If A and B are two angles such that A, B, $\in(0, \pi)$ and they are not supplementary angles such that…
If A and B are two angles such that A, B, $\in(0, \pi)$ and they are not supplementary
angles such that $\sin A-\sin B=0$, then
$A-B=\frac{\pi}{3}$
$A-B=\frac{\pi}{2}$
$A=B$
$A \neq B$
Solution
$\sin A-\sin B=0$
$\sin A=\sin B$ and we know that $\sin A=\sin (\pi-A)=\sin B$
$\therefore \mathrm{A}=\mathrm{B}$ or $\pi-\mathrm{A}=\mathrm{B}$
$\therefore \mathrm{A}=\mathrm{B}$ or $\mathrm{A}+\mathrm{B}=\pi$
Since the angles are not supplementary we say $\mathrm{A}=\mathrm{B}$.