What is the Cosine Rule in Trigonometry and when is it used?

The Cosine Rule, also known as the Law of Cosines, is a formula used in trigonometry to find a side or angle in a triangle when certain other measurements are known. It is particularly useful in non-right-angled triangles. The rule states that for any triangle with sides a, b, and c, and angle C opposite side c, the formula is c² = a² + b² - 2ab * cos(C). This rule is often used in the Cambridge IGCSE Mathematics curriculum to solve problems involving triangles where the standard Pythagorean theorem does not apply.

How do you apply the Cosine Rule to find an unknown side in a triangle?

To find an unknown side using the Cosine Rule, you need to know the lengths of the other two sides and the measure of the included angle. For example, if you have a triangle with sides a and b, and angle C between them, the formula c² = a² + b² - 2ab * cos(C) can be rearranged to solve for side c. This is particularly useful in Cambridge IGCSE Mathematics when dealing with non-right-angled triangles.

How can the Cosine Rule be used to find an unknown angle in a triangle?

To find an unknown angle using the Cosine Rule, you need the lengths of all three sides of the triangle. For example, if you know sides a, b, and c, you can find angle C using the rearranged formula: cos(C) = (a² + b² - c²) / (2ab). This allows you to calculate the angle by taking the inverse cosine of the result. This application is essential in the Cambridge IGCSE Mathematics curriculum for solving problems involving angles in non-right-angled triangles.

What are some common mistakes to avoid when using the Cosine Rule?

Common mistakes when using the Cosine Rule include incorrectly identifying the sides and angles, not using the correct formula for the given problem, and errors in calculation, especially with the cosine function. It's important to ensure that the angle used is the one opposite the side being calculated. Additionally, students should be careful with units and ensure their calculator is set to the correct mode (degrees or radians) as required by the problem. These considerations are crucial for success in Cambridge IGCSE Mathematics.

Can the Cosine Rule be used in right-angled triangles, and if so, how?

While the Cosine Rule can technically be applied to right-angled triangles, it is not necessary because the Pythagorean theorem is simpler and more direct for these cases. However, if you choose to use the Cosine Rule, remember that the cosine of a 90-degree angle is zero, simplifying the formula to the Pythagorean theorem: c² = a² + b². Understanding when and how to apply these rules is part of the Cambridge IGCSE Mathematics curriculum.

Why is "Subtracting instead of adding $b^{2} + c^{2}$" a common mistake in Cosine Rule?

Why it happens: Misremembering the sign pattern. How to avoid it: Cosine rule is Pythagoras + correction: a^2 = b^2 + c^2 - 2bc A. The two sides are added; the correction is subtracted.

Why is "Rearranging incorrectly to find $\cos A$" a common mistake in Cosine Rule?

Why it happens: Confusing which terms move. How to avoid it: Memorise the rearranged form: A = b^2 + c^2 - a^22bc. Notice the side opposite A goes on top with a MINUS. Source: 0580/42 — recurring

Why is "Using cosine rule when sine rule is faster" a common mistake in Cosine Rule?

Why it happens: Defaulting to one rule. How to avoid it: Have a side + its opposite angle? → sine rule. Have SAS or SSS? → cosine rule.

Why is "Forgetting to square-root after computing $a^{2}$" a common mistake in Cosine Rule?

Why it happens: Stopping a step too early. How to avoid it: If you computed a^2, your final step is a = √(a^2).

TrigonometryCambridge IGCSE Maths 0580 Extended

Cosine Rule

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Cosine Rule — Cambridge IGCSE 0580 Maths Extended (2026)

$a^{2} = b^{2} + c^{2} - 2bc\cos A$ . The 'no opposite pair' rule. Use for SAS to find the third side, or SSS to find an angle.

What you’ll learn

Mapped to the Cambridge IGCSE 0580 syllabus (2025-2027).

E8.5 — Use the cosine rule to solve problems involving the lengths and angles of any triangle.

The cosine rule

$a^{2} = b^{2} + c^{2} - 2bc\cos A$ . The angle $A$ is opposite the side $a$ .

For any triangle with sides $a, b, c$ opposite angles $A, B, C$ : $a^{2} = b^{2} + c^{2} - 2bc\cos A.$

The same rule cycles round the labels: $b^{2} = a^{2} + c^{2} - 2ac\cos B,$ $c^{2} = a^{2} + b^{2} - 2ab\cos C.$

When to use.

SAS (two sides + the included angle): find the third side.
SSS (three sides): find an angle (rearrange the rule for $\cos$ ).

Each side is named with the lower-case letter of the angle directly opposite it.

Worked (SAS). Triangle with $b = 7\,\text{cm}$ , $c = 9\,\text{cm}$ , included angle $A = 50\degree$ . Find $a$ .

$a^{2} = 49 + 81 - 2(7)(9)\cos 50\degree$ .
$a^{2} = 130 - 126\cos 50\degree$ .
$a^{2} \approx 130 - 81.0 = 49.0$ .
$a \approx 7.00\,\text{cm}$ .

$a^{2} = b^{2} + c^{2} - 2bc\cos A$ .
$A$ opposite $a$ .
Use SAS to find the missing third side.
Calculator in DEG mode.

Finding an angle (SSS)

Rearrange the cosine rule: $\cos A = \dfrac{b^{2} + c^{2} - a^{2}}{2bc}$ .

When all three sides are known, you can find any angle. Rearrange the cosine rule to isolate $\cos A$ : $\cos A = \dfrac{b^{2} + c^{2} - a^{2}}{2bc}.$

Worked. Triangle with $a = 5\,\text{cm}$ , $b = 7\,\text{cm}$ , $c = 8\,\text{cm}$ . Find angle $A$ .

$\cos A = \dfrac{49 + 64 - 25}{2 \times 7 \times 8} = \dfrac{88}{112} = 0.7857$ .
$A = \cos^{-1}(0.7857) \approx 38.2\degree$ .

No ambiguous case. Unlike the sine rule, the cosine rule has NO ambiguity — $\cos^{-1}$ returns a unique angle in $(0\degree, 180\degree)$ . If $\cos A < 0$ , the angle is OBTUSE; if $\cos A > 0$ , it's ACUTE; if $\cos A = 0$ , it's exactly $90\degree$ .

With all three sides known, rearrange the cosine rule to isolate the angle.

$\cos A = \dfrac{b^{2} + c^{2} - a^{2}}{2bc}$ .
Negative cosine $\Rightarrow$ obtuse angle.
No ambiguous case.

Sine rule vs cosine rule — quick decision

Opposite angle-side pair? Sine rule. SAS or SSS? Cosine rule.

Decision flow.

What's given	Tool
Right angle + 2 sides	Pythagoras / SOH CAH TOA
Angle and OPPOSITE side, plus one more	Sine rule
Two sides + INCLUDED angle (SAS)	Cosine rule (for third side)
Three sides (SSS)	Cosine rule (for any angle)

Worked combined example. Triangle has $A = 65\degree$ , $b = 12\,\text{cm}$ , $c = 9\,\text{cm}$ .

$A$ is the included angle between $b$ and $c$ → cosine rule.
$a^{2} = 144 + 81 - 2(12)(9)\cos 65\degree = 225 - 91.27 \approx 133.73$ .
$a \approx 11.6\,\text{cm}$ .
Now to find $B$ , use the sine rule: $\dfrac{11.6}{\sin 65\degree} = \dfrac{12}{\sin B}$ .
$\sin B = \dfrac{12 \sin 65\degree}{11.6} \approx 0.9376 \Rightarrow B \approx 69.7\degree$ .
Then $C = 180\degree - 65\degree - 69.7\degree = 45.3\degree$ .

Right-angled? Pythagoras / SOH CAH TOA.
Opposite pair? Sine rule.
SAS or SSS? Cosine rule.
Cosine first, then sine for the next angle is a common one-two.

How it’s examined

Cosine rule appears every Paper 4 (4-6 marks), often combined with sine rule and bearings in a multi-part question. Examiner reports flag wrong-angle-side pairing as the recurring error: students apply $a^{2} = b^{2} + c^{2} - 2bc\cos A$ but use the WRONG angle (not opposite $a$ ).

Sources: Cambridge IGCSE Mathematics 0580 syllabus 2025-2027 (E8.5); 0580/42 Oct/Nov 2024 — Q14 (cosine rule + bearings); 0580 Examiner Reports 2022-2024. Last reviewed 2026-05-05.

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Step-by-step worked examples — Cosine Rule

Step-by-step solutions to past-paper-style questions on cosine rule, written exactly the way a tutor would explain them at the board.

1Find a side using the cosine rule
Extended• Adapted from 0580/42 May/Jun 2024 Q16• find side
Question
In triangle $ABC$ , $b = 8$ cm, $c = 10$ cm, $\angle A = 50\degree$ . Find $a$ .
Step-by-step solution
1. Step 1
  $a^{2} = b^{2} + c^{2} - 2bc\cos A$ .
  $a^{2} = 64 + 100 - 160\cos 50\degree \approx 164 - 102.85$
2. Step 2
  Compute and square-root.
  $a \approx \sqrt{61.15} \approx 7.82\,\text{cm}$
Answer
$a \approx 7.82\,\text{cm}$
2Find an angle using the cosine rule
Extended• find angle
Question
In triangle $ABC$ , $a = 6$ , $b = 7$ , $c = 9$ . Find $\angle C$ .
Step-by-step solution
1. Step 1
  Rearrange: $\cos C = \dfrac{a^{2} + b^{2} - c^{2}}{2ab}$ .
  $\cos C = \dfrac{36 + 49 - 81}{2(6)(7)} = \dfrac{4}{84} \approx 0.0476$
2. Step 2
  Apply $\cos^{-1}$ .
  $C = \cos^{-1}(0.0476) \approx 87.3\degree$
Answer
$\angle C \approx 87.3\degree$
3Use cosine rule then sine rule
Extended• mixed
Question
In triangle $XYZ$ , $x = 5$ , $y = 7$ , $\angle Z = 65\degree$ . Find $z$ and $\angle X$ .
Step-by-step solution
1. Step 1
  Cosine rule for $z$ .
  $z^{2} = 25 + 49 - 70\cos 65\degree \approx 74 - 29.58$
2. Step 2
  Compute.
  $z \approx \sqrt{44.42} \approx 6.66$
3. Step 3
  Sine rule for $\angle X$ .
  $\sin X = \dfrac{x \sin Z}{z} = \dfrac{5 \sin 65\degree}{6.66} \approx 0.680$
4. Step 4
  Apply $\sin^{-1}$ .
  $\angle X \approx 42.8\degree$
Answer
$z \approx 6.66,\ \angle X \approx 42.8\degree$
4Find the largest angle from three sides
Core• Adapted from 0580/22 Oct/Nov 2023 Q17• SSS, find angle
Question
Triangle $ABC$ has sides $a = 5$ cm, $b = 7$ cm, $c = 9$ cm. Find the largest angle.
Step-by-step solution
1. Step 1
  The largest angle is opposite the longest side, $c = 9$ . Apply the cosine rule for $\angle C$ .
  $\cos C = \dfrac{a^{2} + b^{2} - c^{2}}{2ab} = \dfrac{25 + 49 - 81}{2(5)(7)} = \dfrac{-7}{70} = -0.1$
2. Step 2
  Apply $\cos^{-1}$ . A negative cosine signals an obtuse angle.
  $C = \cos^{-1}(-0.1) \approx 95.7\degree$
Answer
$\angle C \approx 95.7\degree$
Examiner tip
0580 examiner reports flag candidates who switch the sign of the numerator. Use the formula in its exact form; the negative value of $\cos C$ is what tells you the angle is obtuse.
5Decide which rule to use
Extended• Adapted from 0580/42 Feb/Mar 2024 Q14• mixed, strategy
Question
In triangle $DEF$ , $DE = 8$ cm, $\angle D = 55\degree$ and $\angle E = 67\degree$ . Find $EF$ .
Step-by-step solution
1. Step 1
  We have two angles and the included side (ASA). Find the third angle first: $\angle F = 180\degree - 55\degree - 67\degree = 58\degree$ .
2. Step 2
  Now we have a side ( $DE$ ) with its opposite angle ( $\angle F$ ). That is the sine-rule signal, not cosine.
  $\dfrac{EF}{\sin 55\degree} = \dfrac{8}{\sin 58\degree}$
3. Step 3
  Solve.
  $EF = \dfrac{8 \sin 55\degree}{\sin 58\degree} \approx 7.73\,\text{cm}$
Answer
$EF \approx 7.73$ cm
Examiner tip
Although this is a cosine-rule subtopic, real exam papers test strategy. Cosine rule is unnecessary when a side and its opposite angle are already paired — picking it wastes time and introduces calculation errors.
6Cosine rule on a coordinate plane
Extended• coordinates
Question
Points $A(1, 2)$ , $B(7, 4)$ and $C(3, 8)$ form a triangle. Find $\angle BAC$ .
Step-by-step solution
1. Step 1
  Find the three side lengths using the distance formula.
  $AB = \sqrt{6^{2} + 2^{2}} = \sqrt{40}$
2. Step 2
  Continue.
  $AC = \sqrt{2^{2} + 6^{2}} = \sqrt{40},\ BC = \sqrt{(-4)^{2} + 4^{2}} = \sqrt{32}$
3. Step 3
  Apply the cosine rule for $\angle BAC$ (opposite $BC$ ).
  $\cos A = \dfrac{AB^{2} + AC^{2} - BC^{2}}{2 \cdot AB \cdot AC} = \dfrac{40 + 40 - 32}{2\sqrt{40}\sqrt{40}} = \dfrac{48}{80} = 0.6$
4. Step 4
  Apply $\cos^{-1}$ .
  $\angle BAC = \cos^{-1}(0.6) \approx 53.1\degree$
Answer
$\angle BAC \approx 53.1\degree$
7Real-world surveying problem
Extended• Adapted from 0580/42 May/Jun 2023 Q15• real-world
Question
Two roads leave a junction $J$ . Town $A$ is $4.5$ km along one road and town $B$ is $6.2$ km along the other. The angle between the roads at $J$ is $112\degree$ . Find the straight-line distance from $A$ to $B$ .
Step-by-step solution
1. Step 1
  Apply the cosine rule with the included angle.
  $AB^{2} = 4.5^{2} + 6.2^{2} - 2(4.5)(6.2)\cos 112\degree$
2. Step 2
  Compute. Note $\cos 112\degree \approx -0.3746$ , so the third term becomes positive.
  $AB^{2} = 20.25 + 38.44 - 55.8(-0.3746) \approx 58.69 + 20.90 = 79.59$
3. Step 3
  Square root.
  $AB \approx \sqrt{79.59} \approx 8.92\,\text{km}$
Answer
$AB \approx 8.92$ km
Examiner tip
When the included angle is obtuse, $\cos$ is negative — the subtraction in the formula becomes an addition. Many candidates mis-handle the sign here and lose two marks.
8A* combine cosine rule with area
Challenge• Adapted from 0580/42 Oct/Nov 2024 Q22• multi-step, area
Question
Triangle $PQR$ has $PQ = 10$ cm, $QR = 14$ cm and area $52$ cm $^{2}$ . Find the length of $PR$ , given that $\angle PQR$ is obtuse.
Step-by-step solution
1. Step 1
  Use the area rule to find $\angle Q$ .
  $52 = \dfrac{1}{2}(10)(14)\sin Q \Rightarrow \sin Q = \dfrac{52}{70} \approx 0.7429$
2. Step 2
  Principal value is $48.0\degree$ but the obtuse alternative is required.
  $Q = 180\degree - 48.0\degree = 132.0\degree$
3. Step 3
  Apply the cosine rule.
  $PR^{2} = 10^{2} + 14^{2} - 2(10)(14)\cos 132.0\degree \approx 100 + 196 - 280(-0.6691) \approx 296 + 187.35 = 483.35$
4. Step 4
  Square root.
  $PR \approx \sqrt{483.35} \approx 21.99\,\text{cm}$
Answer
$PR \approx 22.0$ cm
Examiner tip
A* candidates must use the obtuse value of $\angle Q$ when the question specifies it. Failure to spot the constraint costs the final accuracy mark.
9Bearings application
Challenge• Adapted from 0580/42 May/Jun 2024 Q19• bearings, challenge
Question
A ship leaves port $P$ and sails $35$ km on a bearing of $048\degree$ to $Q$ . From $Q$ it sails $52$ km on a bearing of $147\degree$ to $R$ . Find $PR$ .
Step-by-step solution
1. Step 1
  Find the angle at $Q$ inside the triangle. The back-bearing of $P$ from $Q$ is $228\degree$ . The angle from north at $Q$ to $R$ is $147\degree$ . So $\angle PQR = 228\degree - 147\degree = 81\degree$ .
2. Step 2
  Apply the cosine rule with the included angle at $Q$ .
  $PR^{2} = 35^{2} + 52^{2} - 2(35)(52)\cos 81\degree$
3. Step 3
  Compute. $\cos 81\degree \approx 0.1564$ .
  $PR^{2} \approx 1225 + 2704 - 3640(0.1564) \approx 3929 - 569.30 = 3359.70$
4. Step 4
  Square root.
  $PR \approx \sqrt{3359.70} \approx 57.96\,\text{km}$
Answer
$PR \approx 58.0$ km
Examiner tip
The hardest step is correctly identifying $\angle PQR$ inside the triangle from the two bearings. Sketch the north line at $Q$ and mark both bearings clearly before subtracting.
10Find a side with an acute included angle
Core• SAS, find side
Question
Triangle $PQR$ has $PQ = 9$ cm, $PR = 11$ cm and $\angle QPR = 35\degree$ . Find $QR$ .
Step-by-step solution
1. Step 1
  Apply the cosine rule.
  $QR^{2} = 9^{2} + 11^{2} - 2(9)(11)\cos 35\degree$
2. Step 2
  Compute.
  $QR^{2} = 81 + 121 - 198\cos 35\degree \approx 202 - 162.22 = 39.78$
3. Step 3
  Square root.
  $QR \approx \sqrt{39.78} \approx 6.31\,\text{cm}$
Answer
$QR \approx 6.31$ cm
Examiner tip
For an acute included angle the third term is positive, so $a^{2}$ is smaller than $b^{2} + c^{2}$ . Sanity-check by comparing $QR$ with the longer of the two given sides.

Key Formulae — Cosine Rule

The formulae you need to memorise for cosine rule on the Cambridge IGCSE 0580 paper, with every variable defined in plain English and a note on when to use it.

Cosine rule (side)
$a^{2} = b^{2} + c^{2} - 2bc\cos A$
$a$
side opposite angle $A$
$b, c$
the other two sides
$A$
angle opposite side $a$
When to use
When you know two sides and the included angle (SAS).
Cosine rule (angle)
$\cos A = \dfrac{b^{2} + c^{2} - a^{2}}{2bc}$
When to use
When you know all three sides (SSS) and want an angle.

Key Definitions and Keywords — Cosine Rule

Definitions to memorise and the exact keywords mark schemes credit for cosine rule answers — sharpened from recent examiner reports for the 2026 0580 sitting.

Cosine rule
Examiner keyword
A generalisation of Pythagoras to non-right-angled triangles.
SAS configuration
Examiner keyword
Two sides plus the angle between them — a use-cosine-rule signal.
SSS configuration
Examiner keyword
All three sides known — a use-cosine-rule-for-angle signal.

Common Mistakes and Misconceptions — Cosine Rule

The traps other students keep falling into on cosine rule questions — taken from recent Cambridge IGCSE 0580 examiner reports and mark schemes — and how to avoid them.

✕Subtracting instead of adding $b^{2} + c^{2}$
Why it happens
Misremembering the sign pattern.
How to avoid it
Cosine rule is Pythagoras + correction: $a^{2} = b^{2} + c^{2} - 2bc\cos A$ . The two sides are added; the correction is subtracted.
✕Rearranging incorrectly to find $\cos A$
0580/42 — recurring
Why it happens
Confusing which terms move.
How to avoid it
Memorise the rearranged form: $\cos A = \dfrac{b^{2} + c^{2} - a^{2}}{2bc}$ . Notice the side opposite $A$ goes on top with a MINUS.
✕Using cosine rule when sine rule is faster
Why it happens
Defaulting to one rule.
How to avoid it
Have a side + its opposite angle? → sine rule. Have SAS or SSS? → cosine rule.
✕Forgetting to square-root after computing $a^{2}$
Why it happens
Stopping a step too early.
How to avoid it
If you computed $a^{2}$ , your final step is $a = \sqrt{a^{2}}$ .

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

Past paper style quiz

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Video lesson

Short walkthrough of the concepts students most often get stuck on.

Cosine Rule — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

For any triangle with sides $a, b, c$ opposite angles $A, B, C$ : $a^{2} = b^{2} + c^{2} - 2bc\cos A.$

The same rule cycles round the labels: $b^{2} = a^{2} + c^{2} - 2ac\cos B,$ $c^{2} = a^{2} + b^{2} - 2ab\cos C.$

When to use.

SAS (two sides + the included angle): find the third side.
SSS (three sides): find an angle (rearrange the rule for $\cos$ ).

Each side is named with the lower-case letter of the angle directly opposite it.

Worked (SAS). Triangle with $b = 7\,\text{cm}$ , $c = 9\,\text{cm}$ , included angle $A = 50\degree$ . Find $a$ .

$a^{2} = 49 + 81 - 2(7)(9)\cos 50\degree$ .
$a^{2} = 130 - 126\cos 50\degree$ .
$a^{2} \approx 130 - 81.0 = 49.0$ .
$a \approx 7.00\,\text{cm}$ .

What's given

Tool

Right angle + 2 sides

Pythagoras / SOH CAH TOA

Angle and OPPOSITE side, plus one more

Sine rule

Two sides + INCLUDED angle (SAS)

Cosine rule (for third side)

Three sides (SSS)

Cosine rule (for any angle)

Cosine Rule

Short Study Notes in the form of Slides

Short Study Notes — Cosine Rule

Detailed Notes

What you’ll learn

How it’s examined

1Find a side using the cosine rule

2Find an angle using the cosine rule

3Use cosine rule then sine rule

4Find the largest angle from three sides

5Decide which rule to use

6Cosine rule on a coordinate plane

7Real-world surveying problem

8A* combine cosine rule with area

9Bearings application

10Find a side with an acute included angle

Cosine rule (side)

Cosine rule (angle)

Cosine rule

SAS configuration

SSS configuration

✕Subtracting instead of adding b2+c2b^{2} + c^{2}b2+c2

✕Rearranging incorrectly to find cos⁡A\cos AcosA

✕Using cosine rule when sine rule is faster

✕Forgetting to square-root after computing a2a^{2}a2

Practice questions

Past paper style quiz

Assess your understanding

Video lesson

Cosine Rule — frequently asked questions

Cosine Rule

Short Study Notes in the form of Slides

Short Study Notes — Cosine Rule

Detailed Notes

What you’ll learn

How it’s examined

1Find a side using the cosine rule

2Find an angle using the cosine rule

3Use cosine rule then sine rule

4Find the largest angle from three sides

5Decide which rule to use

6Cosine rule on a coordinate plane

7Real-world surveying problem

8A* combine cosine rule with area

9Bearings application

10Find a side with an acute included angle

Cosine rule (side)

Cosine rule (angle)

Cosine rule

SAS configuration

SSS configuration

✕Subtracting instead of adding b2+c2b^{2} + c^{2}b2+c2

✕Rearranging incorrectly to find cos⁡A\cos AcosA

✕Using cosine rule when sine rule is faster

✕Forgetting to square-root after computing a2a^{2}a2

Practice questions

Past paper style quiz

Assess your understanding

Video lesson

Cosine Rule — frequently asked questions

✕Subtracting instead of adding $b^{2} + c^{2}$

✕Rearranging incorrectly to find $\cos A$

✕Forgetting to square-root after computing $a^{2}$

✕Subtracting instead of adding $b^{2} + c^{2}$

✕Rearranging incorrectly to find $\cos A$

✕Forgetting to square-root after computing $a^{2}$