The segments \(BC\) and \(AC\) have length \(1\).
The angle \(\widehat{BCA}\) is \(30^{\circ}\), see
.
By applying the law of cosines in the triangle \(ABC\) we get
\[AB^2= 1 + 1 -2\cdot \cos(30^{\circ})=2-\sqrt{3}\]
and hence
\[AB = \sqrt{2-\sqrt{3}}\,.\]