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Question
In the given figure, the centre O of the small circle lies on the circumference of the bigger circle. If ∠APB = 75° and ∠BCD = 40°, find :
- ∠AOB,
- ∠ACB,
- ∠ABD,
- ∠ADB.
Solution
Join AB and AD
i. ∠AOB = 2∠APB
= 2 × 75°
= 150°
(Angle at the centre is double the angle at the circumference subtended by the same chord).
ii. In cyclic quadrilateral AOBC,
∠ACB = 180° – ∠AOB
= 180° – 150°
= 30°
(Pair of opposite angles in a cyclic quadrilateral are supplementary)
iii. In cyclic quadrilateral ABDC
∠ABD = 180° – ∠ACD
= 180° – (40° + 30°)
= 110°
(Pair of opposite angles in a cyclic quadrilateral are supplementary
iv. In cyclic quadrilateral AOBD,
∠ADB = 180° – ∠AOB
= 180° – 150°
= 30°
(Pair of opposite angles in a cyclic quadrilateral are supplementary
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