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Difference between revisions of "Cosine theorem"

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The square of a side of a triangle is equal to the sum of the squares of the other two sides, minus double the product of the latter two sides and the cosine of the angle between them:
 
The square of a side of a triangle is equal to the sum of the squares of the other two sides, minus double the product of the latter two sides and the cosine of the angle between them:
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c026/c026660/c0266601.png" /></td> </tr></table>
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$$c^2=a^2+b^2-2ab\cos C.$$
  
Here <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c026/c026660/c0266602.png" /> are the sides of the triangle and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c026/c026660/c0266603.png" /> is the angle between <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c026/c026660/c0266604.png" /> and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c026/c026660/c0266605.png" />.
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Here $a,b,c$ are the sides of the triangle and $C$ is the angle between $a$ and $b$.

Latest revision as of 14:25, 19 March 2014

The square of a side of a triangle is equal to the sum of the squares of the other two sides, minus double the product of the latter two sides and the cosine of the angle between them:

$$c^2=a^2+b^2-2ab\cos C.$$

Here $a,b,c$ are the sides of the triangle and $C$ is the angle between $a$ and $b$.

How to Cite This Entry:
Cosine theorem. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Cosine_theorem&oldid=13758
This article was adapted from an original article by Yu.A. Gor'kov (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article