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  $\angle{C}=\dfrac{\pi}{2}$