Example 2 In a convex quadrilateral $ABCD$, $\angle ABC=60^{\circ}, \angle BAD=\angle BCD=90^{\circ}, AB=2, CD=1$, the diagonals $AC, BD$ intersect at point $O$. As shown in Figure 3-6, find $\sin \angle AOB$.
(1996 Beijing High School Mathematics Competition)
Given that $\angle B A D=\angle B C D=90^{\circ}$, then $A, B, C, D$ are concyclic. Extend $B A, C D$ to intersect at $P$, then $\angle A D P=\angle A B C=60^{\circ}$.

Let $A D=x$, then $A P=\sqrt{3} x, D P=2 x$. By the secant theorem, we have $(2+\sqrt{3} x) \cdot \sqrt{3} x=2 x(1+2 x)$. Solving for $A D=x=2 \sqrt{3}-2$, and $B C=\frac{B P}{2}=4-\sqrt{3}$.
Applying Ptolemy's theorem to $A B C D$, we have
$$
\begin{aligned}
B D \cdot A C & =(4-\sqrt{3})(2 \sqrt{3}-2)+2 \cdot 1=10 \sqrt{3}-12 . \\
\text { Also, } S_{A B C D} & =S_{\triangle A B D}+S_{\triangle B C D} \\
& =(2 \sqrt{3}-2)+\frac{1}{2}(4-\sqrt{3})=\frac{3 \sqrt{3}}{2} .
\end{aligned}
$$

Thus, $\frac{1}{2}(10 \sqrt{3}-12) \cdot \sin \angle A O B=\frac{3 \sqrt{3}}{2}$. Therefore, $\sin \angle A O B=\frac{15+6 \sqrt{3}}{26}$.