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Ask iit jee aieee pet cbse icse state board experts Discussion Response Post to: circumcenter=(x1sin2A+x2sin2B+x3sin3C/sigma sin2A,put y1,y2,y3,in place of x1,x2,x3),how?
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[Post New]14 Dec 2007 04:17:54 IST
    Subject: Re:circumcenter=(x1sin2A+x2sin2B+x3sin3C/sigma sin2A,put y1,y2,y3,in place of x1,x2,x3),ho
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bvsatyaram (211)

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Olaaa!! Perrrfect answer. 37  [50 rates]
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U can find this derivation in almost all standard text books. I hope even ur school text book has this.

It would be too tidious to type in the whole derivation.
Satyaram B V,
General Secretary, Mandakini Hostel,
IIT Madras
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