Find all the angles of $\triangle ABC$ when $a = 25$, $b = 26$ and $c = 27$.
Given: $\;$ $a = 25$, $\;$ $b = 26$, $\;$ $c = 27$
In $\triangle ABC$, $\;$ $\tan \left(\dfrac{A}{2}\right) = \sqrt{\dfrac{\left(s - b\right) \left(s - c\right)}{s \left(s - a\right)}}$, $\;$ $\tan \left(\dfrac{B}{2}\right) = \sqrt{\dfrac{\left(s - c\right) \left(s - a\right)}{s \left(s - b\right)}}$
where $s$ is the semi-perimeter of $\triangle ABC$ $= \dfrac{a + b + c}{2}$
Here, $\;$ $s = \dfrac{25 + 26 + 27}{2} = 39$
$\begin{aligned}
\tan \left(\dfrac{A}{2}\right) & = \sqrt{\dfrac{\left(39 - 26\right) \left(39 - 27\right)}{39 \times \left(39 - 25\right)}} \\\\
& = \sqrt{\dfrac{13 \times 12}{39 \times 14}} = \sqrt{\dfrac{2}{7}} = 0.5345
\end{aligned}$
$\implies$ $\dfrac{A}{2} = \tan^{-1} \left(0.5345\right) = 28.1245^{\circ}$
$\implies$ $A = 2 \times 28.1245^{\circ} = 56.2490^{\circ} = 56^{\circ}14'56.4''$
$\begin{aligned}
\tan \left(\dfrac{B}{2}\right) & = \sqrt{\dfrac{\left(39 - 27\right) \left(39 - 25\right)}{39 \times \left(39 - 26\right)}} \\\\
& = \sqrt{\dfrac{12 \times 14}{39 \times 13}} = 0.5757
\end{aligned}$
$\implies$ $\dfrac{B}{2} = \tan^{-1} \left(0.5757\right) = 29.9290^{\circ}$
$\implies$ $B = 2 \times 29.9290^{\circ} = 59.858^{\circ} = 59^{\circ}51'28.8''$
$\therefore \;$ $C = 180^{\circ} - \left(56^\circ 14' 56.4'' + 59^\circ 51' 28.8''\right) = 180^\circ - 116^\circ 6' 25.2'' = 63^\circ 53' 35''$
$\therefore \;$ $A = 56^\circ 14' 56.4''$, $\;$ $B = 59^\circ 51' 28.8''$, $\;$ $C = 63^\circ 53' 35''$