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Difference between revisions of "Law of Cosines"

(→‎Acute Triangle: Picture)
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==Proofs==
 
==Proofs==
 
===Acute Triangle===
 
===Acute Triangle===
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<asy>picture+pic;
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<asy>
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pair+A,B,C,D,E;
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pair A,B,C,D,E;
 
C=(30,70);
 
C=(30,70);
 
B=(0,0);
 
B=(0,0);
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label("f",(30,35),(1,0));
 
label("f",(30,35),(1,0));
 
label("d",(15,0),(0,-1));
 
label("d",(15,0),(0,-1));
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label("e",(50,0),(0,-1.5));</asy>
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label("e",(50,0),(0,-1.5));
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</asy>
 
Let <math>a</math>, <math>b</math>, and <math>c</math> be the side lengths, <math>C</math> is the angle measure opposite side <math>c</math>, <math>f</math> is the distance from angle <math>C</math> to side <math>c</math>, and <math>d</math> and <math>e</math> are the lengths that <math>c</math> is split into by <math>f</math>.
 
Let <math>a</math>, <math>b</math>, and <math>c</math> be the side lengths, <math>C</math> is the angle measure opposite side <math>c</math>, <math>f</math> is the distance from angle <math>C</math> to side <math>c</math>, and <math>d</math> and <math>e</math> are the lengths that <math>c</math> is split into by <math>f</math>.
  

Revision as of 14:43, 7 October 2007

This is an AoPSWiki Word of the Week for Oct 4-Oct 10

The Law of Cosines is a theorem which relates the side-lengths and angles of a triangle. 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

See also

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