Sine of an Angle and Opposite SideDate: 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 Check out our web site! http://mathforum.org/dr.math/ |
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