In the given figure, AB and CD are straight lines through the centre O of a circle.

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°

 

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