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

(N/A) Let us consider two right-angled triangles $ABC$ and $PQR$ such that $\angle C = 90^{\circ}$ and $\angle R = 90^{\circ}$,where $\sin B = \sin Q$.
We have $\sin B = \frac{AC}{AB}$ and $\sin Q = \frac{PR}{PQ}$.
Given $\sin B = \sin Q$,therefore $\frac{AC}{AB} = \frac{PR}{PQ}$.
This can be rewritten as $\frac{AC}{PR} = \frac{AB}{PQ} = k$ (where $k$ is a constant) .......... $(1)$.
Now,using the Pythagoras theorem in both triangles:
$BC = \sqrt{AB^2 - AC^2}$ and $QR = \sqrt{PQ^2 - PR^2}$.
Taking the ratio of the third sides:
$\frac{BC}{QR} = \frac{\sqrt{AB^2 - AC^2}}{\sqrt{PQ^2 - PR^2}}$.
Substituting $AB = kPQ$ and $AC = kPR$ from $(1)$:
$\frac{BC}{QR} = \frac{\sqrt{(kPQ)^2 - (kPR)^2}}{\sqrt{PQ^2 - PR^2}} = \frac{\sqrt{k^2(PQ^2 - PR^2)}}{\sqrt{PQ^2 - PR^2}} = \frac{k \sqrt{PQ^2 - PR^2}}{\sqrt{PQ^2 - PR^2}} = k$ .......... $(2)$.
From $(1)$ and $(2)$,we have $\frac{AC}{PR} = \frac{AB}{PQ} = \frac{BC}{QR} = k$.
By the $SSS$ similarity criterion,$\Delta ACB \sim \Delta PRQ$.
Since the triangles are similar,their corresponding angles are equal. Therefore,$\angle B = \angle Q$.