Question

​a circle with centre 0 circumscribes a pentagon ABCDE. If AB=BC=CD and angle BCD=126 degree Find:
1) angle AEB
2) angle AED
3) angle AOC

Answer 1

According to question

Given that    AB = BC = CD
and      Angle BCD = 126

Thus according to question

In triangle BOC and COD

OB = OD
OC is common
and 
BC = CD

so triangles are similar

Thus Angle OBC = 126/2 = 63
Similarly
      Angle  OAB = 63
Thus   Angle AEB = 180 - 63 = 117

Answer 2

Answer:

AB = BC = CD, ∠BCD = 126°

To find : (i) ∠AEB, (ii) ∠AED, (iii) ∠AOC

Join AC, BE, AO, OC

∵ AB = BC = CD and ∠BCD = 126°

∴ ∠ABC = 126°

In ∆ABC, ∠ABC = ∠BCD =126° .

∴ ∠BAC=∠BCA=180−126/2=54/2=27

∠AEB and ∠ACB are in the same segment

∴ ∠AEB = ∠ACB = 27°

∵ EBCD is a cyclic quad.

∴ ∠BCD + ∠AED = 180° ⇒ 126° + ∠BED = 180°

⇒ ∠BED = 180°- 126° = 54°

∴ ∠AED = ∠AEB + ∠BED = 27° + 54° = 81°

Arc AEDC subtends ∠AOC at the centre and ∠ABC at the remaining part of the circle Reflex ∠AOC = 2 × ∠ABC = 2 × 126° = 252°

∠AOC = 360° – 252° = 108°

Step-by-step explanation:

if it helpful pls vote