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Please answer the igcse math problem.
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SAB=90-54=36

SBA=90-(360-324)=54

ASB= 180-36-54=90

As \ ab = as \ 6 = SIN54 of SBA at the angle of sin.

Therefore, AS=4.85KM.

Bs \ ab = BS\AB=SIN angle SAB \ 6 = sin36.

The result is BS=3.53KM.

SIN∠SBA=2.5\6 ∠SBA=24.62。

The bearing of the ship is from A=24.62.