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Difference between revisions of "Law of Sines"

m (Law of sines moved to Law of Sines: consistency (I hope))
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<center>[[Image:Lawofsines.PNG]]</center>
 
<center>[[Image:Lawofsines.PNG]]</center>
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{{asy replace}}
  
 
=== Method 2 ===
 
=== Method 2 ===
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The formula for the area of a triangle is:  
 
The formula for the area of a triangle is:  
−
<math> \displaystyle [ABC] = \frac{1}{2}ab\sin C </math>
+
<math>[ABC] = \frac{1}{2}ab\sin C </math>
  
 
Since it doesn't matter which sides are chosen as <math>a</math>, <math>b</math>, and <math>c</math>, the following equality holds:  
 
Since it doesn't matter which sides are chosen as <math>a</math>, <math>b</math>, and <math>c</math>, the following equality holds:  
  
−
<math> \displaystyle \frac{1}{2}bc\sin A = \frac{1}{2}ac\sin B = \frac{1}{2}ab\sin C </math>
+
<math>\frac{1}{2}bc\sin A = \frac{1}{2}ac\sin B = \frac{1}{2}ab\sin C </math>
  
−
Multiplying the equation by <math> \displaystyle \frac{2}{abc} </math> yeilds:  
+
Multiplying the equation by <math>\frac{2}{abc} </math> yeilds:  
  
−
<math> \displaystyle \frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} </math>
+
<math>\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} </math>
  
 
==See also==
 
==See also==

Revision as of 17:16, 24 September 2007

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

Method 1

In the diagram below, circle $O$ circumscribes triangle $ABC$. $OD$ is perpendicular to $BC$. Since $\triangle ODB \cong \triangle ODC$, $BD = CD = \frac a2$ and $\angle BOD = \angle COD$. But $\angle BAC = 2\angle BOC$ making $\angle BOD = \angle COD = \theta$. Therefore, we can use simple trig in right triangle $BOD$ to find that

$\sin \theta = \frac{\frac a2}R \Leftrightarrow \frac a{\sin\theta} = 2R.$

The same holds for b and c thus establishing the identity.

Lawofsines.PNG

This picture could be replaced by an asymptote drawing. It would be appreciated if you do this.

Method 2

This method only works to prove the regular (and not extended) Law of Sines.

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

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

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

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

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

See also

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