In $\triangle ABC$, if $a = 287, \; b = 816, \; c = 865$, find the values of $\tan \left(\dfrac{A}{2}\right)$ and $\tan A$.
In any $\triangle ABC$,
$\tan \left(\dfrac{A}{2}\right) = \sqrt{\dfrac{\left(s - b\right) \left(s - c\right)}{s \left(s - a\right)}}$
$\tan A = \dfrac{2 \tan \left(\dfrac{A}{2}\right)}{1 - \tan^2 \left(\dfrac{A}{2}\right)}$
$s = \dfrac{a + b + c}{2} = $ semi perimeter of $\triangle ABC$
Given: $\;$ $a = 287, \; b = 816, \; c = 865$
$s = \dfrac{287 + 816 + 865}{2} = 984$
$\begin{aligned}
\tan \left(\dfrac{A}{2}\right) & = \sqrt{\dfrac{\left(984 - 816\right) \left(984 - 865\right)}{984 \left(984 - 287\right)}} \\\\
& = \sqrt{\dfrac{168 \times 119}{984 \times 697}} = \sqrt{\dfrac{7 \times 7}{41 \times 41}} = \dfrac{7}{41}
\end{aligned}$
$\begin{aligned}
\tan A & = \dfrac{2 \times \dfrac{7}{41}}{1 - \left(\dfrac{7}{41}\right)^2} \\\\
& = \dfrac{2 \times 7 \times 41}{1632} = \dfrac{287}{816}
\end{aligned}$