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fig1

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.

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  • $\begingroup$ I was wondering what tool did you use for this nice drawing? Thanks. $\endgroup$ – NoChance Jul 9 '19 at 22:23
  • $\begingroup$ @NoChance I used Geogebra. It has its limitations but overall a great free program for CAS and geometry. $\endgroup$ – VinMilligan Jul 10 '19 at 16:16
  • $\begingroup$ @nmasanta thanks for cleaning up my question! $\endgroup$ – VinMilligan Jul 10 '19 at 16:18
  • $\begingroup$ Please write in separate lines.. symbols of what is given and what is required. $\endgroup$ – Narasimham Jul 10 '19 at 16:33
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    $\begingroup$ @FareedAF please re-read the question. I tried to simplify what I'm looking for. Simplify the relations is one part. Essentially I'm trying to isolate the variables into expressions where alpha equals something which only contains terms of beta, and where beta equals something which only contains terms of X. $\endgroup$ – VinMilligan Jul 10 '19 at 18:27
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HINT.-Because of $\angle{HFB}=\angle{DCB}$ one has $$x= DA\tan (\pi-\beta)=\frac{DB}{\tan(\alpha)}$$

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  • $\begingroup$ Thanks @Piquito but can you elaborate further based on the problem? $\endgroup$ – VinMilligan Jul 11 '19 at 18:09
  • $\begingroup$ Look at your figure you have $$\tan(\pi-\beta)=\dfrac{x}{1.6}\iff -\tan(\beta)=\dfrac{x}{1.6}$$ so you do have $$\beta=\arctan(\frac{-x}{1.6})$$ On the other hand, the given values of $\alpha=0.98$ and $\beta=2.42$ gives a relation $$\alpha=t\beta$$ and from this you get $$\alpha=0.405\beta$$ $\endgroup$ – Piquito Jul 13 '19 at 20:29
  • $\begingroup$ Thanks @Piquito but the aim of the problem is to ignore all values of angles and lengths. We're looking for general relationships, not actual values. $\endgroup$ – VinMilligan Jul 15 '19 at 19:31

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