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