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Thread: trignometery bearing problem

  1. #1
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    trignometery bearing problem

    A point P is 12 km due north of another point Q. the bearing of a lighthouse,R, from P is 135 degress and, from Q, it is 120 degress. calculate the distance PR.
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  2. #2
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    Hello, Sohail!

    Did you make a sketch?


    A point P is 12 km due north of another point Q.
    The bearing of a lighthouse R from P is 135, and from Q it is 120.
    Calculate the distance PR.
    Code:
          |
        P * 135
          | *
          |45*
     r=12 |     *
          |       *
          |         *
        Q * 120      *
             *          * q
                *         *
                   *        *
                      *       *
                         *      *
                            *     *
                               *    *
                                  *15*
                                     *  * 
                                          * R

    In \Delta PQR, we have: . \angle P = 45^o,\;\angle Q = 120^o, then \angle R = 15^o
    . . also: .side  r = PQ = 12 and we want side q = PR.

    Law of Sines: . \frac{q}{\sin Q} \,= \,\frac{r}{\sin R}\quad\Rightarrow\quad q \:=\:\frac{r\sin Q}{\sin R}

    Therefore, we have: . q\:=\:PQ\:=\:\frac{12\cdot\sin120^o}{\sin15^o} \:\approx\:40.15 km.

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