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Question involving the mathematics of physics and trigonometry

  1. Sep 3, 2011 #1
    I am having trouble solving this question, please help and show work! Thank you!

    A boat that can travel at 4.0 km/h in still water crosses a river with a current of 2.0 km/h. At what angle must the boat be pointed upstream (that is, relative to its actual path) to go straight across the river?

    A] 27
    B] 30
    C] 60
    D]63
    E] 90
     
  2. jcsd
  3. Sep 3, 2011 #2

    Pengwuino

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    So, that giant template that you deleted when you posted your thread, did you even bother to read it? The template that said you must show your own attempt at the problem?
     
  4. Sep 3, 2011 #3
    Unfortunately, I would've attempted but I have absolutely no clue. Therefore I couldn't have found a way to attempt it. I did read the template and I kept it in mind for future questions, but I couldn't for this one.
     
  5. Sep 3, 2011 #4

    Pengwuino

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    So let's assume you're at the "bottom" side of a river going horizontally. The current is going to the right and you want to move to the "top" side of the river. You'll need to point your boat and row across and to the left so as to compensate for the river trying to push you to the right. Pointing at this angle gives you a triangle across the river with horizontal and vertical components to the velocity of your boat.

    To go straight across, your horizontal component must be 2km/h. The problem says you can row at a total of 4km/h. This would be your hypotenuse. So you have your horizontal component which is 1 leg of the triangle, the total velocity which is the hypotenuse, and this gives you enough information to find the angle for this angle you have to travel at.
     
  6. Sep 3, 2011 #5
    Thank you very much!
     
  7. Sep 5, 2011 #6
    Sorry to bother again, I tried approaching this by finding the other leg (through Pythagorean Theorem) and got 3.46. Then I did a proportion, 90 / 4.0 = x / 3.46 , but I couldn't get an appropriate answer.
     
  8. Sep 5, 2011 #7

    lewando

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    You are trying to find the angle, right? Your method is inappropriate. Use trig.
     
  9. Sep 5, 2011 #8
    So then how would I use trig?
     
  10. Sep 5, 2011 #9

    lewando

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    Don't you have something like this:

    trig.jpg .

    Solving for Θ is pretty straightforward. If you recognize the triangle, it should be obvious, otherwise think opposite/hypotenuse.
     
  11. Sep 5, 2011 #10
    I did that, i did sin -1 (2/4) and i got .5235987756 , but i dont know how to use that to get an answer
     
  12. Sep 5, 2011 #11

    Pengwuino

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    As lewando said, your ratio method doesn't make sense.

    What you calculated is your answer, except it's in radians. Convert back to degrees if you need the answer in degrees.
     
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