Law of cosines

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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
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The law of cosines is a formula that helps in solving triangles when two or three side lengths of a triangle are known. The formula combines the squares of two side lengths of a triangle and some offset, classified by the cosine of a particular angle, to calculate the square of the third side. For this reason, the law of cosines is often thought of as the generalization of the Pythagorean theorem, which only applies to right triangles. The law of cosines adds an extra term to the Pythagorean theorem so that a third side length of a triangle can be determined when there is no right angle.
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particular angle.
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<math> c^{2} = a^{2} + b^{2} - 2ab \cos C </math>
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::<math> c^{2} = a^{2} + b^{2} - 2ab \cos C </math>
When to use it: SAS, SSS.
When to use it: SAS, SSS.
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==Proof==
==Proof==
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===By the Pythagorean Theorem===
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An easy way to think of the law of cosines is as an extension of the Pythagorean theorem:
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::<math>a^{2} + b^{2} = c^{2} </math>
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[[Image:Pythagorean_cosines_proof.jpg]]
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===Using the Distance Formula===
Let <math> \vartriangle ABC </math> be oriented so that <math> C</math> is at the origin, and <math> B</math> is at the point<math> (a,0)</math>.
Let <math> \vartriangle ABC </math> be oriented so that <math> C</math> is at the origin, and <math> B</math> is at the point<math> (a,0)</math>.
[[Image:Law_of_cosines_proof.jpg]]
[[Image:Law_of_cosines_proof.jpg]]
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===Distance Formula===
 
<math> distance = \sqrt {(x_{2}-x_{1})^{2} + (y_{2}-y_{1})^{2}}</math>
<math> distance = \sqrt {(x_{2}-x_{1})^{2} + (y_{2}-y_{1})^{2}}</math>
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Since <math> \cos^{2} C + \sin^{2} C = 1</math>,
Since <math> \cos^{2} C + \sin^{2} C = 1</math>,
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<math> c = \sqrt {(a^{2}+b^{2}-2ab \cos C+b^{2}}</math>
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<math> c = \sqrt {a^{2}+b^{2}-2ab \cos C+b^{2}}</math>
Square both sides for
Square both sides for
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<math> c^{2} = (a^{2}+b^{2}-2ab \cos C+b^{2}</math>
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<math> c^{2} = a^{2}+b^{2}-2ab \cos C+b^{2}</math>
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==Example Triangulation==
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==Example Problem==
Complete the triangle using the law of cosines.
Complete the triangle using the law of cosines.

Revision as of 10:53, 3 June 2011

Image:inprogress.png
Law of Cosines
[[Image:Image:Law_of_cosines_pictiure.jpg|center|400px]]
Field: Geometry
Created By: [[Author:| ]]
Website: [ ]

Law of Cosines

The law of cosines is a trigonometric extension of the Pythagorean Theorem.


Contents

Basic Description

The law of cosines is a formula that helps in solving triangles when two or three side lengths of a triangle are known. The formula combines the squares of two side lengths of a triangle and some offset, classified by the cosine of a particular angle, to calculate the square of the third side. For this reason, the law of cosines is often thought of as the generalization of the Pythagorean theorem, which only applies to right triangles. The law of cosines adds an extra term to the Pythagorean theorem so that a third side length of a triangle can be determined when there is no right angle.
c^{2} = a^{2} + b^{2} - 2ab \cos C

When to use it: SAS, SSS.

A More Mathematical Explanation

Proof

By the Pythagorean Theorem

An easy way to think of the law of cosines is as an extens [...]

Proof

By the Pythagorean Theorem

An easy way to think of the law of cosines is as an extension of the Pythagorean theorem:

a^{2} + b^{2} = c^{2}

Image:Pythagorean_cosines_proof.jpg


Using the Distance Formula

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 = \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 Problem

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

Using inverse trigonometry, we know that

B = 90^\circ

And we can find the last angle measure A by subtracting the other two measures from 180^\circ

180^\circ - 90^\circ - 45^\circ = 45^\circ

A=45^\circ

Image:SAS_solution.jpg




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