Difference between revisions of "The Law of Cosines"

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* [https://mathresearch.utsa.edu/wikiFiles/MAT1093/The%20Law%20of%20Cosines/ The Law of Cosines]. Written notes created by Professor Esparza, UTSA.
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==Law of Cosines==
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[[Image:Law-of-cosines1.svg]]
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The Pythagorean theorem is a special case of the more general theorem relating the lengths of sides in any triangle, the law of cosines:
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:<math>a^2+b^2-2ab\cos(\theta)=c^2</math>
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where <math>\theta</math> is the angle between sides <math>a</math> and <math>b</math> .
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===Does the formula make sense?===
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This formula had better agree with the Pythagorean Theorem when <math>\theta=90^\circ</math> . 
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So try it...
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When <math>\theta=90^\circ</math> , <math>\cos(\theta)=\cos(90^\circ)=0</math>
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The <math>-2ab\cos(\theta)=0</math> and the formula reduces to the usual Pythagorean theorem.
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==Permutations==
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For any triangle with angles <math>A,B,C</math> and corresponding opposite side lengths <math>a,b,c</math> , the Law of Cosines states that
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:<math>a^2=b^2+c^2-2bc\cdot\cos(A)</math>
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:<math>b^2=a^2+c^2-2ac\cdot\cos(B)</math>
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:<math>c^2=a^2+b^2-2ab\cdot\cos(C)</math>
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===Proof===
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[[Image:Law-of-cosines2.svg]]
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Dropping a perpendicular <math>OC</math> from vertex <math>C</math> to intersect <math>AB</math> (or <math>AB</math> extended) at <math>O</math> splits this triangle into two right-angled triangles <math>AOC</math> and <math>BOC</math> , with altitude <math>h</math> from side <math>c</math> .
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First we will find the lengths of the other two sides of triangle <math>AOC</math> in terms of known quantities, using triangle <math>BOC</math> .
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:<math>h=a\sin(B)</math>
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Side <math>c</math> is split into two segments, with total length <math>c</math> .
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:<math>\overline{OB}</math> has length <math>\overline{BC}\cos(B)=a\cos(B)</math>
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:<math>\overline{AO}=\overline{AB}-\overline{OB}</math> has length <math>c-a\cos(B)</math>
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Now we can use the Pythagorean Theorem to find <math>b</math> , since <math>b^2=\overline{AO}^2+h^2</math> .
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:{|
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|<math>b^2</math>
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|<math>=\bigl(c-a\cos(B)\bigr)^2+a^2\sin^2(B)</math>
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|-
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|
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|<math>=c^2-2ac\cos(B)+a^2\cos^2(B)+a^2\sin^2(B)</math>
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|-
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|
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|<math>=a^2+c^2-2ac\cos(B)</math>
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|}
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The corresponding expressions for <math>a</math> and <math>c</math> can be proved similarly.
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The formula can be rearranged:
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:<math>\cos(C)=\frac{a^2+b^2-c^2}{2ab}</math>
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and similarly for <math>cos(A)</math> and <math>cos(B)</math> .
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==Applications==
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This formula can be used to find the third side of a triangle if the other two sides and the angle between them are known. The rearranged formula can be used to find the angles of a triangle if all three sides are known.
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==Resources==
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* [https://mathresearch.utsa.edu/wikiFiles/MAT1093/The%20Law%20of%20Cosines/Esparza%201093%20Notes%204.3A.pdf The Law of Cosines]. Written notes created by Professor Esparza, UTSA.
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* [https://mathresearch.utsa.edu/wikiFiles/MAT1093/The%20Law%20of%20Cosines/Esparza%201093%20Notes%204.3B.pdf The Law of Cosines Continued]. Written notes created by Professor Esparza, UTSA.
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== Licensing ==
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Content obtained and/or adapted from:
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* [https://en.wikibooks.org/wiki/Trigonometry/Law_of_Cosines Law of Cosines, Wikibooks] under a CC BY-SA license

Latest revision as of 17:00, 28 October 2021

Law of Cosines

Law-of-cosines1.svg

The Pythagorean theorem is a special case of the more general theorem relating the lengths of sides in any triangle, the law of cosines:

where is the angle between sides and .

Does the formula make sense?

This formula had better agree with the Pythagorean Theorem when .

So try it...

When ,

The and the formula reduces to the usual Pythagorean theorem.

Permutations

For any triangle with angles and corresponding opposite side lengths , the Law of Cosines states that

Proof

Law-of-cosines2.svg

Dropping a perpendicular from vertex to intersect (or extended) at splits this triangle into two right-angled triangles and , with altitude from side .

First we will find the lengths of the other two sides of triangle in terms of known quantities, using triangle .

Side is split into two segments, with total length .

has length
has length

Now we can use the Pythagorean Theorem to find , since .

The corresponding expressions for and can be proved similarly.

The formula can be rearranged:

and similarly for and .

Applications

This formula can be used to find the third side of a triangle if the other two sides and the angle between them are known. The rearranged formula can be used to find the angles of a triangle if all three sides are known.


Resources

Licensing

Content obtained and/or adapted from: