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RAJASTHAN ­ PET Rajasthan PET Solved Paper-2006

  • question_answer
    If the sides of the \[\Delta \]ABC are changed by a very small quantity such that the radius of circumcircle is remains unchanged, then the value of\[\frac{da}{\cos A}+\frac{db}{\cos B}+\frac{dc}{\cos C}\]is

    A)  0                

    B)  2R

    C)  6R               

    D)  8R

    Correct Answer: A

    Solution :

     According to sine rule, \[\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R\] \[\Rightarrow \] \[a=2R\sin A,b=2R\sin B,c=2R\sin C\] On differentiating, we get \[da=2R\cos AdA,db=2R\cos BdB,\] \[dc=2R\cos CdC\] \[\Rightarrow \]\[\frac{da}{\cos A}=RdA,\frac{db}{\cos B}=RdB,\frac{dc}{\cos C}=RbC\] \[\Rightarrow \]\[\frac{da}{\cos A}+\frac{db}{\cos B}+\frac{dc}{\cos C}\] \[=R(dA+dB+dC)\] Now,  \[A+B+C=k\]                (given) \[\Rightarrow \] \[dA+dB+dC=0\] Then, from Eq. (i), \[\frac{da}{\cos A}+\frac{db}{\cos B}+\frac{dc}{\cos C}=0\]


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