Trigonometry on circle as function of distances to (-r,0 ) and (r,0) I have two point A and B on a circle centered at the origin $ O = (0,0)$ with radius r
And I am only told:


*

*A and B are both on the upper half plane ($ y \ge 0 $ ) 

*the distance $a_1$ from A to the point $(r,0)$ 

*the distance $a_2$ from A to the point $(-r,0)$ 

*the distance $b_1$ from B to the point $(r,0)$ 

*the distance $b_2$ from B to the point $(-r,0)$ 


(PS $r$ itself is not given) 
How can I calculate the $\angle AOB$ ? 
 A: 

This is the picture you should have in mind, where $AE = a1,AD=a2,BE=b1,BD=b2$.The answer to your question is $\lvert {\alpha - \beta}\lvert$.
To find $\alpha$ you can write law of cosines in the $\bigtriangleup{AOE}$ and $\bigtriangleup{AOD}$. Respectively
$$a1^2 = 2r^2-2r^2\cos{\alpha}$$
$$a2^2 = 2r^2+2r^2\cos{\alpha}$$
From these two equations it's easy to find $\alpha = \arccos{\frac{a_{1}^2-a_{2}^2}{a_{1}^2+a_{2}^2}}$ 
The process of finding $\beta$ is identical and $\beta = \arccos{\frac{b_{1}^2-b_{2}^2}{b_{1}^2+b_{2}^2}}$

So answer to your question is $\lvert \arccos{\frac{a_{1}^2-a_{2}^2}{a_{1}^2+a_{2}^2}} - \arccos{\frac{b_{1}^2-b_{2}^2}{b_{1}^2+b_{2}^2}} \lvert$
A: Hint:
let $M=(r,0)$ and $N=(-r,0)$. 
The triangles $AMN$ and $BMN$ are rectangles and :
1) the height $AH$ with respect to the hypotenuse $MN$ is $AH=r \sin(\angle AOM)$
2) the height $BK$ with respect to the hypotenuse $MN$ is $BK= r\sin(\angle BOM)$
3) $\overline{AN}\times \overline{AM}=2r\overline{AH}$ and $\overline{BN}\times \overline{BM}=2r\overline{BK}$
4) $\angle BOA = \angle BOM - \angle AOM$
A: For sake of convenience let A be closer to C(-r,0) and B closer to D(r,0). O is the origin. We know  $\mathbf{\angle AOC, \angle BOD}$ from cosine rule (sides OC = OA = OB = OD = r and $\mathbf{AC = a_2}$ and $\mathbf{BD = b_1}$). Now $\mathbf{\angle AOC + \angle AOB + \angle BOD = 180^0}$, so $\mathbf{\angle AOB = 180^0 - (\angle AOC + \angle BOD)}$   
