Question

In ∆ABC, if sin² A + sin² B = sin² C. show that the triangle is right-angled.

Answer 1

In ∆ ABC,
Given, ${\sin }^{2}A+{\sin }^{2}B={\sin }^{2}C......\left(1\right)$

Suppose $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$ .

⇒ $\sin A=\frac{a}{k},\sin B=\frac{b}{k},\sin C=\frac{c}{k}$

On putting these values in equation (1), we get:

$\frac{a^2}{k^2}+\frac{b^2}{k^2}=\frac{c^2}{k^2}\Rightarrow a^2+b^2=c^2$

Thus, ∆ ABC is right-angled.