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) ∠ABC.
Solution:
(i)
∠CED = 90° (Angle in a semi circle)
In ΔCED, we have:
∠CED +∠EDC + ∠DCE = 180° (Angle sum property of a triangle)
⇒ 90° + 40° + ∠DCE = 180°
⇒ ∠DCE = (180° – 130°) = 50° ...(i)
∴ ∠DCE = 50°
(ii)
As ∠AOC and ∠BOC are linear pair, we have:
∠BOC = (180° – 80°) = 100° ...(ii)
In Δ BOC, we have:
∠OBC + ∠OCB + ∠BOC = 180° (Angle sum property of a triangle)
⇒ ∠ABC + ∠DCE + ∠BOC = 180° [∵ ∠OBC = ∠ABC and ∠OCB = ∠DCE]
⇒ ∠ABC = 180° – (∠BOC + ∠DCE)
⇒ ∠ABC = 180° – (100° + 50°) [From (i) and (ii)]
⇒ ∠ABC = (180° - 150°) = 30°