Calculate angle between a chord and the circle's origin I need to make an arc that has a varied size. For that I need to calculate the starting and ending angle of the chord cutting the arc. How can I do that knowing only the radius and the chord's length.

How do I calculate $\alpha$ and $\beta$?
 A: This is right triangle trig if you drop a perpendicular bisector from the chord to the circle center.
$\beta = 2\cdot \sin^{-1}(\frac{l}{2r})$
and $\alpha = \frac{180 - \beta}{2}$
A: 
There is a circle, where the center is the point $A$ and chord $EF$. Because $AE=AF=R$, where $R$ is the radius of a circle, $\triangle AEF$ is an isosceles triangle, thus $AG$ is height, bisector, and median in $\triangle AEF$. $sin\angle EAG=\frac{EF}{2R}$, from which we get $\angle EAG=arcsin\frac{EF}{2R}$. $\angle EAG$ is half of the $\angle EAF$, so $\angle EAF=2*arcsin\frac{EF}{2R}$.
Note, that $\angle FAD=\angle GAD-\angle GAF$ and $\angle EAC=\angle GAC-\angle GAE$. Because $\angle GAC=\angle GAD$ and $\angle GAE=\angle GAF$ $\Rightarrow$ $\angle EAC=\angle FAD$.  $\angle FAD=\frac{180°-\angle EAF}{2}$, which is to say $\angle FAD=\frac{180°- 2*arcsin\frac{EF}{2R}}{2}$
A: You could also use Cosine rule.
Let chord length be $x$; so after substituting values:
    $$x^2 = r^2 + r^2 - 2(r * r * \cos(∠\beta)$$
Which after simplifying would be:
    $$∠\beta = \cos^{-1}\left(\frac{2r^2 - x^2}{2r^2}\right)$$
To find $\alpha$ you can do:
    $$180^\text{o} = 2\alpha + \beta$$
Which after simplifying is:
    $$\frac{180^\text{o} - \beta}{2} = \alpha$$
