’O’ is the centre ∠AOB = 100°
Thus ∠ACB = 1/2 ∠AOB
[∵ angle made by an arc at the centre is twice the angle made by it on the remaining part]]
= 1/2 x 100° = 50°
∠ACB and ∠ADB are supplementary
[ ∵ Opp. angles of a cyclic quadrilateral]
∴ ∠ADB = 180°-50° = 130°
[OR]
∠ADB is the angle made by the major arc \(\widehat{ACB}\) at D.
∴ ∠ADB = 1/2 ∠AOB [where ∠AOB is the angle;
made by \(\widehat{ACB}\) at the centre]
= 1/2 [360° – 100°] [from the figure]
= 1/2 x 260° = 130°
