# Thread: Exterior angle if Isos triangle

1. ## 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

2. Hello, IDontunderstand!

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

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

Find $\displaystyle \angle BAF$ 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 $\displaystyle x = \angle E.\;\text{Then: }\:\angle CDE = 10x$

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

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

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

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

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

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

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

3. 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.