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Math Help - Find the measure of Angle A?

  1. #1
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    Find the measure of Angle A?

    I have absolutely no idea what to do!

    It is the circled one, #21.
    Attached Thumbnails Attached Thumbnails Find the measure of Angle A?-id006.jpg  
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  2. #2
    Member u2_wa's Avatar
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    Triangle

    Quote Originally Posted by careness View Post
    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
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  3. #3
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    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.

    Last edited by Soroban; March 8th 2009 at 03:24 PM.
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