5. As shown in Figure 1, in the right triangle $\triangle ABC$, the hypotenuse $BC=4$, $\angle ABC=30^{\circ}$, and circles are drawn with $AB$ and $AC$ as diameters, respectively. Then the area of the common part of these two circles is ( )
(A) $\frac{2 \pi}{3}+\frac{\sqrt{3}}{2}$
(B) $\frac{5 \pi}{6}-\frac{2 \sqrt{3}}{3}$
(C) $\frac{5 \pi}{6}-\sqrt{3}$
(D) $\frac{2 \pi}{3}-\sqrt{3}$
5.C.

Let the area of the common part of the two circles be $S$. As shown in Figure 4, it is easy to see that
$$
\begin{array}{l}
=\frac{\pi}{3} \times 1^{2}+\frac{\pi}{6} \times(\sqrt{3})^{2}-\frac{1}{2} \times 2 \times \sqrt{3}=\frac{5 \pi}{6}-\sqrt{3} . \\
\end{array}
$$