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In a triangle $\mathrm{ABC}$, if $\mathrm{a}=3, \mathrm{~b}=4$ and $\sin \mathrm{A}=\frac{3}{4}$, then $\angle \mathrm{B}$ equals
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The correct answer is:
$90^{\circ}$
We have, $\frac{\sin A}{a}=\frac{\sin B}{b}$
or, $\quad \sin B=\frac{b}{a} \sin A$
Since, $a=3, b=4, \sin A=\frac{3}{4}$,
we get, $\sin B=\frac{4}{3} \times \frac{3}{4}=1 \quad \therefore \angle B=90^{\circ}$
or, $\quad \sin B=\frac{b}{a} \sin A$
Since, $a=3, b=4, \sin A=\frac{3}{4}$,
we get, $\sin B=\frac{4}{3} \times \frac{3}{4}=1 \quad \therefore \angle B=90^{\circ}$
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