Example 2 In a convex quadrilateral $ABCD$, $\angle ABC=60^{\circ}, \angle BAD=$ $\angle BCD=90^{\circ}, AB=2, CD=$ 1 , the diagonals $AC$ and $BD$ intersect at point $O$, as shown in Figure 2. Then $\sin \angle AOB=$ $\qquad$
(1996, Beijing Middle School Mathematics Competition)
Analysis: From $\angle B A D=$ $\angle B C D=90^{\circ}$, we know that points $A, B, C, D$ are concyclic. To find $\sin \angle A O B$, we think of Ptolemy's theorem, which requires us to find $B C$ and $A D$.

Solution: Since $\angle B A D=\angle B C D=90^{\circ}$, points $A, B, C, D$ are concyclic. Extend $B A$ and $C D$ 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 $(2+\sqrt{3} x) \sqrt{3} x=2 x(1+2 x)$.
Solving this, we get $A D=x=2 \sqrt{3}-2, B C=\frac{1}{2} B P=4-\sqrt{3}$.
By Ptolemy's theorem,
$$
\begin{aligned}
B D \cdot C A & =(4-\sqrt{3})(2 \sqrt{3}-2)+2 \times 1 \\
& =10 \sqrt{3}-12 . \\
\text { Also, } S_{A B C D}= & S_{\triangle A B D}+S_{\triangle X(C)}=\frac{3 \sqrt{3}}{2},
\end{aligned}
$$

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