Triangle inscribed in two squares

Since the circle's area equals $\pi$ , the radius must be 1. Now let's call the angle FOG $2\alpha$ . The area of the blue triangle can be written as function of $\alpha$ . The height of the triangle equals $cos(\alpha)$ and the base equals $2sin(\alpha)$ . Now the area of the blue triangle is: $\frac{1} {2}\cdot 2sin(\alpha)cos(\alpha)= \frac{1} {2}\cdot sin(2\alpha) = 0.25$ . So $sin(2\alpha) = 0.5$ and $2\alpha = 30$ . This leads to $\angle AOF = 90 - 30 = 60$ .