Question 1037921
The bearing 0 degrees is true north (see 1:20 of <a href="https://www.youtube.com/watch?v=QO0k_k4bbYM">this video</a>)


So what you do is first aim true north and you rotate clockwise 310 degrees until you end up in quadrant II. This is what the drawing should look like


<img src="http://image.prntscr.com/image/6125c668af8d4c2e878d1d7b63048ecd.png">


You'll notice a right triangle forming in Q2 with the angle of 40 degrees touching the x axis. The vertical component of this triangle is what we're after. This is the opposite side to the 40 degree angle. We don't know the opposite side, so we'll just call it x.



The 40 degree angle is from the fact that 270+40 = 310 degrees. The 40 degree angle is the piece of the 310 degree angle where it's in Q2.



The ship has traveled D = r*t = 25*(0.25) = 6.25 km
Note: 15 min = (15/60) hrs = (1/4) hrs = 0.25 hrs


So this triangle has a hypotenuse of 6.25 km


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Now we use trig 



Sine = opposite/hypotenuse


sin(40) = x/6.25


x = 6.25*sin(40)


x = 4.01742256054088



So the carrier is approximately 4.01742256054088 kilometers north of its original starting point.


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So you had the right idea, just the wrong trig function.