62
60
65
64

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Since $\angle B = \angle C = 90^{\circ},$ $AC$ is a diameter of $\circledcirc ABCD.$ By cosine rule on $\triangle ABD,$ we find out that $BD= \sqrt{AB^2 + AD^2 - 2 \cdot AB \cdot AD \cdot \cos \angle A} = \sqrt{2883}.$ By Sine rule on $\triangle ABD,$ $\dfrac{BD}{\sin \angle A} = AC \implies AC = \dfrac{2 \sqrt{2883}}{\sqrt{3}} = \boxed{62}.$