The cosine rule is Pythagoras with a correction term. When the angle is $90^\circ$ the correction vanishes and you get Pythagoras back exactly.
Today's hook, Pythagoras only works at $90^\circ$. The cosine rule tells you precisely how wrong Pythagoras is at any other angle, and the size of the error is $2ab\cos C$.
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Recall, your gut answer first
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In $c^2 = a^2 + b^2 - 2ab\cos C$, substitute $C = 90^\circ$. What happens, and what does that tell you about the relationship between this rule and Pythagoras?
Before you work it out, what is your instinct? Write it down, then check it against the lesson.
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Pythagoras with a correction term
Work through the core explanation before applying it.
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Pythagoras with a correction term
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When $C = 90^\circ$, $\cos C = 0$ and the rule collapses to $c^2 = a^2 + b^2$. For any other angle the term $2ab\cos C$ corrects Pythagoras: it subtracts when $C$ is acute and adds when $C$ is obtuse, because $\cos C$ is then negative.
No side-and-opposite-angle pair means the cosine rule: two sides and the included angle, or all three sides.
The unknown side faces the known angle
In $c^2 = a^2 + b^2 - 2ab\cos C$, the side $c$ is opposite the angle $C$. Line those up before substituting.
A negative cosine means an obtuse angle
If $\cos C$ comes out negative, the angle is between $90^\circ$ and $180^\circ$. That is a valid answer, not an error.
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What you'll master
Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.
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What you'll master
Know
Key facts
The cosine rule is $c^2 = a^2 + b^2 - 2ab\cos C$, with $c$ opposite $C$.
Rearranged for an angle it is $\cos C = \dfrac{a^2 + b^2 - c^2}{2ab}$.
At $C = 90^\circ$ it reduces to Pythagoras.
Use it for two sides and the included angle, or for all three sides.
Understand
Concepts
Why the cosine rule is a generalisation of Pythagoras rather than a separate result.
Why a negative cosine correctly signals an obtuse angle.
Why the cosine rule, unlike the sine rule, never produces an ambiguous answer for an angle.
Can do
Skills
Examine and reproduce the proof of the cosine rule.
Find the third side given two sides and the included angle.
Find any angle given all three sides, including obtuse angles.
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Key terms
Cosine ruleA relationship in any triangle linking all three sides to one angle. Like this: with $a = 5$, $b = 7$ and $C = 60^\circ$, $c^2 = 25 + 49 - 2(5)(7)\cos 60^\circ = 39$.
Included angleThe angle between the two sides you know, which is what the cosine rule needs to find the third side. Like this: given $a$, $b$ and $C$, the angle $C$ sits between $a$ and $b$.
Obtuse angleAn angle between $90^\circ$ and $180^\circ$, whose cosine is negative. Like this: $\cos 120^\circ = -0.5$, so a negative cosine value means the angle is obtuse.
Pythagoras theoremThe right-angled special case of the cosine rule. Like this: when $C = 90^\circ$, $\cos C = 0$ and $c^2 = a^2 + b^2$.
SSSKnowing all three sides, which lets you find any angle with the cosine rule. Like this: sides 4, 5 and 6 determine every angle of the triangle.
SASKnowing two sides and the angle between them, which lets you find the third side. Like this: sides 8 and 10 with a $40^\circ$ angle between them.
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Proving the cosine rule
Work through the core explanation before applying it.
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Proving the cosine rule
core concept
Drop a perpendicular of height $h$ from $B$ to $AC$, meeting it at $H$, and let $AH = x$, so $HC = b - x$.
Pythagoras in the left triangle gives $c^2 = h^2 + x^2$, and in the right triangle $a^2 = h^2 + (b-x)^2$. Also $x = c\cos A$.
Expanding the second equation and substituting gives $a^2 = b^2 + c^2 - 2bc\cos A$. Relabelling the vertices produces the version for any angle.
Same construction, different bookkeeping. The sine rule came from writing $h$ two ways; the cosine rule comes from applying Pythagoras twice to the same two triangles. One diagram gives both results.
Quick check: what does $c^2 = a^2 + b^2 - 2ab\cos C$ become when $C = 90^\circ$?
The cosine rule comes from dropping the same perpendicular as the sine rule and applying Pythagoras to both right triangles. At $C = 90^\circ$ the cosine term vanishes and Pythagoras is recovered, so the cosine rule generalises it.
Pause, copy the construction ($h$, $x$, $b-x$), the two Pythagoras statements, and the note that $C = 90^\circ$ recovers Pythagoras, into your book.
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Finding the third side from two sides and the included angle
core concept
We just saw where the cosine rule comes from. That raises a question: when do you reach for it instead of the sine rule? This card answers it → whenever you have no complete side-and-opposite-angle pair, which is exactly the two-sides-and-included-angle case.
Given sides $a$ and $b$ with the angle $C$ between them, substitute straight into $c^2 = a^2 + b^2 - 2ab\cos C$ and square root at the end.
For example with $a = 8$, $b = 5$ and $C = 60^\circ$: $c^2 = 64 + 25 - 2(8)(5)(0.5) = 89 - 40 = 49$, so $c = 7$.
Take the square root only once, at the very end. Rounding before then loses accuracy quickly.
Line up the letters. The side you are finding must be the one opposite the angle you were given. If your labels do not match the formula, relabel the diagram rather than juggling the formula.
Fill the blank: with $a = 3$, $b = 4$ and $C = 90^\circ$, the cosine rule gives $c = $ .
For two sides and the included angle, substitute into $c^2 = a^2 + b^2 - 2ab\cos C$ and square root at the end. The unknown side must be opposite the known angle, so relabel the diagram if the letters do not line up.
Pause, copy the SAS method with the worked $a=8$, $b=5$, $C=60^\circ$ giving $c=7$, and the warning to square root only at the end, into your book.
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Finding an angle from three sides, including the obtuse case
core concept
We just saw the cosine rule finding a side. That raises a question: can it run backwards, finding an angle when all three sides are known? This card answers it → yes, and unlike the sine rule it never leaves an ambiguity.
Rearranging gives $\cos C = \dfrac{a^2 + b^2 - c^2}{2ab}$. Substitute the three sides and take the inverse cosine.
If the result is negative the angle is obtuse. For sides 5, 6 and 10: $\cos C = \dfrac{25 + 36 - 100}{60} = -0.65$, so $C \approx 130^\circ 32'$.
This is why the cosine rule is safer than the sine rule for finding angles: the inverse cosine returns a unique answer between $0^\circ$ and $180^\circ$, so no second solution can hide.
Find the largest angle first. If you need every angle, use the cosine rule on the angle opposite the longest side. Once the only possible obtuse angle is settled, the sine rule is safe for the rest.
Which case does NOT let you use the cosine rule directly?
For three sides use $\cos C = \frac{a^2+b^2-c^2}{2ab}$ and take the inverse cosine. A negative value means an obtuse angle, which is a valid answer. The cosine rule gives a unique angle, so it cannot be ambiguous.
Pause, copy the rearranged formula, the worked obtuse example with sides 5, 6, 10, and the strategy of finding the largest angle first, into your book.
Worked examples · 3 in a row, reveal as you go
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Work examples end to end
Follow the reasoning through complete worked solutions.
PROBLEM 1 · THIRD SIDE FROM SAS
In triangle $ABC$, $a = 11$ cm, $b = 7$ cm and $C = 62^\circ$. Find $c$ to one decimal place.
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$c^2 = 11^2 + 7^2 - 2(11)(7)\cos 62^\circ$
Substitute; $c$ is opposite the given angle $C$.
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$= 121 + 49 - 154\cos 62^\circ = 170 - 72.31$
Evaluate the correction term.
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$c^2 \approx 97.69$, so $c \approx 9.9$ cm
Square root once, at the end.
PROBLEM 2 · ANGLE FROM SSS
A triangle has sides 5 m, 6 m and 10 m. Find its largest angle, to the nearest minute.
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The largest angle faces the longest side, so find the angle opposite $10$
$\sin B = \dfrac{9\sin 47^\circ}{8.8} \approx 0.748$, so $B \approx 48.4^\circ$
With the pair established the sine rule is quickest.
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Quick-fire practice
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Quick-fire practice
+10 XP
Find $c$ if $a = 6$, $b = 8$ and $C = 40^\circ$.
Find the largest angle of a triangle with sides 7, 9 and 12.
Find $b$ if $a = 15$, $c = 20$ and $B = 110^\circ$.
Explain in one line why $\cos C < 0$ means $C$ is obtuse.
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Revisit the right-angled case
Run the quick drill and copy the summary into your book.
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Revisit the right-angled case
You substituted $C = 90^\circ$ at the start. Write one sentence explaining what the term $2ab\cos C$ is doing geometrically, and say what it means that the term is negative when $C$ is obtuse.
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Multiple choice
Answer the drill bank and rate your confidence.
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Multiple choice
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Short answer
Write full responses, then check them against the model answers.
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Short answer
ApplyBand 43 marks
Q1. In triangle $XYZ$, $x = 13$ cm, $z = 9$ cm and $Y = 56^\circ$. Find $y$ correct to one decimal place. (3 marks)
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ApplyBand 54 marks
Q2. A triangular paddock has sides 45 m, 62 m and 80 m. Find its largest angle to the nearest minute, and justify which angle you chose to calculate. (4 marks)
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UnderstandBand 42 marks
Q3. Explain why the cosine rule can never give an ambiguous answer when finding an angle, while the sine rule can. (2 marks)
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📖 Comprehensive answers (click to reveal)
Practice 1: $c^2 = 36 + 64 - 96\cos 40^\circ \approx 26.5$, $c \approx 5.1$. Practice 2: $\cos C = \frac{49+81-144}{126} \approx -0.111$, $C \approx 96.4^\circ$. Practice 3: $b^2 = 225 + 400 - 600\cos 110^\circ \approx 830.2$, $b \approx 28.8$. Practice 4: cosine is negative only for angles between $90^\circ$ and $180^\circ$.
Q2 (4 marks): The largest angle is opposite the longest side, 80 m, because a larger side subtends a larger angle [1]. $\cos \theta = \frac{45^2 + 62^2 - 80^2}{2(45)(62)} = \frac{2025 + 3844 - 6400}{5580}$ [1] $= \frac{-531}{5580} \approx -0.0952$ [1]. $\theta \approx 95.46^\circ = 95^\circ 28'$ [1].
Q3 (2 marks): The inverse cosine returns exactly one angle between $0^\circ$ and $180^\circ$, and the sign of the cosine distinguishes acute from obtuse, so the answer is unique [1]. The inverse sine returns only the acute angle, but $\sin\theta = \sin(180^\circ - \theta)$, so a second obtuse solution may also fit the triangle [1].
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Review and finish
Take the module quiz if you are ready, then mark the lesson complete.
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Boss battle · Cosine Rule Clash
earn bronze · silver · gold
Find third sides and unknown angles, deciding each time whether the sine or cosine rule is the faster route. Beat the boss to bank a tier, gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.