Difference between revisions of "Law of Cosines"
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==Proofs== | ==Proofs== | ||
===Acute Triangle=== | ===Acute Triangle=== | ||
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{{image}} | {{image}} | ||
| + | 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>. | ||
| − | + | We use the Pythagorean theorem: | |
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| − | |||
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| − | We use the | ||
<cmath>a^2+b^2-2f^2=d^2+e^2</cmath> | <cmath>a^2+b^2-2f^2=d^2+e^2</cmath> | ||
Revision as of 14:10, 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 
, 
and 
opposite angles of measure 
, 
and 
, respectively, the Law of Cosines states:
In the case that one of the angles has measure 
(is a right angle), the corresponding statement reduces to the Pythagorean Theorem.
Proofs
Acute Triangle
An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.
Let , , and be the side lengths, is the angle measure opposite side , 
is the distance from angle to side , and 
and 
are the lengths that is split into by .
We use the Pythagorean theorem:
![\[a^2+b^2-2f^2=d^2+e^2\]](http://latex.artofproblemsolving.com/a/6/0/a600f283e1d0ec9352e0acfe76e17e8eeed3db53.png)
We are trying to get 
on the LHS, because then the RHS would be 
.
We use the addition rule for cosines and get:
![\[\cos{C}=\dfrac{f}{a}*\dfrac{f}{b}-\dfrac{d}{a}*\dfrac{e}{b}=\dfrac{f^2-de}{ab}\]](http://latex.artofproblemsolving.com/c/3/4/c3470947802c3947ed2be19390ced22eee737bc9.png)
We multiply by -2ab and get:
![\[2de-2f^2=-2ab\cos{C}\]](http://latex.artofproblemsolving.com/c/e/1/ce15598b4d95ea94c52106c092dfc81e921ce1a0.png)
Now remember our equation?
![\[a^2+b^2-2f^2+2de=c^2\]](http://latex.artofproblemsolving.com/1/d/6/1d6cea7b27e07bf64f79ab8626eb9a43c06b7bb1.png)
We replace the 
by 
and get:
![\[c^2=a^2+b^2-2ab\cos{C}\]](http://latex.artofproblemsolving.com/c/5/4/c54e32f9760aee6e977d625b529959edba45a4d3.png)
We can use the same argument on the other sides.
Right Triangle
Since 
, 
, so the expression reduces to the Pythagorean Theorem. You can find several proofs of the Pythagorean Theorem here
