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Geometry & measures · Trigonometry (sine & cosine rules)

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The cosine rule

Why a² = b² + c² − 2bc cos A is Pythagoras with a correction term, when to choose it over the sine rule, and the rearrangement that finds an angle from three sides.

Geometry & measures · Trigonometry (sine & cosine rules)

The cosine rule

Why a² = b² + c² − 2bc cos A is Pythagoras with a correction term, when to choose it over the sine rule, and the rearrangement that finds an angle from three sides.

Why it works

It is Pythagoras with a correction term. For a triangle with sides aa, bb, cc and the angle AA opposite aa:

a2=b2+c2−2bccos⁡A.a^2 = b^2 + c^2 - 2bc\cos A.

When A=90°A = 90°, cos⁡90°=0\cos 90° = 0 and the correction vanishes, leaving a2=b2+c2a^2 = b^2 + c^2 — ordinary Pythagoras. When AA is acute, cos⁡A>0\cos A > 0, so we subtract something and aa comes out shorter than Pythagoras would predict; when AA is obtuse, cos⁡A\cos A is negative, the correction adds, and aa comes out longer. That built-in sign behaviour is why the rule handles obtuse triangles without any special case — unlike the sine rule.

Choose it when the sine rule can't start. Use the cosine rule when you have either
  • two sides and the angle between them (SAS) → find the third side, or
  • all three sides (SSS) → find any angle.
Both are exactly the situations where no side is paired with its opposite angle, so the sine rule has nothing to work with.

Keep the letters consistent. The side you are finding, aa, must be the one opposite the angle AA that you use — the two "wrapping" sides bb and cc are the ones enclosing that angle. Label the diagram first.

Rearranged for an angle:

cos⁡A=b2+c2−a22bc.\cos A = \frac{b^2 + c^2 - a^2}{2bc}.

Then A=cos⁡−1(… )A = \cos^{-1}(\dots). This form is safe for obtuse angles: if the fraction comes out negative, the inverse cosine returns an angle over 90°90° automatically — no ambiguity to resolve, which is a real advantage over the sine rule.

Order of operations bites here. b2+c2−2bccos⁡Ab^2 + c^2 - 2bc\cos A means "work out 2bccos⁡A2bc\cos A as one quantity, then subtract". Don't compute b2+c2−2bcb^2 + c^2 - 2bc and then multiply by cos⁡A\cos A. And don't square-root until the very end.