Revision notes for CIE IGCSE Maths The Cosine Rule. Open the guide for explanations and worked examples. Written against the CIE IGCSE Maths (0580) specification, so the content matches what's examinable rather than general Maths background.
Revision notes for CIE IGCSE Maths The Cosine Rule. Open the guide for explanations and worked examples. Written against the CIE IGCSE Maths (0580) specification, so the content matches what's examinable rather than general Maths background.
The Cosine Rule works in any triangle, not just right-angled triangles. It is especially useful when Pythagoras and basic trigonometry do not apply directly.
Included angle and opposite side
The included angle is the angle between two known sides. The opposite side is the side directly across from an angle.
In the diagram below, angle CCC is between sides aaa and bbb, and side ccc is opposite angle CCC.

Spotting the included angle
Suppose a triangle has sides 8 cm and 11 cm with a 70° angle between them.
The 70° angle is the included angle, because it sits between the two known sides.
The side opposite the 70° angle is the side you can find directly using the Cosine Rule.
Quick check
If the angle is drawn between the two given sides, think Cosine Rule for a side.
The Cosine Rule
If side ccc is opposite angle CCC, and the two sides around angle CCC are aaa and bbb, then:
c2=a2+b2−2abcosCc^2 = a^2 + b^2 - 2ab\cos Cc2=a2+b2−2abcosCThis looks a bit like Pythagoras, but with an extra cosine part. In fact, when C=90∘C = 90^\circC=90∘, cos90∘=0\cos 90^\circ = 0cos90∘=0, so it becomes Pythagoras.
Main idea
Use c2=a2+b2−2abcosCc^2 = a^2 + b^2 - 2ab\cos Cc2=a2+b2−2abcosC when you know two sides and the angle between them, and you want the third side.
Finding a missing side
A triangle has two sides of length 12 cm and 9 cm. The angle between them is 105°. Find the opposite side xxx, to 1 decimal place.
Identify the two known sides and the included angle: the sides are 12 cm and 9 cm, and the included angle is 105°.
Substitute into the Cosine Rule:
x2=122+92−2×12×9×cos105∘x^2 = 12^2 + 9^2 - 2 \times 12 \times 9 \times \cos 105^\circx2=122+92−2×12×9×cos105∘Work out the value of x2x^2x2:
x2≈280.9x^2 \approx 280.9x2≈280.9Square root and round to 1 decimal place: x≈16.8x \approx 16.8x≈16.8 cm.
Forgetting the square root
The formula gives x2x^2x2 first, not xxx. Always square root at the end when finding a length.
Obtuse angle sanity check
If the included angle is bigger than 90°, the opposite side is usually quite long. That is because the triangle is opening out wider.
Sometimes you are given all three sides and asked for an angle. Then you rearrange the Cosine Rule.
Starting from:
c2=a2+b2−2abcosCc^2 = a^2 + b^2 - 2ab\cos Cc2=a2+b2−2abcosCRearrange to get:
cosC=a2+b2−c22ab\cos C = \frac{a^2 + b^2 - c^2}{2ab}cosC=2aba2+b2−c2Then use inverse cosine, written cos−1\cos^{-1}cos−1, to get the angle.
Inverse cosine
The inverse cosine, cos−1\cos^{-1}cos−1, is the calculator function that finds the angle when you know its cosine value.
Finding an angle from three sides
In triangle ABCABCABC, AB=6AB = 6AB=6 m, BC=8BC = 8BC=8 m and AC=9AC = 9AC=9 m. Find angle ABCABCABC to the nearest degree.
Angle ABCABCABC is at BBB, so the two sides touching the angle are AB=6AB = 6AB=6 and BC=8BC = 8BC=8.
The side opposite angle BBB is AC=9AC = 9AC=9.
Substitute into the rearranged Cosine Rule:
cosB=62+82−922×6×8\cos B = \frac{6^2 + 8^2 - 9^2}{2 \times 6 \times 8}cosB=2×6×862+82−92Simplify the fraction:
cosB=1996\cos B = \frac{19}{96}cosB=9619Use inverse cosine:
B=cos−1(1996)≈78.6∘B = \cos^{-1}\left(\frac{19}{96}\right) \approx 78.6^\circB=cos−1(9619)≈78.6∘Round to the nearest degree: angle ABCABCABC is 79°.
Using the wrong opposite side
When finding an angle, the side opposite the angle is the one being subtracted in the numerator: a2+b2−c2a^2 + b^2 - c^2a2+b2−c2. Label the opposite side before substituting.
Calculator mode
Make sure your calculator is in degrees, not radians. Your answer should be in degrees for these questions.
Use the Cosine Rule when you have either:
If you have a right-angled triangle, try Pythagoras or SOHCAHTOA first. If you have a matching opposite side and angle pair, the Sine Rule may be more suitable.
Some harder questions give side lengths like x+1x + 1x+1 or 2x−12x - 12x−1. You still use the Cosine Rule, but you will need to expand brackets and solve an equation.
Quadratic equation
A quadratic equation is an equation involving x2x^2x2, such as x2−x−20=0x^2 - x - 20 = 0x2−x−20=0.
Using the Cosine Rule with algebra
A triangle has sides x+1x + 1x+1 and 2x−12x - 12x−1 with an included angle of 60°. The opposite side is 63\sqrt{63}63. Find xxx.
Since the opposite side is 63\sqrt{63}63, its square is 63.
Substitute into the Cosine Rule:
63=(x+1)2+(2x−1)2−2(x+1)(2x−1)cos60∘63 = (x + 1)^2 + (2x - 1)^2 - 2(x + 1)(2x - 1)\cos 60^\circ63=(x+1)2+(2x−1)2−2(x+1)(2x−1)cos60∘Use cos60∘=12\cos 60^\circ = \frac{1}{2}cos60∘=21:
63=(x+1)2+(2x−1)2−(x+1)(2x−1)63 = (x + 1)^2 + (2x - 1)^2 - (x + 1)(2x - 1)63=(x+1)2+(2x−1)2−(x+1)(2x−1)Expand and simplify:
63=3x2−3x+363 = 3x^2 - 3x + 363=3x2−3x+3Rearrange and factorise:
3x2−3x−60=0⇒x2−x−20=0⇒(x−5)(x+4)=03x^2 - 3x - 60 = 0 \Rightarrow x^2 - x - 20 = 0 \Rightarrow (x - 5)(x + 4) = 03x2−3x−60=0⇒x2−x−20=0⇒(x−5)(x+4)=0The solutions are x=5x = 5x=5 or x=−4x = -4x=−4. Reject x=−4x = -4x=−4 because it would give negative side lengths, so x=5x = 5x=5.
Accepting an impossible value
Lengths cannot be negative. If an algebra question gives two possible values, check that all side lengths are positive.
You may need to use the triangle area formula first.
Area of a triangle using sine
If two sides are aaa and bbb with included angle CCC, then the area is:
Area=12absinC\text{Area} = \frac{1}{2}ab\sin CArea=21absinCThis can help you find the included angle. Then you can use the Cosine Rule to find the missing side and finish the perimeter.
Finding a perimeter using area first
Triangle ABCABCABC has AB=18.5AB = 18.5AB=18.5 m, AC=12.8AC = 12.8AC=12.8 m, and area 70 m². The angle at AAA is acute. Find the perimeter to 3 significant figures.
Use the area formula with the two sides around angle AAA:
70=12×18.5×12.8×sinA70 = \frac{1}{2} \times 18.5 \times 12.8 \times \sin A70=21×18.5×12.8×sinARearrange to find sinA\sin AsinA:
sinA=14018.5×12.8≈0.591\sin A = \frac{140}{18.5 \times 12.8} \approx 0.591sinA=18.5×12.8140≈0.591Use inverse sine, choosing the acute angle:
A≈sin−1(0.591)≈36.2∘A \approx \sin^{-1}(0.591) \approx 36.2^\circA≈sin−1(0.591)≈36.2∘Use the Cosine Rule to find BCBCBC:
BC2=18.52+12.82−2×18.5×12.8×cos36.2∘BC^2 = 18.5^2 + 12.8^2 - 2 \times 18.5 \times 12.8 \times \cos 36.2^\circBC2=18.52+12.82−2×18.5×12.8×cos36.2∘This gives BC≈11.1BC \approx 11.1BC≈11.1 m, so the perimeter is approximately 42.4 m to 3 significant figures.
Sine ambiguity
When you use inverse sine, there may be two possible angles between 0° and 180°. Use the diagram or wording, such as “acute” or “obtuse”, to choose the correct one.
A question may ask for:
Keep full calculator values until the final line. Rounding too early can change the final answer.
In the exam
Mark the angle you are using, then mark the side opposite it.
Write the Cosine Rule formula before substituting numbers, so the method is clear.
Do not round too early; give the final answer in the form requested by the question.
Check yourself
If you know two sides and the angle between them, which version of the Cosine Rule do you use?
When finding an angle, which side goes in the “minus” part of the numerator?
Why must you check side lengths when solving an algebraic Cosine Rule question?
Test yourself with 5 quick questions on this guide. Answer them all correctly to complete it.
Test yourself on this topic, or move on to the next guide.
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