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Thread: Help

  1. #1
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    Help

    Hi, i would be verry happy if someone could help me with this tasks. Its number 5 i cant figure out.
    AB=AC, AD=AE, BAD=30 degree , what is the angel x?

    So plz help. zorry for my bad english...
    thanks /Jonas

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  2. #2
    Member
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    See attachment before reading.

    PART 1: angles around D

    $\displaystyle (150^0 - \alpha) + \beta + x = 180^0$

    $\displaystyle x = 180^0 - (150^0 - \alpha) - \beta = 30^0 + \alpha - \beta$

    PART 2: angles in triangle DCE


    $\displaystyle (180^0 - \beta) + \alpha + x = 180^0$

    $\displaystyle x = 180^0 - (180^0 - \beta) - \alpha = \beta - \alpha$

    PART 3: x = x

    $\displaystyle x = x $

    $\displaystyle 30^0 + \alpha - \beta = \beta - \alpha$

    $\displaystyle 2\beta = 30^0 + 2\alpha \therefore \beta = 15^0 + \alpha$

    PART 4: Returning to angles around D

    $\displaystyle (150^0 - \alpha) + \beta + x = 180^0$

    $\displaystyle (150^0 - \alpha) + \alpha + 15^0 + x = 180^0$

    $\displaystyle 165^0 + x = 180^0$

    $\displaystyle x = 15^0$

    Feel free to ask any question if something is unclear
    Attached Thumbnails Attached Thumbnails Help-triangle.jpg  
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  3. #3
    Super Member

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    Lexington, MA (USA)
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    Hello, J0naz!

    I have a solution, but I'm sure someone will find a shorter one.


    Given: .$\displaystyle AB = AC,\;AD = AE,\;\angle BAD = 30^o$

    Find: .$\displaystyle \angle x \,=\,\angle EDC$
    Code:
                                  A
                                  o
                               *   * *
                            *  30  *   *
                         *           *     *
                      *               *       *     E
                   *                   *       α o
                *                       *      *    * 
             * θ                         *α * x        *
        B o   *   *   *   *   *   *   *   o   *   *   *   o C
                                          D

    Since $\displaystyle \Delta ABC$ is isosceles, $\displaystyle \angle B = \angle C = \theta$


    In $\displaystyle \Delta ABD\!:\;\angle B + \angle BAD + \angle ADB \:=\:180^o\quad\Rightarrow\quad \theta + 30^o + \angle ADB \:=\:180^o$
    . . Hence: .$\displaystyle \angle ADB \,=\,150^o -\theta $ .[1]


    In $\displaystyle \Delta ABC\!:\;\angle A + \angle B + \angle C \:=\:180^o \quad\Rightarrow\quad (\angle DAC + 30^o) + \theta + \theta \:=\:180^o $

    . . Hence: .$\displaystyle \angle DAC \:=\:150^o - 2\theta$


    In isosceles triangle $\displaystyle ADE\!:\;\angle ADE = \angle AED = \alpha$

    Then: .$\displaystyle (150^o-2\theta) + \alpha + \alpha \:=\:180^o\quad\Rightarrow\quad \alpha \:=\:\theta + 15^o$ .[2]


    At vertex $\displaystyle D$, we have: .$\displaystyle \angle ADB + \alpha + \angle x \:=\:180^o$

    Substitute [1] and [2]: .$\displaystyle (150^o - \theta) + (\theta + 15^o) + \angle x \:=\:180^o$

    . . Therefore: .$\displaystyle \angle x \:=\:15^o$

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  4. #4
    Newbie
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    Feb 2009
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    ok,

    thank you guys for the help.
    But i think unforturnally this one was a little bit over my skill.
    I understand your explanation in the beginning, but then its to mutch for me. I have just started the math course so hopfully i will understand it later on.
    But thank you for your time anyway, maybe I come back another day with new questions.

    Have a nice day
    /Jonas
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