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Math Help - problem with bearings

  1. #1
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    problem with bearings

    From home George jogs 3.5km NE and then 8.2km SE. Find the bearing of his final position from the starting position.
    The problem about all bearing questions is that i don't know how to find the conventional like sometimes u have to subtract 360 degrees and then add on 270. Its CONFUSINGG Please helpp

    And a diagram would be reallly goood please a diagram would help. Thanks a lot
    Last edited by mr fantastic; April 25th 2009 at 06:25 PM. Reason: Deleted potentially offensive word and excessive smilies
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  2. #2
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    Quote Originally Posted by jonomantran View Post
    From home George jogs 3.5km NE and then 8.2km SE. Find the bearing of his final position from the starting position.
    The problem about all bearing questions is that i don't know how to find the conventional like sometimes u have to subtract 360 degrees and then add on 270. Its CONFUSINGG Please helpp

    And a diagram would be reallly goood please a diagram would help. Thanks a lot
    Check your dictionary.
    Almost all will have a picture showing the compass points.

    From your question, the NE bearing is 90 degrees to the SE bearing, thus you have a right angle.The distance back to the initial position is the square root of ( 3.5^2 + 8.2^2).
    Even without a diagram you can visualize that the return bearing will be in the NorthWest quadrant.
    It actually will be the NW bearing + the Arctan(3.5/8.2).

    After you lookup and determine where the compass points are, you can simply graph (or draw) the lengths given [of course you should draw it to a reduced scale so that it will fit on a sheet of paper, as opposed to drawing the line 3.5 km long. The real problem with drawing the line to scale is the number of pencils required, but you are free to persue that method.]

    3.5^2
    3*3 = 9 and 4*4 = 16
    3.5^2 is approximately halfway between 9 and 16 or about 12.

    8.2^2
    8*8 = 64 and 9*9 = 81, difference 81-64 = 17
    0.2 or 1/5 * the difference of 17 is about 3
    so 8.2^2 is about 67 = 64+3

    67+12 = 79 which is about equal to 81.
    Square root of 81 is 9

    The returning distance is going to be about 9 km.

    As for the bearing, 3.5 divided by 8.2 is roughly 1/2
    The Arctan of 1 is 45 degrees, so half of that is 22 degrees.
    That means the return bearing will be NW or 45 degrees + 22 degrees
    or ROUGHLY 67 degrees West of North.

    You can use a calculator to determine more precise values for the answer.
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  3. #3
    No one in Particular VonNemo19's Avatar
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    no problem!

    All you gotta do is put this problem on the grid in standard position and convert to rectangular coordinates:

    NE is the same thing as saying 45 degrees Right?

    So you got an angle and the hypotenuese of a right triangle. So if this guy left the origin and went along the hypoteneuse For 3.5, how far along the x- axis is he? Then you could figure the heght above the x-axis he is by figuring out what the opposite side is... Then your going to have (x,y).

    Do that again with the guy making a 90degree turn SE and then find the distance from the origin. use the distance formula. It should be real easy because one of your points will be (0,0).

    then all ya gotta do is find the angle off of the x-axis and you got your "bearing"!
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  4. #4
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    Hello, jonomantran!

    Why are you confused?
    You know which way North is, and which way East is.
    And I assume you know which way NorthEast is . . .



    From home George jogs 3.5km NE and then 8.2km SE.
    Find the bearing of his final position from the starting position.
    Code:
          N       B
          |       o - - - -
          |     *   * 45
          |45* 3.5   *
          | *           * 8.2
        A o               *
                *           *
                      *       *
                            *   *
                                  o C
    Note that \angle B = 90^o

    George starts at A, facing North.
    He turns 45 clockwise and jobs 3.5 km to point B.
    Then he turns 90 clockwise and jogs 8.2 km to point C.

    In right triangle ABC\!:\;\;\tan A \:=\:\frac{8.2}{3.5} \:=\:2.342857143
    . . Hence: . \angle A \:\approx\:67^o


    Therefore, the bearing of AC is: . \angle NAC \:=\:45^o + 67^o \;=\;112^o

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