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Math Help - 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)

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

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

    (6)

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

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

    30=m\widehat{AC}+25

    5=m\widehat{AC}

    (8)

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

    (3x-5)+(3x)+(5x)+(2x)=360

    13x-5=360

    13x=365

    x=\frac{365}{13}

    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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