Law of Cosines

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The Law of Cosines is a theorem which relates the side-lengths and angles of a triangle. It can be derived in several different ways, the most common of which are listed in the "proofs" section below. It can be used to derive the third side given two sides and the included angle. All triangles with two sides and an include angle are congruent by the Side-Angle-Side congruence postulate.


For a triangle with edges of length $a$, $b$ and $c$ opposite angles of measure $A$, $B$ and $C$, respectively, the Law of Cosines states:

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

\[b^2 = a^2 + c^2 - 2ac\cos B\]

\[a^2 = b^2 + c^2 - 2bc\cos A\]

In the case that one of the angles has measure $90^\circ$ (is a right angle), the corresponding statement reduces to the Pythagorean Theorem.


Proof 1

Acute Triangle

[asy] pair A,B,C,D,E; C=(30,70); B=(0,0); A=(100,0); D=(30,0); size(100); draw(B--A--C--B); draw(C--D); label("A",A,(1,0)); dot(A); label("B",B,(-1,-1)); dot(B); label("C",C,(0,1)); dot(C); draw(D--(30,4)--(34,4)--(34,0)--D); label("f",(30,35),(1,0)); label("d",(15,0),(0,-1)); label("e",(50,0),(0,-1.5)); [/asy]

Let $a$, $b$, and $c$ be the side lengths, $C$ is the angle measure opposite side $c$, $f$ is the distance from angle $C$ to side $c$, and $d$ and $e$ are the lengths that $c$ is split into by $f$.

We use the Pythagorean theorem:


We are trying to get $a^2+b^2-2f^2+2de$ on the LHS, because then the RHS would be $c^2$.

We use the addition rule for cosines and get:

\[\cos{C}=\dfrac{f}{a}\cdot \dfrac{f}{b}-\dfrac{d}{a}\cdot \dfrac{e}{b}=\dfrac{f^2-de}{ab}\]

We multiply by $-2ab$ and get:


Now remember our equation?


We replace the $-2f^2+2de$ by $-2ab\cos{C}$ and get:


We can use the same argument on the other sides.

Right Triangle

Since $C=90^{\circ}$, $\cos C=0$, so the expression reduces to the Pythagorean Theorem. You can find several proofs of the Pythagorean Theorem here.

Obtuse Triangle

The argument for an obtuse triangle is the same as the proof for an acute triangle.

Proof 2 (Vector Dot Product)

Consider $\triangle{ABC}$. Let $\vec{AB}=\vec{c}, \vec{AC}=\vec{b},\vec{BC}=\vec{a}$.

Because of the identity $|\vec{a}|^2=\vec{a}\cdot\vec{a}$,we can complete our proof as the following.

Draw the diagram. Note that $\vec{c}+\vec{a}=\vec{b}$. Then $\vec{b}-\vec{c}=\vec{a}$ and $\vec{a}\cdot \vec{a}=a^2$. $(\vec{b}-\vec{c})^2=b^2+c^2-2\cdot b\cdot c\cdot \cos{A}=|\vec{a}|^2$. Now, we have finished the proof because the two quantities are equal.

Credits to China High School Math textbook $\emph{Mathematics Vol 5B Textbook}$ by People's Education Press (this textbook is currently discontinued but it has been used for hinting the proof. The proof is done by myself. But the letting and the process of guiding students to verify the identity $|\vec{a}|^2=\vec{a}\cdot\vec{a}$ is written in the textbook.




1. If the sides of a triangle have lengths 2, 3, and 4, what is the radius of the circle circumscribing the triangle?

$\mathrm{(A) \ } 2 \qquad \mathrm{(B) \ } 8/\sqrt{15}  \qquad \mathrm{(C) \ } 5/2 \qquad \mathrm{(D) \ } \sqrt{6} \qquad \mathrm{(E) \ }  (\sqrt{6} + 1)/2$


2. In the quadrilateral $ABCD$, $\angle{ADC}=90^\circ$, $AB=2$, $BD=5$.

(1) Find $\cos{\angle{ADB}}$.

(2) If $DC=2\sqrt{2}$, find $BC$.

(2018 China Gaokao Syllabus I #17)

Solution link: P.S.: Since the solution is on a forum, please read all the way to thread #3 for the solutions!


A tripod has three legs each of length $5$ feet. When the tripod is set up, the angle between any pair of legs is equal to the angle between any other pair, and the top of the tripod is $4$ feet from the ground. In setting up the tripod, the lower 1 foot of one leg breaks off. Let $h$ be the height in feet of the top of the tripod from the ground when the broken tripod is set up. Then $h$ can be written in the form $\frac m{\sqrt{n}},$ where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $\lfloor m+\sqrt{n}\rfloor.$ (The notation $\lfloor x\rfloor$ denotes the greatest integer that is less than or equal to $x.$)



A tetrahedron $ABCD$ is inscribed in the sphere $S$. Find the locus of points $P$, situated in $S$, such that

$\frac{AP}{PA_{1}} + \frac{BP}{PB_{1}} + \frac{CP}{PC_{1}} + \frac{DP}{PD_{1}} = 4,$

where $A_{1}, B_{1}, C_{1}, D_{1}$ are the other intersection points of $AP, BP, CP, DP$ with $S$.


See Also