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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.

Contents

Theorem

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.

Proofs

Acute Triangle

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)...


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:

a^2+b^2-2f^2=d^2+e^2

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}*\dfrac{f}{b}-\dfrac{d}{a}*\dfrac{e}{b}=\dfrac{f^2-de}{ab}

We multiply by -2ab and get:

2de-2f^2=-2ab\cos{C}

Now remember our equation?

a^2+b^2-2f^2+2de=c^2

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

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

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.

Problems

Introductory

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{(...

(Source)

Intermediate

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.)

(Source)

Olympiad

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.

(Source)

See Also

Looking for a challenging algebra text? Preparing for MATHCOUNTS or the AMC exams?
Check out Art of Problem Solving's Introduction to Algebra by Richard Rusczyk.
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