Bash and Effectiveness

The Law of Cosines is like the bigger, better brother to the Law of Sines. It is used quite often in computational contests (well, at least more than Law of Sines) and is key to bashing an unknown side in terms of a variable, finding a length directly, or finding the cosine of an angle. You can then use this derived length in whatever tools you need (e.g. similar triangles, pythagorean theorem, etc.).

Formula & Derivation

Let's lay out some notation first. In any triangle $\triangle ABC$, we label each side with the lowercase letter of the angle it sits across from, so side $a$ is opposite angle $A$, side $b$ is opposite $B$, and side $c$ is opposite $C$.

Each side is named after the angle it faces.

The Law of Cosines states:

There's nothing special about $C$; by symmetry, you get one version at each angle:

And if it's the angle you want rather than a side, just solve for the cosine:

Why "generalized Pythagorean theorem"? Set $C = 90^\circ$. Then $\cos C = 0$, the last term cancels to 0, and you're left with $c^2 = a^2 + b^2$. The $-2ab\cos C$ term is indeed the correction for the angle not being right. And note: if $C$ is obtuse then $\cos C < 0$, so the term adds to $c$, which is basically what you'd expect from opening the angle wider.

Derivation

The trick is the one that we have stressed many times: drop an altitude and split one annoying triangle into two right triangles, where trigonometry and Pythagoras actually apply.

Drop a perpendicular from $B$ onto side $b = CA$, call the foot $H$.

The altitude $h$ splits $\triangle ABC$ into two right triangles, $\triangle BHC$ and $\triangle BHA$.

Look at the right triangle $\triangle BHC$. Its hypotenuse is $a$, and the angle at $C$ is our angle $C$, so from trigonometry:

Since the full base $CA$ has length $b$, the leftover segment is:

Now apply the Pythagorean theorem to the other right triangle, $\triangle BHA$ — its legs are $AH$ and $h$, and its hypotenuse is $c$:

Expand:

Group the two $a^2$ terms and cancel them with $\sin^2 C + \cos^2 C = 1$:

(If $C$ is obtuse, $H$ lands outside segment $CA$ and $\cos C$ comes out negative, so the formula still holds.)

Techniques

The Law of Cosines is useful in a variety of ways, one obvious case being finding a missing side length given two side lengths and a known angle between them. However, there is quite some nuance in how it's used at more advanced levels of competition. For instance:

• You may get a side length in terms of a variable and use that new side length as new information to solve further. If this is unclear, consider the examples we present, and as always, practice!
• Solve for the cosine of the angle. You may have similar triangles, and knowing the cosine of that angle can be used in other triangles where that angle is also present.
• Noticing congruent, complementary, or supplementary angles and using how their cosines relate to the cosine of the original angle (e.g. $\cos(180^\circ - \theta) = -\cos\theta$).

Always match $a$ with $A$, $b$ with $B$, and $c$ with $C$. The side opposite to the angle is the one being solved for in $a^2=b^2+c^2-2bc\cos A$.

Elementary Examples

Example 1: Solving for a side

Find the missing side. In $\triangle ABC$, two sides of length $8$ and $5$ meet at a $60^\circ$ angle at vertex $C$. Find the length of the third side $c = AB$.

The $60^\circ$ angle is enclosed by the two known sides, sitting opposite the unknown side $c$ — exactly the configuration the Law of Cosines handles:

Hence $c = \sqrt{49} = 7$.

Example 2: Cosine of an angle in terms of a variable

Express the angle. In $\triangle ABC$, the side $BC$ has unknown length $x$, while $AB = 3$ and $AC = 5$. Write $\cos B$ as a function of $x$.

We want the angle at $B$. It is enclosed by the sides $AB = 3$ and $BC = x$, and it sits opposite the side $AC = 5$. Solving the Law of Cosines for the cosine gives what we want:

Worked Example

2026 AIME I · Problem 5. Points $A$ and $B$ lie in a plane with $AB = 1$. Rotate $A$ counterclockwise about $B$ through an acute angle $\theta$ to land at $A'$; then rotate $B$ clockwise about $A'$ through the same angle $\theta$ to land at $B'$. Given that $AB' = \tfrac{4}{3}$, write $\cos\theta = \tfrac{m}{n}$ in lowest terms and find $m+n$.

Solid: the parallelogram $BAA'B'$. Dashed: its diagonals, meeting at the midpoint $O$.

Rotation preserves distance to its center, so $BA' = BA = 1$, and then $A'B' = A'B = 1$. Hence three of the four connecting segments equal $1$, with $AB' = \tfrac{4}{3}$ given.

Because the two rotations turn $\vec{BA}$ into $\vec{B'A'}$, the points form a parallelogram $BAA'B'$ whose diagonals are $BA' = 1$ and $AB' = \tfrac{4}{3}$. Diagonals bisect each other at $O$, so

The rotation angle is exactly $\angle ABA' = \theta$, which is the angle at $B$ in $\triangle ABO$ (since $O$ lies on $BA'$). That triangle has sides $AB = 1$ and $BO = \tfrac{1}{2}$ enclosing $\theta$, opposite the side $AO = \tfrac{2}{3}$. Apply the Law of Cosines:

Therefore $\cos\theta = \tfrac{5}{4} - \tfrac{4}{9} = \tfrac{45 - 16}{36} = \tfrac{29}{36}$, giving $m + n = 29 + 36 = \boxed{65}$.

Practice Problems