Question

In a triangle $\Delta ABC$ it is given ${\sin }^{2}A+{\sin }^{2}B={\sin }^{2}C$ then prove that the triangle is right-angled.

Answer 1

Given,

${\sin }^{2}A+{\sin }^{2}B={\sin }^{2}C------------\left(1\right)$

From since rule of triangle we get,

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

or, $\sin A=\frac{a}{2R},\sin B=\frac{b}{2R};\sin C=\frac{c}{2R}$

Using these in equation $\left(1\right)$ we get,

$\frac{a^2}{4R^2}+\frac{b^2}{4R2}=\frac{C^2}{4R^2}$4$

or, $a^2+b^2=c^2$,

This gives, $△ABC$ is might-angled triangle.