If $b = 90$, $c = 70$ and $A = 72^\circ 48' 30''$, find $B$ and $C$.
Given: $\;$ $b = 90$, $\;$ $c = 70$, $\;$ $A = 72^\circ 48' 30''$
By definition,
$\begin{aligned}
\tan \left(\dfrac{B - C}{2}\right) & = \dfrac{b - c}{b + c} \cot \left(\dfrac{A}{2}\right) \\\\
& = \dfrac{90 - 70}{90 + 70} \cot \left(\dfrac{72^\circ 48' 30''}{2}\right) \\\\
& = \dfrac{20}{160} \cot \left(36^\circ 24' 15''\right) \\\\
& = \dfrac{1}{8} \times 1.3562 = 0.1695
\end{aligned}$
$\therefore \;$ $\dfrac{B- C}{2} = \tan^{-1} \left(0.1695\right) = 9.6202^\circ$
$\implies$ $B - C = 2 \times 9.6202^\circ = 19.2404^\circ = 19^\circ 14' 25.44''$ $\;\;\; \cdots \; (1)$
Now, in $\triangle ABC$, $\;$ $A + B + C = 180^\circ$
$\implies$ $B + C = 180^\circ - A$
i.e. $\;$ $B + C = 180^\circ - 72^\circ 48' 30'' = 107^\circ 11' 30''$ $\;\;\; \cdots \; (2)$
Adding equations $(1)$ and $(2)$ we have,
$2 B = 126^\circ 25' 55.44''$ $\implies$ $B = 63^\circ 12' 57.78''$
Substituting the value of $B$ in equation $(2)$ gives,
$C = 107^\circ 11' 30'' - 63^\circ 12' 57.78'' = 43^\circ 58' 32''$
$\therefore \;$ $B = 63^\circ 12' 57.78''$, $\;$ $C = 43^\circ 58' 32''$