Question

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

Answer 1

In $∆$ABC, $\angle$C = 90$°$
sinA = $\frac{\mathrm{BC}}{\mathrm{AB}}$ and
sinB = $\frac{\mathrm{AC}}{\mathrm{AB}}$

As, sinA = sinB
$\Rightarrow$$\frac{\mathrm{BC}}{\mathrm{AB}}$ = $\frac{\mathrm{AC}}{\mathrm{AB}}$
$\Rightarrow$BC = AC
So, $\angle$A = $\angle$B (Angles opposite to equal sides are equal)