Mathematics
In the given figure, AB is a diameter of a circle with centre O and chord ED is parallel to AB and ∠EAB = 65°. Calculate : (i) ∠EBA (ii) ∠BED (iii) ∠BCD
Circles
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Answer
(i) We know that,
Angle in semi-circle is a right angle.
∴ ∠AEB = 90°.
In △AEB,
⇒ ∠AEB + ∠EBA + ∠EAB = 180°
⇒ 90° + ∠EBA + 65° = 180°
⇒ 155° + ∠EBA = 180°
⇒ ∠EBA = 180° - 155° = 25°.
Hence, ∠EBA = 25°.
(ii) ∠BED = ∠EBA = 25° Alternate interior angles ED ∥ AB, EB as a transversal
Hence, ∠BED = 25°.
(iii) As, AB ∥ ED
∴ ∠DEB = ∠EBA = 25° [Alternate angles]
BCDE is a cyclic quadrilateral.
∴ ∠DEB + ∠BCD = 180° [Sum of opposite angles in a cyclic quadrilateral = 180°.]
⇒ 25° + ∠BCD = 180°
⇒ ∠BCD = 180° - 25° = 155°.
Hence, ∠BCD = 155°.
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