Overview

Use the Law of Cosines when a triangle is not right but you need a missing side or angle.

Key Ideas

  • a2=b2+c22bccosAa^2 = b^2 + c^2 - 2bc\cos A and cyclic versions.
  • cosA=b2+c2a22bc\cos A = \frac{b^2 + c^2 - a^2}{2bc}.
  • If A=90A=90^\circ, the formula reduces to the Pythagorean theorem.

Core Skills

Choose the Correct Opposite Side

Always match aa with AA, bb with BB, and cc with CC. The side opposite the angle is the one being solved for in a2=b2+c22bccosAa^2=b^2+c^2-2bc\cos A.

Find a Side from Two Sides and an Angle

When two sides and their included angle are known, compute the third side directly using the first form.

Find an Angle from Three Sides

Use cosA=(b2+c2a2)/(2bc)\cos A = (b^2+c^2-a^2)/(2bc), then take arccos\arccos if needed.

Worked Example

Triangle has sides a=7a=7, b=8b=8, c=9c=9. Find cosA\cos A.

cosA=82+9272289=64+8149144=2/3\cos A = \frac{8^2 + 9^2 - 7^2}{2\cdot 8\cdot 9} = \frac{64+81-49}{144} = 2/3.

More Examples

Example 1: Find a Side

Triangle has sides b=6b=6, c=10c=10, and included angle A=60A=60^\circ. Find aa.

a2=62+1022610cos60=36+10060=76a^2 = 6^2 + 10^2 - 2\cdot 6\cdot 10\cdot \cos 60^\circ = 36+100-60=76, so a=76a=\sqrt{76}.

Example 2: Detect an Obtuse Angle

Triangle has sides a=9a=9, b=5b=5, c=7c=7. Is A\angle A obtuse?

Compute b2+c2a2=25+4981=7b^2 + c^2 - a^2 = 25 + 49 - 81 = -7, so cosA<0\cos A < 0 and AA is obtuse.

Strategy Checklist

  • Identify the included angle when two sides are known.
  • Use the cosine form to check for obtuse or acute angles.
  • Reduce to Pythagorean theorem when A=90A=90^\circ.

Common Pitfalls

  • Using the wrong opposite side for angle AA.
  • Dropping the minus sign on 2bccosA2bc\cos A.
  • Forgetting to square all side lengths in the equation.

Practice Problems

StatusSourceProblem NameDifficultyTags
AMC 12AHard
Show TagsCircles, Geometry, Tangency
AMC 12AHard
Show TagsCyclic Quadrilateral, Geometry, Law of Cosines
AIME IMedium
Show TagsAlgebra, Geometry, Triangle
AMC 12BHard
Show TagsEquilateral, Geometry, Medians, Triangle

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