Question

chords AC and DE intersect at B. If m(arc DC) =110, m(arc AE) =94° , find angle ABE​

Answer 1

Given : chords AC and DE intersect at B  

m(arc DC) =110, m(arc AE) =94°  

To Find : ∠ABE

Solution:

∠ABE = ∠BCE + ∠BEC  ( exterior angle of triangle = Sum of opposite interior angle )

∠BCE = ∠ACE   as B lies on AC

∠ACE  = (1/2) m(arc AE)

=> ∠ACE  =  (1/2) 94°

= > ∠ACE  =47°

= >  ∠BCE = 47°

∠BEC  = ∠DEC  as B lies on DE

∠DEC =   (1/2) m(arc DC)

=> ∠DEC =     (1/2) 110°

=> ∠DEC =     55°

=> ∠BEC  = 55°

∠ABE = ∠BCE + ∠BEC

=> ∠ABE = 47°  + 55°

=> ∠ABE = 102°

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