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 18 mi/hr. If the ship maintains its speed and course, what is the rate of change of the tracking angle theta between the shore and the line between the radar station and the ship at 12:30 PM? (Hint: Use the law of sines.)

Use the image provided here for clarification purposes. It helps so much to make a drawing in this sort of questions. The tracking angle here is drawn as [tex]\theta[/tex].

Law of sines is applied to the vertical side (ship's trayectory) and the horizontal side (shore line). Here, time must be expressed in hours:

[tex]\frac{sin \theta}{18t}=\frac{sin(\pi /2 - \theta)}{3}=\frac{cos\theta}{3}[/tex]

[tex]tan(\theta)=\frac{18t}{3} = 6t[/tex].

[tex]\theta=arctan(6t)[/tex].

So, the rate of change of our tracking angle, [tex]\theta[/tex], can be obtained by deriving this expression:

[tex]\frac{d \theta}{dt}=6\frac{1}{1+(6t)^2}[/tex].

At 12:30 PM, i.e., half an hour after departure, t = 1/2, so:

[tex]\frac{d \theta}{dt}(t=1/2)=6\frac{1}{1+(3)^2}=6/10=0.6[/tex] rads/h.

Please note this is the actual rate of change of the tracking angle at this very moment, but not for the rest of moments. It is a function, variable with time, applied to a specific instant. For a value in degrees,

[tex]0.6\frac{180deg}{\pi}=34.37 deg/h[/tex]