How to solve for the angle of an arc in this circumstance? This is a shape I came upon during work (3D modelling), and I have been trying to figure out how to better describe it.
The shape is comprised of four equal arcs of a circle, such that the radius of each arc is the distance between their midpoints, and the origin of each arc is the midpoint of the arc opposite:

How to I determine the angle of the arc, $\theta$? I know from playing around graphically that $\theta\approx48.59°$, but I would love to know how to actually solve for it. It should be noted that $\theta$ is a constant, and is independent of the radius $r$.
 A: Consider the triangle formed by the centers of two adjacent arcs and their intersection point. It has sides $r,r,\frac{r}{\sqrt{2}}$(To see that the third side is $\frac{r}{\sqrt{2}}$, let the tangents to to the arcs at their respective midpoints meet at $A,B,C,D$, it is easy to see that they form a square and the side of the square is $r$. The line joining the two centers of the arcs is the line joining the midpoint of two adjacent sides of the square and its length is obviously $\frac{r}{\sqrt{2}}$!) and one of the two equal angles of the triangle is $\frac{\theta}{2}+45^{\circ}$(This is by symmetry). In the isosceles triangle in consideration, simple trigonometry yields $\cos(\frac{\theta}{2}+45^{\circ})=\frac{1}{2\sqrt{2}}$. Now $$-\sin(\theta)=\cos(\theta+90^{\circ})=\cos(2\cdot(\frac{\theta}{2}+45^{\circ}))=2(\frac{1}{2\sqrt{2}})^2-1=\frac{-3}{4}$$ $$\implies \sin(\theta)=\frac{3}{4} \implies \theta=\arcsin(\frac{3}{4})\approx 48.59^{\circ}$$ Just to make things clearer, here is the diagram: $IEG$ is the triangle in consideration.

A: Claim:
$$\theta=(2\arccos\frac{1}{2\sqrt2}-90)^\circ\approx 48.6^\circ.$$
Proof:

Firstly, it is clear that $ABCD$ is a square.
By resolving, we find that $HE=r\cos\frac{\theta}{2}$, and so $EF=r-r\cos\frac{\theta}{2}$.
By symmetry, this means that $GH=EF=r-r\cos\frac{\theta}{2}$. This in turn means that the length of each side of the square - given by $AB$ and is also given by $EG$- can be written as  $$AB=EG=r-2(r-r\cos\frac{\theta}{2})=2r\cos\frac{\theta}{2}-r.$$
However, we can also write $AB$ in a different way:
$$AB=2AE=2r\sin\frac{\theta}{2}.$$
Equating our two expressions for $AB$:
$$\begin{align}&2r\sin\frac{\theta}{2}=2r\cos\frac{\theta}{2}-r\\
\implies &\cos\frac{\theta}{2}-\sin\frac{\theta}{2}=\frac{1}{2}\\
\implies &\frac{1}{\sqrt2}\left(\cos\frac{\theta}{2}-\sin\frac{\theta}{2}\right)=\frac{1}{2\sqrt2}\\
\implies&\cos\left(\frac{\theta}{2}+45^\circ\right)=\frac{1}{2\sqrt2}\\
\implies&\frac{\theta}{2}+45^\circ=\arccos\frac{1}{2\sqrt2}\\
\implies & \theta=(2\arccos\frac{1}{2\sqrt2}-90)^\circ.\end{align}$$
