<PRE><B>Question 1101656</B>
 A ship leaves a port at 1:00 pm and sails in the direction N34°W at a rate of 24 mi/hr. Another ship leaves port at 1:30 pm and sails in the direction N56°E at a rate of 20 mi/hr.
Ship # 1:: N34W has a bearing of 90+34 = 124 degrees; speed = 24 m)h; time = 2 hrs
Ship # 2:: N56E has a bearing of 90-56 = 34 degrees; speed = 20 mph; time = 1.5 hrs
 
a. Approximately how far apart are the ships at 3:00 pm?
Ship # 1 position:: (x = 3*24*cos(134) ; y = 3*24*sin(134)) = (-50.02,51.79)
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Ship # 2 position:: (x = 1.5*20*cos(34) ; y = 1.5*20*sin(34)) = (24.87,16.78)
distance = sqrt[74.89^2+35.1^2] = 82.71 miles
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b. What is bearing, to the nearest degree, from the first ship to the second?
slope = (51.79-16.78)/(-50.02-24.87) = (35.01/-74.89) = -0.4675
bearing = arctan(-0.4675) = 180-25.06 = 154.94 degrees
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Cheers,
Stan H.
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