The cosine rule

For any triangle, with the side $a$ opposite the angle $A$ :

a^2 = b^2 + c^2 - 2bc\cos A

It looks like Pythagoras' theorem with a correction on the end, and that is exactly what it is.

Before you start

You need Pythagoras' theorem and the cosine ratio, $\cos\theta = \dfrac{\text{adjacent}}{\text{hypotenuse}}$ in a right-angled triangle.

Predict first

Before reading on, predict: if $A = 90^\circ$ , what does $\cos A$ equal, and what does the rule become? Work that out first — the answer says why this is Pythagoras with something added rather than a separate rule to memorise.

The proof

Take a triangle with corners $A$ , $B$ and $C$ , and sides $a$ , $b$ , $c$ opposite them. Suppose $A$ is acute for now; the last section deals with an obtuse $A$ .

Drop a perpendicular from $C$ to the side $AB$ , meeting it at $D$ . Call the height $h$ , and call the length $AD$ the distance $x$ .

That splits the triangle into two right-angled triangles, and the proof is Pythagoras applied to each of them.

In the left triangle, $ACD$ :

the hypotenuse is $AC = b$ ;
the adjacent side to angle $A$ is $AD = x$ .

So the cosine ratio gives

\cos A = \frac{x}{b} \qquad\text{so}\qquad x = b\cos A

and Pythagoras gives

h^2 = b^2 - x^2

In the right triangle, $BCD$ :

the hypotenuse is $BC = a$ ;
the vertical side is $h$ ;
the horizontal side is $DB$ , which is the whole of $AB$ minus $AD$ , so $c - x$ .

Pythagoras gives

a^2 = h^2 + (c - x)^2

Substitute. Replace $h^2$ with $b^2 - x^2$ and expand the bracket:

a^2 = (b^2 - x^2) + (c^2 - 2cx + x^2)

The $x^2$ terms cancel:

a^2 = b^2 + c^2 - 2cx

And $x = b\cos A$ , so

a^2 = b^2 + c^2 - 2bc\cos A

which is the cosine rule.

Why it becomes Pythagoras when the angle is right

When $A = 90^\circ$ , the cosine of a right angle is $0$ . The correction term $-2bc\cos A$ is therefore zero, and what is left is

a^2 = b^2 + c^2

So Pythagoras' theorem is the case of the cosine rule where the angle happens to be a right angle. Nothing new was needed for it; it falls out.

The obtuse case, and why it works

If $A$ is obtuse the perpendicular from $C$ lands outside the side $AB$ , so the picture is different: the foot $D$ sits beyond $B$ rather than between $A$ and $B$ .

The algebra still works, with two changes. The length $DB$ becomes $x - c$ rather than $c - x$ , and the angle you measure inside the right-angled triangle at $A$ is $180^\circ - A$ , whose cosine is $-\cos A$ . Both changes flip a sign, and they flip it in the same place, so the final statement is unchanged.

That is worth checking rather than assuming. A proof that quietly only covered acute angles would not be a proof of the cosine rule, and the sign that looks like a detail is the whole reason the rule is stated with a minus.

Check your understanding

Use the rule with $A = 90^\circ$ to show it becomes Pythagoras' theorem.
In a triangle, $b = 7$ , $c = 9$ and $A = 60^\circ$ . Find $a$ .
The rule is often rearranged to $\cos A = \dfrac{b^2 + c^2 - a^2}{2bc}$ . What is that version for?

Answers

$\cos 90^\circ = 0$ , so $-2bc\cos A = 0$ and $a^2 = b^2 + c^2$ .
$a^2 = 49 + 81 - 2(7)(9)(0.5) = 130 - 63 = 67$ , so $a \approx 8.19$ .
Finding an angle when all three sides are known. The original form gives a side from two sides and the angle between them; this form solves for the angle instead.

A common error

Watch the angle you use. The rule needs the angle between the two sides you know — the one opposite the side you are finding. Reaching for an angle that is not between them is the most common way to get a wrong answer, and the rule gives no warning: it will happily compute a number for the wrong triangle.