Trig in a hexagon

By cosine law ,

$y^2=2^2+2^2-2(2)(2)(\cos~120)=8-8\left(-\dfrac{1}{2}\right)=8+\dfrac{8}{2}=12$ $\color{#3D99F6}\large \implies$ $\color{#D61F06}\large \boxed{y=\sqrt{12}}$

By pythagorean theorem ,

$x^2=1^2+(\sqrt{12})^2=1+12=13$ $\color{#3D99F6}\large \implies$ $\color{plum}\large \boxed{x=\sqrt{13}}$ $\large \boxed{answer}$

Notice that triangle ACD is equilateral. This means that angle AXB is also 60 degrees. By applying the Law of Cosines to triangle AXB, we find that the length of segment AB is equal to $\sqrt{4^2+1^2- 2 \cdot 4 \cdot 1 \cdot cos(60)}$ = $\sqrt{13}$

We use simple coordinate geometry. Let $ACDEFG$ be the regular hexagon traversed anti-clockwise. Let $A=(0,0)$ and $C = (2,0)$ . Since $\vec(CD)$ makes $60^{\circ}$ with $AC$ , the x-axis, we have $D= (2 + 2\cos 60^{\circ} , 2\sin 60^{\circ}) = (3, \sqrt{3}).$

Also, $DB = 1$ and $DB$ makes $120^{\circ}$ with the x-axis. This gives $B = (3 + \cos 120^{\circ} , \sqrt{3} + \sin 120^{\circ}) = (\frac{5}{2} , \frac{3\sqrt 3}{2}).$

Thus $AB = \sqrt{(\frac{5}{2})^2 + (\frac{3\sqrt 3}{2})^2} = \boxed{\sqrt{13}}.$