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Question
AB is the diameter of the circle with centre O. OD is parallel to BC and ∠AOD = 60°. Calculate the numerical values of :
- ∠ABD,
- ∠DBC,
- ∠ADC.
Solution
Join BD.
i. `∠ABD = 1/2 ∠AOD = 1/2xx 60^circ = 30^circ`
(Angle at the first is double the angle at the circumference subtended by the same chord)
ii. ∠BDA = 90°
(Angle in a semicircle)
Also, ΔOAD is equilateral (∵ ∠OAD = 60°)
∴ ∠ODB = 90° – ∠ODA
= 90° – 60°
= 30°
Also, OD || BC
∴ ∠DBC = ∠ODB = 30° (Alternate angles)
iii. ∠ABC = ∠ABD + ∠DBC
= 30° + 30°
= 60°
In cyclic quadrilateral ABCD,
∠ADC = 180° – ∠ABC
= 180° – 60°
= 120°
(Pair of opposite angles in a cyclic quadrilateral are supplementary)
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