Please help! A port and a radar station are 3 mi apart on a straight shore runni

Question

Please help! A port and a radar station are 3 mi apart on a straight shore running east and west. A ship leaves the port at noon traveling at a rate of 20 mi/hr. If the ship maintains its speed and course, what is the rate of change of the tracking angle 0 between the 1 shore and the line between the radar station and the ship at i 12:30 PM? The rate of change of the tracking angle theta between the shore and the line between the radar station and the ship at is 12:30 PM is Q rad/hr. (Round to tour decimal places as needed.)

Explanation / Answer

solution

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angle ABC = theta, let us denote theta with x.

so angle ACB = 180 - x- 45 = 135-x

fron sin rule

AB/sin C = AC/sin B

so 2/sin(135-x) = AC / sin x

2sinx = AC(cosx+sinx)/sqrt(2)

2sqrt(2) sinx = AC(cosx+sinx)

differentiating with respect to t

2sqrt(2) cosx dx/dt = AC(cosx - sinx) dx/dt + (cosx+sinx)dAC/dt

dAC/dt = 20 mi/hr

so 2sqrt(2) cosx dx/dt = AC(cosx - sinx) dx/dt + (cosx+sinx)20

at time 12:30, s = 20/3 miles = AC

2sqrt(2) sinx = AC(cosx+sinx) = 20(cosx + sinx)/3

6sqrt(2) = 20 (cotx + 1)

cotx=-0.576

so x = 67.01

so cosx = 0.39

sin x= 0.92

substituting

2sqrt(2)(0.92) dx/dt = 20(-1.38) dx/dt + 1.31(20)

so dx/dt = 0.87

rate of change of theta = 0.87

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