There's something I'm missing here and I need to reach out for some help. So if you're a trig expert, I need your advice.
Referring to the current diagram (fig 1), I'm trying to find alpha in terms of beta, and then beta in terms of $X$. Pay no attention to the values of the angles since point C can move freely.
$AC = l$
$BC = m$
$DC = X$
FB = EA = 1
The angle $\beta ~ \beta = 2~\beta$
$AB = \frac{1}{2}$
We know this:
$m ~\sin(\alpha) = l ~\sin(2~\beta - \frac{\pi}{2}) + \frac{1}{2}$
$X = \frac{1}{2}~\cos(\alpha) ~\sin(2~\beta) ~\cos(\alpha - 2~\beta)$
Essentially, what I'm looking for are expressions where:
$\alpha$ is an expression which only contains the variable $\beta$
and where:
$\beta$ is an expression which only contains the variable $X$
The variables $l$ and $m$ should not be included in such expressions.
I've been racking my brains over this construction for weeks and I'm not academically trained in these matters, so I'm sure someone knows this easily. Thank you for any help.
HINT.-Because of $\angle{HFB}=\angle{DCB}$ one has $$x= DA\tan (\pi-\beta)=\frac{DB}{\tan(\alpha)}$$