# Thread: Find the measure of Angle A?

1. ## Find the measure of Angle A?

I have absolutely no idea what to do!

It is the circled one, #21.

2. ## Triangle

Originally Posted by careness
I have absolutely no idea what to do!

It is the circled one, #21.
let angle A = x

angle AED = 180-2x (AED is isosceles triangle)
angle EDB = 180-4x (EDB is isosceles triangle)
angle BDC =180-(180-4x+x)=3x
angle C = 3x ,BCD is isosceles triangle
ABC is isosceles triangle so angle B=3x
hence angle A+B+C=180, x+2(3x)=180
Solve to get an angle A

3. Hello, careness

Edit: I read my diagram incorrectly!

21. Given: .$\displaystyle AB = AC,\;BC = BD = DE = AE$

Find $\displaystyle m\angle A$
Code:
              A
*
θ
* *

*   *

E o     *

*2θ    θ*

*         o D
2θ
*           *
180-3θ
*             *

B o   *   *   *   o C

Draw segments $\displaystyle BD, DE.$
. . Let $\displaystyle \theta \,=\,\angle A$
Since $\displaystyle \Delta AED$ is isosceles: .$\displaystyle \angle ADE = \theta$

$\displaystyle \angle BED$ is an exterior angle of $\displaystyle \Delta AED$
. . Hence: .$\displaystyle \angle BED = 2\theta$
Since $\displaystyle \Delta BED$ is isosceles: .$\displaystyle \angle EBD = 2\theta$

At point D, we have: .$\displaystyle \angle ADE + \angle EDB +\angle BDC \:=\:180^o$
. . $\displaystyle \theta + (180^o-4\theta) + \angle BDC \:=\:180^o \quad\Rightarrow\quad \angle BDC \:=\:3\theta$
Since $\displaystyle \Delta BCD$ is isosceles: .$\displaystyle \angle C \:=\:3\theta$ .[1]

But $\displaystyle C$ is a base angle of an isosceles triangle with vertex angle $\displaystyle \theta.$
.Hence: .$\displaystyle \angle C \:=\:\frac{180^o-\theta}{2}$ .[2]

Equate [2] and [1]: .$\displaystyle \frac{180^o-\theta}{2} \:=\:3\theta$

. . $\displaystyle 180^o-\theta \:=\:6\theta \quad\Rightarrow\quad 7\theta \:=\:180^o \quad\Rightarrow\quad \theta \:=\:\frac{180^o}{7}$

$\displaystyle \angle A$ is the central angle of a regular heptagon.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Consider this diagram . . .
Code:
                A
*
/θ\
/   \
/     \
/       \
/         \  D
/           *
/        *    \
/     *         \
/  *              \
B * - - - - - - - - - *C

Given: .Isosceles triangle $\displaystyle ABC\!:\;AB = AC$
. . . . . .$\displaystyle BC = BD = DA$

Then: .$\displaystyle \theta \:=\:\frac{180^o}{5} \:=\:72^o$, the central angle of a regular pentagon.