In the figure given alongside, AB and CD are straight lines through the centre O of a circle. If ∠AOC = 80° and ∠CDE = 40°, find the number of degrees in :

(i) ∠DCE,

(ii) ∠ABC.

(i) From figure,

∠CED = 90° [As angle in semi-circle = 90°]

In △CED,

⇒ ∠CED + ∠CDE + ∠DCE = 180° [By angle sum property of triangle]

⇒ 90° + 40° + ∠DCE = 180°

⇒ ∠DCE + 130° = 180°

⇒ ∠DCE = 180° - 130° = 50°.

Hence, ∠DCE = 50°.

(ii) We know that,

An exterior angle is equal to sum of two opposite interior angles.

In ∆BOC,

⇒ ∠AOC = ∠OCB + ∠OBC

⇒ ∠OBC = ∠AOC - ∠OCB

⇒ ∠OBC = ∠AOC - ∠DCE

⇒ ∠OBC = 80° - 50° = 30°.

From figure,

∠ABC = ∠OBC = 30°.

Hence, ∠ABC = 30°.