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Question
In the figure, O is the centre of the circle, ∠AOE = 150°, ∠DAO = 51°. Calculate the sizes of the angles CEB and OCE.
Solution
∠ADE = `1/2` Reflex (∠AOE)
= `1/2 (360^circ - 150^circ)`
= 105°
(Angle at the center is double the angle at the circumference subtended by the same chord)
∠DAB + ∠BED = 180°
(Pair of opposite angles in a cyclic quadrilateral are supplementary)
`=>` ∠BED = 180° – 51° = 129°
∴ ∠CEB = 180° – ∠BED (Straight line)
= 180° – 129°
= 51°
Also, by angle sum property of ∆ADC,
∠OCE = 180° – 51° – 105° = 24°
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