Find the measure of Angle A?

**careness** · March 5th 2009, 09:56 PM

I have absolutely no idea what to do!

It is the circled one, #21.

**u2_wa** · March 5th 2009, 10:43 PM

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

**Soroban** · March 7th 2009, 08:19 AM

Hello, careness

Edit: I read my diagram incorrectly!

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

Find $m\angle A$

Code:

              A
              *
              θ
             * *

            *   *

         E o     *

          *2θ    θ*

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

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

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

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

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

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

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

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

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

Consider this diagram . . .

Code:

                A
                *
               /θ\
              /   \
             /     \
            /       \
           /         \  D
          /           *
         /        *    \
        /     *         \
       /  *              \
    B * - - - - - - - - - *C

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

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

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