Find the length of AB in Triangle ABC In $\Delta ABC , m \angle A = 2 m \angle C$ , side $BC$ is 2 cm longer than side $AB$ . $AC = 5 $What is $AB$ ?


Well I thought you can use trigonometry or Complete Pythagoras theorem , but I don't really know how to apply it
Let $|AB|=c$, $|BC|=a=c+2$,
$|AC|=b=5$, 
$\angle BCA=\gamma$,
$\angle CAB=\alpha=2\gamma$

By the sine rule we have

\begin{align} 
\frac{\sin\alpha}{a}
&amp;=
\frac{\sin\beta}{b}
=\frac{\sin\gamma}{c}
,\\
\frac{\sin2\gamma}{c+2}
&amp;=
\frac{\sin(\pi-3\gamma)}{5}
=\frac{\sin\gamma}{c}
.
\end{align}  

By the rules based on 
componendo and dividendo,

\begin{align} 
\frac{\sin2\gamma}{c+2}
&amp;=
\frac{\sin\gamma}{c}
=\frac{\sin2\gamma-\sin\gamma}{c+2-c}
=\frac{\sin2\gamma-\sin\gamma}{2}
,\\
\frac{\sin(3\gamma)}{5}
&amp;=
\frac{\sin2\gamma-\sin\gamma}{2}
,
\end{align}

\begin{align} 
2\sin(3\gamma)
-
5\sin2\gamma+5\sin\gamma
&amp;=0
,\\
8\sin\gamma\cos^2\gamma-2\sin\gamma
-10\sin\gamma\cos\gamma+5\sin\gamma
&amp;=0
,\\
8\cos^2\gamma
-10\cos\gamma+3
&amp;=0
,
\end{align}  

so $\cos\gamma$ 
must be either $\tfrac12$ or $\tfrac34$.
It follows that possible values for $\gamma$
are $60^\circ$ or $\arccos\tfrac34\approx41.41^\circ$. 

But since $\beta=180^\circ-3\gamma$, 
$\gamma=60^\circ$ results in $\beta=0$, a degenerate case,
so the only suitable choice is

\begin{align} 
\cos\gamma&amp;=\tfrac34
,\\
\frac{\sin\gamma}c&amp;=
\frac{\sin2\gamma-\sin\gamma}2
,\\
c&amp;=\frac{2\sin\gamma}{\sin2\gamma-\sin\gamma}
,\\
&amp;=
\frac{2\sin\gamma}{2\sin\gamma\cos\gamma-\sin\gamma}
=
\frac{2}{2\cos\gamma-1}
=
\frac{2}{2\cdot\tfrac34-1}
=4.
\end{align}

Thus $\triangle ABC$ has sides $4,5$ and $6$cm.
Let $AB=x$ and $AD$ be bisector of $\Delta ABC$.

Thus, $$\Delta ABD\sim\Delta CBA,$$ which gives
$$\frac{BD}{x}=\frac{x}{x+2}$$ or
$$BD=\frac{x^2}{x+2},$$ which gives
$$DC=x+2-\frac{x^2}{x+2}=\frac{4x+4}{x+2}$$ and since
$$\frac{AB}{AC}=\frac{BD}{DC},$$ we obtain:
$$\frac{x}{5}=\frac{\frac{x^2}{x+2}}{\frac{4x+4}{x+2}},$$
which gives $x=4$.