• 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

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         $A=\frac{\pi }{2}$