Question

If $\angle Aand\angle B$ are acute angles such that sin A = sin B then prove that $\angle A=\angle B$.

Answer 1

$In△ABC,\angle C={90}^{o}sinA=\frac{BC}{AB}andsinB=\frac{AC}{AB}AssinA=sinB\Rightarrow \frac{BC}{AB}=\frac{AC}{AB}\Rightarrow BC=ACSo,\angle A=\angle B$

(Angles opposite to equal sides are equal)