SOLUTION: A coal carrier departs from the terminal at a bearing of 310 degrees. If it travels at a speed of 25 km per hour, how far north of the terminal is the carrier after 15 minutes?

Question 1037921: A coal carrier departs from the terminal at a bearing of 310 degrees. If it travels at a speed of 25 km per hour, how far north of the terminal is the carrier after 15 minutes?
Cos40= N/375
375(Cos40)=N
375(0.7660)=N
287.4=N
This has to be soo wrong. Help please.
Thank you
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
The bearing 0 degrees is true north (see 1:20 of this video)

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

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

---------------------------------------------------------

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.

--------------------------------------------------------

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