Difference between revisions of "Law of Sines"
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<center>[[Image:Lawofsines.PNG]]</center> | <center>[[Image:Lawofsines.PNG]]</center> | ||
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=== 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> | + | <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> | + | <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> | + | Multiplying the equation by <math>\frac{2}{abc} </math> yeilds: |
− | <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, .
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:
See also
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