A Question about 2018, and Angles

Let $O$ be center of this semicircle (with radius $r=1009$ ). Given that inscribed angle $\angle{XAY} = 20^{\degree} \Rightarrow$ $\widehat{XY} = 40^{\degree}$ and:

$m = |\widehat{XY}| = r \cdot \angle{XOY} = 1009 \cdot (40 \cdot \frac{\pi}{180}) = \frac{2018\pi}{9}$ .

Next, draw radius $OO'$ such that $\overline{ OO'} \perp \overline{XY}$ at the point $Z$ and $\overline{OO'}$ bisects $\angle{XOY}.$ Using either one of right triangles $\triangle{OZX}, \triangle{OZY}$ we find that:

$n = |\overline{XY}| = 2r \cdot \sin \angle{XOZ} = 2r \cdot \sin \angle{YOZ} = 2(1009) \sin(20^{\degree}) \approx 690.2$

Thus, $m-n = \frac{2018\pi}{9} - 690.2 \approx \boxed{14.22}$ .

$NOTE:$ The angle $\angle{YAB}$ is entirely extraneous to our solution!