Difference between revisions of "Law of Cosines"

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The '''Law of Cosines''' is a theorem which relates the side-[[length]]s and [[angle]]s of a [[triangle]]. For a triangle with [[edge]]s of length <math>a</math>, <math>b</math> and <math>c</math> opposite [[angle]]s of measure <math>A</math>, <math>B</math> and <math>C</math>, respectively, the Law of Cosines states:
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The '''Law of Cosines''' is a theorem which relates the side-[[length]]s and [[angle]]s of a [[triangle]]. It can be derived in several different ways, the most common of which are listed in the "proofs" section below.
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==Theorem==
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For a triangle with [[edge]]s of length <math>a</math>, <math>b</math> and <math>c</math> opposite [[angle]]s of measure <math>A</math>, <math>B</math> and <math>C</math>, respectively, the Law of Cosines states:
  
 
<math>c^2 = a^2 + b^2 - 2ab\cos C</math>
 
<math>c^2 = a^2 + b^2 - 2ab\cos C</math>
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The argument for an obtuse triangle is the same as the proof for an acute triangle.
 
The argument for an obtuse triangle is the same as the proof for an acute triangle.
  
==See also==
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==Problems==
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===Introductory===
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If the sides of a triangle have lengths 2, 3, and 4, what is the radius of the circle circumscribing the triangle?
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<center><math>
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\mathrm{(A) \ } 2
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\qquad \mathrm{(B) \ } 8/\sqrt{15}
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\qquad \mathrm{(C) \ } 5/2
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\qquad \mathrm{(D) \ } \sqrt{6}
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\qquad \mathrm{(E) \ }  (\sqrt{6} + 1)/2
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</math></center>
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([[University of South Carolina High School Math Contest/1993 Exam/Problem 29|Source]])
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===Intermediate===
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A tripod has three legs each of length <math>5</math> 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 <math>4</math> feet from the ground. In setting up the tripod, the lower 1 foot of one leg breaks off. Let <math> h </math> be the height in feet of the top of the tripod from the ground when the broken tripod is set up. Then <math> h </math> can be written in the form <math> \frac m{\sqrt{n}}, </math> where <math> m </math> and <math> n </math> are positive integers and <math> n </math> is not divisible by the square of any prime. Find <math> \lfloor m+\sqrt{n}\rfloor. </math> (The notation <math> \lfloor x\rfloor </math> denotes the greatest integer that is less than or equal to <math> x. </math>)
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([[2006 AIME I Problems/Problem 14|Source]])
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===Olympiad===
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A tetrahedron <math>ABCD </math> is inscribed in the sphere <math>S </math>.  Find the locus of points <math>P </math>, situated in <math>S </math>, such that
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<center>
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<math> \frac{AP}{PA_{1}} + \frac{BP}{PB_{1}} + \frac{CP}{PC_{1}} + \frac{DP}{PD_{1}} = 4, </math>
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</center>
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where <math>A_{1}, B_{1}, C_{1}, D_{1} </math> are the other intersection points of <math>AP, BP, CP, DP </math> with <math>S </math>.
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([[1973 IMO Shortlist Problems/Bulgaria 1|Source]]
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==See Also==
 
* [[Law of Sines]]
 
* [[Law of Sines]]
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* [[Law of Tangents]]
 
* [[Trigonometry]]
 
* [[Trigonometry]]
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[[Category:Theorems]]
 
[[Category:Trigonometry]]
 
[[Category:Trigonometry]]

Revision as of 22:04, 31 January 2008

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.

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

[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:

\[a^2+b^2-2f^2=d^2+e^2\]

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?

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

(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

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