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Math Help - bearings

  1. #1
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    bearings

    Not sure if I'm in the right forum as I'm new to this, but here goes.

    I have a car traveling in a straight line from point C (-100,100) to point D (200,-100). I need to find the direction of travel as a bearing to the nearest degree. Any help?
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  2. #2
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    Re: bearings

    slope diagram = -1
    y-x/x+100= -1 simplifies to y= x

    0,0 is the origin on a graph and the center 0f a 360 degree compass pointing true north. your course is 45 degrees W of north or 315 degrees true. Iwent the wrong way.change to 45 degrees south of east or135 degrees true
    Last edited by bjhopper; May 10th 2012 at 02:32 PM. Reason: added correction sentence
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  3. #3
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    Re: bearings

    Quote Originally Posted by sarahheath1985 View Post
    Not sure if I'm in the right forum as I'm new to this, but here goes.
    I have a car traveling in a straight line from point C (-100,100) to point D (200,-100). I need to find the direction of travel as a bearing to the nearest degree. Any help?
    I do not understand reply #2.
    The vector of direction is <3,-2>\text{ or }3i-2j~.

    Thus the direction angle is \arctan\left(\frac{-2}{3}\right). You need to change that to degrees.
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  4. #4
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    Re: bearings

    Yes Igoofed.the slope is -2/3 and angle = 33.7 and course is 33.7 degrees south of east or 123.7 degrees true
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  5. #5
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    Re: bearings

    so,

    what is the general formula? say the coordinates change to (-200,100) at C and (400,-200) at D?
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  6. #6
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    Re: bearings

    There is no general formula. You are given two poimts on a coordinate grid. From these you determine the slope of the line connecting them.In this new case slope equals -1/2and angle equals arc tangent of -1/2= 26.6 deg. Coursefrom C to D is 26.6 deg south of east or 116.6 deg true
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  7. #7
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    Re: bearings

    The "slope" of a line from (x_0, y_0) to (x_1, y_1) is the fraction \frac{y_1- y_0}{x_1- x_} and the angle the line makes with the x- axis is arctan\left(\frac{y_1- y_0}{x_1- x_0}\right). However, that is measured counterclockwise from the x-axis while the "bearing" (as on a map) is measured clockwise from the y-axis ("north"). To convert to a "bearing", subtract 90 degrees and multiply by -1:
    -\left(arctan\left(\frac{y_1- y_0}{x_1- x_0}\right)- 90\right) degrees.
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  8. #8
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    Re: bearings

    The slope angle is always a positive acute angle. A sign is added only to show how it slants relative to the horizontal and does not apply to the angle itself.
    Last edited by bjhopper; May 12th 2012 at 03:13 AM. Reason: missing words
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