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Thread: how do u find the angle of these points?

  1. Dec 31st, 2003, 12:17 AM #1
    voidflux
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    how do u find the angle of these points?

    Hello everyone, I was reading next semesters programming book and I can't even figure out how to do this problem on paper, so It doesn't look like i'll be programming it anytime soon. Here's my question, They say, design and implement a class called circle_location to keep track of the postion of a single point that travels around a circle. An object of this clas records the position of the point as an angle, measured in a clockwise direction from the top of the circle. Programming this isn't my problem, but I'm confused on how i'm suppose to figure out the angle of the point. So say I have a point at (0,0) and (3,3) this would make a line then I could draw a vertical line down to make a right angle, but how would I find the angle of that? with just given 2 points? Or am i misunderstanding what they really want me to do?
    Thanks!
    C¤ry Sanchez
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  2. Dec 31st, 2003, 02:08 AM #2
    kedaman
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    google on trigonometry
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    writing software in C++ is like driving rivets into steel beam with a toothpick.
    writing haskell makes your life easier:
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  3. Jan 1st, 2004, 01:31 PM #3
    jemidiah
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    You can use tan-1 (abs(y1 - y2))/(abs(x1 - x2))

    With the originating point being (x2, y2) and the other being (x1, y1). Now, you'll have to use whether (y1 - y2) and (x1 - x2) or positive or negative to tell you which quadrant it's in.

    - , + = quadrant 4
    - , - = quadrant 3
    + , - = quadrant 2
    + , + = quadrant 1

    I'm sure there are other ways (like somehow using the parametric representation of a circle and solving for T), but that *should* work with some fiddling around.
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  4. Jan 2nd, 2004, 11:52 AM #4
    voidflux
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    Thanks jemidiah, That will work great!
    C¤ry Sanchez
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    @ Penn State
    IBM.zSeries Intern
    Mandriva 2007
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  6. Jan 2nd, 2004, 09:56 PM #5
    jemidiah
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    One thing I remembered: when x1 - x2 = 0, you'd get a divide by zero error (duh), so the angle'll be 90
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  7. Jan 4th, 2004, 05:14 AM #6
    sw_is_great
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    yeh its infinity and tan 90 is infinity
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  8. Jan 4th, 2004, 01:27 PM #7
    TheManWhoCan
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    I haven't nit-picked for a while, so here goes:

    tan90o is infinite not infinity, they are different.
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  9. Jan 4th, 2004, 03:43 PM #8
    kedaman
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    I thought tan 90° was not defined, likewise cot 0° csc 0° and sec 90°
    Use
    writing software in C++ is like driving rivets into steel beam with a toothpick.
    writing haskell makes your life easier:
    reverse (p (6*9)) where p x|x==0=""|True=chr (48+z): p y where (y,z)=divMod x 13
    To throw away OOP for low level languages is myopia, to keep OOP is hyperopia. To throw away OOP for a high level language is insight.
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  11. Jan 4th, 2004, 05:56 PM #9
    jemidiah
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    They're all divide by zero errors, so you could debate about anything you want (search the forums if you really want to see war )
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  12. Jan 4th, 2004, 06:41 PM #10
    kedaman
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    yeah but i dont like war.. or nit picking for that matter I just thought some general info on how to use trigonometry would do good.
    Use
    writing software in C++ is like driving rivets into steel beam with a toothpick.
    writing haskell makes your life easier:
    reverse (p (6*9)) where p x|x==0=""|True=chr (48+z): p y where (y,z)=divMod x 13
    To throw away OOP for low level languages is myopia, to keep OOP is hyperopia. To throw away OOP for a high level language is insight.
    Reply With Quote Reply With Quote
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