The Law of Cosines

Q: What is the Law of Cosines?

c² = a² + b² − 2ab cos C, where C is the angle between sides a and b and c is the side opposite it.

Q: When do I use the Law of Cosines instead of the Law of Sines?

When you know two sides and the angle between them (SAS), or all three sides (SSS). In both cases no side is paired with its opposite angle.

Q: How is it related to the Pythagorean theorem?

When C is 90°, cos C = 0 and the correction term disappears, leaving c² = a² + b².

Q: How do I find an angle from three sides?

Rearrange to cos C = (a² + b² − c²)/(2ab), then take the inverse cosine. The answer is unambiguous, because cosine separates acute from obtuse angles.

Q: Why is there no ambiguous case with cosine?

Cosine is positive for acute angles and negative for obtuse ones, so each value between −1 and 1 belongs to exactly one angle between 0° and 180°.

Quick answer

When two sides and the angle between them are known, or all three sides, the Law of Sines cannot start, because no side comes with its opposite angle. The Law of Cosines handles both: c² = a² + b² − 2ab cos C. It is the Pythagorean theorem plus a correction that vanishes at a right angle, shrinks the third side for an acute angle and stretches it for an obtuse one.

What you'll learn

Prove the Law of Cosines using coordinates
Find the third side from two sides and the included angle
Find an angle from three sides, and choose between the two laws

When the Law of Sines cannot start

The Law of Sines needs a side together with the angle across from it. Two common situations never supply that pair:

SAS — two sides and the angle between them
SSS — all three sides

For both, there is a second law:

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

Here $C$ is the angle between sides $a$ and $b$ , and $c$ is the side across from it. The same pattern holds for any angle: $a^2 = b^2 + c^2 - 2bc\cos A$ , and so on.

The Pythagorean theorem, corrected

Look at the formula next to $c^2 = a^2 + b^2$ . It is the Pythagorean theorem with one extra term, $-2ab\cos C$ , and that term depends only on the angle.

Angle $C$	$\cos C$	Effect on $c$
$90°$	$0$	no correction: $c^2 = a^2 + b^2$
acute	positive	subtracts, so $c$ is shorter
obtuse	negative	adds, so $c$ is longer

That matches what a hinge does. Close the angle between two fixed sides and the far ends move together. Open it and they spread apart.

Why the correction is 2ab cos C

Put the triangle on coordinates. Place $C$ at the origin and $B$ on the positive $x$ -axis, so $B = (a, 0)$ . Point $A$ is $b$ units from the origin at angle $C$ , so by the unit circle definition it sits at

A = (b\cos C,\; b\sin C)

The Law of Cosines on coordinates

Now $c$ is the distance from $A$ to $B$ . By the distance formula:

c^2 = (b\cos C - a)^2 + (b\sin C)^2

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

Group the two terms containing $b^2$ :

c^2 = a^2 + b^2\left(\cos^2 C + \sin^2 C\right) - 2ab\cos C

The Pythagorean identity says the bracket is exactly $1$ :

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

Nothing in the argument needed $C$ to be acute. For an obtuse angle, $A$ lands to the left of the $y$ -axis, $\cos C$ is negative, and every step goes through the same.

SAS: finding the third side

Two sides $a = 7$ and $b = 10$ meet at $C = 50°$ . Find $c$ .

c^2 = 7^2 + 10^2 - 2(7)(10)\cos 50° = 149 - 140\cos 50°

c^2 \approx 149 - 89.99 = 59.01 \qquad c \approx 7.68

Multiply $140\cos 50°$ before subtracting. That product is one term.

SSS: finding an angle

Rearrange the law to isolate the cosine:

\cos C = \frac{a^2 + b^2 - c^2}{2ab}

A triangle has sides $5$ , $8$ and $7$ . Find the angle opposite the side of length $7$ .

\cos C = \frac{25 + 64 - 49}{2(5)(8)} = \frac{40}{80} = \frac{1}{2} \qquad C = 60°

Unlike sine, cosine never leaves two choices. It is positive for acute angles and negative for obtuse ones, so each value belongs to exactly one angle between $0°$ and $180°$ .

Which law to use

Known	Start with
two angles and a side (AAS, ASA)	Law of Sines
two sides and a non-included angle (SSA)	Law of Sines, checking for two triangles
two sides and the included angle (SAS)	Law of Cosines
three sides (SSS)	Law of Cosines

After the Law of Cosines supplies one missing piece, either law can finish the triangle.

Worked examples

Common mistakes

Practice problems

Find $c$ when $a = 5$ , $b = 8$ and $C = 60°$ .

Answer

$7$

Full solution

$c^2 = 25 + 64 - 80\cos 60° = 89 - 40 = 49$ , so $c = 7$ .
A triangle has sides $3$ , $5$ and $7$ . Find the angle opposite the side of length $5$ .

Answer

About $38.21°$

Full solution

$\cos B = \tfrac{9 + 49 - 25}{2(3)(7)} = \tfrac{33}{42} \approx 0.7857$ , so $B \approx 38.21°$ .
Find $c$ when $a = 12$ , $b = 15$ and $C = 110°$ .

Answer

About $22.18$

Full solution

$c^2 = 144 + 225 - 360\cos 110° \approx 369 + 123.13 = 492.13$ , so $c \approx 22.18$ .
A triangle has sides $7$ , $8$ and $9$ . Find its largest angle.

Answer

About $73.40°$

Full solution

The largest angle is across from $9$ : $\cos C = \tfrac{49 + 64 - 81}{2(7)(8)} = \tfrac{32}{112} \approx 0.2857$ , so $C \approx 73.40°$ .
Two hikers leave a trailhead on paths $40°$ apart. One walks $3$ miles and the other $5$ miles. How far apart are they?

Answer

About $3.32$ miles

Full solution

$d^2 = 9 + 25 - 30\cos 40° \approx 34 - 22.98 = 11.02$ , so $d \approx 3.32$ miles.
For each set of given parts, name the law to start with: (a) $A$ , $B$ , $c$ ; (b) $a$ , $b$ , $C$ ; (c) $a$ , $b$ , $c$ ; (d) $a$ , $b$ , $A$ .

Answer

(a) Sines. (b) Cosines. (c) Cosines. (d) Sines, checking for two triangles.

Full solution

(a) Two angles give the third, and then every side has its opposite angle.

(b) Two sides and the included angle: SAS.

(c) Three sides: SSS.

(d) Side $a$ comes with its opposite angle $A$ , but this is SSA, which can give two triangles.
Explain why an obtuse angle $C$ makes $c^2$ larger than $a^2 + b^2$ .

Answer

For an obtuse angle $\cos C < 0$ , so $-2ab\cos C$ is positive and adds to $a^2 + b^2$ .

Full solution

$a$ and $b$ are positive lengths, so the sign of $-2ab\cos C$ is the opposite of the sign of $\cos C$ .

An obtuse angle has a negative cosine, which makes the correction positive. The side across from an obtuse angle is longer than the hypotenuse of a right triangle with the same two legs.
Can a triangle have sides $4$ , $5$ and $10$ ? Use the Law of Cosines to decide.

Answer

No.

Full solution

The angle across from $10$ would need $\cos C = \tfrac{16 + 25 - 100}{40} = -1.475$ .

No angle has a cosine below $-1$ , so no such triangle exists. This is the triangle inequality in another form: $4 + 5$ is less than $10$ .
Use the coordinate proof to explain where the term $b^2$ in $a^2 + b^2$ comes from.

Answer

From $b^2\cos^2 C + b^2\sin^2 C$ , which the Pythagorean identity collapses to $b^2$ .

Full solution

Expanding the distance from $A(b\cos C, b\sin C)$ to $B(a, 0)$ produces $b^2\cos^2 C$ from the horizontal part and $b^2\sin^2 C$ from the vertical part.

Together they are $b^2(\cos^2 C + \sin^2 C) = b^2$ .
With $a = 7$ , $b = 10$ and $C = 50°$ , Theo computes $c^2 = (49 + 100 - 140)\cos 50°$ and gets $c \approx 2.41$ . Find his error.

Hint

Which operation comes first in $149 - 140\cos 50°$ ?

Answer

He subtracted before multiplying. The correct value is $c \approx 7.68$ .

Full solution

In $a^2 + b^2 - 2ab\cos C$ , the cosine multiplies only $2ab$ . So $140\cos 50° \approx 89.99$ is computed first, then subtracted from $149$ .

$c^2 \approx 149 - 89.99 = 59.01$ , so $c \approx 7.68$ .

A sanity check catches Theo’s answer. With a $50°$ angle between sides of $7$ and $10$ , the third side cannot be as short as $2.41$ — the ends of the two sides are much farther apart than that.

Frequently asked questions

What is the Law of Cosines?

c² = a² + b² − 2ab cos C, where C is the angle between sides a and b and c is the side opposite it.

When do I use the Law of Cosines instead of the Law of Sines?

When you know two sides and the angle between them (SAS), or all three sides (SSS). In both cases no side is paired with its opposite angle.

How is it related to the Pythagorean theorem?

When C is 90°, cos C = 0 and the correction term disappears, leaving c² = a² + b².

How do I find an angle from three sides?

Rearrange to cos C = (a² + b² − c²)/(2ab), then take the inverse cosine. The answer is unambiguous, because cosine separates acute from obtuse angles.

Why is there no ambiguous case with cosine?

Cosine is positive for acute angles and negative for obtuse ones, so each value between −1 and 1 belongs to exactly one angle between 0° and 180°.

What to learn next

The Dot Product and the Angle Between VectorsMultiply two vectors into a number: u · v = u₁v₁ + u₂v₂ equals |u||v|cos θ, which measures the angle between them, tests for right angles and gives projections.

Standards alignment

This lesson covers the following Common Core State Standards for Mathematics.

CCSS.MATH.CONTENT.HSG.SRT.D.10Similarity, Right Triangles, and Trigonometry(+) Prove the Laws of Sines and Cosines and use them to solve problems.
CCSS.MATH.CONTENT.HSG.SRT.D.11Similarity, Right Triangles, and Trigonometry(+) Understand and apply the Law of Sines and the Law of Cosines to find unknown measurements in right and non-right triangles (e.g., surveying problems, resultant forces).

For teachersPrint problems onlyTick to print a blank worksheet; leave it clear to print the lesson with the full answer key.

Quick answer

What you'll learn

Frequently asked questions

What is the Law of Cosines?

When do I use the Law of Cosines instead of the Law of Sines?

How is it related to the Pythagorean theorem?

How do I find an angle from three sides?

Why is there no ambiguous case with cosine?

What to learn next

Standards alignment

More Geometry lessons