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Question
In the given figure, AB = BC = CD and ∠ABC = 132°.
Calcualte:
- ∠AEB,
- ∠AED,
- ∠COD.
Solution
In the figure, O is the centre of circle, with AB = BC = CD.
∠ABC = 132°
i. In cyclic quadrilateral ABCE
∠ABC + ∠AEC = 180° ...[Sum of opposite angles]
`=>` ∠132° + ∠AEC = 180°
`=>` ∠AEC = 180° – 132°
`=>` ∠AEC = 48°
Since, AB = BC, ∠AEB = ∠BEC ...[Equal chords subtends equal angles]
∴ `∠AEB = 1 /2 ∠AEC`
= `1/2 xx 48^circ`
= 24°
ii. Similarly, AB = BC = CD
∠AEB = ∠BEC = ∠CED = 24°
∠AED = ∠AEB + ∠BEC + ∠CED
= 24° + 24° + 24°
= 72°
iii. Arc CD subtends ∠COD at the centre and ∠CED at the remaining part of the circle.
∴ COD = 2∠CED
= 2 × 24°
= 48°
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