If $a = 2b$ and $C = 120^\circ$, find the values of $A$, $B$ and the ratio of $c$ to $a$.
In $\triangle ABC$,
$\tan \left(\dfrac{A - B}{2}\right) = \dfrac{a - b}{a + b} \cot \left(\dfrac{C}{2}\right)$
$\begin{aligned}
i.e. \; \tan \left(\dfrac{A - B}{2}\right) & = \dfrac{2b - b}{2b + b} \cot \left(\dfrac{120^\circ}{2}\right) \\\\
& = \dfrac{1}{3} \cot 60^\circ \\\\
& = \dfrac{1}{3 \sqrt{3}} = 0.1925
\end{aligned}$
$\implies$ $\dfrac{A - B}{2} = \tan^{-1} \left(0.1925\right) = 10.8962^\circ$
i.e. $\;$ $A - B = 2 \times 10.8962^\circ = 21.7924^\circ$ $\;\;\; \cdots \; (1)$
In $\triangle ABC$, $\;$ $A + B + C = 180^\circ$
i.e. $\;$ $A + B = 180^\circ - C = 180^\circ - 120^\circ = 60^\circ$ $\;\;\; \cdots \; (2)$
Adding equations $(1)$ and $(2)$ we have,
$2 A = 60^\circ + 21.7924^\circ = 81.7924^\circ$
$\implies$ $A = 40.8962^\circ = 40^\circ 53' 46.32''$
Substituting the value of $A$ in equation $(2)$ we have,
$B = 60^\circ - A = 60^\circ - 40^\circ 53' 46.32'' = 19^\circ 6' 14''$
$\therefore \;$ $A = 40^\circ 53' 46.32''$, $\;$ $B = 19^\circ 6' 14''$
Now, in $\triangle ABC$, by cosine rule we have,
$\cos C = \dfrac{a^2 + b^2 - c^2}{2 a b}$
i.e. $\;$ $\cos 120^\circ = \dfrac{a^2 + \dfrac{a^2}{4} - c^2}{2 \times a \times \dfrac{a}{2}}$
i.e. $\;$ $-\dfrac{1}{2} = \dfrac{\dfrac{5 a^2}{4} - c^2}{a^2}$
i.e. $\;$ $- a^2 = \dfrac{5 a^2}{2} - 2 c^2$
i.e. $\;$ $\dfrac{7 a^2}{2} = 2 c^2$
i.e. $\;$ $\dfrac{c^2}{a^2} = \dfrac{7}{4}$
$\implies$ $\dfrac{c}{a} = \dfrac{\sqrt{7}}{2}$