Find the length of AB in Triangle ABC

Question

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

Answer 1

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 $x=4$.

Answer 2

Hint: with the Theorem of cosines we get

$$c^2=(c+2)^2+25-2(c+2)\cdot 5\cos(\gamma)$$ and with the Theorem of sines we get

$$\frac{\sin(3\gamma)}{\sin(\gamma)}=\frac{c}{5}$$ With these equations you can eliminate $\gamma$ and then you can compute $c$

Answer 3

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

\begin{align} 2\sin(3\gamma) - 5\sin2\gamma+5\sin\gamma &=0 ,\\ 8\sin\gamma\cos^2\gamma-2\sin\gamma -10\sin\gamma\cos\gamma+5\sin\gamma &=0 ,\\ 8\cos^2\gamma -10\cos\gamma+3 &=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&=\tfrac34 ,\\ \frac{\sin\gamma}c&= \frac{\sin2\gamma-\sin\gamma}2 ,\\ c&=\frac{2\sin\gamma}{\sin2\gamma-\sin\gamma} ,\\ &= \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.