Get an angle from the center of circle, based on other angle in the circle

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 :

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

• 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. – Jean Marie Jan 28 at 20:38
• @JeanMarie I believe that is what the OP meant by "P is always looking at the center of the circle" – R. Burton Jan 28 at 20:42
• Indeed, the red line is the angle bissector. In my words, I would say, the red line cuts the green angle in half :) – Xys Jan 28 at 20:47
• Apply the sine law to the triangle with sides $d$, $r$ and the green one. – Aretino Jan 28 at 21:15
• They call it the aperture. – Yves Daoust Jan 28 at 21:57

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 $$a$$.

• 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) ? – Xys Jan 28 at 23:40
• The final formula given by @Aretino can be expressed under the form: $b/2=$acos$\left(\cos\alpha\sqrt{1-{d^2\over r^2}\sin^2\alpha}-{d\over r}\sin^2\alpha\right)$ – Jean Marie Jan 29 at 0:01
• @JeanMarie Thanks ! – Xys Jan 29 at 9:45
• 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. – Xys Jan 29 at 11:41
• 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. – Aretino Jan 29 at 13:26

Take a look at the figure below that you will easily recognize :

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 $$b/2<\pi/2$$.

• $\sin(\pi-b/2)=\sin(b/2)$. – Aretino Jan 28 at 21:47
• @Aretino Thanks ! – Jean Marie Jan 28 at 21:54
• Does the final solution change if $POH$ does not bisect angle $b$? – let's have a breakdown Jan 28 at 22:01
• @Chase Ryan Taylor Yes ; another answer is that we need a supplementary information to be able to conclude. – Jean Marie Jan 28 at 22:40
• 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) ? – Xys Jan 28 at 23:42