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Math Help - solutions of a triangle

  1. #1
    Member grgrsanjay's Avatar
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    Cool solutions of a triangle

    prove that: \:\dfrac{c}{a+b}\:\ = \:\dfrac{1-tan\:\dfrac{A}{2}\:\tan\:\dfrac{B}{2}\ }{1 +tan\:\dfrac{A}{2}\:\tan\:\dfrac{B}{2} }

    where \:\boxed{a,b,c} are the sides opposite to the angles \:\boxed{A,B,C} respectively in a triangle
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  2. #2
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    Let  P , Q , R be the points of tangency of the incircle to the sides  BC , CA , AB respectively .


    Then  CP = CQ = (a+b-c)/2 = r \cot(\frac{C}{2}) where  r is the inradius .


    Also ,  c = AR + BR = r [ \cot(\frac{A}{2})  + \cot(\frac{B}{2})  ]


    But  \cot(\frac{C}{2})  = \tan( \frac{A+B}{2} ) = \tan(\frac{A}{2}) \tan(\frac{B}{2})  ~ \frac{ \cot(\frac{A}{2})  + \cot(\frac{B}{2})  }{ 1 - \tan(\frac{A}{2}) \tan(\frac{B}{2}) }


    Therefore ,  \frac{ a+b-c}{c} = 2r \tan(\frac{A}{2}) \tan(\frac{B}{2}) ~ \frac{ \cot(\frac{A}{2})  + \cot(\frac{B}{2})  }{ 1 - \tan(\frac{A}{2}) \tan(\frac{B}{2}) } ~ \frac{1}{ r[  \cot(\frac{A}{2})  + \cot(\frac{B}{2}) ]}

     =  \frac{2\tan(\frac{A}{2}) \tan(\frac{B}{2}) }{ 1 - \tan(\frac{A}{2}) \tan(\frac{B}{2}) }


     \frac{ a+b}{c} = \frac{1 + \tan(\frac{A}{2}) \tan(\frac{B}{2}) }{1-\tan(\frac{A}{2}) \tan(\frac{B}{2}) }

    or  \boxed{ \displaystyle{ \frac{c}{a+b} = \frac{ 1 -\tan(\frac{A}{2}) \tan(\frac{B}{2}) }{1+ \tan(\frac{A}{2}) \tan(\frac{B}{2}) } }}
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  3. #3
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    Quote Originally Posted by simplependulam to grgrsanjay;
    Let  \:\boxed{P , Q , R } be the points of tangency of the incircle to the sides  BC , CA , AB respectively .

    what does tangency to a circle mean actually???
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  4. #4
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    Quote Originally Posted by grgrsanjay View Post
    what does tangency to a circle mean actually???
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