5. In $\triangle A B C$, $A B=A C=a$, a regular triangle $B C D$ is constructed outward on side $B C, A=$
$\qquad$ when, the area of quadrilateral $A B C D$ is maximized.
5. $150^{\circ}$.

Let $\angle B A D=\theta, A D$ and $B C$ intersect at point $O, A D$ is clearly the perpendicular bisector of $B C$, and $\angle B D A=30^{\circ}$, so $\angle A B D=150^{\circ}-\theta$. In $\triangle A B D$, $\frac{A D}{\sin \left(150^{\circ}-\theta\right)}=\frac{A B}{\sin 30^{\circ}}=2 a$, thus $A D=2 a \sin \left(150^{\circ}-\theta\right)$, $B C=2 B D=2 a \sin \theta, S_{A B C D}=\frac{1}{2} B C \cdot O A+\frac{1}{2} B C \cdot O D=\frac{1}{2} B C \cdot A D=\frac{1}{2} \cdot 2 a \sin \theta$. $2 a \sin \left(150^{\circ}-\theta\right)=a^{2}\left[\frac{\sqrt{3}}{2}+\cos \left(A-150^{\circ}\right)\right]$. Therefore, when $A=150^{\circ}$, $S_{A B C D}$ is maximized.