jmc

Let $x$ denote the measure in degrees of $\angle BOC$. Because $\angle BOD$ and $\angle COA$ are right angles, $\angle COD$ and $\angle BOA$ each measure $90-x$ degrees. Therefore $\angle AOD=x+(90-x)+(90-x)$ degrees. Solving $$3.5x=x+90-x+90-x$$we find $x=180/4.5=40$. Therefore $\angle AOD=180^{\circ }-40^{\circ }=\boxed{140\text{ degrees}}$.