Given that
[tex]\begin{gathered} \text{arcAB}=\text{arcDC} \\ \text{Thus,} \\ AB+BC+DC+DA=360^0(sum\text{ of angles of a circumference)} \end{gathered}[/tex]
Given that
[tex]\begin{gathered} arcDA=2BC \\ 104^0+104^0+2BC+BC=360^0_{} \\ 3BC=360^0-208^0 \\ BC=\frac{152^0}{3}=50.7^0 \\ \text{Hence} \\ AC=50.7^0+104=154.7^0\approx155^0 \end{gathered}[/tex]
