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Thread: 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).

    $\displaystyle X_0 = 0$
    $\displaystyle Y_0 = 0 $

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

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

    Since $\displaystyle X_0 = 0 $ & $\displaystyle Y_0 = 0 $ The distance AB is $\displaystyle \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 : $\displaystyle \frac{X_2 - X_0} {Y_2 - Y_0} $

    As a check:


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

    $\displaystyle 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).

    $\displaystyle x=30sin(120)=25.98$

    $\displaystyle y=30cos(120)=-15$

    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 $\displaystyle x=25.98+50sin(80)=75.22$

    $\displaystyle y=-15+50cos(80)=-6.32$

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

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

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

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

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

    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 $\displaystyle A$ 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 $\displaystyle B.$
    Calculate distance $\displaystyle AB$ and bearing of $\displaystyle {\color{blue}A}$ from $\displaystyle {\color{blue}B}.$ .
    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 $\displaystyle P$:
    . . $\displaystyle AP = 30,\angle NAP = 120^o,\;\angle SAP = \angle APQ = 60^o$

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


    In $\displaystyle \Delta APB$, use the Law of Cosines:

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

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

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



    In $\displaystyle \Delta APB$, use the Law of Cosines.

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

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

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


    Therefore, the bearing of $\displaystyle A$ from $\displaystyle B$ is: .$\displaystyle 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 $\displaystyle P$:
    . . $\displaystyle AP = 30,\angle NAP = 120^o,\;\angle SAP = \angle APQ = 60^o$

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


    In $\displaystyle \Delta APB$, use the Law of Cosines:

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

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

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



    In $\displaystyle \Delta APB$, use the Law of Cosines.

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

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

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


    Therefore, the bearing of $\displaystyle A$ from $\displaystyle B$ is: .$\displaystyle 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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