# Thread: Help

1. ## 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

2. 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

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

4. ## 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