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प्रश्न
In the given figure, AC is the diameter of the circle with centre O. CD and BE are parallel. Angle ∠AOB = 80° and ∠ACE = 10°.
Calculate:
- Angle BEC,
- Angle BCD,
- Angle CED.
उत्तर
i. ∠BOC = 180° – 80° = 100° (Straight line)
And ∠BOC = 2∠BEC
(Angle at the centre is double the angle at the circumference subtended by the same chord)
`=> ∠BEC = (100^circ)/2 = 50^circ`
ii. DC || EB
∴ DCE = ∠BEC = 50° (Alternate angles)
∴ ∠AOB = 80°
`=> ∠ACB = 1/2 ∠AOB = 40^circ`
Angle at the center is double the angle at the circumference subtended by the same chord)
We have,
∠BCD = ∠ACB + ∠ACE + ∠DCE
= 40° + 10° + 50°
= 100°
iii. ∠BED = 180° – ∠BCD
= 180° – 100°
= 80°
(Pair of opposite angles in a cyclic quadrilateral are supplementary)
`=>` ∠CED + 50° = 80°
`=>` ∠CED = 30°
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