【解析】:(1)设CD的长为x米,则 $\tan \alpha = \frac { x } { 3 5 } , \tan \beta = \frac { x } { 8 0 } , \ \because \frac { \pi } { 2 } > \alpha \geq 2 \beta > 0$
$$
\therefore \tan \alpha \geq \tan 2 \beta , \ \therefore \tan \alpha \geq { \frac { 2 \tan \beta } { 1 - \tan ^ { 2 } \beta } } , \ \therefore { \frac { x } { 3 5 } } \geq { \frac { 2 { \frac { x } { 8 0 } } } { 1 - { \frac { x ^ { 2 } } { 6 4 0 0 } } } } = { \frac { 1 6 0 x } { 6 4 0 0 - x ^ { 2 } } } ,
$$
解得 $0 < x \leq 2 0 \sqrt { 2 } \approx 2 8 . 2 8$ ,∴CD的长至多为28.28米
> ②设 $D B = a , D A = b , D C = m \ , \angle A D B = 1 8 0 ^ { \circ } - \alpha - \beta = 1 2 3 . 4 3 ^ { \circ }$
则 $\frac { a } { \sin \alpha } = \frac { A B } { \sin \angle A D B }$ ,解得 $a = { \frac { 1 1 5 \sin 3 8 . 1 2 ^ { \circ } } { \sin 1 2 3 . 4 3 ^ { \circ } } } \approx 8 5 . 0 6$
$: m = { \sqrt { 8 0 ^ { 2 } + a ^ { 2 } - 1 6 0 a \cos 1 8 . 4 5 ^ { \circ } } } \approx 2 6 . 9 3 , \ : C D$ 的长为 26.93米