3. In a convex quadrilateral $ABCD$, $\angle ABC = 60^{\circ}, \angle BAD = \angle BCD = 90^{\circ} \cdot AB = 2, CD = 1$, the diagonals $AC$ and $BD$ intersect at point $O$, as shown in the right figure. Then $\sin \angle AOB =$ $\qquad$
3. $\frac{15+6 \sqrt{3}}{26}$.

It is known that points $A$, $B$, $C$, and $D$ are concyclic.
Extend $BA$ and $CD$ to intersect at $P$. Then,
$$
\angle ADP = \angle ABC = 60^{\circ}.
$$

Let $AD = x$,
then $AP = \sqrt{3} x$, $DP = 2x$.
By the secant theorem, we have
$$
\begin{array}{l}
(2 + \sqrt{3} x) \cdot \sqrt{3} x \\
= 2x(1 + 2x).
\end{array}
$$

Solving for $AD = x = 2\sqrt{3} - 2$,
$$
\begin{aligned}
BC = & \frac{1}{2} BP = \frac{1}{2}[2 + 2\sqrt{3}(\sqrt{3} - 1)] \\
= & 4 - \sqrt{3}, \\
S_{ABCD} = & S_{\triangle ABD} + S_{\triangle BCD} \\
= & \frac{1}{2} \cdot 2 \cdot (2\sqrt{3} - 2) \\
& + \frac{1}{2} \cdot 1 \cdot (4 - 1 - 1) \\
= & \frac{5\sqrt{3}}{2}.
\end{aligned}
$$

According to Ptolemy's theorem (if Ptolemy's theorem is not used, the length of $AC$ can be calculated using the cosine rule in $\triangle ADC$, and the length of $BD$ can be calculated using the Pythagorean theorem in the right $\triangle ABD$. Applying Ptolemy's theorem can simplify the radical operations.), we have
$$
\begin{array}{l}
BD \cdot CA = (4 - \sqrt{3})(2\sqrt{3} - 2) + 2 \times 1 \\
\quad = 10\sqrt{3} - 12, \\
\therefore \frac{1}{2}(10\sqrt{3} - 12) \sin \angle AOB = \frac{3}{2} \sqrt{3} \\
\therefore \sin \angle AOB = \frac{3\sqrt{3}}{10\sqrt{3} - 12} \\
= \frac{3\sqrt{3}(\sqrt{300} + 12)}{156} = \frac{15 + 6\sqrt{3}}{26}.
\end{array}
$$