Exterior angle if Isos triangle

**IDontunderstand** · October 13th 2009, 01:01 PM

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

**Soroban** · October 14th 2009, 05:25 AM

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: . $\angle BAF + 90^o + 30^o \:=\:180^o \quad\Rightarrow\quad \boxed{\angle BAF \:=\:60^o}$

**IDontunderstand** · October 14th 2009, 01:23 PM

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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