Given, a parallelogram ABCD
In the ΔOBC, we have
y+30∘=100∘ [exterior angle property of triangle]
⇒y=70∘
By the angle sum property of a triangle,
we have, x+y+30=180∘
⇒x+70∘+30∘=180∘⇒x=180∘−100∘=80∘
Now, since AD∥BC and BD is transversal, therefore
∠ADO=∠OBC [alternate interior angles]
⇒z=30∘