two sides and the included angle
  • Combine both rules with bearings in a multi-step problem
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Key terms
Cosine ruleA relationship holding in every triangle that links all three sides to one angle. Like this: in a triangle with sides $8$ and $11$ meeting at $47°$, the third side satisfies $c^2 = 8^2 + 11^2 - 2(8)(11)\cos 47°$.
Included angleThe angle formed where two named sides meet. Like this: if the sides are $a$ and $b$, the included angle is $C$, and it is the only angle both of those sides touch.
SASSide, angle, side, meaning two sides and the angle between them are known. Like this: given $a = 5$, $C = 120°$ and $b = 8$, you have SAS and the cosine rule finds $c$ in one step.
SSSThree sides known and no angle. Like this: given $7$, $9$ and $13$, no angle is available to start the sine rule, so the cosine rule in its angle form is the only way in.
Area ruleThe area of any triangle from two sides and the angle between them, $A = \tfrac{1}{2}ab\sin C$. Like this: sides $14$ and $9$ meeting at $63°$ enclose an area of $\tfrac{1}{2}(14)(9)\sin 63°$, with no perpendicular height ever measured.
2

Choose your rule

Decide between sine and cosine from what you are given, and read the sign of the cosine.

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Choosing your rule, and reading the sign of the cosine
core concept

Before you write anything, name what you were given. Three shapes of question cover almost everything in this focus area:

SAS, unknown side: $\quad c^2 = a^2 + b^2 - 2ab\cos C$
SSS, unknown angle: $\quad \cos C = \dfrac{a^2 + b^2 - c^2}{2ab}$
SAS, area: $\quad A = \tfrac{1}{2}ab\sin C$

The angle form carries something the sine rule could not give you. Cosine is positive for acute angles and negative for obtuse ones, so the sign of your answer settles the matter before you touch the inverse cosine. A negative value means the angle is obtuse and the calculator will return the obtuse angle directly. There is nothing to subtract from $180°$ and nothing ambiguous to resolve.

Which rule, decided in one line. If you can see a side and the angle opposite it, use the sine rule. If you cannot, you were almost certainly given SAS or SSS, and that is the cosine rule. For an area, look for two sides with the angle between them, and use the area rule rather than hunting for a height.

The cosine rule is $c^2 = a^2 + b^2 - 2ab\cos C$, where $C$ is the angle included between sides $a$ and $b$, and $c$ is the side opposite $C$. Rearranged for an angle, $\cos C = (a^2 + b^2 - c^2)/(2ab)$, and a negative result means $C$ is obtuse. The area rule is $A = \tfrac{1}{2}ab\sin C$, using the same two sides and the same included angle.

Pause, copy the cosine rule in both forms, the area rule, and one line saying that a negative cosine means an obtuse angle, into your book.

Quick check: In a triangle $b = 5$, $c = 8$ and $A = 60°$. Find $a$, to 2 decimal places.

3

Work three cosine rule examples

Follow an unknown side from SAS, an unknown angle from SSS and an area from two sides and the angle between.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · UNKNOWN SIDE, SAS

In triangle $ABC$, $b = 8\text{ cm}$, $c = 11\text{ cm}$ and $A = 47°$. Find $a$, correct to 2 decimal places.

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$a^2 = b^2 + c^2 - 2bc\cos A$
Sides $b$ and $c$ meet at $A$, so $A$ is the included angle and $a$ is the side opposite it. Every letter in the rule shifts together.
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PROBLEM 2 · UNKNOWN ANGLE, SSS

A triangle has sides $a = 7$, $b = 9$ and $c = 13$. Find its largest angle, correct to the nearest minute.

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$c = 13 = \max(a,b,c) \;\Rightarrow\; C \text{ is the largest angle}$
The largest angle always faces the longest side, so you can name which angle you are after before calculating anything.
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PROBLEM 3 · AREA FROM TWO SIDES AND THE ANGLE BETWEEN

A triangular garden bed has sides of $14\text{ m}$ and $9\text{ m}$ meeting at an angle of $63°$. Find its area, correct to 2 decimal places.

1
$A = \tfrac{1}{2}ab\sin C$
The two given sides are $a$ and $b$, and $63°$ is the angle they form, so it is the included angle the rule asks for.
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True or false: If the cosine rule gives a negative value for $\cos C$, then angle $C$ is obtuse.

4

Avoid the three traps

Spot the missing square root, the wrong included angle and the dropped negative.

Common errors · the 3 traps that cost marks

Trap 01
Answering with $c^2$ instead of $c$
The cosine rule delivers the square of the side. Writing $64.97$ when the answer is $8.06$ loses the mark outright, and the number looks plausible enough that it rarely gets a second glance. Circle the square root before you start substituting.
Trap 02
Using an angle that is not the included one
Both the cosine rule and the area rule need the angle between the two sides you are using. Given sides $a$ and $b$ with angle $A$, neither rule applies as written. Redraw and relabel so the angle sits between your two sides, or use the sine rule instead.
Trap 03
Subtracting from 180 out of habit
Last lesson trained you to convert a sine rule angle to its obtuse partner. The cosine rule needs none of that. It returns the correct angle, acute or obtuse, from the sign of the cosine alone, so subtracting from $180°$ here turns a right answer into a wrong one.

Fill the gaps: With $b = 8$, $c = 11$ and $A = 47°$: $b^2 + c^2 =$ , the correction term $2bc\cos A =$ (2 dp), so $a =$ (2 dp).

5

Drill the cosine rule

Run the quick-fire calculations until rule choice is automatic.

Quick-fire practice · 5 calculations

1

$b = 6$, $c = 10$, $A = 55°$. Find $a$ to 2 decimal places.

2

$a = 5$, $b = 8$, $C = 120°$. Find $c$ to 2 decimal places.

3

$a = 6$, $b = 7$, $c = 9$. Find angle $A$ to the nearest degree.

4

A triangle has sides $11\text{ cm}$ and $15\text{ cm}$ meeting at $82°$. Find its area to 2 decimal places.

5

A triangular block of land has sides $45\text{ m}$, $60\text{ m}$ and $80\text{ m}$. Find its largest angle to the nearest minute, then its area to the nearest square metre.

Match what you were given to the rule that starts the question:

  • Two sides and the angle between them, find the third side
  • All three sides, find an angle
  • Two sides and the angle between them, find the area
  • A side, the angle opposite it, and one more angle
  • Sine rule
  • Cosine rule, angle form
  • Cosine rule, side form
  • Area rule, one half ab sin C
6

Revisit your thinking

Return to your opening answer and name what has changed.

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Revisit your thinking

Back to the paddock. Two fences of $240\text{ m}$ and $180\text{ m}$ meet at $110°$, and there is no matched pair anywhere, so the sine rule never starts.

Third side: $d^2 = 240^2 + 180^2 - 2(240)(180)\cos 110°$. Since $\cos 110° = -0.342020$ is negative, that last term is added, giving $d^2 = 90000 + 29550.54 = 119550.54$ and $d \approx \mathbf{346\text{ m}}$ of wire.

Area: $A = \tfrac{1}{2}(240)(180)\sin 110° = 21600(0.939693) \approx \mathbf{20297\text{ m}^2}$, which is about $2.03$ hectares.

Notice the third side came out longer than either fence. An obtuse included angle opens the triangle up, and the negative cosine is what tells you so.

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7

Practise the cosine rule

Answer the question bank, then write full short-answer responses.

01
Multiple choice
+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.

02
Short answer
ApplyBand 33 marks

Q1. Two straight roads leave an intersection at an angle of $68°$. One cyclist rides $4.2\text{ km}$ along the first road and a second cyclist rides $5.7\text{ km}$ along the other. Find the distance between them, correct to the nearest $10\text{ m}$. (3 marks)

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ApplyBand 44 marks

Q2. A triangular sail has sides of $3.4\text{ m}$, $4.1\text{ m}$ and $5.2\text{ m}$. Find the angle between the two shorter sides, correct to the nearest minute, and then find the area of the sail, correct to 2 decimal places. (4 marks)

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AnalyseBand 55 marks

Q3. A ship leaves port $P$ and sails $24\text{ km}$ on a bearing of $042°$T to a point $Q$. It then sails $31\text{ km}$ on a bearing of $128°$T to a point $R$. Find the distance $PR$ correct to the nearest $0.1\text{ km}$, and the bearing of $R$ from $P$ correct to the nearest degree. (5 marks)

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📖 Comprehensive answers (click to reveal)

Drill 1: $a^2 = 36 + 100 - 120\cos 55° = 136 - 68.83 = 67.17$, so $a = 8.20$

Drill 2: $\cos 120° = -0.5$, so $c^2 = 25 + 64 + 40 = 129$ and $c = 11.36$. The correction term is added because the cosine is negative.

Drill 3: $\cos A = \tfrac{49 + 81 - 36}{2(7)(9)} = \tfrac{94}{126} = 0.746032$, so $A = 41.75° = 42°$ to the nearest degree

Drill 4: $A = \tfrac{1}{2}(11)(15)\sin 82° = 82.5 \times 0.990268 = 81.70\text{ cm}^2$

Drill 5: The largest angle faces the $80\text{ m}$ side: $\cos\theta = \tfrac{2025 + 3600 - 6400}{2(45)(60)} = \tfrac{-775}{5400} = -0.143519$, so $\theta = 98.2515° = 98°\,15'$. Then $A = \tfrac{1}{2}(45)(60)\sin 98.2515° = 1350 \times 0.989618 = 1336\text{ m}^2$ to the nearest square metre.

Q1 (3 marks): The two roads and the line joining the cyclists form a triangle with two sides and the angle between them known, so it is the cosine rule, not the sine rule [1]. $d^2 = 4.2^2 + 5.7^2 - 2(4.2)(5.7)\cos 68° = 50.13 - 17.94 = 32.19$ [1]. So $d = 5.6739\text{ km} = \mathbf{5.67\text{ km}}$ to the nearest $10\text{ m}$ [1].

Q2 (4 marks): The angle between the two shorter sides is the angle opposite the $5.2\text{ m}$ side. $\cos\theta = \tfrac{3.4^2 + 4.1^2 - 5.2^2}{2(3.4)(4.1)} = \tfrac{1.33}{27.88} = 0.047704$ [1]. So $\theta = 87.2657°$, and $0.2656 \times 60 = 15.9$, giving $\mathbf{87°\,16'}$ [1]. That angle is included between the two sides used, so the area rule applies directly [1]: $A = \tfrac{1}{2}(3.4)(4.1)\sin 87.2657° = 6.97 \times 0.998832 = \mathbf{6.96\text{ m}^2}$ [1].

Q3 (5 marks): The back-bearing from $Q$ to $P$ is $042° + 180° = 222°$, and the bearing from $Q$ to $R$ is $128°$, so angle $PQR = 222° - 128° = 94°$ [1]. That is two sides with the angle between them, so $PR^2 = 24^2 + 31^2 - 2(24)(31)\cos 94° = 1537 + 103.80 = 1640.80$ [1], giving $PR = 40.5068 = \mathbf{40.5\text{ km}}$ [1]. For the bearing, the sine rule gives $\sin(\angle QPR) = \tfrac{31\sin 94°}{40.5068} = 0.763440$, so $\angle QPR = 49.77°$ [1]. Angle $PQR = 94°$ is obtuse and a triangle can hold only one obtuse angle, so $\angle QPR$ must be the acute value and no ambiguity arises. The bearing of $R$ from $P$ is $042° + 49.77° = 91.77°$, which is $\mathbf{092°\text{T}}$ to the nearest degree [1].