Question 144, As shown in the figure, plane \(ABDE \perp\) plane \(\mathrm{ABC}\), \(\triangle \mathrm{ABC}\) is an isosceles right triangle, \(A C = B C = 4\), quadrilateral \(\mathrm{ABDE}\) is a right trapezoid, \(\mathrm{BD} / / \mathrm{AE}, \mathrm{BD} \perp \mathrm{AB}, \mathrm{BD} = 2, \mathrm{AE} = 4\), points \(O\) and \(\mathrm{M}\) are the midpoints of \(\mathrm{CE}\) and \(\mathrm{AB}\) respectively. Find the sine value of the angle formed by line \(\mathrm{CD}\) and plane \(\mathrm{ODM}\).
Question 144, Solution: As shown in the figure, according to the conditions, we can supplement the figure to form a cube EFGH$A N B C$ with a side length of 4. Taking the midpoint of $A C$ as $K$, then $O K$ is parallel and equal to $B D$, so quadrilateral $O D B K$ is a rectangle. Taking the midpoint of $A K$ as $\mathrm{B}_{2}$, then $\mathrm{ML} / / \mathrm{BK} / / 0 \mathrm{D}$.

Since $S_{\triangle O L C}=\frac{L C \cdot O K}{2}=\frac{3 \times 2}{2}=3$, and the distance from $D$ to the plane $O L C$ is 4, thus $\mathrm{CO}-\mathrm{OLD}=\frac{3 \times 4}{3}=4$. Also, $\mathrm{OL}=\sqrt{5}, \mathrm{OD}=2 \sqrt{5}, \mathrm{LD}^{2}=\sqrt{29}$, so according to Heron's formula, $\mathrm{S}_{\triangle \mathrm{OLD}}=2 \sqrt{6}$. Therefore, we have
$$
4=V_{C-O L D}=\frac{2 \sqrt{6} \cdot d(C, O L D)}{3} \Rightarrow d(C, O L D)=\sqrt{6}
$$

Noting that $C D=2 \sqrt{5}$, the sine value of the angle formed by the line $C D$ and the plane $O D M$ is $\frac{\sqrt{6}}{2 \sqrt{5}}=\frac{\sqrt{30}}{10}$.