Principle Of Mathematical Induction

Question
CBSEENMA11014348

Prove that for any triangle ABC

     2 straight R space equals space fraction numerator straight a over denominator sin space straight A end fraction equals straight b over sinB equals fraction numerator straight c over denominator sin space straight C end fraction

where R is the radius of the circumcircle.

Solution

Proof:
                 
Let space space angle straight A be acute in figure (i), obtuse in figure (ii) and right angle in figure (iii)
Join BO and produce it to meet the circle in A' and join CA',
then triangle BCA' is a right angle triangle in (i) and (ii), but C and A' coincide in (iii).
In figure (i),
Since angles in the same segment are equal 
∴          space space angle BAC space equals space angle BA apostrophe straight C
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#6 {main}</pre>                                         [By sine formula]
                       fraction numerator 2 straight R over denominator 1 end fraction space equals space fraction numerator straight a over denominator sin space straight A end fraction                                             [∵ BCA' = 90 degree]
In figure (ii),  fraction numerator BA apostrophe over denominator sin space 90 degree end fraction equals fraction numerator BC over denominator sin space straight A apostrophe end fraction
rightwards double arrow                 fraction numerator BA apostrophe over denominator sin space 90 degree end fraction equals fraction numerator BC over denominator sin left parenthesis straight pi minus straight A right parenthesis end fraction
                                     (∵ Sum of opposite angles of a cyclic quadrilateral is 180 degree)
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#6 {main}</pre>
In figure (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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                          2 straight R equals fraction numerator straight a over denominator sin space straight A end fraction                           [∵ angle straight A space equals space 90 degree i.e. angle is semi-circle]
In all cases, we have
                      2 straight R space equals space fraction numerator straight a over denominator sin space straight A end fraction
But,             fraction numerator straight a over denominator sin space straight A end fraction equals fraction numerator straight b over denominator sin space straight B end fraction equals fraction numerator straight c over denominator sin space straight C end fraction
Hence,                  2 straight R space equals space fraction numerator straight a over denominator sin space straight A end fraction equals fraction numerator straight b over denominator sin space straight B end fraction space equals space fraction numerator straight c over denominator sin space straight C end fraction
                       
                         
                      

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