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Thread: Tricky Right Triangle Problem (Picture Included)

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    Tricky Right Triangle Problem (Picture Included)

    Tricky Right Triangle Problem (Picture Included)-triangleproblem.png
    It states that Segments AD, DF, FE, EC, and CB are all equal. Triangle ACB is a Right Triangle. That is all the information given. What is Angle A?
    An explanation would be great because I can't seem to understand it!
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    Lightbulb Re: Tricky Right Triangle Problem (Picture Included)

    Try A=40.6463194 (unchecked)
    Last edited by MaxJasper; Oct 12th 2012 at 03:02 PM.
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    Re: Tricky Right Triangle Problem (Picture Included)

    Hello, JacobFischer!

    Click image for larger version. 

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    Segments AD, DF, FE, EC, and CB are all equal.
    Triangle ACB is a Right Triangle.
    What is angle A?

    Let $\displaystyle \theta$ = angle A.
    We will calculate all the angles in the diagram.
    Write these figures in your diagram as we go along.

    Since $\displaystyle \Delta ADF$ is isosceles, $\displaystyle \angle DFA = \theta.$

    Since $\displaystyle \angle EDF$ is an exterior angle, $\displaystyle \angle EDF = 2\theta.$

    Since $\displaystyle \Delta EDF$ is isosceles, $\displaystyle \angle DEF = 2\theta.$

    Then $\displaystyle \angle DFE \:=\:180^o - 2\theta - 2\theta \:=\:180^o - 4\theta$

    Hence, $\displaystyle \angle EFC \:=\:180^o - \theta - (180^o-\theta)\:=\:3\theta$

    Since $\displaystyle \Delta FEC$ is isosceles, $\displaystyle \angle ECF = 3\theta.$

    Then $\displaystyle \angle FEC \:=\:180^o - 3\theta - 3\theta \:=\:180^o - 6\theta$

    Hence, $\displaystyle \angle BEC \:=\:180^o - 3\theta - (180^o - 6\theta) \:=\:4\theta$

    Since $\displaystyle \Delta ECB$ is isosceles, $\displaystyle \angle EBC = 4\theta$

    Hence, $\displaystyle \angle ECB \:=\:180^o - 4\theta - 4\theta \:=\:180^o - 8\theta$ .[1]


    But we know that: $\displaystyle \angle ECB + \angle ECF \:=\:90^o$
    . . That is: .$\displaystyle \angle ECB + 3\theta \:=\:90^o$ .[2]

    Substitute [1] into [2]: .$\displaystyle (180^o - 8\theta) + 3\theta \:=\:90^o $

    . . and we have: .$\displaystyle \text{-}5\theta \:=\:\text{-}90^o \quad\Rightarrow\quad \boxed{\theta \:=\:18^o}$
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