If $b = 91$, $c = 125$ and $\tan \left(\dfrac{A}{2}\right) = \dfrac{17}{6}$, find the value of $a$.
By definition, $\;$ $\tan\left(\dfrac{A}{2}\right) = \sqrt{\dfrac{\left(s - b\right) \left(s - c\right)}{s \left(s - a\right)}}$ $\;\;\; \cdots \; (1)$
where $\;$ $s = \dfrac{a + b + c}{2}$ $\;$ is the semi-perimeter of $\triangle ABC$.
Given:
$b = 91$, $\;$ $c = 125$, $\;$ $\tan \left(\dfrac{A}{2}\right) = \dfrac{17}{6}$ $\;\;\; \cdots \; (2)$
Here, $\;$ $s = \dfrac{a + 91 + 125}{2} = \dfrac{a}{2} + 108$ $\;\;\; \cdots \; (3)$
Now, from equation $(1)$ we have,
$\tan^2 \left(\dfrac{A}{2}\right) = \dfrac{\left(s - b\right) \left(s - c\right)}{s \left(s - a\right)}$
i.e. $\left(\dfrac{17}{6}\right)^2 = \dfrac{\left(\dfrac{a}{2} + 108 - 91\right) \left(\dfrac{a}{2} + 108 - 125\right)}{\left(\dfrac{a}{2} + 108\right) \left(\dfrac{a}{2} + 108 - a\right)}$ $\;\;$ [by equations $(2)$ and $(3)$]
i.e. $\;$ $\dfrac{289}{36} = \dfrac{\left(\dfrac{a}{2} + 17\right) \left(\dfrac{a}{2} - 17\right)}{\left(\dfrac{a}{2} + 108\right) \left(108 - \dfrac{a}{2}\right)}$
i.e. $\;$ $\dfrac{289}{36} = \dfrac{\dfrac{a^2}{4} - 289}{11664 - a^2}$
i.e. $\;$ $289 \left(11664 - \dfrac{a^2}{4}\right) = 36 \left(\dfrac{a^2}{4} - 289\right)$
i.e. $\;$ $3370896 - \dfrac{289 a^2}{4} = \dfrac{36 a^2}{4} - 10404$
i.e. $\;$ $3381300 = \dfrac{325 a^2}{4}$
i.e. $\;$ $a^2 = 41616$
$\implies$ $a = 204$