2. As shown in Figure 4, in the convex quadrilateral $ABCD$, $AB=1$, $BC=2$, $CD=\sqrt{3}$, $\angle ABC=120^{\circ}$, $\angle BCD=90^{\circ}$.
Then $AD=$ $\qquad$
2. $\sqrt{7}$.

As shown in Figure 9, take the midpoint $E$ of $BC$, and connect $ED$, $EA$. Draw $BF \perp EA$ at $F$.
By the Pythagorean theorem, we know $DE=2$, so $\angle CED=60^{\circ}$.
Since $BA=BE$, we have $\angle BEA=30^{\circ}$.
Therefore, $\angle AED=90^{\circ}$.
Also, $EA=2EF=2\sqrt{1-\frac{1}{4}}=\sqrt{3}$, then by the Pythagorean

theorem, $AD=\sqrt{AE^2+ED^2}=\sqrt{7}$.