Given a triangle with vertices A, B and C, find a formula for the radius R of thecircumcircle
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angle BOC is twice angle BAC since they intercept the same chord BC but BOC is a central angle Euclid's Elements, Book III, Proposition 20 and central and inscribed angles. In triangle BOC, BM = MC. Which leads to angle BOM = angle MOC.
Which leads to saying that angle BOM = angle MOC = angle A.
In the triangle BOM,
sin(BOM) = BM / OB
But BM =BC/2
Which leads to:
sin(BOM) = BC / (2*OB)
angle BOM = angle A and OB is the radius R of the circle.
Hence sin(A) = BC / (2R)
or 2R = BC / sin(A)
and now using the sine law, we have
2R = BC / sin(A) = AC / sin(B) = AB / sin(C)
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