Ta có:

`{(\hat(BAC) = 90^@ - 60^@ = 30^@),(\hat(DAB) = 90^@ - 30^@ = 60^@):}`

`<=>\hat(CAD) = 60^@ - 30^@ = 30^@`

Ta lại có:

`(DC)/(sin CAD) = (AC)/(sin 30^@)`

`<=>AC = (4sin30^@)/(sin30^@) = 4`

Nên:

`sin\hat(ACD) = (AB)/(AC)`

`<=>AB = 4sin60^@ = 2sqrt3~~3,5(m)`

`<=>BC = AC . cos60^@ = 2(m)`

`<=>DB = 4 + 2  =6(m)`

Vậy: `...`