Question

In the given figure, O is the centre of the circle, ∠AOB = 100° and ∠AFB = 70°, then ∠AEB equals

Answer 1

Given : O is the centre of the circle, ∠AOB = 100° and ∠AFB = 70°

To Find : ∠AEB

Solution:

∠AOB = 100°

∠ACB = ∠ADB = (1/2)∠AOB   ( angle by same arc AB at center and  remaining circle)

∠ACB = ∠ADB = (1/2)100°

∠ACB = ∠ADB = 50°

∠ECB + ∠ACB = 180°  Linear pair

=> ∠ECB + 50° = 180°  

=> ∠ECB  = 130°

∠ECF = ∠ECB  =  130°

Similarly ∠EDF  = 130°

∠CFD = ∠AFB  ( vertically opposite angles)

=> ∠CFD = 70°

in Quadrilateral CFDE

∠CFD + ∠EDF + ∠DEC + ECF  = 360°

=> 70° + 130° + ∠DEC +130° = 360°

=> ∠DEC = 30°

∠AEB = ∠DEC

=> ∠AEB = 30°

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