∠xCE = 20o, ∠xCD = 40o => ∠ECD = 40o - 20o = 20o
∠DCB = 90o - ∠xCD = 90o - 40o = 50o
DB = CB x tanDCB = 15 x tan50 = ~17,88 (m)
EB = CB x tanECB = 15 x tan(50 + 20) = 15 x tan70 = ~41,21 (m)
=> ED = EB - DB = 41,21 - 17,88 = 23,33 (m)
Vậy khoảng cách giữa 2 thuyền là 23,33 (m)