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Math Help - Trig Word Problem

  1. #1
    Junior Member casey_k's Avatar
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    Question Trig Word Problem

    Two forest fire towers, A and B are 20.3km apart. The bearing from A to B is N70degreesE. The ranger in each tower observes a fire and radios the fire's bearing from the tower. The bearing from tower A is N25degreesE. From tower B, the bearing is N15degreesW. How far is the fire from each tower?
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  2. #2
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    Hello, casey_k!

    We need a good sketch for this problem . . .


    Two forest fire towers, A and B are 20.3 km apart.
    The bearing from A to B is N70^oE

    The ranger in each tower observes a fire and radios the fire's bearing from the tower.
    The bearing from tower A is N25^oE.
    From tower B, the bearing is N15^oW.

    How far is the fire from each tower?
    Code:
          N               P
          |         F     |
          |         *     |
          |        / \    |
          |       /40\15|
          |      /     \  |
          |     /       \ |
          |    /         \|
          |25/       95 * B
          |  /        *   |
          | / 45 *    70|
          |/  *  20.3     |
        A *               Q
    We are given: . AB \,=\, 20.3,\;\;\angle NAB \:=\: 70^o \:=\: \angle ABQ

    The fire is at F.
    \angle NAF = 25^o \quad\Rightarrow\quad \angle FAB = 45^o

    Since \angle PBF = 15^o,\;\angle ABQ = 70^o, then \angle FBA = 95^o . . . then: . \angle F = 40^o


    In \Delta AFB. use the Law of Sines:

    . . \frac{AF}{\sin95^o} \:=\:\frac{20.3}{\sin40^o} \quad\Rightarrow\quad AF \:\approx\:31.5\text{ km}

    . . \frac{BF}{\sin45^o} \:=\:\frac{20.3}{\sin40^o} \quad\Rightarrow\quad BF \:\approx\:22.3\text{ km}


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