Difference between revisions of "Law of Sines"

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(Added proof for the non-extended law of sines.)
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Given a [[triangle]] with sides of length a, b and c, opposite [[angle]]s of measure A, B and C, respectively, and a [[circumcircle]] with radius R,  <math>\frac{a}{\sin{A}}=\frac{b}{\sin{B}}=\frac{c}{\sin{C}}=2R</math>.
 
Given a [[triangle]] with sides of length a, b and c, opposite [[angle]]s of measure A, B and C, respectively, and a [[circumcircle]] with radius R,  <math>\frac{a}{\sin{A}}=\frac{b}{\sin{B}}=\frac{c}{\sin{C}}=2R</math>.
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==Proof of the Law of Sines==
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The formula for the area of a triangle is:
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<math> \displaystyle [ABC] = \frac{1}{2}ab\sin C </math>
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Since it doesn't matter which sides are chosen as <math>a</math>, <math>b</math>, and <math>c</math>, the following equality holds:
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<math> \displaystyle \frac{1}{2}bc\sin A = \frac{1}{2}ac\sin B = \frac{1}{2}ab\sin C </math>
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Multiplying the equation by <math> \displaystyle \frac{2}{abc} </math> yeilds:
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<math> \displaystyle \frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} </math>
  
 
==See also==
 
==See also==

Revision as of 19:20, 29 June 2006

Given a triangle with sides of length a, b and c, opposite angles of measure A, B and C, respectively, and a circumcircle with radius R, $\frac{a}{\sin{A}}=\frac{b}{\sin{B}}=\frac{c}{\sin{C}}=2R$.

Proof of the Law of Sines

The formula for the area of a triangle is: $\displaystyle [ABC] = \frac{1}{2}ab\sin C$

Since it doesn't matter which sides are chosen as $a$, $b$, and $c$, the following equality holds:

$\displaystyle \frac{1}{2}bc\sin A = \frac{1}{2}ac\sin B = \frac{1}{2}ab\sin C$

Multiplying the equation by $\displaystyle \frac{2}{abc}$ yeilds:

$\displaystyle \frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}$

See also

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