Principle Of Mathematical Induction

Question
CBSEENMA11014337

Prove in any triangle ABC

(i) sin straight A over 2 equals square root of fraction numerator left parenthesis straight s minus straight b right parenthesis space left parenthesis straight s minus straight c right parenthesis over denominator bc end fraction end root (ii) sin straight B over 2 equals square root of fraction numerator left parenthesis straight s minus straight c right parenthesis left parenthesis straight s minus straight a right parenthesis over denominator ac end fraction end root (iii) sin straight C over 2 equals square root of fraction numerator left parenthesis straight s minus straight a right parenthesis left parenthesis straight s minus straight b right parenthesis over denominator ab end fraction end root

Solution

(i)  We have
          2 space sin squared straight A over 2 space equals space 1 minus cosA space equals space 1 minus open square brackets fraction numerator straight b squared plus straight c squared minus straight a squared over denominator 2 bc end fraction close square brackets
                         equals space fraction numerator 2 bc minus straight b squared minus straight c squared plus straight a squared over denominator 2 bc end fraction equals fraction numerator straight a squared minus left parenthesis straight b minus straight c right parenthesis squared over denominator 2 bc end fraction
                          equals fraction numerator left parenthesis straight a minus straight b plus straight c right parenthesis space left parenthesis straight a plus straight b minus straight c right parenthesis over denominator 2 bc end fraction equals fraction numerator left parenthesis 2 straight s minus 2 straight b right parenthesis left parenthesis 2 straight s minus 2 straight c right parenthesis over denominator 2 bc end fraction
rightwards double arrow            sin squared straight A over 2 space equals space fraction numerator left parenthesis straight s minus straight b right parenthesis space left parenthesis straight s minus straight c right parenthesis over denominator bc end fraction
or            sin straight A over 2 equals square root of fraction numerator left parenthesis straight s minus straight b right parenthesis space left parenthesis straight s minus straight c right parenthesis over denominator bc end fraction end root
Similarly, we can prove that
(ii)            sin straight B over 2 equals square root of fraction numerator left parenthesis straight s minus straight a right parenthesis left parenthesis straight s minus straight c right parenthesis over denominator ac end fraction end root
(iii)            sin straight C over 2 equals square root of fraction numerator left parenthesis straight s minus straight a right parenthesis left parenthesis straight s minus straight b right parenthesis over denominator ab end fraction end root

          
  

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