It is given that ∠ABC=680 and AB=BC.
In △ABC, the sum of three angles is 1800 that is:
∠CAB+∠ABC+∠ACB=1800
But ∠CAB=∠ACB (Angle opposite to equal sides are equal)
Therefore,
∠ACB+∠ACB+680=1800⇒2∠ACB+680=1800⇒2∠ACB=1800−680⇒2∠ACB=1120⇒∠ACB=11202=560
Thus, ∠ACB=560.
Now, ∠AOB=2∠ACB (Angle at centre is twice the angle at the circumference)
Therefore,
∠AOB=2×560=1120
Also, we know that ∠AOB+∠ADB=1800, thus
∠AOB+∠ADB=1800⇒1120+∠ADB=1800⇒∠ADB=1800−1120=680
Hence, ∠ADB=680.