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Math Help - 3D Trig. Probs

  1. #1
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    3D Trig. Probs

    g&t homework, pleasse help draw the diagrams first
    thanks.

    1. the pilot of an aircraft flying due South, at an altitude of 750m, sees an airfield directly ahead, angle of depression 17. at the same time the pilot notices a town due east, angle of depression 10. how far is town from airfield?

    2.from the top of a cliff 120m above sea level a boat is observed due North, angle of depression 20. the boat travels due east and, 10min later, its angle of depression from the top of the cliff is 10. how fast is the boat moving?

    3. a river runs east to west. a tree stands on the edge of one bank and from a point C, on the opposite bank and due south of the tree, the angle of elevation of the top of the tree is 28. from point D, 65m due east of C the angle of elevation of the top of the tree is 20
    calculate a) height of tree
    b)width of river at C.
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  2. #2
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    Hello, Fibonacci!!

    1. The pilot of an aircraft flying due South, at an altitude of 750m,
    sees an airfield directly ahead, angle of depression 17.
    At the same time the pilot notices a town due east, angle of depression 10.
    How far is town from airfield?
    Once you "see" the situation, make separate diagrams.

    Code:
        P * - - - - - - - - - - - - S
          |   * 17
          |       *
      750 |           *
          |               *
          |               17 *
        Q * - - - - - - - - - - - * A
                750/tan17
    The pilot is at P at an altitude of 750 m: . PQ = 750
    The airfield is at A.\;\;\angle SPA \:=\:\angle PAQ \:=\:17^o
    \tan17^o \:=\:\frac{750}{AQ}\quad\Rightarrow\quad AQ \:=\:\frac{750}{\tan17^o}



    Code:
        P * - - - - - - - - - - - - E
          |   * 10
          |       *
      750 |           *
          |               *
          |               10 *
        Q * - - - - - - - - - - - * T
                750/tan10
    The pilot is at P: . PQ = 750
    The town is at T.\;\;\;\angle EPT = \angle PTQ = 10^o
    \tan10^o \:=\:\frac{750}{QT}\quad\Rightarrow\quad QT \:=\:\frac{750}{\tan10^o}




    Looking straight down at the ground, we have this diagram:
    Code:
                     750/tan10
              P * - - - - - - - - * T
                |              *
                |           *
     750/tan17 |        *
                |     *
                |  *
              A *

    I assume you can find the distance AT now.

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