Question

If $a^2+b^2+c^2=8R^2$, then prove that the triangle is a right-angled triangle.

Answer 1

Proof:

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

Using $\mathrm{Sin}$ law, $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$

$\Rightarrow {\left(2R\sin A\right)}^2+{\left(2R\sin B\right)}^2+{\left(2R\sin C\right)}^2=8R^2$

$\Rightarrow 4R^2{\sin }^2\left(A\right)+4R^2{\sin }^2\left(B\right)+4R^2{\sin }^2\left(C\right)=8R^2$

$\Rightarrow 4R^2\left({\sin }^2\left(A\right)+{\sin }^2\left(B\right)+{\sin }^2\left(C\right)\right)=8R^2$

$\Rightarrow {\sin }^2\left(A\right)+{\sin }^2\left(B\right)+{\sin }^2\left(C\right)=2$

$\Rightarrow \frac{1-\cos \left(2A\right)}{2}+\frac{1-\cos \left(2B\right)}{2}+\frac{1-\cos \left(2C\right)}{2}=2$ $$\begin{gathered} \left[\because \cos \left(2\theta \right)=1-2{\sin }^2\left(\theta \right) \\ \Rightarrow {\sin }^2\left(\theta \right)=\frac{1-\cos \left(2\theta \right)}{2}\right] \end{gathered}$$

$\Rightarrow 3-(\cos 2A+\cos 2B+\cos 2C)=4$

$\Rightarrow \cos 2A+\cos 2B+\cos 2C=-1$

$\Rightarrow -1-4\cos A\cos B\cos C=-1(Cos2A+Cos2B+Cos2C=-1-4\cos A\cos B\cos C,\mathrm{when}\mathrm{A}+\mathrm{B}+\mathrm{C}=180)$

$\Rightarrow \cos A\cos B\cos C=0$

For the above equation to be $0$, either of $\cos \left(A\right)$ or $\cos \left(B\right)$ or $\cos \left(C\right)$ should be equal to $0$.

We know that, $\cos \left(90°\right)=0$.

Thus, either of angles $A,B$ or $C$ is $90°$(right-angled).

Hence, it is proved that the triangle is a right-angled triangle.