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I'm revising for an exam tomorrow, but got stuck on the omission of a derivation from this:


This is an image of a triangulation sensor, I worked out how they derived r, but I've no idea where to begin with the derivation for beta. Could anyone shed some light on this for me?


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@Chimoo: It would be of help if you could paste the question here or try to draw the question here. – user17762 May 30 '11 at 17:48
put the screenshot in for you – Chimoo May 30 '11 at 17:57
@Chimoo:This may be off-topic could you please tell me the steps how to manage to get that screen shot to that form after print-screen? – Quixotic May 30 '11 at 20:45
you just paste it into paint and cut it down in there. save as png or jpg and upload it using the button provided in the edit box – Chimoo May 30 '11 at 21:08
The expression for $\beta$ looks so close to something like $$\tan(a+b)=\frac{\tan a+\tan b}{1-\tan a\tan b}$$ or $$\tan(a-b)=\frac{\tan a-\tan b}{1+\tan a\tan b}$$ and the trailing $-1$ seems so incongruous that I think there's probably a typesetting mistake in there somewhere, but I'm not really sure. – Isaac May 30 '11 at 22:29
up vote 1 down vote accepted

Hint: Use sine theorem. I'll post the complete answer when I get home.

I seem to be getting some problems with that $-1$ constant after $\arctan$. Are you sure those are the right formulas?

The idea to get angle $\beta$ in terms of $\alpha_1,\alpha_2$ is the following: Apply sine theorem in the two halves of triangle and obtain $\displaystyle \frac{\sin(\beta+\alpha_1+\alpha_2)}{\sin \beta}=\frac{\sin \alpha_2}{\sin\alpha_1}$. From here, using usual trigonometric formulas, you can extract $\tan \beta$, and then try and prove it has a closed form as presented in your formulas.

Here are some more details. We have $\frac{\sin \beta \cos(\alpha_1+\alpha_2)+\cos\beta\sin(\alpha_1+\alpha_2)}{\sin\beta}=\frac{\sin\alpha_2}{\sin\alpha_1}$. Then $\cos(\alpha_1+\alpha_2)+\cot\beta\sin(\alpha_1+\alpha_2)=\frac{\sin\alpha_2}{\sin\alpha_1}$. From here, you get a formula for $\cot \beta$, then you invert and get $\tan \beta$. I didn't manage to get your closed form, but if there is a way, this is the path. :)

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they are copied from the paper. perhaps it is wrong, whats the solution without it? – Chimoo May 30 '11 at 21:07
thanks for answering, I don't understand how you would go from that representation to the $tan$ representation given? – Chimoo May 31 '11 at 7:34

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