#### Question

In the given figure, AB = BC = CD and ∠ABC = 132 . Calcualte:

(i) ∠AEB, (ii) ∠AED, (iii) ∠COD.

#### Solution

In the figure, O is the centre of circle, with AB = BC = CD.

Also, ∠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 /2xx 48° `

= 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°

Is there an error in this question or solution?

Solution In the Given Figure, Ab = Bc = Cd and ∠Abc = 132 . Calcualte: (I) ∠Aeb, (Ii) ∠Aed, (Iii) ∠Cod. Concept: Arc and Chord Properties - the Angle that an Arc of a Circle Subtends at the Center is Double that Which It Subtends at Any Point on the Remaining Part of the Circle.