JEE Main & Advanced Sample Paper JEE Main Sample Paper-43

  • question_answer
    Direction: Let ABC be a triangle, R be the circumradius of the triangle. Also, given \[{{R}^{2}}=\frac{1}{8}\,({{a}^{2}}+{{b}^{2}}+{{c}^{2}})\],  then
    Hence, the ABC can be

    A)  equilateral

    B)  isosceles and scalene

    C)  cannot say                             

    D)  right angled

    Correct Answer: D

    Solution :

     By above solution \[\sum{\cos \,2A=-1}\] \[\Rightarrow \]            \[\cos \,2A+\cos \,2B+\,(\cos \,2C+1)=0\] \[\Rightarrow \]            \[2\,\cos \,(A+B)\,\cos \,(A-B)\,+2\,{{\cos }^{2}}\,C=0\] but                   \[A+B=\pi -C\] \[\Rightarrow \]            \[2\,\cos \,(\pi -C)\cdot \cos (A+B)+2\,{{\cos }^{2}}C=0\] \[\Rightarrow \]            \[-2\cos \,C\cdot \cos \,(A-C)+2\,{{\cos }^{2}}C=0\] \[\Rightarrow \]            \[-2\,\cos \,C\,[\cos \,(A-C)-\cos \,C]=0\] \[\Rightarrow \]            Either \[\cos \,C=0,\,\,i.e.,\,\,C=\frac{\pi }{2}\] \[\Rightarrow \]       \[\cos \,(A-B)\,-\,\cos \,C=0\] \[\Rightarrow \]       \[2\,\,\sin \,\left( \frac{C+A-B}{2} \right)\sin \,\left( \frac{C-A+B}{2} \right)=0\] i.e.,      \[C+A-B=0\]  or         \[C-A+B=0\] i.e.,      \[B=\frac{\pi }{2}\]                or         \[A=\frac{\pi }{2}\]

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