In the given figure, O is the centre of a circle and chords AC and BD intersect at E. If ∠AEB = 110° and ∠CBE = 30° then ∠ADB = ?
Given: ∠AEB = 110° and ∠CBE = 30°
∠AEC = ∠AEB + ∠BEC = 180°
∠AEB + ∠BEC = 180°
110° + ∠BEC = 180°
∠BEC = 180° – 110°
∠BEC = 70°
In ΔBEC
By angle sum property
∠CBE + ∠BEC + ∠ECB = 180°
30° + 70° + ∠ECB = 180°
∠ECB = 180° – 30° – 70°
∠ECB = 80°
Here,
∠ECB = ∠ADB (angles in the same segment)
∴ ∠ECB = ∠ADB = 80°
∴ ∠ADB = 80°