If ∠A and ∠B are acute angles such that sinA


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



$\ln \triangle \mathrm{ABC}, \angle \mathrm{C}=90^{\circ}$

$\sin \mathrm{A}=\frac{\mathrm{BC}}{\mathrm{AB}}$ and

$\sin B=\frac{A C}{A B}$

As, sinA = sinB

$\Rightarrow \frac{\mathrm{BC}}{\mathrm{AB}}=\frac{\mathrm{AC}}{\mathrm{AB}}$

">BC = AC
So, ">A = ">B             (Angles opposite to equal sides are equal)

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