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Math Help - three-dimensional applications

  1. #1
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    three-dimensional applications

    A hot-air balloon, B, is observed simultaneously from two points, P and Q, on horizontal ground. From P the bearing of B is 60 degrees at an angle of elevation of 45 degrees. From Q the bearing of B is 330 degrees at an angle of elevation of 60 degrees. The distance BQ is 800 m.

    a) Draw a sketch showing the positions of P, Q and B.
    b) Calculate the height of the balloon above the ground.
    c) Calculate the bearing of Q from P.

    Once again I need your help.
    Time: 2 days.
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  2. #2
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    Quote Originally Posted by Eftychia View Post
    A hot-air balloon, B, is observed simultaneously from two points, P and Q, on horizontal ground. From P the bearing of B is 60 degrees at an angle of elevation of 45 degrees. From Q the bearing of B is 330 degrees at an angle of elevation of 60 degrees. The distance BQ is 800 m.

    a) Draw a sketch showing the positions of P, Q and B.
    b) Calculate the height of the balloon above the ground.
    c) Calculate the bearing of Q from P.

    Once again I need your help.
    Time: 2 days.
    1. Draw a sketch.

    2. You are dealing actually with 3 right triangles. As the 1st step you can calculate the lengths of the sides of the grey triangle QFB:

    |\overline{QF}|=800 \cdot \cos(60^\circ)

    |\overline{BF}|=800 \cdot \sin(60^\circ)

    3. Calculate the lengths of the sides of the yellow triangle PFB.

    4. Triangle PQF is a right triangle too (why?). Since you know \overline{PF} and \overline{QF} you can calculate the length of \overline{PQ}.

    5. Calculate the values of the angles \angle(QPF) and \angle(FQP) and then consequently the corresponding bearings.
    Attached Thumbnails Attached Thumbnails three-dimensional applications-balloneinpeilen.png  
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  3. #3
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    I think the sketch should look like this.
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  4. #4
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    Quote Originally Posted by emakarov View Post
    I think the sketch should look like this.
    As far as I'm informed bearings are measured clockwise wrt N = 0 = 360.

    Therefore the situation looks like this (see attachment).
    I've added the North-direction at P and Q. Add a 3rd parallel line to the North-direction passing through F and you'll see why \angle(PFQ) = 90^\circ.
    Attached Thumbnails Attached Thumbnails three-dimensional applications-peilung_fusspkt.png  
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