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Question
In the given figure, A, B, C and D are four points on a circle. AC and BD intersect at a point E such that ∠BEC = 130° and ∠ECD = 20°. Find ∠BAC.
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Solution
In ΔCDE,
∠CDE + ∠DCE = ∠CEB ...(Exterior angle)
⇒ ∠CDE + 20° = 130°
⇒ ∠CDE = 110°
However, ∠BAC = ∠CDE ...(Angles in the same segment of a circle)
⇒ ∠BAC = 110°
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