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Principle Of Mathematical Induction

Question
CBSEENMA11014338

Prove in any triangle ABC

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#6 {main}</pre> (ii) cos straight B over 2 equals square root of fraction numerator straight s left parenthesis straight s minus straight b right parenthesis over denominator ac end fraction end root  (iii) <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
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#6 {main}</pre>


Solution

We have
           2 cos squared straight A over 2 equals 1 plus cosA space equals space 1 plus fraction numerator straight b squared plus straight c squared minus straight a squared over denominator 2 bc end fraction equals fraction numerator 2 bc plus straight b squared plus straight c squared minus straight a squared over denominator 2 bc end fraction
                          equals fraction numerator left parenthesis straight b plus straight c right parenthesis squared minus straight a squared over denominator 2 bc end fraction equals fraction numerator left parenthesis straight b plus straight c minus straight a right parenthesis space left parenthesis straight b plus straight c plus straight a right parenthesis over denominator 2 bc end fraction equals fraction numerator left parenthesis 2 straight s minus 2 straight a right parenthesis space left parenthesis 2 straight s right parenthesis over denominator 2 bc end fraction
rightwards double arrow                 cos squared straight A over 2 space equals space fraction numerator straight s left parenthesis straight s minus straight a right parenthesis over denominator bc end fraction
or                     cos straight A over 2 space equals space square root of fraction numerator straight s left parenthesis straight s minus straight a right parenthesis over denominator bc end fraction end root
Similarly, we can show (ii) and (iii)