The Cosine Rule

Suppose you know your distances to two landmarks and the angle between them as you swing from one to the other — how far apart are the landmarks themselves? A ship altering course, a drone crossing a field, a builder sizing a roof truss: whenever you have two sides and the angle wedged between them, the cosine rule hands you the third distance.

The sine rule is a wonderful tool — but it has one demand: you must already know a side paired with the angle opposite it. Plenty of real triangles refuse to hand you such a pair. Picture a triangle where you know two sides and only the angle wedged between them. No side lines up with a known opposite angle, so the sine rule simply cannot start.

This is where the cosine rule steps in. It is a generalisation of Pythagoras that works in any triangle, right-angled or not:

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

Use it in two situations:

you know two sides and the included angle (b, c and A) and want the third side a;
you know all three sides and want an angle — rearrange to \cos A = \dfrac{b^2 + c^2 - a^2}{2bc} then take the inverse cosine.

Notice that when A = 90^\circ we have \cos A = 0, so the last term vanishes and the rule collapses to a^2 = b^2 + c^2 — Pythagoras' theorem. The cosine rule is Pythagoras with a correction term stitched on for triangles that aren't right-angled.

In any triangle, with each side opposite the angle of the same letter:

to find a side from two sides and the included angle, use a^2 = b^2 + c^2 - 2bc\cos A;
to find an angle from all three sides, rearrange to \cos A = \dfrac{b^2 + c^2 - a^2}{2bc};
when the included angle is 90^\circ, \cos A = 0 and it reduces to Pythagoras' theorem a^2 = b^2 + c^2.

The included angle

Here is a scalene triangle. Step through it to see how sides b and c meet at vertex A — the angle A between them is the one the cosine rule uses to reach the opposite side a.

Worked example 1 — find the third side

Two roads leave a village at an angle of 50^\circ to each other. One friend walks 7\text{ km} along the first road; another walks 9\text{ km} along the second. How far apart are they now?

You know two sides — b = 7 and c = 9 — and the angle A = 50^\circ squeezed between them. That's a two-sides-and-the-included-angle setup, so drop it straight into the rule:

a^2 = b^2 + c^2 - 2bc\cos A = 7^2 + 9^2 - 2(7)(9)\cos 50^\circ

a^2 = 49 + 81 - 126 \times 0.6428 = 130 - 81.0 = 49.0

a = \sqrt{49.0} \approx 7.0\text{ km}

Do the whole b^2 + c^2 - 2bc\cos A as one calculation before taking the square root — the -2bc\cos A is a single term, not something to square on its own.

Worked example 2 — find an angle from three sides

A triangular garden bed has sides a = 8\text{ m}, b = 5\text{ m} and c = 7\text{ m}. Find the angle A opposite the longest side.

Three sides, no angles — so use the rearranged form. Put the side you want the opposite angle for (a) with a minus sign:

\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{5^2 + 7^2 - 8^2}{2(5)(7)}

\cos A = \frac{25 + 49 - 64}{70} = \frac{10}{70} \approx 0.1429

A = \cos^{-1}(0.1429) \approx 81.8^\circ

A neat feature: unlike the sine rule, the cosine has no ambiguous case for angles. Because \cos is negative for obtuse angles and positive for acute ones, \cos^{-1} pins down exactly one angle between 0^\circ and 180^\circ.

Worked example 3 — the right-angle special case

Take sides b = 6 and c = 8 with an included angle of exactly 90^\circ. The cosine rule gives

a^2 = 6^2 + 8^2 - 2(6)(8)\cos 90^\circ = 36 + 64 - 96 \times 0 = 100,

a = \sqrt{100} = 10.

The correction term switched itself off, and we're left with a^2 = b^2 + c^2 — the familiar 6-8-10 Pythagorean triangle. The cosine rule didn't replace Pythagoras; it contains it.

Three slip-ups catch people out with the cosine rule:

The angle must be the included one. To find a side you need the two sides and the angle between them. If the angle you're given isn't wedged between the two known sides, the formula won't work — you'd want the sine rule instead. To find an angle you need all three sides.
Mind the order of operations. Compute a^2 = b^2 + c^2 - 2bc\cos C as one expression. The -2bc\cos C is a single term; do not square (b+c), and do not forget to multiply by \cos C. And remember the final answer for a side needs a square root at the end.
Keep the calculator in degrees. If it's set to radians, \cos 50 gives a wildly wrong number. A quick check: \cos 60^\circ should read exactly 0.5.

Look again at a^2 = b^2 + c^2 - 2bc\cos A. The first part, b^2 + c^2, is pure Pythagoras. The tail -2bc\cos A is a correction term that accounts for the angle not being 90^\circ. Open the angle wider than a right angle and \cos A turns negative, so we add to b^2 + c^2 and the far side stretches longer. Close it below 90^\circ and \cos A is positive, so we subtract and the side shrinks. Set it to exactly 90^\circ and \cos A = 0 — the correction disappears and Pythagoras stands alone.

That one elegant generalisation, teamed up with the sine rule, lets you solve absolutely any triangle from a handful of measurements. Together they are the mathematical backbone of navigation, surveying, and 3D computer graphics — every time a game engine lights a polygon or a ship plots a course, cosine-rule triangles are doing quiet work underneath.