36. In quadrilateral $A B C D$, $A B=4, B C=1, C D=7$. If $\angle D A B=\angle B C D=120^{\circ}$. Then, $D A=(\quad)$.
(A) $2+3 \sqrt{5}$
(B) $2 \sqrt{3}+\sqrt{46-8 \sqrt{3}}$
(C) $2 \sqrt{3}+\sqrt{46+8 \sqrt{3}}$
(D) $-2+3 \sqrt{5}$
$(\mathrm{E}) *$
36. (D).

As shown in Figure 3, we have
$$
\begin{array}{l}
A B^{2}+A D^{2}+A B \cdot 4 I \\
=B C^{2}+C D^{2}+B C \cdot C D \text {. } \\
4^{2}+A D^{2}+4 \cdot D \\
=1^{2}+7^{2}+7 \text {. } \\
\end{array}
$$

That is,
then $A D^{2}+4 A D=41$.
Thus, $A D=-2+3 \sqrt{5}$.