#### 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:

(i) ∠AOB, (ii) ∠ACB, (iii) ∠ABD, (iv) ∠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)

Is there an error in this question or solution?

Solution 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: (I) ∠Aob, (Ii) ∠Acb, (Iii) ∠Abd, (Iv) ∠Adb Concept: Arc and Chord Properties - the Angle that an Arc of a Circle Subtends at the Center is Double that Which It Subtends at Any Point on the Remaining Part of the Circle.