4 A port and a radar station are 3 miles apart on a straight

4. A port and a radar station are 3 miles apart on a straight shore running east and west. [so that we all have same picture please place the port at the origin of your Cartesian plane and the Radar station to the east of the port). A ship leaves the port at 1pm traveling northeast at a rate of 25 mph. If the ship maintains its speed and course, what is the rate of change of the tracking angle ( ) between the shore and the line between the radar station and the ship at 1:30pm?

Solution

HEre we have that as there is a difference of 30 minutes between two times, and ship runs at a speed of 25 mph, that means in 30 minutes, it is 12.5 miles away from the port in northeast direction surely at 45 degree angle. So if we draw a line from radar station to this shore line, it forms a triangle there in which the distance of drawn line is missing.

So on applying cosine formula here, we get

cos A = (b^2+c^2-a^2)/2bc

or cos 45 = (12.5^2+3^2-a^2)/2(12.5)(3)

0.707 =(156.25 +9 - a^2)/75

or 0.707(75) = 164.25-a^2

or 53.03 =164.25-a^2

or a^2=164.25 - 53.03 = 111.22

or a=root(111.22) =10.55

Now on again using the sine rule here, we get

sin 45/a = sin B/3

or 0.707/10.55 = sin B/3

or sin B = 0.707(3)/10.55 = 2.1213/10.55 =0.2010

or B = arcsin(0.2010)=11.6 degree.

So required rate of change of the tracking angle is 11.6 degree per hour.

Answer