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In the Figure below, the line segments $OA$ and $OB'$ make angle $\theta$ and $-\theta$ respectively with the positive x axis. Similiarly $AB$ is orthogonal to the x axis with $D$ the point of intersection and let $d$ be the distance between point $O$ and $D$.

$AB$ is then tilted about point $D$ by an unknown angle $\beta$ (in ccw direction) to form $A'B'$ as shown in the figure.

enter image description here

My question is given that the variables $\theta,\, d, \, y_1$ and $y_2$ are known (experimentally), how do I find the value of $\beta$?

By projecting $y_1$ and $y_2$ on the x axis and using simple algebraic and trigonometric manipulation I obtain following 2 equations for $y_1$ and $y_2$ as a function of $\beta$.

$$y_1(\beta) = \dfrac{d}{\sin(\beta) + \dfrac{\cos(\beta)}{\tan(\theta)}} \, $$ $\qquad$ and

$$y_2(\beta) = \dfrac{d}{-\sin(\beta) + \dfrac{\cos(\beta)}{\tan(\theta)}}$$

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2 Answers 2

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If $y_1,y_2,d$ and $\theta$ are known, then from

$$y_1(\beta) = \dfrac{d}{\sin(\beta) + \dfrac{\cos(\beta)}{\tan(\theta)}} \, $$ $\qquad$ and

$$y_2(\beta) = \dfrac{d}{-\sin(\beta) + \dfrac{\cos(\beta)}{\tan(\theta)}}$$

you get $$\frac{1}{y_1}+\frac{1}{y_2}=\frac{2\cos{\beta}}{d\tan{\theta}}$$ and hence $$\frac{d\cdot \tan{\theta}}{2}\cdot \left(\frac{1}{y_1}+\frac{1}{y_2}\right)=\cos{\beta}$$ from which you can determine $\beta$.

Alternatively, (see Blue's comment) you take differences to obtain

$$\frac{1}{y_1}-\frac{1}{y_2}=\frac{2\sin(\beta)}{d}$$ from which the result follows.

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    $\begingroup$ +1. Alternatively, you can write $$\frac{1}{y_1}-\frac{1}{y_2}=\frac{2\sin\beta}{d}$$ $\endgroup$
    – Blue
    Nov 5, 2019 at 21:17
  • $\begingroup$ I just updated the comment because I, too, ignored a factor (namely $d$). So, the solution is independent of $\theta$, but still depends upon $d$. $\endgroup$
    – Blue
    Nov 5, 2019 at 21:22
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    $\begingroup$ oops, got it! thanks again. By the way, I've made use of some of your work in your blog in my class as I am teaching Trigonometry at the moment. Just wanted to thank you while we're here ;) $\endgroup$
    – b00n heT
    Nov 5, 2019 at 21:25
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    $\begingroup$ It's nice to know that someone's seeing that stuff and making good use of it. ;) $\endgroup$
    – Blue
    Nov 5, 2019 at 21:27
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    $\begingroup$ thanks!!! the inverse subtraction or addition really does the trick... $\endgroup$
    – unfinished_sentenc
    Nov 5, 2019 at 22:11
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If the 2 equations in the original posts are true then the problem is solved. Credits to b00n heT and Blue.

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  • $\begingroup$ On this website, once you are happy with one of the given answer, you can accept it by clicking on the tick which you will find to the left of the answer itself :) $\endgroup$
    – b00n heT
    Nov 5, 2019 at 22:32
  • $\begingroup$ Thank you, and thanks for the nice problem! I'll make good use of it $\endgroup$
    – b00n heT
    Nov 5, 2019 at 22:52

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