Three Angles and a Ratio

If $O$ is the center of the circle in which quadrangle $ABCD$ is inscribed, then $\angle BOC=120^\circ$ and $\angle OBC=30^\circ$ . If we pick the radius of the circle to be 1, then $BC=2\times cos(30^\circ)=\sqrt{3}$ . Therefore $AB=\frac{\sqrt{3}}{2}$ . Half of it is $\frac{\sqrt{3}}{4}$ , which gives us $\angle ABO=arccos(\frac{\sqrt{3}}{4})\approx64.34^\circ$ . Since $\angle DBC=180^\circ-60^\circ-75^\circ=45^\circ$ . The angle we are searching for $\alpha=\angle ABD$ has to satisfy $\alpha+45^\circ=64.34^\circ+30^\circ$ so $\alpha=49.34^\circ$ .