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I know the radius "r" of a circle. I have a point "P", always in the circle and always "looking" at the center of the circle, with a certain angle, or overture "a". I know the distance between "P" and the center of the circle. I would like to know the angle or overture "b" from the center, so that "b" covers the same arc of the circle as "a".

Here's a schema explaining the problem :

enter image description here

The goal is to retrieve the angle "b" from all the other parameters. Thanks a lot in advance !

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  • $\begingroup$ You can retrieve b from a and p in the particular case where the red line is the angle bissector ; otherwise, it is not possible. $\endgroup$ – Jean Marie Jan 28 at 20:38
  • $\begingroup$ @JeanMarie I believe that is what the OP meant by "P is always looking at the center of the circle" $\endgroup$ – R. Burton Jan 28 at 20:42
  • $\begingroup$ Indeed, the red line is the angle bissector. In my words, I would say, the red line cuts the green angle in half :) $\endgroup$ – Xys Jan 28 at 20:47
  • $\begingroup$ Apply the sine law to the triangle with sides $d$, $r$ and the green one. $\endgroup$ – Aretino Jan 28 at 21:15
  • $\begingroup$ They call it the aperture. $\endgroup$ – Yves Daoust Jan 28 at 21:57
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Let $\alpha=a/2$ and $\beta=b/2$. Applying the sine law to the triangle with sides $d$, $r$ we get: $$ {r\over\sin\alpha}={d\over\sin(\beta-\alpha)}, $$ which after expanding $\sin(\beta-\alpha)$ becomes: $$ \sin\beta=\tan\alpha\cos\beta+{d\over r}\tan\alpha. $$ This equation can be solved, for example, plugging it into $\sin^2\beta+\cos^2\beta=1$ and solving for $\cos\beta$: $$ \cos\beta=\cos\alpha\sqrt{1-{d^2\over r^2}\sin^2\alpha}-{d\over r}\sin^2\alpha, $$ where I discarded the negative solution as $0\le\beta\le\pi/2$.

EDIT.

Here's a graph of $b$ vs. $d/r$, comparing (for $a=180°$) the exact solution above (black curve) with the approximate solution $b=(1+d/r)a$ (red curve). The difference is less pronounced for smaller values of $a$.

enter image description here

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  • $\begingroup$ Thanks a lot for your answer ! Unfortunately, I'm not sure to understand.. Is there a function that express b directly ? Like b = f(a) ? $\endgroup$ – Xys Jan 28 at 23:40
  • $\begingroup$ The final formula given by @Aretino can be expressed under the form: $b/2=$acos$\left(\cos\alpha\sqrt{1-{d^2\over r^2}\sin^2\alpha}-{d\over r}\sin^2\alpha\right)$ $\endgroup$ – Jean Marie Jan 29 at 0:01
  • $\begingroup$ @JeanMarie Thanks ! $\endgroup$ – Xys Jan 29 at 9:45
  • $\begingroup$ It seems to me that if d=r, then b=2a. Of course if d=0, then b=a. Then Isn't the solution just : b = (1 + d/r) a ? I can't prove it, but the values seems to confirm my little theorem. $\endgroup$ – Xys Jan 29 at 11:41
  • $\begingroup$ The solution cannot be written in the simple form you propose: just try both formulas for some values of $d/r$ to be convinced. Of course you could use your formula if an approximate result is enough. $\endgroup$ – Aretino Jan 29 at 13:26
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Take a look at the figure below that you will easily recognize :

enter image description here

Let us use 2 properties : a) the sine law in triangle POQ :

$$\dfrac{r}{\sin(a/2)}=\underbrace{\dfrac{c}{\sin(\pi-b/2)}}_{= \ \dfrac{c}{\sin(b/2)}}\tag{1}$$

b) orthogonal projection on axis $POH$ expressing that $PH=PO+OH$ :

$$c \cos(a/2)=d+r\cos(b/2)\tag{2}$$

It suffices now to extract the unknown $c$ from (2) and to plug it into (1) giving :

$$\sin(a/2)(d+r \cos(b/2))=r \cos(a/2)\sin(b/2)\tag{3}$$

As you want to express $b$ as a function of $a$, a good option here is to take the classical formulas (https://en.wikipedia.org/wiki/Tangent_half-angle_formula) :

$$\cos(b/2)=\frac{1-t^2}{1+t^2}, \ \ \sin(b/2)=\frac{2t}{1+t^2}, \ \text{with} \ \ t:=\tan(b/4)$$

in order to transform (3) into a quadratic in $t$. Solving it will give you two roots $t_1$ and $t_2$, out of which you will extract the solutions, under the constraint that $b/2<\pi/2$.

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  • $\begingroup$ $\sin(\pi-b/2)=\sin(b/2)$. $\endgroup$ – Aretino Jan 28 at 21:47
  • $\begingroup$ @Aretino Thanks ! $\endgroup$ – Jean Marie Jan 28 at 21:54
  • $\begingroup$ Does the final solution change if $POH$ does not bisect angle $b$? $\endgroup$ – let's have a breakdown Jan 28 at 22:01
  • $\begingroup$ @Chase Ryan Taylor Yes ; another answer is that we need a supplementary information to be able to conclude. $\endgroup$ – Jean Marie Jan 28 at 22:40
  • $\begingroup$ Thanks a lot for your answers guys ! As I said to Aretino, unfortunately, I'm not sure to understand.. Is there a function that express b directly ? Like b = f(a) ? $\endgroup$ – Xys Jan 28 at 23:42

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