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Bearings (compass headings) question high school

  1. Jul 9, 2012 #1
    A man walks 3 km due east and then 5 km northeast. Find his distance and bearing from his original position.

    ive drawn the triangle, but have no idea how to get an angle out of it, i've looked at the solution and they have somehow got the angle 135, though I understand their steps after they got the angle, just don't know how they got it.
     
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  3. Jul 9, 2012 #2

    jedishrfu

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    Re: bearings question high school

    welcome to the forums phospho!

    you need to show some work before we can help.

    I'd start by drawing a picture. Also you know the angle when he starts walking NE as 45 degrees from the horizontal right?

    So that means the interior angle is? ...answer left to the student...
     
  4. Jul 9, 2012 #3

    LCKurtz

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    Re: bearings question high school

    Do you get an angle about 32 degrees and distance ##\sqrt{89}##? I don't see where 135 degrees has anything to do with this problem either, at least not if you know how to use vectors.

    [Edit] Never mind, my calculations are wrong so just ignore this post, especially since you aren't to use vectors anyway.
     
    Last edited: Jul 9, 2012
  5. Jul 9, 2012 #4
    Re: bearings question high school

    I have no idea about exterior and interior angles, do you have a link which explains them?

    the bold part made sense of how they got 135, i was then able to solve the question... I feel like such an idiot.

    this kind of leads on to vectors, they don't want us to use vectors as of yet.
     
  6. Jul 9, 2012 #5
    Re: bearings question high school

    I don't see how you would get an angle greater than 45. I made a drawing and the angle has to be less than 45 degrees.
     
  7. Jul 9, 2012 #6

    jedishrfu

    Staff: Mentor

    Re: bearings question high school

    If you draw the path of the man the angle between walking east and then going NE is 135 ie 180-45) given that angle and the lengths of the sides of the triangle created you have enough info the compute the angle and actual distance from the staring point.

    So the 135 degrees is intermediate info needed to solve the problem.
     
  8. Jul 9, 2012 #7

    HallsofIvy

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    Re: bearings question high school

    So you understand the 135 degrees? Good. Now you have a triangle with sides of length 3 and 5 and angle between them 135 degrees. Now use the "cosine law": if a triangle has sides of length a, b, and c, and the angle opposite side c has measure c, then [itex]c^2= a^2+ b^2- 2ab cos(C)[/itex]. (If C= 90 degrees, so this is a right triangle, cos(C)= 0 and this is the Pythagorean theorem. If C= 180 degrees, so we really have a straight line, cos(180)= -1 and this is [itex]c^2= a^2+ b^2+ 2ab= (a+ b)^2[/itex], c= a+ b. If C= 0 degrees, so we are "doubling back", cos(0)= 1 and this is [itex]c^2= a^2+ b^2- 2ab= (a- b)^2[/itex], c= |a- b|.)
     
    Last edited: Jul 9, 2012
  9. Jul 9, 2012 #8
    Re: bearings question high school

    Well, that's why i'm trying to fill in the gaps that I am missing.
     
  10. Jul 9, 2012 #9

    HallsofIvy

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    Re: bearings question high school

    I have edited my original response.
     
  11. Jul 9, 2012 #10
    Re: bearings question high school

    thanks, completely understand what you did there, but did not seem to figure out the angle 135... oh dear.

    thanks again everyone.
     
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