In RT triangle ABC, BC = ab/tan30 = 10 √ 3m.
In the RT triangle ABD, BD = ab/tan45 = 10m.
Because the angle DBC = 15+75 = 90.
So in BDC of RT triangle, DC = √ [(10 √ 3) 2+10 2] m = 20m.
Therefore, the ship speed is 20m/min.
2. According to the meaning of the question, the angle EBD = 90-75 = 15, and because DC in the RT triangle BDC = 2bd.
So angle BDC = 60, angle BDE =120, bed = 45.
EB=BD*sinBDE/sinBED=5√6m
So 5√6m off the island.