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Math Help - height of a mountain

  1. #1
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    height of a mountain


    From a point on ground level, you measure the angle of elevation to the top of a mountain to be 35. Then you walk 160 m farther away from the mountain and find that the angle of elevation is now 15. Find the height of the mountain.

    I keep getting 30, but it is not correct. I can't figure out where the 15 degrees goes. Do I have it drawn correctly thus far? Thanks much.
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  2. #2
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    Hello, daydreembelievr!

    Your diagram is wrong . . .


    From a point on ground level, the angle of elevation to the top of a mountain is 35.
    160 m farther away from the mountain, the angle of elevation is 15.
    Find the height of the mountain.
    Code:
        P *
          | *  *
          |   *     *
         h|     *        *
          |       *           *
          |     35 *         15  *
          * - - - - - * - - - - - - - - *
          Q     x     A       160       B

    The height of the mountain is: h = PQ.

    The first observation is made at A\!:\;\;x = QA,\;\angle PAQ = 35^o

    The second observation is made at B\!:\;\;AB = 160,\;\angle PBQ = 15^o


    In right triangle PQB\!:\;\;\tan15^o \:=\:\frac{h}{x+160} \quad\Rightarrow\quad x \:=\:\frac{h}{\tan15^o} - 160 .[1]

    In right triangle PQA\!:\;\;\tan35^o \:=\:\frac{h}{x}\quad\Rightarrow\quad x \:=\:\frac{h}{\tan35^o} .[2]


    Equate [1] and [2]: . \frac{h}{\tan15^o}-160 \:=\:\frac{h}{\tan35^o} \quad\hdots\quad \text{and solve for }h

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  3. #3
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    unsure

    Sooo helpful. I forgot to make it x+160 instead of just x. But now when I try to solve it I keep getting 25.9 which is close to what I was getting before...but I know that isn't the answer (I have the answer key). Am I doing something wrong?

    h/0.7002 = h/0.26795 - 160

    h=(0.7002)h/0.26795 - 160
    h= 0.7002h/0.26795 - 112.032
    0.26795h=0.7002h-112.032
    h-25.9

    thanks sooo much.
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  4. #4
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    Hello, daydreembelievr!


    We have: . \frac{h}{\tan15^o}-160 \:=\:\frac{h}{\tan35^o}

    Multiply by \tan15^o\tan35^o\!:\quad h\tan35^o - 160\tan15^o\tan35^o \:=\:h\tan15^o

    . . . . . . h\tan35^o - h\tan15^o \:=\:160\tan15^o\tan35^o

    Factor: . h(\tan35^o - \tan15^o) \:=\:160\tan15^o\tan35^o \quad\Rightarrow\quad h \:=\:\frac{160\tan15^o\tan35^o}{\tan35^o-\tan15^o}

    Therefore: . h \;=\;69.44737423...

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