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Adding Vector Components Help

  1. Jun 25, 2008 #1
    1. The problem statement, all variables and given/known data
    A person follows the path outlined below. 100m east, then 300m south then 150m 30 degrees south of west, then 200m 60 degrees north of east.

    2. Relevant equations
    I used component vectoring and have not been coming up with the resultant answer that the instructor has outlined.

    3. The attempt at a solution
    I label each vector as A, B, C, and D. Then figure the components of each: Ax=100 Ay=0, Bx=0 By=-300, Cx-129.9 Cy=-75, and Dx=-100 and Dy 173.2. This gives me Rx- -129.9 and Ry= 201.8. I then use pythagorean theorum to do radicand (-129.9 squared)+(-201.8 squared) which gives me 154.4. Yet ther instructor shows his answer as 240. I also then do angle theta =Tan-1(-201.8/-129.9) which gives me tan-1(1.553) wihch gives me the answer 57.2, but the instructor gives me the answer 237 degrees.

    Any help would be greatly appreciated as I feel im getting hte hang of it except for this type of question.

    Thanks in advance,

  2. jcsd
  3. Jun 25, 2008 #2


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    I think I see the problem. What quadrant is vector D in? (Incidentally, I don't see how your instructor gets their answer, either. I'm wondering if they misinterpreted one of the directions, for C or D...)
  4. Jun 25, 2008 #3
    Crud, sorry it should read 60 degrees north of West. My fault, hope this can help you figure it out. Thanks for taking a look at it so far though:) Any help would be greatly appreciated.

  5. Jun 26, 2008 #4


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    In that case, this is correct:

    But I will ask you how you got the hypotenuse to be smaller than one of the sides...

    Mayhap there was a calculator entry error: you should get your instructor's result now...

    Inverse trig functions constantly reveal the danger of trusting a calculator... Your tangent is calculated correctly. But what quadrant is your resultant vector in? How many angles can have the same tangent value? What other angle has a tangent of 1.553?
  6. Jun 26, 2008 #5
    Thanks, I got it now. If I had just looked at my diagram it would have become clear to me. Being in quadrant III means it would be what my calculator got (57.2) but I should have added 180 degrees, correct? Sorry about not following your explanation completely but I have never had trig or geometry. I am learning them as I go through this class:)

  7. Jun 26, 2008 #6


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    When you have picked up enough trigonometry, you'll learn that there are (generally) two angles in the cycle from 0º to 360º which have the same values for sine, cosine, and tangent. If you reverse the question to ask what angle has a particular value of one of these functions, you have two answers (usually) then. But a mathematical function can only give one "output" value for a chosen "input" value and a calculator can only work with functions.

    So the "inverse trig functions" -- arcsine, arccosine, arctangent, etc. -- are created by restricting their "outputs": arcsine and arctangent only give values from -90º to +90º and arccosine only gives values from 0º to 180º. However, real problems can lead to angles in any of the four quadrants, so the user of calculators or software has to know how to interpret the "output" from a requested function. In your problem, the resultant vector points into the third quadrant, but arctangent can only provide results in the fourth or first. When that happens, knowing that the output is in the "wrong" quadrant, you have to take the value of the arctangent and add 180º to it. (There are similar, but different, manipulations for arcsine and arccosine.)
  8. Jun 26, 2008 #7
    Thanks, I am getting the hang of it now.

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