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

Let us consider two right triangles $ABC$ and $PQR$ where $\sin B=\sin Q$(see $Fig.$)

We have $\quad \sin B =\frac{A C}{A B}$

and $\sin Q =\frac{ PR }{ PQ }$

Then $\quad \frac{A C}{A B}=\frac{P R}{P Q}$

Therefore, $\frac{A C}{P R}=\frac{A B}{P Q}=k,$ say ...........$(1)$

Now, using Pythagoras theorem,

$BC =\sqrt{ AB ^{2}- AC ^{2}}$

and $QR =\sqrt{ PQ ^{2}- PR ^{2}}$

So, $\quad \frac{ BC }{ QR }=\frac{\sqrt{ AB ^{2}- AC ^{2}}}{\sqrt{ PQ ^{2}- PR ^{2}}}=\frac{\sqrt{k^{2} PQ ^{2}-k^{2} PR ^{2}}}{\sqrt{ PQ ^{2}- PR ^{2}}}=\frac{k \sqrt{ PQ ^{2}- PR ^{2}}}{\sqrt{ PQ ^{2}- PR ^{2}}}=k$ ..........$(2)$

From $( 1 )$ and $( 2 ),$ we have

$\frac{A C}{P R}=\frac{A B}{P Q}=\frac{B C}{Q R}$

Then,$\Delta ACB \sim \Delta PRQ$ and therefore, $\angle B =\angle Q$.