AD=BC=8 ft
AB=CD=2 ft
Then,
[tex]\begin{gathered} AC=BD=\sqrt[]{2^2+8^2} \\ =\sqrt[]{68} \\ OA=OB=OC=OD=\frac{\sqrt[]{68}}{2}\text{ ft} \end{gathered}[/tex]
In triangle BDC,
[tex]\begin{gathered} \tan \angle BDC=\frac{8}{2} \\ =4 \\ \angle BDC=75.963 \\ \angle DBC=14.04 \end{gathered}[/tex][tex]\begin{gathered} OE=\frac{8}{2} \\ =4\text{ ft} \\ \angle BDC=\angle OCD=75.96 \end{gathered}[/tex]
In triangle ODC,
[tex]\begin{gathered} \angle ODC+\angle OCD+\angle OCD=180 \\ \\ \angle\text{DOC}=180-(2\cdot75.96) \\ \angle DOC=28 \\ \angle X=\angle DOC=28 \end{gathered}[/tex]
So, the correct option is option C.
