Question

In In $\Delta ABC,$ if $8R^2=a^2+b^2+c^2,$ then the triangle is

• A. Right angled
• B. Equilateral
• C. Acute angled
• D. Obtuse angled

Answer 1

The correct option is A

Right angled

$8R^2=a^2+b^2+c^2=4R^2({\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C)\Rightarrow {\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C=2\Rightarrow ({\cos }^{2}A-{\sin }^{2}C)+{\cos }^{2}B=0\Rightarrow \cos (A-C)\cos (A+C)+{\cos }^{2}B=0\Rightarrow \cos (A-C)\cos (A+C)+{\cos }^2(\pi -(A+C\left)\right)=0\Rightarrow \cos (A+C)\left(\cos \right(A-C)+\cos (A+C\left)\right)=0\Rightarrow \cos (\pi -B)\left(2\cos A\cos C\right)=0\Rightarrow -2\cos A\cos C\cos B=0$
$\therefore \cos A=0\text{or}\cos B=0\text{or}\cos C=0\Rightarrow a=\frac{\pi }{2}\text{or}B=\frac{\pi }{2}\text{or}C=\frac{\pi }{2}$