If $A = 30^{o}$ , $a = 7, b = 8$ in $△ A B C$ , then prove that $B$ has two solutions.

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Solution

As from sine rule,
$\frac{a}{sin A} = \frac{b}{sin B}$
$\Rightarrow sin B = \frac{b}{a} sin A = \frac{8}{7} \times \frac{1}{2} = \frac{4}{7}$
As B is between $(0, π)$ , we have two solution
$B = {sin}^{- 1} (\frac{4}{7})$ and $π - {sin}^{- 1} (\frac{4}{7})$ .

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