Law of cosines

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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
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<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
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<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.

c^{2} = a^{2} + b^{2} - 2ab \cos C

When to use it: SAS, SSS.

Contents

Proof

Let \vartriangle ABC be oriented so that C is at the origin, and B is at the point (a,0).

Image:Law_of_cosines_proof.jpg

Distance Formula

distance = \sqrt {(x_{2}-x_{1})^{2} + (y_{2}-y_{1})^{2}}

c is the distance from A to B.

Substituting the appropriate points into the distance formula gives us

c = \sqrt {(a-b \cos C)^{2} + (0-b \sin C)^{2}}

Squaring the inner terms, we have

c = \sqrt {(a^{2}-2ab \cos C+b^{2} \cos^{2} C) + (b^{2} \sin^{2} C)}

Since \cos^{2} C + \sin^{2} C = 1,

c = \sqrt {(a^{2}+b^{2}-2ab \cos C+b^{2}}

Square both sides for

c^{2} = (a^{2}+b^{2}-2ab \cos C+b^{2}

Example Triangulation

Complete the triangle using the law of cosines.


c^{2} = a^{2} + b^{2} - 2ab \cos C

Image:SAS triangle.jpg

Solution

To find the side length c,


c^{2} = 6^{2} + (6 \sqrt{2})^{2} -2 (6) (6 \sqrt{2}) \cos 45^\circ

Simplify for

c^{2} =36 + 36 (2) - 72 \sqrt{2}) \cos 45^\circ

Since \cos 45^\circ = \frac{1}{\sqrt{2}}, substitution gives us

c^{2} =36 + 36 (2) - 72 \sqrt{2} (\frac{1}{\sqrt{2}})

Simplify for

c^{2} =36 + 72 - 72


c^{2} =36

Taking the square root of both sides gives us

c =6

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


b^{2} = a^{2} + c^{2} - 2ab \cos B

Substituting in the appropriate side lengths gives us

(6 \sqrt{2})^{2} = 6^{2} + 6^{2} - 2(6)(6) \cos B

Simplify for

36 (2) = 36 + 36 - 72 \cos B

72 = 72 - 72 \cos B

Subtracting 72 from both sides gives us

0 = - 72 \cos B

Dividing both sides by -72 gives us

0 = \cos B

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