Question
In $\triangle A B C, a=3, b=4$ and $\sin A=\frac{3}{4}$, find $\angle B$

Answer

By sine rule,
$ \frac{a}{\sin A}=\frac{b}{\sin B}$
$\therefore \frac{3}{\frac{3}{4}}=\frac{4}{\sin B}$
$\therefore \sin B=1=\sin \frac{\pi}{2}$
$\therefore \angle B=\frac{\pi}{2} $

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