Question

If in a $△ABC,a^2+b^2+c^2=8R^2,$ where $R=$ circumradius,then the triangle is

• A. equilateral
• B. isosceles
• C. right angled
• D. none of these

Answer 1

Answer: (C)

right angled

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