If in a $$\triangle ABC,{a}^{2}+{b}^{2}+{c}^{2}=8{R}^{2},$$ where $$R=$$ circumradius,then the triangle is
  • A
    equilateral
  • B
    isosceles
  • C
    right angled
  • D
    none of these

Solution

Since, $${a}^{2}+{b}^{2}+{c}^{2}=8{R}^{2}$$
Using sine rule, we have
$${\left(2R\sin{A}\right)}^{2}+{\left(2R\sin{B}\right)}^{2}+{\left(2R\sin{C}\right)}^{2}=8{R}^{2}$$
$$\Rightarrow 4{R}^{2}\left[{\left(\sin{A}\right)}^{2}+{\left(\sin{B}\right)}^{2}+{\left(\sin{C}\right)}^{2}\right]=8{R}^{2}$$
$$\Rightarrow {\sin}^{2}A+{\sin}^{2}B+{\sin}^{2}C=2$$
$$\Rightarrow 1-{\cos}^{2}A+{\sin}^{2}B+1-{\cos}^{2}C=2$$
$$\Rightarrow -{\cos}^{2}A+{\sin}^{2}B-{\cos}^{2}C=2-2=0$$
$$\Rightarrow {\cos}^{2}A-{\sin}^{2}B+{\cos}^{2}C=0$$
$$\Rightarrow \cos{\left(A+B\right)}\cos{\left(A-B\right)}+{\cos}^{2}C=0$$
Since, $$A+B+C=\pi \Rightarrow A+B=\pi-C$$
$$\Rightarrow \cos{\left(\pi-C\right)}\cos{\left(A-B\right)}+{\cos}^{2}C=0$$
$$\Rightarrow  -\cos{C}\cos{\left(A-B\right)}+{\cos}^{2}C=0$$
$$\Rightarrow  \cos{C}\left[-cos{\left(A-B\right)}+\cos{C}\right]=0$$
$$\Rightarrow \cos{C}=0, -cos{\left(A-B\right)}+\cos{C}=0$$
$$\Rightarrow C=\dfrac{\pi}{2}$$ or $$-cos{\left(A-B\right)}+\cos{\left(\pi-\left(A+B\right)\right)}=0$$
$$\Rightarrow -cos{\left(A-B\right)}-cos{\left(A+B\right)}=0$$
Using transformation angle formula, we have
$$\Rightarrow 2\cos{A}\cos{B}=0$$
$$\therefore \cos{A}=0,\cos{B}=0$$
Hence $$\angle{A}=\dfrac{\pi}{2} $$or $$\angle{B}=\dfrac{\pi}{2}, $$ or  $$\angle{C}=\dfrac{\pi}{2}$$