Question

The diagram shows a pentagon ABCDE inscribed in a circle center O . Given AB=BC=CD and angle ABC =132. Find angle AEB, angleAED and angle COD.(plz show full solution and any useless answers like typing anything just for points will be reported)

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

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Answer 2

Answer:

Step-by-step explanation:

Triangle AOB ~∆BOC (BO =CO :radius)

(BO=AO:radius)

(BA=BC given)

SSS rule

So angle ABC =angle OBC

angle ABO =132÷2= 48°

48°+AOE = 180° ( angles on a straight line)

AOE= 132°

In ∆AOE

132+OAE+OEA =180°

OEA =OAE (OA and OE are radius so the adjacent angles are also equal)

OAe =OEA = 48÷2=24°

Or AEB can also be written as OEA

Now apply congruency on ∆BOC AND ∆COD

YOU WILL FIND THAT COD = 48°

HOPE IT HELPS YOU!!