In your diagram, all three sides of $\triangle AEC$ are known. Using the law of cosines you get $2\alpha = \cos^{-1}(7/8)$ and $\alpha \approx 14.47751219^{\circ}$.
Continuing to use the law of cosines we get $AD=\dfrac{3\sqrt{10}}{2}$ and $AB=3\sqrt{6}$.
Finally, using the law of cosines on $\triangle ABD$ we get $\alpha \approx 71.170769^{\circ}$.