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Math Help - Need some help ! newbie here !

  1. #1
    Newbie jhaycianofficial's Avatar
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    Need some help ! newbie here !

    hi i'm Jhaycian Velasquez 19 years of age an engineering student )

    can u please help me out of this give me some figure

    two observers who are 2 miles apart on a horizontal plane observe a balloon in a same vertical plane with them

    two observers who are 2 miles apart on a horizontal plane observe a balloon in a same vertical plane with themselves. the angles of elevation are 50 degrees and 65 degrees respectively. find the height of the balloon,(a)if it is between the observers;(b)if it is on the same side of both of them.

    ) i hope u will help me )
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  2. #2
    Super Member

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    Re: Need some help ! newbie here !

    Hello, Jhaycian!

    Two observers are 2 miles apart on a horizontal plane.
    They observe a balloon in a same vertical plane with themselves.
    The angles of elevation are 50 degrees and 65 degrees respectively.
    Find the height of the balloon

    (a) if it is between the observers.

    Code:
                C
                *
               *| *
              * |   *
             *  |h    *
            *   |       *
           *65  |      50 *
        A *-----*-----------* B
          :  x  D    2-x    :
    The observers are at A and B: AB = 2.
    The balloon is at C; its height is h = CD.
    \angle A = 65^o,\;\angle B = 50^o.
    Let x = AD\quad\Rightarrow\quad 2-x = DB.

    In \Delta CDA\!:\;\tan65^o \,=\,\frac{h}{x} \quad\Rightarrow\quad x \:=\:\frac{h}{\tan65^o} .[1]

    In \Delta CDB\!:\;\tan50^i \:=\:\frac{h}{2-x} \quad\Rightarrow\quad x \:=\:\frac{2\tan50^o - h}{\tan50^o} .[2]

    Equate [1] and [2]: . \frac{h}{\tan65^o} \:=\:\frac{2\tan50^i - h}{\tan50^o}

    . . . . . . . . . . . . . . h\tan50^o \:=\:2\tan50^o\tan65^o - h\tan65^o

    . . . . . . . \h\tan50^o + h\tan65^o \;=\;2\tan50^o\tan65^o

    . . . . . . h(\tan50^o + \tan65^o) \;=\;2\tan50^o\tan65^o

    Therefore: . h \;=\;\frac{2\tan50^o\tan65^o}{\tan50^o+\tan65^o} \;\;\approx\;\;1.532\text{ miles.}




    (b) if it is on the same side of both of them.

    This is the diagram for this scenario.
    Code:
                          * C
                        **|
                      * * |
                    *  *  |
                  *   *   |h
                *    *    |
              *     *     |
            *50    *65    |
        A *-------*-------* D
          :   2   B   x   :
    Give it a try . . .
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  3. #3
    Newbie jhaycianofficial's Avatar
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    manila
    Posts
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    Re: Need some help ! newbie here !

    Quote Originally Posted by Soroban View Post
    Hello, Jhaycian!


    Code:
                C
                *
               *| *
              * |   *
             *  |h    *
            *   |       *
           *65  |      50 *
        A *-----*-----------* B
          :  x  D    2-x    :
    The observers are at A and B: AB = 2.
    The balloon is at C; its height is h = CD.
    \angle A = 65^o,\;\angle B = 50^o.
    Let x = AD\quad\Rightarrow\quad 2-x = DB.

    In \Delta CDA\!:\;\tan65^o \,=\,\frac{h}{x} \quad\Rightarrow\quad x \:=\:\frac{h}{\tan65^o} .[1]

    In \Delta CDB\!:\;\tan50^i \:=\:\frac{h}{2-x} \quad\Rightarrow\quad x \:=\:\frac{2\tan50^o - h}{\tan50^o} .[2]

    Equate [1] and [2]: . \frac{h}{\tan65^o} \:=\:\frac{2\tan50^i - h}{\tan50^o}

    . . . . . . . . . . . . . . h\tan50^o \:=\:2\tan50^o\tan65^o - h\tan65^o

    . . . . . . . \h\tan50^o + h\tan65^o \;=\;2\tan50^o\tan65^o

    . . . . . . h(\tan50^o + \tan65^o) \;=\;2\tan50^o\tan65^o

    Therefore: . h \;=\;\frac{2\tan50^o\tan65^o}{\tan50^o+\tan65^o} \;\;\approx\;\;1.532\text{ miles.}





    This is the diagram for this scenario.
    Code:
                          * C
                        **|
                      * * |
                    *  *  |
                  *   *   |h
                *    *    |
              *     *     |
            *50    *65    |
        A *-------*-------* D
          :   2   B   x   :
    Give it a try . . .
    Thanks a lot for help me out
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