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Homework Help: Map question involving vectors (find the angle)

  1. Sep 15, 2018 at 11:09 PM #1
    1. The problem statement, all variables and given/known data
    Instructions for finding a buried treasure include the following: Go 66.0 paces at 256deg, turn to 140deg and walk 125 paces, then travel 100 paces at 169deg. The angles are measured counterclockwise from an axis pointing to the east, the +x direction. Determine the resultant displacement from the starting point. Enter the distance (without units) and the angle relative to the positive x-axis.



    2. Relevant equations


    3. The attempt at a solution
    I already figured out the displacement which is 213 paces but i thought the angle could be found using the two components (x and y), 35.4 and -210. Please help because I'm wrong!!
     
  2. jcsd
  3. Sep 15, 2018 at 11:27 PM #2

    Delta²

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    so if you do ##\tan\theta=\frac{-35.4}{210}## the angle you get isn't the correct answer? I suspect that the answer is given in positive number , which other positive angle has the same tangent as that negative angle?
     
  4. Sep 16, 2018 at 7:52 AM #3
    I tried that as well but it’s apparently still wrong
     
  5. Sep 16, 2018 at 8:36 AM #4

    Delta²

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    I didn't check your answer for the displacement, is 213 paces correct?
     
  6. Sep 16, 2018 at 8:37 AM #5
    Yes it’s 213 paces
     
  7. Sep 16, 2018 at 8:38 AM #6

    Delta²

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    What's the answer key for the angle?
     
  8. Sep 16, 2018 at 8:39 AM #7
    It unfortunately doesn’t say, its an online where it tells me whether I’m right or wrong
     
  9. Sep 16, 2018 at 8:45 AM #8

    Delta²

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    So neither -9.568 degrees or 350.432 degrees is the correct answer?
     
  10. Sep 16, 2018 at 8:48 AM #9
    Could you explain to me why it would be 350.432 degrees?
     
  11. Sep 16, 2018 at 8:51 AM #10

    Delta²

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    Trigonometry formulas say that the tangent of angle ##-\theta## is equal to the tangent of angle ##2\pi-\theta## or ##360-\theta## in degrees.
     
  12. Sep 16, 2018 at 10:28 AM #11
    Well i just tried 3.50 x 10^2 degrees (since i can only carry 3 significant digits) and unfortunately still a no :(,
     
  13. Sep 16, 2018 at 10:35 AM #12

    Delta²

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    Well I don't know what else, maybe you calculated the x and y as y,x, try ##\tan\theta=-\frac{210}{35.4}## which leads to ##\theta=-80.4## or ##\theta=279.6##
     
  14. Sep 16, 2018 at 10:37 AM #13
    Thank you so much for your help I really appreciate it!
     
  15. Sep 17, 2018 at 2:54 PM #14

    robphy

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    For the counterclockwise angle from the positive-x-axis, [itex]\theta=\tan^{-1} \left(\frac{R_y}{R_x}\right) [/itex],
    with the rule of thumb to add [itex]180^\circ[/itex] if [itex]R_x<0[/itex] (since [itex]\tan^{-1} [/itex] returns a value between [itex]-90^\circ[/itex] and [itex]+90^\circ[/itex]).
     
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