Difference between revisions of "Law of Sines"
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The same holds for b and c thus establishing the identity. | The same holds for b and c thus establishing the identity. | ||
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=== Method 2 === | === Method 2 === | ||
This method only works to prove the regular (and not extended) Law of Sines. | This method only works to prove the regular (and not extended) Law of Sines. |
Revision as of 18:16, 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
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
![$\sin \theta = \frac{\frac a2}R \Leftrightarrow \frac a{\sin\theta} = 2R.$](http://latex.artofproblemsolving.com/0/3/f/03f858d898d88e6bb7665ea3514b180f2e8ebcf5.png)
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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