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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 : ∠BCD.
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उत्तर
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°
संबंधित प्रश्न
In the figure, m∠DBC = 58°. BD is the diameter of the circle. Calculate:
1) m∠BDC
2) m∠BEC
3) m∠BAC
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Calculate:
- ∠BDC
- ∠BEC
- ∠BAC
