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Sine of an Angle and Opposite Side

Date: 5/17/96 at 16:54:58
From: Anonymous
Subject: Relationship of sine of an angle to its opposite side

Hello.  I am a 10th grader in high school and I am currently in 
Algebra II.  Our math teacher gave us a problem for bonus, but she 
didn't know the complete answer.  Do you think you could help?

Given triangle ABC inscribed in a circle.  Prove why (sin A)/a equals 
the diameter of the circle.

Date: 5/28/96 at 22:55:37
From: Doctor Pete
Subject: Re: Relationship of sine of an angle to its opposite side

Let the center of the circumscribed circle of triangle ABC be O.  
Draw OA to intersect at D, so AD is a diameter.  Then angle 
BOD = 2BAD.  This is because AO = BO, so ABO is isoceles, and 
therefore angle ABO = BAO; since BOD = ABO+BAO, BOD = 2BAD. 

Similarly, COD = 2CAD, so BOC = BOD+COD = 2(BAD+CAD) = 2BAC.  
Then the angle bisector of BOC also bisects BC.
Since triangle BOC is isoceles; let the midpoint of BC be E.  
Hence sin(A)/a = sin(BAC)/BC = sin(BOC/2)/(2BE) = sin(BOE)/(2BE) 
= (BE/BO)/(2BE) = 1/(2BO). But BO is the radius R of O, so 
a/sin(A) = D, where D is the diameter of the circumscribed circle.

Note that sin(A)/a cannot give the diameter (think of the units); if 
one has two similar triangles ABC and A'B'C', sin(A) = sin(A'), so you 
would expect the radii to be *directly* proportional to the length of 
a side, not *inversely* proportional.

Also, you can see this gives a proof of the Law of Sines, which states
sin(A)/a = sin(B)/b = sin(C)/c.

-Doctor Pete,  The Math Forum
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Associated Topics:
High School Trigonometry

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