Angle in figure consisting of a square surrounded by semi circles

Question

I'd like to know how to get the angle in the following problem:

It is a square with side equal to 1. The radius of each semi circle is equal to the side of the square. How can this angle be determined?

Answer 1

It's $30^\circ$. Let the point of intersection of "upper" arcs $BD$ and $AC$ be called $E$, and of upper $BD$ with lower arc $AC$ be called $F$. You should recognize that $\triangle ABE$ is equilateral (why?). What about $\triangle ADF$? Now finish.

Answer 2

Drop a vertical line from the top intersection to the base of the square. You now have a right triangle whose hypotenuse is $1$ and side adjacent is $0.5$. $0.5$ is the cosine of the angle between upper line and base. $\cos^{-1}(0.5)$ is $60^\circ$

In a similar fashion drop a vertical line from the lower intersection. Now you have aright triangle whose hypotenuse is $1$ and the side opposite is $0.5$. $0.5$ is the sine of the angle between hypotenuse and base. $\sin^{-1}(0.5)$ is $30^\circ$.

To get angle between these lines subtract $30^\circ$ from $60^\circ$.

Answer 3

This looks very similar to the construction of an angle of 60 degrees. You should be able to use that to find the angle.

http://www.mathopenref.com/constangle60.html