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Thread: 3 more circle questions

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    3 more circle questions

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  2. #2
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    Quote Originally Posted by dgenerationx2 View Post
    in attachments
    (3)

    $\displaystyle m\widehat{BDC}=360-70=290$

    $\displaystyle m\angle A=\frac{1}{2}(m\widehat{BDC}-m\widehat{BC})=\frac{1}{2}(290-70)=\frac{1}{2}(220)=110^{\circ}$

    (6)

    $\displaystyle m\angle AEC=\frac{1}{2}(m\widehat{AC}+m\widehat{BD})$

    $\displaystyle 15=\frac{1}{2}(m\widehat{AC}+25)$

    $\displaystyle 30=m\widehat{AC}+25$

    $\displaystyle 5=m\widehat{AC}$

    (8)

    $\displaystyle m\widehat{AC}+m\widehat{BC}+m\widehat{BD}+m\wideha t{AD}=360^{\circ}$

    $\displaystyle (3x-5)+(3x)+(5x)+(2x)=360$

    $\displaystyle 13x-5=360$

    $\displaystyle 13x=365$

    $\displaystyle x=\frac{365}{13}$

    $\displaystyle m\angle DEB=\frac{1}{2}(m\widehat{DB}+m\widehat{AC})=\frac {1}{2}(5x+3x-5)=\frac{1}{2}\left(8\left(\frac{365}{13}\right)-5\right)=109.8^{\circ}$ which is not one of your choices.
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