Difference between revisions of "Law of Sines"
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==Proof == | ==Proof == | ||
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− | In the diagram below, circle <math> O </math> [[circumscribe]]s triangle <math> ABC </math>. <math> OD </math> is perpendicular to <math> BC </math>. Since <math> \triangle ODB \cong \triangle ODC </math>, <math> BD = CD = \frac a2 </math> and <math> \angle BOD = \angle COD </math>. But <math> \angle BAC = 2\angle BOC </math> making <math> \angle BOD = \angle COD = \theta </math>. Therefore, we can use simple trig in right triangle <math> BOD </math> to find that | + | In the diagram below, circle <math> O </math> [[circumscribe]]s triangle <math> ABC </math>. <math> OD </math> is [[perpendicular]] to <math> BC </math>. Since <math> \triangle ODB \cong \triangle ODC </math>, <math> BD = CD = \frac a2 </math> and <math> \angle BOD = \angle COD </math>. But <math> \angle BAC = 2\angle BOC </math> making <math> \angle BOD = \angle COD = \theta </math>. Therefore, we can use simple trig in [[right triangle]] <math> BOD </math> to find that |
<center><math> \sin \theta = \frac{\frac a2}R \Leftrightarrow \frac a{\sin\theta} = 2R. </math> </center> | <center><math> \sin \theta = \frac{\frac a2}R \Leftrightarrow \frac a{\sin\theta} = 2R. </math> </center> |
Revision as of 12:43, 20 October 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, .
Contents
[hide]Proof
Method 1
In the diagram below, circle circumscribes triangle . is perpendicular to . Since , and . But making . Therefore, we can use simple trig in right triangle to find that
The same holds for b and c thus establishing the identity.
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:
Since it doesn't matter which sides are chosen as , , and , the following equality holds:
Multiplying the equation by yeilds: