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 :

1. ∠AOB,
2. ∠ACB,
3. ∠ABD,
4. ∠ADB.

Answer 1

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