Principle Of Mathematical Induction

Question
CBSEENMA11014349

Prove that in any triangle ABC

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Solution

Proof:
               We have straight R space equals space fraction numerator straight a over denominator 2 space sin space straight A end fraction space equals space fraction numerator straight a over denominator 2 begin display style 2 over bc end style square root of straight s left parenthesis straight s minus straight a right parenthesis space left parenthesis straight s minus straight b right parenthesis space left parenthesis straight s minus straight c right parenthesis end root end fraction equals fraction numerator abc over denominator 4 straight capital delta end fraction equals straight R. straight H. straight S.

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