In $\triangle ABC$, if $a = 9$, $b = 12$ and $A = 30^\circ$, find $c$.
Given: $\;$ $a = 9$, $\;$ $b = 12$, $\;$ $A = 30^\circ$
Now, $\;$ $b \sin A = 12 \times \sin 30^\circ = \dfrac{12}{2} = 6 < a$
$\implies$ There are two values of $B$, $\;$ $0^\circ < B < 90^\circ$ (first quadrant) and $90^\circ < B < 180^\circ$ (second quadrant)
By sine rule, $\;$ $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$
$\implies$ $\sin B = \dfrac{b \sin A}{a}$
i.e. $\;$ $\sin B = \dfrac{12 \times \sin 30^\circ}{9} = \dfrac{12}{9 \times 2} = \dfrac{2}{3}$
i.e. $\;$ $B = \sin^{-1} \left(\dfrac{2}{3}\right)$
i.e. $\;$ $B = 41.8103^\circ = 41^\circ 48' 37.08''$ $\;$ or $\;$ $B = 138.1897^\circ = 138^\circ 11' 22.92''$
By projection formula, $\;$ $c = a \cos B + b \cos A$
When $\;$ $B = 41^\circ 48' 37.08''$,
$c = 9 \cos \left(41^\circ 48' 37.08''\right) + 12 \cos \left(30^\circ\right)$
i.e. $c = 9 \times 0.7454 + 12 \times 0.8660 = 17.10$
When $\;$ $B = 138^\circ 11' 22.92''$,
$c = 9 \cos \left(138^\circ 11' 22.92''\right) + 12 \cos \left(30^\circ\right)$
i.e. $c = 9 \times \left(- 0.7454\right) + 12 \times 0.8660 = 3.68$