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Angle on coordinate axis, related acute angle

  1. Mar 8, 2013 #1
    1. The problem statement, all variables and given/known data
    Tan A = -2/6
    a) Draw two possible locations on the coordinate axis for the terminal arm of angle A
    b) Find two possible values for the measure of angle A and the related acute angle.


    2. Relevant equations
    c^2 = a^2 + b^2
    SOHCAHTOA


    3. The attempt at a solution
    I know that the two possible locations are in quadrants II and IV. But I'm not sure how they're supposed to be drawn? I attached an image of how I think it's supposed to be, but I'm not sure if I did it right.
    Also, I'm not sure what angle A represents? Is it the angle IN the triangle? or the angle outside of it?
     
  2. jcsd
  3. Mar 8, 2013 #2
    Picture
     

    Attached Files:

  4. Mar 8, 2013 #3

    eumyang

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    Unfortunately, the triangles are not drawn correctly. The way you have it, tan A = -6/2.

    For angles in standard position, the initial side lies on the positive x-axis. You measure them by going counterclockwise until you reach the terminal side. So in quadrant I, the angle is in the triangle, but for the other quadrants, the angle is NOT in the triangle.
     
  5. Mar 8, 2013 #4
    Okay thanks, so would the triangles look more like this? (image attached)

    And does that mean the related acute angle is the angle IN the triangle, and the angle A (in the 2nd quadrant) would be 180 - (related acute angle)? and the angle A (in the 4th quadrant) would be 360 - (related acute angle)?
     

    Attached Files:

  6. Mar 8, 2013 #5

    eumyang

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    Still incorrect, I'm afraid. Take the first diagram you uploaded and just switch the 6's and 2's (but leave the negatives alone). Remember that tan θ = y/x.

    Yes.
     
  7. Mar 8, 2013 #6

    SteamKing

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    The triangle in Quadrant IV is flipped.
     
  8. Mar 8, 2013 #7
    Oh okay, I think I understand. Like this?

    I'm just not sure why that happens? Why can x be -6 or 6 and y be -2 or 2?
     

    Attached Files:

  9. Mar 8, 2013 #8
    The b in the 4th quadrant is in the wrong spot, meant to put it along x-axis
     
  10. Mar 8, 2013 #9
    6* not b
     
  11. Mar 8, 2013 #10

    eumyang

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    Yes (after taking into account the misplacement of 6 in Q IV), but I don't understand your question. The way your latest diagram is drawn , in Quadrant II x = -6 and y = 2, and in Quadrant IV, x = 6 and y = -2. :confused:
     
  12. Mar 8, 2013 #11
    That is the correct triagnles though? That's what I mean, I just don't understand why the values are switched? (from x being negative to positive, and same with y)
     
  13. Mar 8, 2013 #12

    eumyang

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    You have [itex]\tan A = -\frac{2}{6}[/itex], and you know that [itex]\tan \theta = \frac{y}{x}[/itex]. If you have a fraction that is negative, then either the numerator or the denominator is negative, but not both. So these are equivalent:
    [itex]-\frac{2}{6} = \frac{-2}{6} = \frac{2}{-6}[/itex]
    Therefore, we draw triangles in Quadrants II and IV. In Quadrant II, x is negative and y is positive. In Quadrant IV, x is positive and y is negative.

    Does that answer your question? If not, then I still don't know what you are asking.
     
  14. Mar 8, 2013 #13
    Oh okay. Yes, that explains it. Thank you!
     
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