Math Core

The law of sines needs a side and the angle across from it. When you know two sides and the angle between them (SAS), or all three sides (SSS), you have no such pair. The law of cosines handles both cases. It is the Pythagorean theorem, upgraded to work in every triangle.

From Pythagoras to every triangle

In a right triangle with the right angle at $C$ , you know $c^2 = a^2 + b^2$ . If you open angle $C$ wider than $90^\circ$ , side $c$ gets longer than the Pythagorean theorem predicts. If you close it tighter, side $c$ gets shorter. The law of cosines measures that correction exactly.

Place $C$ at the origin with side $b$ along the positive $x$ -axis, so $A = (b, 0)$ . Vertex $B$ is a distance $a$ from the origin at angle $C$ , so $B = (a \cos C,\ a \sin C)$ . Side $c$ is the distance from $A$ to $B$ :

\begin{aligned} c^2 &= (a \cos C - b)^2 + (a \sin C)^2 \\ &= a^2 \cos^2 C - 2ab \cos C + b^2 + a^2 \sin^2 C \\ &= a^2(\cos^2 C + \sin^2 C) + b^2 - 2ab \cos C \\ &= a^2 + b^2 - 2ab \cos C. \end{aligned}

The last step uses the Pythagorean identity $\sin^2 C + \cos^2 C = 1$ .

The law of cosines

In any triangle $ABC$ ,

\begin{aligned} a^2 &= b^2 + c^2 - 2bc \cos A \\ b^2 &= a^2 + c^2 - 2ac \cos B \\ c^2 &= a^2 + b^2 - 2ab \cos C \end{aligned}

Each version starts with the side across from the angle in the cosine. Solved for the angle:

\cos A = \frac{b^2 + c^2 - a^2}{2bc}.

The correction term $-2ab\cos C$ behaves just as the picture suggests. If $C = 90^\circ$ , then $\cos C = 0$ and you get the Pythagorean theorem back. If $C$ is obtuse, $\cos C$ is negative, so the term adds length. If $C$ is acute, $\cos C$ is positive and the term takes length away.

SAS: find the third side

Worked example: Two sides and the included angle

In triangle $ABC$ , $b = 8$ , $c = 11$ and $A = 52^\circ$ . Find $a$ to the nearest tenth.

Two sides and the included angle (SAS).

The unknown side $a$ is across from the known angle $A$ :

\begin{aligned} a^2 &= 8^2 + 11^2 - 2(8)(11)\cos 52^\circ \\ &= 185 - 176 \cos 52^\circ \\ &\approx 185 - 108.36 = 76.64 \end{aligned}

So $a \approx \sqrt{76.64} \approx 8.8$ .

Common mistake

Follow the order of operations. In $185 - 176\cos 52^\circ$ , multiply $176$ by $\cos 52^\circ$ first, then subtract. Computing $(185 - 176)\cos 52^\circ$ is a very common calculator slip. And don't forget the final square root: the formula gives $a^2$ , not $a$ .

SSS: find an angle

When all three sides are known, use the solved form to get the cosine of any angle. Because $\cos^{-1}$ returns angles from $0^\circ$ to $180^\circ$ , it correctly reports obtuse angles. Unlike the law of sines, there is no ambiguity.

Worked example: Three sides

A triangle has sides $a = 7$ , $b = 9$ and $c = 12$ . Find its largest angle to the nearest tenth of a degree.

Three sides known (SSS). The largest angle, C, is across from the longest side.

The largest angle is across from the longest side, so find $C$ :

\cos C = \frac{7^2 + 9^2 - 12^2}{2(7)(9)} = \frac{49 + 81 - 144}{126} = \frac{-14}{126} \approx -0.1111.

So $C = \cos^{-1}(-0.1111) \approx 96.4^\circ$ . The negative cosine told you right away that $C$ is obtuse.

Tip

The numerator $a^2 + b^2 - c^2$ tells you the type of triangle before you compute anything. If $c$ is the longest side: $c^2 < a^2 + b^2$ means acute, $c^2 = a^2 + b^2$ means right, and $c^2 > a^2 + b^2$ means obtuse.

Solving the whole triangle

After the law of cosines gives you a complete pair, you can switch to the law of sines for the rest. To stay safe, use the law of sines for the smallest remaining angle. The smallest angle of a triangle is always acute, so the calculator's $\sin^{-1}$ value is guaranteed to be correct. Then get the last angle from the angle sum.

Worked example: SAS, all the way

Solve the triangle with $a = 10$ , $b = 14$ and $C = 100^\circ$ . Round to the nearest tenth.

Side $c$ .

c^2 = 10^2 + 14^2 - 2(10)(14)\cos 100^\circ = 296 - 280\cos 100^\circ \approx 296 + 48.62 = 344.62,

so $c \approx 18.6$ . (Since $\cos 100^\circ$ is negative, the term added length.)

Angle $A$ . Side $a = 10$ is the shortest, so $A$ is the smallest angle and must be acute:

\sin A = \frac{10 \sin 100^\circ}{c} \approx \frac{9.848}{18.564} \approx 0.5305, \qquad A \approx 32.0^\circ.

Angle $B$ . $B \approx 180^\circ - 100^\circ - 32.0^\circ = 48.0^\circ$ .

Distances in the real world

Worked example: Two hikers

Two trails leave a trailhead with a $38^\circ$ angle between them. One hiker walks $4.2$ km along the first trail and another walks $5.5$ km along the second. How far apart are they, to the nearest tenth of a kilometer?

The two walks and the $38^\circ$ angle between them are SAS:

d^2 = 4.2^2 + 5.5^2 - 2(4.2)(5.5)\cos 38^\circ \approx 17.64 + 30.25 - 36.41 = 11.48,

so $d \approx 3.4$ km.

Choosing a law

You know	Start with
AAS or ASA	law of sines
SSA	law of sines (check the ambiguous case)
SAS	law of cosines for the third side
SSS	law of cosines for an angle (the largest is a good first choice)

Practice

Practice 1

In triangle $ABC$ , $a = 5$ , $b = 7$ and $C = 60^\circ$ . Find $c$ to the nearest tenth.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 2

A triangle has sides $a = 5$ , $b = 7$ and $c = 8$ . Find angle $B$ , in degrees.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 3

In triangle $ABC$ , $b = 6$ , $c = 9$ and $A = 120^\circ$ . Find $a$ to the nearest tenth.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 4

A triangle has sides $4$ , $6$ and $9$ . Find its largest angle to the nearest tenth of a degree.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 5

In triangle $ABC$ you know $b$ , $c$ and angle $A$ . Which equation finds side $a$ ?

Practice 6

A triangle has sides $6$ , $7$ and $10$ . What kind of triangle is it?

Practice 7

A parallelogram has sides of length $8$ and $12$ , and one of its angles is $65^\circ$ . Find the length of the longer diagonal to the nearest tenth.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 8

A plane flies $150$ miles in a straight line, then turns $35^\circ$ to the right and flies another $90$ miles. How far is it from its starting point, to the nearest mile?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.