Given $A = 10^\circ$, $a = 2308.7$, $b = 7903.2$, find the smaller value of $c$.
Given: $\;$ $A = 10^\circ$, $\;$ $a = 2308.7$, $\;$ $b = 7903.2$
Now, $\;$ $b \sin A = 7903.2 \times \sin \left(10^\circ\right) = 1371.9955 < a$
$\implies$ There are two values of $B$, $\;$ $0^\circ < B < 90^\circ$ (first quadrant) and $90^\circ < B < 180^\circ$ (second quadrant)
By sine rule in $\triangle ABC$, $\;$ $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$
$\implies$ $\sin B = \dfrac{b \sin A}{a} = \dfrac{7903.2 \times \sin \left(10^\circ\right)}{2308.7} = 0.5943$
$\implies$ $B = \sin^{-1} \left(0.5943\right)$
i.e. $\;$ $B = 36^\circ 27' 45.72''$ $\;$ or $\;$ $B = 143^\circ 32' 14''$
Now, in $\triangle ABC$, $A + B + C = 180^\circ$ $\implies$ $C = 180^\circ - \left(A + B\right)$
When $\;$ $B = 36^\circ 27' 45.72''$, $\;$ $C = 180^\circ - \left(10^\circ + 36^\circ 27' 45.72''\right) = 133^\circ 32' 14''$
When $\;$ $B = 143^\circ 32' 14''$, $\;$ $C = 180^\circ - \left(10^\circ + 143^\circ 32' 14''\right) = 26^\circ 27' 46''$
$\therefore \;$ Smaller value of $c$ is obtained for $C = 26^\circ 27' 46''$
By sine rule we have,
$c = \dfrac{a \sin C}{\sin A} = \dfrac{2308.7 \times \sin \left(26^\circ 27' 46''\right)}{\sin \left(10^\circ\right)} = \dfrac{2308.7 \times 0.4456}{0.1736} = 5926.02$