5. As shown in Figure 1, given
$$
\begin{array}{l}
\angle B=30^{\circ}, \angle C= \\
45^{\circ}, \angle B D C=150^{\circ},
\end{array}
$$

and $B D=C D=5$. Then $A B=(\quad)$.
(A) 7.5
(B) $5 \sqrt{2}$
(C) $5 \sqrt{3}$
(D) $5+\sqrt{5}$
5. C.
Connect $A D$. It is easy to know that
$$
\angle B D C=\angle B A C+\angle B+\angle C=2 \angle B A C \text {, }
$$

and $B D=C D$.
Therefore, points $A, B, C$ lie on the circle $\odot D$ with radius $B D$.
Thus, $A B=2 B D \cos B=5 \sqrt{3}$.