Question

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

Answer 1

$\sin B-\sin Q=0\Rightarrow 2\cos \left(\frac{B+Q}{2}\right)\sin \left(\frac{B-Q}{2}\right)=0$

$\Rightarrow \cos \left(\frac{B+Q}{2}\right)=0$ or $\sin \left(\frac{B-Q}{2}\right)=0$

$\Rightarrow \frac{B+Q}{2}=\frac{\pi }{2}$ $\frac{B-Q}{2}=0\Rightarrow \angle B=\angle Q$

$\Rightarrow B+Q=\pi$ But $A+B+C=\pi \Rightarrow A=0$ Not possible