Question:
In ∆ABC, if sin2 A + sin2 B = sin2 C. show that the triangle is right-angled
Solution:
In ∆ ABC,
Given, $\sin ^{2} A+\sin ^{2} B=\sin ^{2} C \ldots \ldots$ (1)
Suppose $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$.
$\Rightarrow \sin A=\frac{a}{k}, \quad \sin B=\frac{b}{k}, \quad \sin C=\frac{c}{k}$
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.