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Math Help - Get angle from coordinates of a corner?

  1. #1
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    Get angle from coordinates of a corner?

    Hi, please bear with me as I have not taken trig in over 10 year.. I am PRETTY sure this is a trigonometry problem.

    Here is an image that describes my problem:


    Note: the lengths and angles indicated may not be 100% exact, but they should be close. I measured them as well as I was able. The lengths on the labelled sides ARE the same on all the shown triangles, and do not change. The values for X and Y ARE known, and X does not change either.

    I need an expression that describes the value for Z in terms of X and Y.

    Any help on this would be appreciated, I don't really even know where to start!
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  2. #2
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    Quote Originally Posted by Malkalypse View Post
    Hi, please bear with me as I have not taken trig in over 10 year.. I am PRETTY sure this is a trigonometry problem.

    Here is an image that describes my problem:


    Note: the lengths and angles indicated may not be 100% exact, but they should be close. I measured them as well as I was able. The lengths on the labelled sides ARE the same on all the shown triangles, and do not change. The values for X and Y ARE known, and X does not change either.

    I need an expression that describes the value for Z in terms of X and Y.

    Any help on this would be appreciated, I don't really even know where to start!
    i don't see Z labeled on the actual figure, i have no idea what length or angle it is you want to find
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  3. #3
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    Sorry if I was not clear about that.. Z here describes the angle that is labelled (where it says Z1, Z2, and Z3)

    Basically, I need an expression for the angle of the indicated corner as shown in this link:
    http://www.versatileartist.com/temp/pca_help.avi

    to be more specifically clear about the original image: X is the distance directly right from the bottom corner of the triangle, Y is the distance directly up from it, and Z is the angle of the top corner. X does not change.
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  4. #4
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    Hello, Malkalypse!

    If I interpret the problem correctly, angle z = angle CBA.

    I assume that the 14.776,\:14.967,\;x = 4.848 are fixed
    . . and that y may vary.
    Code:
                                B
                                *
                            *  *|
                         *  z * |
            14.776    *      *  |
                   *        *   |
                *          *    |
             *            *     |
        C *              *      |y
           *            *       |
            *          *_____   | 
             *        *√x+y   |
      14.967  *      *          |
               *   *            |
                * *             |
                 *- - - - - - - *
                 A       x      O

    In right triangle AOB, we have: . x = AO,\:y = BO
    . . Hence: . AB = \sqrt{x^2+y^2}

    From the Law of Cosines: . \cos z \;=\;\frac{CB^2 + AB^2 - CA^2}{2\cdot CB\cdot AB}

    . . so we have: . \cos z \;=\;\frac{14.776^2 + (\sqrt{x^2+y^2})^2 - 14.967^2}{2(14.776)(\sqrt{x^2+y^2})}


    Since x = 4.848

    . . we have: . \cos z \;=\;\frac{y^2 - 17.822191}{29.552\sqrt{y^2 + 23.503104}}

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  5. #5
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    Hm, doesn't seem to work for some reason.. maybe I'm doing it wrong.

    Also, I realized that what I really need is angle CBO.. though I should be able to get that by adding CBA and ABO...

    My distances were also inaccurate (I wanted to make sure all my information was correct so I checked again). AC is 14.9921 and CB is 14.8591.

    However, even taking all this into account, I can tell that the answers I'm getting aren't even close.I don't know if it's maybe an order of operations issue, but I seem to be ending up with numbers less than 1, which isn't right.

    ---
    You know what, I just realized that I am rustier than I even thought. When you say "cos z = ..." ... does that mean "the cosine of z = ..." or that "z as a cosine = ..." or just "z = ..." and is that answer in radians or degrees? :P
    Sorry for asking stupid questions, but I really need to figure this out.
    Last edited by Malkalypse; June 3rd 2007 at 04:50 PM.
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Malkalypse View Post
    You know what, I just realized that I am rustier than I even thought. When you say "cos z = ..." ... does that mean "the cosine of z = ..." or that "z as a cosine = ..." or just "z = ..." and is that answer in radians or degrees? :P
    Sorry for asking stupid questions, but I really need to figure this out.
    it means the cosine of z
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  7. #7
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    Okay, so how do I convert it to degrees?

    Sorry if I seem like an idiot, but it really has been a long time since I have done anything like this. In the meantime, I'm trying to get a very stubborn program to use this information, and it would be a lot easier if I were sure that the the information I am putting in is correct :/
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Malkalypse View Post
    Okay, so how do I convert it to degrees?

    Sorry if I seem like an idiot, but it really has been a long time since I have done anything like this. In the meantime, I'm trying to get a very stubborn program to use this information, and it would be a lot easier if I were sure that the the information I am putting in is correct :/
    are you using a calculator? if so, just put it in degrees mode when doing the computations. if you're using some other program, well, i don't know. it depends on how the program works
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  9. #9
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    Quote Originally Posted by Malkalypse View Post
    Okay, so how do I convert it to degrees?
    Convert what to degrees? Radians?
    Then use the formula,

    \frac{180}{\pi} \cdot x
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