Difference between revisions of "Law of Sines"
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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 = | + | 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> 2\angle BAC = \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 19:56, 28 January 2009
The Law of Sines is a useful identity in a triangle, which, along with the law of cosines and the law of tangents can be used to determine sides and angles. The law of sines can also be used to determine the circumradius, another useful function.
Contents
[hide]Theorem
In triangle , where is the side opposite , opposite , opposite , and where is the circumradius:
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 and , 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 yields:
Problems
Introductory
- If the sides of a triangle have lengths 2, 3, and 4, what is the radius of the circle circumscribing the triangle?
(Source)
Intermediate
- Triangle has sides , , and of length 43, 13, and 48, respectively. Let be the circle circumscribed around and let be the intersection of and the perpendicular bisector of that is not on the same side of as . The length of can be expressed as , where and are positive integers and is not divisible by the square of any prime. Find the greatest integer less than or equal to .
(Source)
Advanced
Let be a convex quadrilateral with , $AC \neqBD$ (Error compiling LaTeX. Unknown error_msg), and let be the intersection point of its diagonals. Prove that if and only if .
(Source)