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Math Help - BEARING/trig problem

  1. #1
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    BEARING/trig problem

    From a point P, the ships R, V are observed bearings 325degreesT and 47degreesT respectively, whilst from a point Q 10km due north of P the ships R, V bear 310degreesT and 79degreesT respectively.
    i) Show that PQ = 10sin50/sin15 and obtain similar expression for PV

    dont realli get how to draw the diagram at all , doesnt make sense to me =/// !!

    any help is GREATLY appreciated !!!
    thankyou !
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  2. #2
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    Quote Originally Posted by iiharthero View Post
    From a point P, the ships R, V are observed bearings 325degreesT and 47degreesT respectively, whilst from a point Q 10km due north of P the ships R, V bear 310degreesT and 79degreesT respectively.
    i) Show that PQ = 10sin50/sin15 and obtain similar expression for PV

    dont realli get how to draw the diagram at all , doesnt make sense to me =/// !!

    any help is GREATLY appreciated !!!
    thankyou !
    All the angles are measured with respect to the North direction, clockwise!

    See attachment.
    Attached Thumbnails Attached Thumbnails BEARING/trig problem-kreuzpeilung.png  
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  3. #3
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    Hello, iiharthero!

    From a point P, the ships R, V are observed at 325 and 47, resp,
    while from a point Q 10km due north of P, the ships R, V bear 310 and 79, resp.

    i) Show that PR \:=\: \frac{10\sin50^o}{\sin15^o} and obtain similar expression for PV.

    This is the diagram . . .

    Code:
                                                o V
                                            * *
                          |              *  *
        R o               |           *   *
            * *           |        *    *
              *   *   50 | 79 *     *
                *     *   |  *      *
                  *       oQ      *
                    *     |     *
                      *35|47*
                        * | *
                          o
                          P
    where PQ = 10


    Consider \Delta PQR and all its angles.
    Code:
        R o
            * *
              *15*           |
                *     *   50 |
                  *       *   |
                    *    130 o Q
                      *       |
                        *     |10
                          *35|
                            * |
                              o
                              P

    Law of Sines: . \frac{PR}{\sin130^o} \:=\:\frac{10}{\sin15^o} \quad\Rightarrow\quad PR \:=\:\frac{10\sin130^o}{\sin15^o}

    Since \sin130^o = \sin50^o, we have: . PR \:=\:\frac{10\sin50^o}{\sin15^o}

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