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

Question

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?

Thanks

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    You are right, I found a different paper and uploaded their result, which replaces the$-1$with an alpha 1. Does this help?2011-05-31

1 Answers 1

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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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    thanks for answering, I don't understand how you would go from that representation to the $tan$ representation given?2011-05-31