Question

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

Calcualte:

1. ∠AEB,
2. ∠AED,
3. ∠COD.

Answer 1

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°