Difference between revisions of "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== | ||
| + | The formula for the area of a triangle is: | ||
| + | <math> \displaystyle [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: | ||
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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> | ||
| + | |||
| + | Multiplying the equation by <math> \displaystyle \frac{2}{abc} </math> yeilds: | ||
| + | |||
| + | <math> \displaystyle \frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} </math> | ||
==See also== | ==See also== | ||
Revision as of 20:20, 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, 
.
Proof of the Law of Sines
The formula for the area of a triangle is:
![$\displaystyle [ABC] = \frac{1}{2}ab\sin C$](http://latex.artofproblemsolving.com/5/6/d/56d1e1798c8fe2f875e8d9483406d06e30c6113c.png)
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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