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प्रश्न
In the given figure, AB is a diameter of the circle. Chord ED is parallel to AB and ∠EAB = 63°.
Calculate:
- ∠EBA,
- ∠BCD.
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उत्तर
i. ∠AEB = 90°
(Angle in a semicircle is a right angle)
Therefore ∠EBA = 90° – ∠EAB
= 90° – 63°
= 27°
ii. AB || ED
Therefore ∠DEB = ∠EBA = 27° (Alternate angles)
Therefore BCDE is a cyclic quadrilateral
Therefore ∠DEB + ∠BCD = 180°
[Pair of opposite angles in a cyclic quadrilateral are supplementary]
Therefore ∠BCD = 180° – 27° = 153°
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