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Math Help - Plz solve these questions.(Trignometry statement questions)

  1. #1
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    Question Plz solve these questions.(Trignometry statement questions)

    Last edited by prashantvrm; July 14th 2008 at 08:12 PM.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by prashantvrm View Post
    you seem to be missing information from number 2), i think you meant to state the angle of elevation, but didn't. anyway, we can still do this problem in principle. the other problems are similar, try attacking them in the same way. that is, draw diagrams and figure out what trig ratio you need to solve for the unknown. i would hope you know what is meant by "angle of elevation" and "angle of depression."

    anyway, in this problem, let the angle of elevation be x and the distance from the foot of the pole to the point be d (see the diagram below).

    we can use the tangent trig ratio here.


    problem 2:

    \tan x = \frac {\text{opposite}}{\text{adjacent}} = \frac {12}d

    \Rightarrow d = \frac {12}{\tan x}

    now, if we know what x is, we can find the distance. otherwise, leave it like that. try the others, be sure to draw diagrams so you know how to think about the problem
    Attached Thumbnails Attached Thumbnails Plz solve these questions.(Trignometry statement questions)-elevation.gif  
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  3. #3
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    Hello, prashantvrm!

    4. The angle of elevation to the top of a chimney from a point on the ground is 30.
    After walking 50 m towards the chimney, the angle of elevation becomes 45.
    Find the height of the chimney.
    Code:
                                      * C
                                  * * |
                              *   *   |
                          *     *     | h
                      *       *       |
                  *         *         |
              * 30       * 45       |
          * - - - - - - * - - - - - - * 
          A     50      B      x      D
    Let CD = h, the height of the chimney.

    When the observer stands at A,\:\angle CAD = 30^o.

    At B (where AB = 50), \angle CBD = 45^o.

    Let x = BD.


    In right triangle CDA\!:\;\frac{h}{x+50} \:=\:\tan30^o \:=\:\frac{1}{\sqrt{3}} \quad\Rightarrow\quad \sqrt{3}h \:=\:x + 50\;\;{\color{blue}[1]}

    In right triangle CDB\!:\;\frac{h}{x} \:=\:\tan45^o \:=\:1 \quad\Rightarrow\quad x \:=\:h\;\;{\color{blue}[2]}


    Substitite [2] into [1]: . \sqrt{3}h \:=\:h + 50 \quad\Rightarrow\quad \sqrt{3}h - h \:=\:50

    Factor: . (\sqrt{3}-1)h \:=\:50 \quad\Rightarrow\quad h \:=\:\frac{50}{\sqrt{3}-1} \;\approx\;\boxed{68.3\text{ m}}

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  4. #4
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    Hello again, prashantvrm!

    3. The top of a tree is broken and the broken part makes a 30 angle with the ground.
    The distance from the tip of the tree to the base of the tree is 40 m.
    Find the height of the tree.
    Code:
        A *
          |   *
          |       *
          |           *
          |           30 *
        C * - - - - - - - - - * B
                  40
    This is a 30-60 right triangle.

    We know the ratios of ths sides . . . . AC : BC : AB \:=\:1 : \sqrt{3} : 2


    Hence, we have: . AC : BC : AB \;=\;\frac{40\sqrt{3}}{3} : 40 : \frac{80\sqrt{3}}{3}


    The height of the tree is: . AC + AB \;=\;\frac{40\sqrt{3}}{3} + \frac{80\sqrt{3}}{3} \;=\;\frac{120\sqrt{3}}{3} \;=\;\boxed{40\sqrt{3}\text{ m}}

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  5. #5
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    hai

    just find the distance and u can find the answer .....
    its just too easy.......
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