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A circle having centre as O' and radius r' touches the incircle of Delta ABC externally at. F, where F is on BC and also touches its circumcircle internally at G. It O is the circumcentre of Delta ABC and I is its incentre, then |
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Answer» OO'=R-r' `OE=DE=|BF-BD|=|s-b-(a)/(2)|=(|c-b|)/(2)` OD = EF = R cos A From `Delta OEO'`, using Pythagoras theorem, we get `(R-r')^(2)=(R cos A+r')^(2)+((c-b)/(2))^(2)` `rArr r' =(BC)/(4R)"tan"^(2)(A)/(2)` `= (Delta)/(a)"tan"^(2)(A)/(2)` |
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