3 more circle questions

Quote:

Originally Posted by dgenerationx2

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.