In figure, for DABC, chord AB @ chord BC,
Please answer with explanation.....
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In figure, for DABC, chord AB @ chord BC,
Please answer with explanation.....
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Answer:
∠BEA = 18°
Step-by-step explanation:
Here,
chord AB = chord BC, ∠ABC=72°, BD is the bisector of ∠ABC and ∠BAC=∠ACB.
Now,
∠ABC + ∠BAC + ∠ACB = 180° [sum of interior angles of a triangle is 180°]
or, 72° + ∠ACB + ∠ACB = 180° [since ∠BAC = ∠ACB]
or, 2 ∠ACB = 180°-72°
or, 2 ∠ACB = 108°
i.e. ∠ACB = 54°
∠ADB = ∠ACB [being inscribed angles made on same base are equal]
= 54°
And,
∠ABD + ∠DBC = ∠ABC
or, ∠DBC + ∠DBC = 72° [since BD is the bisector of ∠ABC]
or, 2 ∠DBC = 72°
i.e. ∠DBC = 36°
Now,
∠BED + ∠DBE = ∠ADB [being exterior angle of a triangle is equal to sum of two non-adjacent opposite interior angle]
or, ∠BEA + ∠DBC = 54° [since ∠BED and
∠BEA are same while ∠DBE and DBC are also same]
or, ∠BEA + 36° = 54°
or, ∠BEA = 54° - 36°
i.e. ∠BEA = 18°