Finding an angle involving two triangles inscribed in a circle

Question

Since $\overline{OA} \parallel \overline{BC}$, we have that So I know that $\angle COA=\angle BCO=40^\circ$. I'm trying to figure out how to make use of the facts that these triangles are inscribed in circles... help appreciated!

Answer 1

$\angle O =\angle OCB = 40^°$ (alternate interior angles, $OA \parallel BC$)

$\angle B =\frac{\angle O}{2} = 20^°$ ($\angle$ of center=2$\angle$ circumference subtended same segment)

$\angle ODA=120^°$(sum of $\angle$ 's in a $\triangle$)

Answer 2

We have $$\angle OAD=\angle{ABC}=\dfrac12\angle{AOC}=20^{\circ}$$ because of Angle at the Centre Is Twice the Angle at the Circumference.

So, in $\Delta ODA$, we have $$\angle AOC=40^\circ\\\angle OAD=20^\circ$$ So, $$\angle ODA=180^\circ-40^\circ-20^\circ=120^\circ$$