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Math Help - Trigonometric Addition Formulae

  1. #1
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    Trigonometric Addition Formulae



    Two telephone poles, AB and CD, are situated 9m apart. They are held by wires that are fixed to the ground at E. Calculate exact value of cos AEC.


    --

    You can't use a calculator....
    The answer is 1/route 442
    I need help on how to arrive at the answer, without the aid of a calculator.
    I know i will have to use something like:
    "cos(alpha + beta)" = cos alpha x cos beta - sin alpha x sin beta

    I know it sounds rude as it's my 1st post, but if your kind enough to help.
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  2. #2
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    Quote Originally Posted by r_maths View Post
    Two telephone poles, AB and CD, are situated 9m apart. They are held by wires that are fixed to the ground at E. Calculate exact value of cos AEC.
    --
    You can't use a calculator....
    The answer is 1/route 442
    I need help on how to arrive at the answer, without the aid of a calculator.
    I know i will have to use something like:
    "cos(alpha + beta)" = cos alpha x cos beta - sin alpha x sin beta

    I know it sounds rude as it's my 1st post, but if your kind enough to help.
    Hello,

    I've modified your sketch a little bit. You can easily calculate the cosine values in the red and blue right triangles.

    EB
    Attached Thumbnails Attached Thumbnails Trigonometric Addition Formulae-telefonmasten.gif  
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  3. #3
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    thanks, il try now, if i need more help with this, please be on to help.
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  4. #4
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    Hello, r_maths!

    Welcome aboard!


    Code:
                                  * C
                                * |
        A *                   *   |
          | *               *     |
          |   *           *       | 5
        4 |     *       *         |
          |       * θ *           |
        B * - - - - * - - - - - - * D
              6     E      3      :

    Two telephone poles, AB and CD, are situated 9m apart.
    They are held by wires that are fixed to the ground at E.
    Calculate exact value of \cos(\angle AEC).

    We want to find \cos\theta which is in \Delta AEC.

    Look at the upper portion of the diagram.
    Code:
                                  * C
                                  | 1
        A *-----------------------+ F
          |           9           |
          |                       |
        4 |                       | 4
          |                       |
        B * - - - - + - - - - - - * D
              6     E      3      :

    Draw AC.
    We have a right triangle CFA with AF = 9,\:CF = 1
    . . Hence: . AC \:=\:\sqrt{9^2 + 1^2} \:=\:\sqrt{82}


    Back to the original diagram . . .

    We have \Delta AEC with AE = \sqrt{52},\:CE = \sqrt{34},\:AC = \sqrt{82}

    Using the Law of Cosines, we have:
    . . \cos\theta \:=\:\frac{AE^2 + CE^2 - AC^2}{2(AE)(CE)} \:=\:\frac{(\sqrt{52})^2 + (\sqrt{34})^2 - (\sqrt{82})^2}{2(\sqrt{52})(\sqrt{34})} = \:\frac{52 + 34 - 82}{2(2\sqrt{13})(\sqrt{34})}

    Therefore: . \cos\theta \:=\:\frac{4}{4(\sqrt{13})(\sqrt{34})} \:=\:\frac{1}{\sqrt{442}}

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  5. #5
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    Quote Originally Posted by r_maths View Post
    thanks, il try now, if i need more help with this, please be on to help.
    Hello,

    I haven't got much time left. So I send you (only for confirmation of course) a possible way to do your problem:

    From the drawing you can read:
    \cos(\alpha)=\frac{4}{\sqrt{52}}
    \sin(\alpha)=\frac{6}{\sqrt{52}}

    \cos(\beta)=\frac{5}{\sqrt{34}}
    \sin(\beta)=\frac{3}{\sqrt{34}}

    Now use the formula you've mentioned above:

    \cos(\alpha+\beta)=\frac{4}{\sqrt{52}} \cdot \frac{5}{\sqrt{34}} - \frac{6}{\sqrt{52}} \cdot \frac{3}{\sqrt{34}}= \frac{20}{\sqrt{52 \cdot 34}} - \frac{18}{\sqrt{52 \cdot 34}}=\frac{2}{\sqrt{52 \cdot 34}}=\frac{1}{\sqrt{442}}

    EB
    Attached Thumbnails Attached Thumbnails Trigonometric Addition Formulae-telefonmasten.gif  
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  6. #6
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    ^
    Soroban, thanks,
    i was having difficulties with earboth solution. i got (19/442) or maybe its my bad maths skills.

    thanks!

    EDIT: thanks earboth
    Last edited by r_maths; January 4th 2007 at 12:25 PM.
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  7. #7
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    Quote Originally Posted by r_maths View Post
    ^
    omg, thanks,
    i was having difficulties with earboth solution. i got (19/442) or maybe its my bad maths skills.

    thanks!
    Hello,

    you didn't subtract but you added the product of sine-values in your formula. That's all.

    EB
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  8. #8
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    Thumbs up

    yep, solved. thanks
    im going to be a regular here if im going to past my exams
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