Example 1 As shown in Figure 4-2, in the convex quadrilateral $ABCD$, $\angle ABC=60^{\circ}, \angle BAD=\angle BCD=90^{\circ}, AB=2$, $CD=1$, the diagonals $AC, BD$ intersect at point $O$. Find $\sin \angle AOB$.
(1996 Beijing High School Mathematics Competition)
Extend $BA, CD$ to intersect at $P$, let $BC=x$, then $PB=2x, PC=\sqrt{3}x$. Applying Stewart's Theorem to $\triangle PBC$ and point $A$ on $PB$, we have
$$
\begin{aligned}
CA^{2} & =PC^{2} \cdot \frac{AB}{PB}+BC^{2} \cdot \frac{PA}{PB}-AB \cdot PA \\
& =(\sqrt{3} x)^{2} \cdot \frac{2}{2 x}+x^{2} \cdot \frac{2 x-2}{2 x}-2(2 x-2) \\
& =x^{2}-2 x+4 .
\end{aligned}
$$

From the similarity of right triangles $\triangle ADP \sim \triangle CBP$, we have $PD \cdot PC = PA \cdot PB$, i.e., $(\sqrt{3} x-1) \cdot \sqrt{3} x = (2 x-2) \cdot 2 x$, solving for $BC=x=4-\sqrt{3}$.

Thus, $CA^{2}=15-6 \sqrt{3}$. In the right triangle $\triangle BCD$, $BD^{2}=x^{2}+1=20-8 \sqrt{3}$, hence $BD \cdot AC=$ $\sqrt{4(5-2 \sqrt{3})} \cdot \sqrt{3(5-2 \sqrt{3})}=10 \sqrt{3}-12$. 

And $S_{ABCD}=S_{\triangle ABD}+S_{\triangle BCD} \doteq(2 \sqrt{3}-2)+\frac{1}{2}(4-\sqrt{3})=\frac{3 \sqrt{3}}{2}$, so $\frac{1}{2}(10 \sqrt{3}-12) \cdot \sin \angle AOB=\frac{3 \sqrt{3}}{2}$, which gives $\sin \angle AOB=\frac{15+6 \sqrt{3}}{26}$ as the desired result.