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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 = 75o and BCD = 40o, find :

(i) AOB,

(ii) ACB,

(iii) ABD,

(iv) ADB.

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Solution

Join A and B as shown in the diagram.

(i) \angle AOB = 2 \angle APB = 2 \times 75^o = 150^o (angle subtended by an chord on the center is double that subtended by the same chord on the circumference)c712

(ii) \angle ACB = 180^o-\angle AOB^o = 180^o-150^o = 30^o (AOBC is a cyclic quadrilateral)

(iii) \angle ABD = 180^o - \angle ACD = 180^o-(40^o+30^o) = 110^o (ABDC is a cyclic quadrilateral)

(iv) \angle ADB = 180^o - \angle AOB = 180^o-150^o=  30^o (ADBO is a cyclic quadrilateral)

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