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Homework Help: Can you have negative angles?

  1. Oct 27, 2011 #1
    1. The problem statement, all variables and given/known data
    Can you have a negative angle, or do you just assume absolute numbers?

    Your coach tells you to turn south for 10.0m and then turn and run east 15.0m. How far must the ball be thrown from where you started to where you ended?

    2. Relevant equations
    a^2 + b^2 = c^2
    tan(theta) = opp/adj

    3. The attempt at a solution
    Since, you are headed south I am making the y component -10.0m. This means when I find the angle it is negative.
    tan(theta) = 15.0m/-10.0m
    theta = -56 degrees

    Do I just assume absolute when I report it out?
  2. jcsd
  3. Oct 27, 2011 #2


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    You can have negative angles. The standard convention is that positive angles correspond to the counter-clockwise direction. An angle of +56 degrees from the x-axis would correspond to a point in the first quadrant while an angle of -56 degrees from the x-axis would correspond to a point in the fourth quadrant. You can't just ignore the sign.

    Your answer just needs to be clear as to what you mean. Instead of saying -56 degrees, you could equivalently say 56 degrees below the x-axis, or, even better in this problem, say 56 degrees south of east. Somehow the information reflected in the minus sign needs to make it into your answer.
  4. Oct 28, 2011 #3
    But, the 56 degrees is reflective of the right triangle I made. So, wouldn't I have to write 34 degrees south of east? I think that is what is confusing me the most.
  5. Oct 28, 2011 #4


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    If by "reflective of the right triangle" you mean it is below the x-axis (east west) then, no, the angle reflecting 56 degrees above the x-axis is again 56 degrees below the x-axis. That is "56 degrees southof east". Using the convention that angles are measured counterclock wise from the x-axis (east-west), that would be -56 degrees or, since an entire circle is 360 degrees, 360- 56= 304 degrees.
  6. Oct 28, 2011 #5
    So, what if I wanted to give the angle in compass directions instead of polar? Isn't it always measured from the x-axis of that quadrant?
  7. Oct 28, 2011 #6


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    I didn't notice you had switched the x and y coordinates in your calculation. You're right. The answer would be 34 degrees south of east.

    If you align the positive y-axis with north, the coordinates of the destination are (x,y) = (15 m, -10 m). If θ is the angle relative to the +x-axis, you always have tan θ=y/x, which in thise case gives you tan θ=-10/15, which corresponds to an angle of θ=-33.7°. If you use positive and negative numbers, it's probably best to stick with this method because the math will always work out correctly.

    If instead you decide to analyze a right triangle, you'd probably be better off sticking with positive numbers because lengths are always positive. So for the triangle you drew with the angle φ between the -y-axis and the hypotenuse, you'd say the opposite leg is 15 m long and the adjacent leg is 10 m long. So tan φ=15/10 or θ=56°.

    What's important with either method is that you make clear which angle you're specifying. Your answer could be "34 degrees south of east" or "56 degrees east of south" or "34 degrees below the x-axis" or "56 degrees to the right of the -y axis". In contrast, if you simply said "34 degrees" or "56 degrees", some might assume you're following the standard convention of measuring angles from the x-axis and conclude incorrectly the point is somewhere in the first quadrant.
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