Question

In $△ABC,a=2,b=3$ and $\sin A=\frac{2}{3}$, then find $\angle B$

Answer 1

In $△ABC,a=2,b=3,\sin A=\frac{2}{3}$

$\frac{a}{b}=\frac{\sin A}{\sin B}⟹\sin B=\frac{b}{a}\times \sin A=\frac{3}{2}\times \frac{2}{3}=1⟹\angle B={90}^{\circ }$