2. Solved Papers
  3. RAJASTHAN ­ PET

RAJASTHAN ­ PET Rajasthan PET Solved Paper-2011

  • question_answer
    In a\[\Delta ABC,\]if\[\frac{\cos A}{a}=\frac{\cos B}{b}=\frac{\cos C}{c}\]and the side\[a=2,\]then area of the triangle is

    A)  1              

    B)  2

    C)  \[\sqrt{3}/2\]          

    D) \[\sqrt{3}\]

    Correct Answer: D

    Solution :

     \[\frac{\cos A}{a}=\frac{\cos B}{b}=\frac{\cos C}{c}\] \[\Rightarrow \]\[\frac{\cos A}{K\sin A}=\frac{\cos B}{K\sin B}=\frac{\cos C}{K\sin C}\](by sin rule) \[\Rightarrow \] \[\cot A=\cot B=\cot C\] \[\Rightarrow \] \[A=B=C=60{}^\circ \] \[\Rightarrow \]\[\Delta \]ABC is an equilateral Hence, \[\Delta =\frac{\sqrt{3}}{4}{{a}^{2}}=\frac{\sqrt{3}}{4}\times 4=\sqrt{2}\]     \[(\because a=2)\]


You need to login to perform this action.
You will be redirected in 3 sec spinner