In the figure, O is the centre of the circle, ∠AOE = 150°, ∠DAO = 51°. Calculate the sizes of the angles CEB and OCE.
Exercise 17 (A) | Q 31 | Page 260
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∠ADE =$\frac{1}{2} \operatorname{Re}$ flex $(\angle AOE )=\frac{1}{2}\left(360^{\circ}-150^{\circ}\right)=105^{\circ}$
(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° - =180° -129° = 51°
Also, by angle sum property of ∆ADC,
∠OCE =180° - 51° -105° = 24°
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