Law of Cosines

Generalises the Pythagorean theorem to any triangle: c² = a² + b² − 2ab·cos(C). Essential for solving triangles without a right angle.

c² = a² + b² − 2ab cos C — drag the angle to see how c responds

c² = 5² + 7² − 35.0 = 39.0→c = 6.245

C = 60°

Acute: cos C > 0, so −2ab cos C is negative — c is shorter than the Pythagorean prediction.

Definition

The Law of Cosines generalises the Pythagorean theorem to any triangle. For a triangle with sides $a$ , $b$ , $c$ and angles $A$ , $B$ , $C$ (where angle $C$ is opposite side $c$ ):

$c^2 = a^2 + b^2 - 2ab\cos C$

Similarly: $a^2 = b^2 + c^2 - 2bc\cos A$ and $b^2 = a^2 + c^2 - 2ac\cos B$ .

When $C = 90°$ : $\cos 90° = 0$ , so $c^2 = a^2 + b^2$ — recovering the Pythagorean theorem. The law of cosines is its generalisation.

When to use it:

Given two sides and the included angle (SAS)
Given all three sides (SSS) — to find an angle

Finding a missing side (SAS)

A triangle has $a = 5$ , $b = 7$ , and $C = 60°$ . Find $c$ .

$c^2 = 5^2 + 7^2 - 2(5)(7)\cos 60° = 25 + 49 - 70 \cdot \frac{1}{2} = 74 - 35 = 39$

$c = \sqrt{39} \approx 6.24$

Try it

A triangle has sides $a = 8$ , $b = 6$ , $c = 10$ . Find the angle $C$ opposite side $c$ .

Solution

Rearrange the law of cosines for $\cos C$ :

$\cos C = \frac{a^2 + b^2 - c^2}{2ab} = \frac{64 + 36 - 100}{2(8)(6)} = \frac{0}{96} = 0$

$C = \arccos(0) = 90°$

This confirms the $6$ - $8$ - $10$ triangle is a right triangle (a multiple of the $3$ - $4$ - $5$ triple).

Related concepts

Needs first

Pythagorean Theorem

Triangles

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