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Math Help - Exterior angle if Isos triangle

  1. #1
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    Exterior angle if Isos triangle

    In the Accompanying diagram AB=AC=CD=DE and <E=1/10 <CDE <BAF in degrees Picture Attached




    So this is what I figure Tirangle CDE is isos So <CDE= x and <E and <C=10x so 21x=180 but cannot use a calculator so figure number should be nicer then what get and only have 5 minutes to solve I get x to equal 8.571428571

    Thanks
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  2. #2
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    Hello, IDontunderstand!

    You should be able to "walk" across the diagram
    . . and determine all the angles.


    In the accompanying diagram:
    AB=AC=CD=DE
    \angle E \:=\:\tfrac{1}{10}\angle CDE

    Find \angle BAF in degrees.
    Code:
        B o
           *  *
            * 45 *
             *        *
              *           *     C
               *         45  o
                *           *   * *
                 *        *  120 *15*
                  * 90 *           *     *
               60 *  * 30       30 *   15 *
          o - - - - o - - - - - - - - - o - - - - o
          F         A                   D         E

    Let x = \angle E.\;\text{Then: }\:\angle CDE = 10x

    Since \Delta CDE is isosceles: . \angle DCE = x

    We have: . 10x + x + x \:=\:180^o \quad\Rightarrow\quad 12x \:=\:180^o \quad\Rightarrow\quad x \:=\:15^o

    Hence: . \angle CDE = 150^o \quad\Rightarrow\quad \angle CDA = 30^o

    Since \Delta ACD is isosceles: . \angle CDA = \angle CAD = 30^o \quad\Rightarrow\quad \angle ACD = 120^o


    About vertex C, we have: . \angle BCA + 120^o + 15^o \:=\:180^o \quad\Rightarrow\quad \angle BCA = 45^o

    Since \Delta BAC is isosceles: . \angle BCA = \angle CBA = 45^o \quad\Rightarrow\quad \angle BAC = 90^o


    About vertex A, we have: . \angle BAF + 90^o + 30^o \:=\:180^o \quad\Rightarrow\quad \boxed{\angle BAF \:=\:60^o}

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  3. #3
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    Thank you I feel dumb what I tried to do was say that <E is 10 times <CDE because I wanted to multiply each side by 10.
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