GeometryRight Triangles and TrigonometryG-SRT.D.11

The Law of Cosines

c² = a² + b² − 2ab cos C extends the Pythagorean theorem.

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The Law of Sines from The Law of Sines needs an angle with its opposite side. When you know two sides and the angle between them (SAS), or all three sides (SSS), you need a different tool, and it turns out to be the Pythagorean theorem with a correction term.

Why it works

Place $C$ at the origin with side $b$ along the $x$-axis: $C(0, 0)$, $A(b, 0)$. Point $B$ is $a$ units from $C$ at angle $C$, so $B = (a \cos C, a \sin C)$. By the distance formula, $$c^2 = (a \cos C - b)^2 + (a \sin C)^2.$$ Expanding, $$c^2 = a^2 \cos^2 C - 2ab \cos C + b^2 + a^2 \sin^2 C.$$ Since $\cos^2 C + \sin^2 C = 1$ (the Pythagorean theorem in a right triangle with hypotenuse 1), this becomes $c^2 = a^2 + b^2 - 2ab \cos C$. When $C = 90^\circ$, $\cos C = 0$ and it is exactly $a^2 + b^2 = c^2$.

Key idea

Law of Cosines: $c^2 = a^2 + b^2 - 2ab \cos C$, where $C$ is the angle between sides $a$ and $b$. Use it for SAS (to find the third side) and for SSS (solved for $\cos C$, to find an angle).
Drag $B$. When the angle at $C$ is $90^\circ$, $c^2 = a^2 + b^2$ exactly; when it is larger, $c$ grows past that; when smaller, $c$ shrinks.

Worked example: Two sides and the included angle

In $\triangle ABC$, $a = 5$, $b = 8$ and $m\angle C = 60^\circ$. Find $c$.
  1. $c^2 = 5^2 + 8^2 - 2(5)(8)\cos 60^\circ$.
  2. Multiply first: $2(5)(8)\cos 60^\circ = 80 \cdot \frac{1}{2} = 40$.
  3. $c^2 = 25 + 64 - 40 = 49$, so $c = 7$.

Answer: $c = 7$

Worked example: Three sides, find an angle

A triangle has sides $3$, $5$ and $7$. Find its largest angle.
  1. The largest angle is opposite the longest side, $7$. Solve for its cosine: $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$, so $\cos C = \frac{9 + 25 - 49}{30} = -\frac{15}{30} = -\frac{1}{2}$.
  2. A negative cosine means an obtuse angle: $C = \cos^{-1}\left(-\frac{1}{2}\right) = 120^\circ$.

Answer: $120^\circ$

Worked example: With a calculator

Two sides of a triangular lot are $40$ m and $55$ m, meeting at $95^\circ$. How long is the third side?
  1. $c^2 = 40^2 + 55^2 - 2(40)(55)\cos 95^\circ$, which is $1600 + 3025 - 4400(-0.0872)$, or about $4625 + 383.5 = 5008.5$.
  2. $c \approx \sqrt{5008.5} \approx 70.8$ m.

Answer: About $70.8$ m

Common mistake

Combining $a^2 + b^2 - 2ab$ first and then multiplying by $\cos C$. Order of operations says multiply $2ab \cos C$ first, then subtract. In the first example, $(25 + 64 - 80) \cdot \frac{1}{2} = 4.5$ is wrong; $25 + 64 - 40 = 49$ is right. Check: the answer should be close to $a^2 + b^2$ when $C$ is near $90^\circ$.

Practice

Pass with 3 right answers, the last 2 in a row. Each right answer earns 3 XP.

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Sources and further reading

  • OpenStax Algebra and Trigonometry 2e, 10.2 Non-right Triangles: Law of Cosines
  • Common Core State Standards for Mathematics, High School Geometry, G-SRT.D.10 and G-SRT.D.11
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