Question

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

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

Answer 1

The correct option is A right angled

Given $a^2+b^2+c^2=8R^2$

from sine rule $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$

$\Rightarrow {\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C=2\Rightarrow {\sin }^{2}A+{\sin }^{2}B+{\sin }^2(A+B)=2\Rightarrow {\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}A{\cos }^{2}B+{\cos }^{2}A{\sin }^{2}B+2\sin A\sin B\cos A\cos B=2\Rightarrow 1-{\cos }^{2}A+1-{\cos }^{2}B+{\sin }^{2}A{\cos }^{2}B+{\cos }^{2}A{\sin }^{2}B+2\sin A\sin B\cos A\cos B=2\Rightarrow 2-{\cos }^{2}A({\sin }^{2}B-1)+{\cos }^{2}B({\sin }^{2}A-1)+2\sin A\sin B\cos A\cos B=2\Rightarrow -2{\cos }^{2}A{\cos }^{2}B+2\sin A\sin B\cos A\cos B=0\Rightarrow \cos A\cos B(\cos A\cos B-\sin A\sin B)=0\Rightarrow \cos A\cos B\cos (A+B)=0\Rightarrow A=\frac{\pi }{2}orB=\frac{\pi }{2}orC=\frac{\pi }{2}$
Hence it is a right angled triangle.