Question

In the given figure, AB and CD are straight lines through the centre O of a circle. If ∠AOC = 80° and ∠CDE = 40°, find (i) ∠DCE, (ii) ∠OBC.

Answer 1

(i)
∠CED = 90° (Angle in a semi-circle)
In ΔCED,
∠CED +∠EDC + ∠DCE = 180° (Angle sum property of a triangle)
⇒ 90° + 40° + ∠DCE = 180°
⇒ ∠DCE = (180° – 130°) = 50° ............(i)
∴ ∠DCE = 50°
(ii)
∠AOC and ∠BOC form a linear pair.
∠BOC = (180° – 80°) = 100° .............(ii)
In ΔBOC,
∠OBC + ∠OCB + ∠BOC = 180° (Angle sum property of a triangle)
∠OBC + ∠DCE + ∠BOC = 180° [Since ∠OCB = ∠DCE]
∠OBC = 180° – (∠BOC + ∠DCE)
= 180° – (100° + 50°) [from (i) and (ii)]
= (180° - 150°) = 30°