Can i get help on this question. It is a 6 marker
Another hint then: Knowing that angle ABO is a right angle, then you know that 2x + angle CBO + 90 degrees = 180 degrees. (Again, prove it.) So what is angle CBO in terms of x?
Here's a site with a list of the circle theorems. Maybe having all of them in one spot might help.
-Dan
Do you understand that $y=m(\angle BAD)=\frac{1}{2}\left( {arc(BCD) - arc(BD)} \right)~?$
The quadrilateral $ABOD$ contains two right angles & the sum of all its angles is $2\pi$. Therefore $m(\angle BOD)=\pi-y.$
Look at the figure: $m(\angle COD)=\pi-2x=m(arc DC)$; moreover since $2x=0.5(arc BC)$ then $4x=m(\angle COB)$.
$m(\angle BOC)+m(\angle COD)+m(\angle DOB)=2\pi$ now you finish and sow us your work.