JCECE Engineering JCECE Engineering Solved Paper-2009

  • question_answer
    In a triangle\[ABC\], if\[\tan \frac{A}{2}=\frac{5}{6}\]and\[\tan \frac{B}{2}=\frac{20}{37}\], then \[a+c\]is equal to

    A) \[b\]                                     

    B) \[2b\]

    C) \[3b\]                                   

    D) \[4b\]

    Correct Answer: B

    Solution :

    Since,\[\frac{A}{2}+\frac{B}{2}=\frac{\pi }{2}-\frac{C}{2}\] \[\Rightarrow \]\[\frac{\tan \frac{A}{2}+\tan \frac{B}{2}}{1-\tan \frac{A}{2}\tan \frac{B}{2}}=\tan \left( \frac{\pi }{2}-\frac{C}{2} \right)\] \[\Rightarrow \]\[\frac{\frac{5}{6}+\frac{20}{37}}{1-\frac{5}{6}\times \frac{20}{37}}=\cot \frac{C}{2}\] \[\Rightarrow \]\[\frac{305}{122}=\cot \frac{C}{2}\] \[\Rightarrow \]\[\tan \frac{C}{2}=\frac{2}{5}\] Now,     \[\tan \frac{A}{2}\tan \frac{C}{2}=\frac{5}{6}\times \frac{2}{5}\] \[\Rightarrow \]\[\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}\sqrt{\frac{(s-a)(s-b)}{s(s-c)}}=\frac{1}{3}\] \[\Rightarrow \]               \[\frac{s-b}{s}=\frac{1}{3}\] \[\Rightarrow \]               \[3s-3b=s\] \[\Rightarrow \]               \[2s=3b\Rightarrow a+b+c=3b\] \[\Rightarrow \]               \[a+c=2b\]


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