3. As shown in Figure $2, A B$ is the diameter of a semicircle, chords $A D$ and $B C$ intersect at point $P$, it is known that $\angle D P B=$ $60^{\circ}, D$ is the midpoint of $\overparen{B C}$. Then $\tan \angle A D C$ equals ( ).
(A) $\frac{1}{2}$
(B) 2
(C) $\sqrt{3}$
(D) $\frac{\sqrt{3}}{3}$
3. D.

Connect $A C$ (figure omitted). Let $\angle D A C=\angle D A B=\alpha, \angle A B C=\beta$. Then $\alpha+\beta=60^{\circ}, 2 \alpha+\beta=90^{\circ}$. Solving these, we get $\alpha=30^{\circ}$. Therefore, $\tan \angle D A C=\frac{\sqrt{3}}{3}$.