2 times chord length equals supplementary arc length This is a problem that has bugged me for a long time. It involves a diameter and a circle with radius 1. I wanted to find the point at which the distance between the point and one end of the diameter was $\frac{1}{2}$ the length of the arc connecting the point to the other end of the diameter. 

I have used a program to find the angle's approximate measure ($118.014011183903^\circ$), and the arc's approximate length ($2.05973305864452$), but would like to know a mathematical way to solve for either one of these, as each leads to the other.
 A: If you bisect the chord you get two right triangles which show $\frac x2=\sin (\frac {\pi-\theta}2)$ and from the arc you have $2x=2 \pi \theta$.  These are two equations in two unknowns, but the mix of polynomial and trig suggests that there will not be an algebraic solution.  Your solution is quite believable because if $x$ were the arc instead of the chord it would be obvious that $\theta=\frac {2\pi} 3$ and your solution is a little smaller because the chord represents a little more angle than the arc.  I think you have done everything that there is to do.  Congratulations.
A: Here is a different approach: In the triangle containing the $x$ you can apply the law of cosines to arrive at $x^2=2+2cos\theta$. (I also applied the identity cos(A-B)=cosAcosB+sinAsinB). The circle has radius $1$ and thus circumference $2\pi$ and thus $2x=\theta$. Now eliminating $\theta$ from those two equations ultimately gives $\theta^2$=4(2+2cos\theta$ and putting this in DESMOS confirms your angle approximation: https://www.desmos.com/calculator . There is no exact answer possible
A: Consider unit radius:
$$ 2 \sin^{-1} \frac{x/2}{1} + 2 \,x = \pi$$
$$ \theta\,\text{marked on your drawing}\, = 2 \,x \, \text {radians}$$ 
$$\rightarrow \theta \approx 2.06 \text{ radians} $$.
