| 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 |
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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