∴BM∥AF.

∴∠ABM=∠BAF=30

In △BAM, AM= 12AB=5, BM=53.

The passing point c is CN⊥AH in n and BD in k.

At Rt△BCK, ∠ CBK = 90-60 = 30.

Let CK=x, then BK=3x.

In Rt△ACN,

* An island C was observed at A in the northeast,

∴∠CAN=45，

∴AN=NC.

∴AM+MN=CK+KN

And NM=BK, BM=KN.

∴ x+53 = 5+3x。 The solution is x=5.

∫5 nautical miles > 4.8 nautical miles,

Fishing boats are not in danger of entering the farm.

A: This fishing boat is not in danger of entering the farm;

Solution 2: Let point C be CE⊥BD and foot be E, as shown in the figure:

∴CE∥GB∥FA.

∴∠BCE=∠GBC=60，ACE=∠FAC=45

∴∠BCA=∠BCE-∠ACE=60 -45 = 15

∠ BAC =∠ FAC-∠ Fab = 45-30 = 15。

∴∠BCA=∠BAC，

∴BC=AB= 10

In Rt△BCE, CE=BC? cos∠BCE=BC？ COS60 = 10× 12 = 5 (nautical mile)

∫5 nautical miles > 4.8 nautical miles, ∴ fishing boats are not in danger of entering the farm.

A: This fishing boat is not in danger of entering the farm.