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Law of Cosines

Law of Cosines

Lesson · Intermediate

Geometry: Plane Geometry

Triangle ABC with sides a, b, and c

For a triangle with sides $a,b,c$ :

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

or

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

Consider the Unit Circle in Trigonometry

Let

BC=a,\qquad CA=b,\qquad AB=c

Draw the altitude

AD\perp BC

and let

BD=x,\qquad DC=a-x,\qquad AD=h

Triangle ABC with altitude AD

By the Pythagorean Theorem in $\triangle ABD$ ,

c^2=x^2+h^2

In $\triangle ACD$ ,

b^2=(a-x)^2+h^2

Subtract the first equation from the second:

b^2-c^2=(a-x)^2-x^2

Expanding,

b^2-c^2=a^2-2ax+x^2-x^2

so

b^2-c^2=a^2-2ax

Hence

2ax=a^2+c^2-b^2

and therefore

x=\frac{a^2+c^2-b^2}{2a}

Now, in the right triangle $ABD$ ,

\cos B=\frac{BD}{AB}=\frac{x}{c}

Substituting the value of $x$ ,

\cos B=\frac{a^2+c^2-b^2}{2ac}

Therefore,

\boxed{\cos B=\frac{a^2+c^2-b^2}{2ac}}

Rearranging,

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

so

\boxed{b^2=a^2+c^2-2ac\cos B}

Similarly, by relabeling the sides and angles,

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

and hence

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

Likewise,

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

Triangle $ABC$ has

AB=13,\qquad AC=15,\qquad BC=14

If $M$ is the midpoint of $BC$ , find $AM$ .

Since $M$ is the midpoint,

BM=MC=7

Let

AM=m,\qquad \angle AMB=\theta

Triangle ABC with median AM

Then

\angle AMC=180^\circ-\theta

Apply the Law of Cosines to $\triangle ABM$ :

13^2=m^2+7^2-14m\cos\theta

So

169=m^2+49-14m\cos\theta

Apply the Law of Cosines to $\triangle ACM$ :

15^2=m^2+7^2-14m\cos(180^\circ-\theta)

Since

\cos(180^\circ-\theta)=-\cos\theta

we get

225=m^2+49+14m\cos\theta

Add the two equations:

169+225=2m^2+98

Thus

\begin{aligned} 394&=2m^2+98\\ 2m^2&=296\\ m^2&=148 \end{aligned}

Therefore,

\boxed{AM=2\sqrt{37}}

A triangle has side lengths

x,\qquad x+1,\qquad x+2

For which positive $x$ is the largest angle obtuse?

The largest angle is opposite $x+2$ .

It is obtuse exactly when

(x+2)^2>x^2+(x+1)^2

Expanding,

x^2+4x+4>2x^2+2x+1

so

x^2-2x-3<0

Factor:

(x-3)(x+1)<0

Thus

-1<x<3

Since $x>0$ ,

\boxed{0<x<3}

If

A=90^\circ

then

\cos A=0

Therefore,

\boxed{a^2=b^2+c^2}

So the Pythagorean Theorem is a special case of the Law of Cosines.

Let $a$ be the largest side.

From

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

we obtain:

\boxed{ \begin{aligned} a^2&<b^2+c^2 &&\Rightarrow \text{ acute}\\ a^2&=b^2+c^2 &&\Rightarrow \text{ right}\\ a^2&>b^2+c^2 &&\Rightarrow \text{ obtuse} \end{aligned} }

If a parallelogram has sides $a,b$ and diagonals $p,q$ ,

\boxed{p^2+q^2=2a^2+2b^2}

Let

AB=CD=a,\qquad BC=AD=b

\angle DAB=\theta,\qquad \angle ABC=180^\circ-\theta

Parallelogram ABCD with diagonals p and q

By the Law of Cosines,

p^2=a^2+b^2-2ab\cos\theta

q^2=a^2+b^2-2ab\cos(180^\circ-\theta)

Since

\cos(180^\circ-\theta)=-\cos\theta

we have

q^2=a^2+b^2+2ab\cos\theta

Adding

p^2+q^2=2a^2+2b^2

Therefore,

\boxed{p^2+q^2=2a^2+2b^2}

AMC 12A 2024 (Problem 19)
AMC 12A 2022 (Problem 12)
AMC 12A 2025 (Problem 8)
AMC 12A 2025 (Problem 24)
AMC 12B 2025 (Problem 21)
AMC 12B 2025 (Problem 25)

Regular Polygons

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