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Math Help - Trig

  1. #1
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    Trig

    A lighthouse B is 8km from a point A and its bearing from A is 70 degrees E of N. A ship sails from A for 6km on a course due NE and then alters its course to 50 degrees E of N. Find from the tables its distances N and E of A when it is due North of B.

    Im not sure how to do this questions. Thanks
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  2. #2
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    Hello deltaxray
    Quote Originally Posted by deltaxray View Post
    A lighthouse B is 8km from a point A and its bearing from A is 70 degrees E of N. A ship sails from A for 6km on a course due NE and then alters its course to 50 degrees E of N. Find from the tables its distances N and E of A when it is due North of B.

    Im not sure how to do this questions. Thanks
    I'm not sure what 'from the tables' means. But here's how I should do it.

    B is 8 \sin 70^o km East of A and 8\cos 70^o km North of A. (Check this out on a diagram. I'll leave you to work these values out.)

    When the ship has completed the first 6 km leg, it is 6\sin45^o km East and 6\cos45^o km North of A. (Work these out as well.)

    If it then sails a distance x km on a bearing 050^o, it will travel a further distance x\sin50^o km East and x \cos 50^o km North.

    If the ship is now due North of B, its distance East of A is the same as the distance that B is East of A. So:
    6\sin 45^o + x\sin50^o = 8\sin 70^o
    From this we can work out the value of x and hence how far North it is of A. (You've already worked out how far East it is.)

    Can you complete the working now?

    Grandad
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  3. #3
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    Hello, deltaxray!

    Did you make a sketch?


    A lighthouse B is 8 km from point A
    and its heading from A is N\,70^o\,E

    A ship sails from A for 6km on a course due NE to point C
    and then alters its course to N\,50^o\,E to point D.

    Find its distances north and east of A when it is due North of B.
    Code:
                        :               o D
                        :           *   |
                        : 50   *       |
                        :   * 40       |
                      C o - - - - - - - o E
                      * |       3.28    |
          :         *   |               o B
          :     6 *     |         *     |
          : 45 *       |   *           |
          :   *       * |  8            |
          : *   *  20  |               |
          o - - - - - - o - - - - - - - o
          A    4.24     F     3.28      G
          : - - - - - -7.52 - - - - - - :

    We have: . AB \,=\,8,\;\angle BAG \,=\,20^o

    . . . Also: . AC \,=\,6,\;\angle CAF \,=\,45^o

    . . . .And: . \angle DCE \,=\,40^o

    We want: . \begin{Bmatrix}AG & \text{east} \\ DG & \text{north} \end{Bmatrix}



    In right triangle BGA\!:\;\;\cos20^o \,=\,\frac{AG}{8} \quad\Rightarrow\quad AG \:=\:8\cos20^o \:=\:7.517590966

    . . Hence: . \boxed{AG \:\approx\:7.52}



    DG \:=\:DE + EG

    In right triangle CFA\!:\;\sin45^o \,=\,\frac{CF}{6} \quad\Rightarrow\quad CF \,=\,6\sin45^o \,=\,4.242670687
    . . Hence: . CF \:\approx\:4.24 \:=\:EG
    . . . Also: . AF \:\approx\:4.24

    Then: . FG \:=\:AG - AF \:=\:7.52 - 4.24 \:=\:3.28 \:=\:CE


    In right triangle DEC\!:\;\tan40^o \,=\,\frac{DE}{CD} \quad\Rightarrow\quad DE \:=\:3.28\tan40^o \:=\:2.75224679
    . . So: . DE\:\approx\:2.75

    Hence: . DG \;=\;DE + EG \:=\:2.75 + 4.24 \quad\Rightarrow\quad\boxed{DG \:=\:6.99}


    Therefore, the ship is 6.99 km north and 7.52 km east of point A.

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