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Math Help - Bearings question with trig.

  1. #1
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    Question Bearings question with trig.

    Hey Im really not sure how to do bearings at all.
    For homework i have this question:

    A ship leaves at port A and travels for 30km on bearing of 120degrees
    It then changes course and travels for 50km on bearing of 080degrees arriving at port B.
    Calculate distance AB and bearing A from B

    thanks
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  2. #2
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    Quote Originally Posted by mitchoboy View Post
    ...
    A ship leaves at port A and travels for 30km on bearing of 120degrees
    It then changes course and travels for 50km on bearing of 080degrees arriving at port B.
    Calculate distance AB and bearing A from B
    ...

    Typically bearings are given from a reference [North or South] and deflecting East or West.

    From the information given, assume the reference for zero bearing is due North or the y-axis; and assume that port A is at the origin (0,0).

     X_0 = 0
     Y_0 = 0

     X_1 = X_0 + \sin \left(120\right) \times\ 30 = 0.866 \times 30 = 25.981
     Y_1 = Y_0 + \cos \left(120\right) \times\ 30 = 0.500 \times 30 = 15.000

     X_2 = X_1 + \sin\left( 80\right) \times\ 50 = X_1 + 0.985 \times 50 = X_1 + 49.240 = 75.221
     Y_2 = Y_1 + \cos \left(80\right) \times\ 50 = Y_1 + 0.174 \times 50 = Y_1 + 8.682 = 23.682

    Since  X_0 = 0 &  Y_0 = 0 The distance AB is  \sqrt{X_2^2 + Y_2^2}


    The bearing is the arctangent of the difference between final coordinates and the initial coordinates:

    The tangent of the bearing AB is :  \frac{X_2 - X_0} {Y_2 - Y_0}

    As a check:


     X_2 = \sin \left ({bearing AB}\right) \times \left ({distance AB}\right)

     Y_2 = \cos \left ({bearing AB}\right) \times \left({distance AB}\right)
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  3. #3
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    Just for kicks, let's do it this way. Let's use a coordinate system so you can see what is going on. Like in surveying.

    Let's say the coordinate of A is (0,0).

    Then, to the turning point, it is 30 km at an azimuth of 120 degrees.

    (Technically, this is an azimuth, not a bearing. But, it doesn't really matter).

    x=30sin(120)=25.98

    y=30cos(120)=-15

    Those are the coordinates of the turning point, (25.98,-15)

    Next, the boat turns 80 degrees from north and goes 50 km to go to B.

    B's coordinates are x=25.98+50sin(80)=75.22

    y=-15+50cos(80)=-6.32

    The coordinates of B are (75.22, -6.32).

    Now, to find the distance back to A where it started, just use ol' Pythagoras.

    \sqrt{(75.22)^{2}+(-6.32)^{2}}=75.485

    To find the bearing back to A, one way of many:

    270+cos^{-1}(\frac{75.22}{75.485})=274.8 \;\ deg

    Here is a diagram so you can see. It is rather sloppy done in paint, but I hope it will suffice.
    Attached Thumbnails Attached Thumbnails Bearings question with trig.-bearing.gif  
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  4. #4
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    Hello, mitchoboy!

    Bearings are measured clockwise from North.
    And a good diagram is essential.


    A ship leaves at port A and travels for 30 km on bearing of 120.
    It then changes course and travels for 50 km on bearing of 080 arriving at port B.
    Calculate distance AB and bearing of {\color{blue}A} from {\color{blue}B}. .
    Is this correct?
    Code:
          N
          |
          |
        A o                     R
          | *     *             |
          |60*           *     |
          |     *       Q       o B
          |    30 *     |     *
          S         *60|80* 50
                      * | *
                        o
                        P

    The ship starts at A and sails 30 km to point P:
    . . AP = 30,\angle NAP = 120^o,\;\angle SAP = \angle APQ = 60^o

    Then it turns and sails 50 km to point B:
    . . PB = 50,\;\angle QPB = 80^o \quad\Rightarrow\quad \angle APB = 140^o


    In \Delta APB, use the Law of Cosines:

    . . AB^2 \:=\:AP^2 + PB^2 - 2(AP)(BP)\cos(\angle APB)

    . . AB^2 \:=\:30^2+50^2 - 2(30)(50)\cos140^o \:=\:5698.133329

    Therefore: . \boxed{AB \;\approx\;75.5\text{ km}}



    In \Delta APB, use the Law of Cosines.

    . . \cos A \:=\:\frac{75.5^2 + 30^2 - 50^2}{2(75,5)(30)} \:=\:0.90413245

    Hence: . \angle A \;\approx\;25.2^o

    Then: . \angle BAS \:=\:25.2^o + 60^o \:=\:85.2^o \:=\:\angle ABR


    Therefore, the bearing of A from B is: . 360^o - 85.2^o \:=\:\boxed{274.8^o}

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  5. #5
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    Quote Originally Posted by Soroban View Post
    Hello, mitchoboy!

    Bearings are measured clockwise from North.
    And a good diagram is essential.

    Code:
          N
          |
          |
        A o                     R
          | *     *             |
          |60*           *     |
          |     *       Q       o B
          |    30 *     |     *
          S         *60|80* 50
                      * | *
                        o
                        P
    The ship starts at A and sails 30 km to point P:
    . . AP = 30,\angle NAP = 120^o,\;\angle SAP = \angle APQ = 60^o

    Then it turns and sails 50 km to point B:
    . . PB = 50,\;\angle QPB = 80^o \quad\Rightarrow\quad \angle APB = 140^o


    In \Delta APB, use the Law of Cosines:

    . . AB^2 \:=\:AP^2 + PB^2 - 2(AP)(BP)\cos(\angle APB)

    . . AB^2 \:=\:30^2+50^2 - 2(30)(50)\cos140^o \:=\:5698.133329

    Therefore: . \boxed{AB \;\approx\;75.5\text{ km}}



    In \Delta APB, use the Law of Cosines.

    . . \cos A \:=\:\frac{75.5^2 + 30^2 - 50^2}{2(75,5)(30)} \:=\:0.90413245

    Hence: . \angle A \;\approx\;25.2^o

    Then: . \angle BAS \:=\:25.2^o + 60^o \:=\:85.2^o \:=\:\angle ABR


    Therefore, the bearing of A from B is: . 360^o - 85.2^o \:=\:\boxed{274.8^o}
    thankyou for your answer. but how did you get apq as 60 degrees?
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