Math Core

The law of sines needs a side paired with its opposite angle. But what if you know two sides and the angle between them (SAS), or all three sides (SSS)? Then there's no complete pair, and you need a different tool: the law of cosines, which is the Pythagorean theorem upgraded to work for every triangle.

The formula

The law of cosines

In any triangle $ABC$ ,

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

The same pattern works for each side:

$a^2 = b^2 + c^2 - 2bc \cos A \qquad b^2 = a^2 + c^2 - 2ac \cos B$

The pattern to remember: the side on the left and the angle at the end are opposite each other, and the other two sides appear everywhere else.

Why it works

Put the triangle on a coordinate plane with $C$ at the origin and side $a$ along the positive $x$ -axis, so $B = (a, 0)$ . Point $A$ is a distance $b$ from the origin, at angle $C$ above the axis. By the definitions of sine and cosine, its coordinates are $(b \cos C, \; b \sin C)$ .

Place C at the origin with side a along the x-axis.

Now $c$ is the distance from $A$ to $B$ . Use the distance formula (which is itself the Pythagorean theorem):

\begin{aligned} c^2 &= (b \cos C - a)^2 + (b \sin C - 0)^2 \\ &= b^2 \cos^2 C - 2ab \cos C + a^2 + b^2 \sin^2 C \\ &= b^2(\cos^2 C + \sin^2 C) + a^2 - 2ab \cos C \\ &= a^2 + b^2 - 2ab \cos C \end{aligned}

The last step uses $\sin^2 C + \cos^2 C = 1$ from the sine and cosine lesson. (Here $\cos^2 C$ is shorthand for $(\cos C)^2$ .)

Connection to the Pythagorean theorem

If $C = 90^\circ$ , then $\cos C = 0$ and the formula becomes $c^2 = a^2 + b^2$ : the Pythagorean theorem is the special case. The term $-2ab \cos C$ is the correction for a triangle that isn't right.

If $C$ is acute, $\cos C > 0$ , so you subtract, and $c^2 < a^2 + b^2$ .
If $C$ is obtuse, $\cos C < 0$ (your calculator will show a negative number), so you add, and $c^2 > a^2 + b^2$ .

That's exactly the acute/obtuse test from the first lesson of this unit, now with a reason behind it.

SAS: finding the third side

Worked example: Two sides and the included angle

In triangle $ABC$ , $a = 8$ , $b = 11$ and $C = 40^\circ$ . Find $c$ to the nearest tenth.

\begin{aligned} c^2 &= 8^2 + 11^2 - 2(8)(11)\cos 40^\circ \\ &= 64 + 121 - 176 \cos 40^\circ \\ &\approx 185 - 134.82 = 50.18 \\ c &\approx \sqrt{50.18} \approx 7.1 \end{aligned}

Worked example: Two ships

Two ships leave the same port. One sails $30$ miles on a straight course, and the other sails $50$ miles on a course that makes a $120^\circ$ angle with the first. How far apart are the ships?

\begin{aligned} d^2 &= 30^2 + 50^2 - 2(30)(50)\cos 120^\circ \\ &= 900 + 2500 - 3000(-0.5) \\ &= 3400 + 1500 = 4900 \\ d &= 70 \text{ miles} \end{aligned}

Because $120^\circ$ is obtuse, its cosine is negative, and the ships end up farther apart than the Pythagorean theorem would suggest ( $\sqrt{3400} \approx 58.3$ ).

SSS: finding an angle

Solve the formula for the cosine:

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

Then use $\cos^{-1}$ . Unlike $\sin^{-1}$ , the inverse cosine gives angles all the way from $0^\circ$ to $180^\circ$ , so a negative cosine correctly gives an obtuse angle. There's no ambiguous case.

Worked example: All three sides

A triangle has sides $a = 7$ , $b = 9$ and $c = 13$ . Find the largest angle, to the nearest tenth of a degree.

The largest angle is opposite the longest side, so find $C$ .

$\cos C = \frac{7^2 + 9^2 - 13^2}{2(7)(9)} = \frac{49 + 81 - 169}{126} = \frac{-39}{126} \approx -0.3095$

$C = \cos^{-1}(-0.3095) \approx 108.0^\circ$

The negative cosine told us right away that the triangle is obtuse.

Tip

When solving an SSS triangle completely, find the largest angle first with the law of cosines. The other two angles are then guaranteed to be acute, so you can safely finish with the law of sines (or the law of cosines again) without worrying about the ambiguous case.

Common mistake

Evaluate in the right order. In $185 - 176 \cos 40^\circ$ , multiply $176 \cos 40^\circ$ first, then subtract. A common mistake is to compute $185 - 176 = 9$ and then multiply by $\cos 40^\circ$ . If your calculator allows it, type the whole expression at once.

Which law should I use?

You know	Use first
a right triangle	trig ratios and the Pythagorean theorem
AAS or ASA	law of sines
SSA	law of sines (check the ambiguous case)
SAS	law of cosines
SSS	law of cosines