# Law of cosines

### From Math Images

(→Example Triangulation) |
(→Solution) |
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Now we can orient the triangle differently to get get a new version of the law of cosines so we can find angle measure <math>B</math>, | Now we can orient the triangle differently to get get a new version of the law of cosines so we can find angle measure <math>B</math>, | ||

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<math> b^{2} = a^{2} + c^{2} - 2ab \cos B </math> | <math> b^{2} = a^{2} + c^{2} - 2ab \cos B </math> | ||

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+ | Substituting in the appropriate side lengths gives us | ||

+ | |||

+ | <math> (6 \sqrt{2})^{2} = 6^{2} + 6^{2} - 2(6)(6) \cos B </math> | ||

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+ | Simplify for | ||

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+ | <math> 36 (2) = 36 + 36 - 72 \cos B </math> | ||

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+ | <math>72 = 72 - 72 \cos B </math> | ||

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+ | Subtracting <math>72</math> from both sides gives us | ||

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+ | <math>0 = - 72 \cos B </math> | ||

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+ | Dividing both sides by <math>-72</math> gives us | ||

+ | |||

+ | <math>0 = \cos B </math> |

## Revision as of 11:21, 30 May 2011

The law of cosines is a formula that helps in triangulation when two or three side lengths of a triangle are known. The formula relates all three side lengths of a triangle to the cosine of a particular angle.

When to use it: SAS, SSS.

## Contents |

## Proof

Let be oriented so that is at the origin, and is at the point.

### Distance Formula

is the distance from to .

Substituting the appropriate points into the distance formula gives us

Squaring the inner terms, we have

Since ,

Square both sides for

## Example Triangulation

Complete the triangle using the law of cosines.

### Solution

To find the side length ,

Simplify for

Since , substitution gives us

Simplify for

Taking the square root of both sides gives us

Now we can orient the triangle differently to get get a new version of the law of cosines so we can find angle measure ,

Substituting in the appropriate side lengths gives us

Simplify for

Subtracting from both sides gives us

Dividing both sides by gives us