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In the given figure, O is the center of the circle. AC and BD intersect at P. If ∠AOB = 100° and ∠DAP = 30°, what is ∠APB?

(a) 77° 

(b) 80° 

(c) 85° 

(d) 90°

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Answer : (b) 80º 

∠ADB = \(\frac{1}{2}\) × ∠AOB = 50° 

In ∆ DPA, ∠ADP + ∠DAP + ∠DPA = 180° 

⇒ ∠DPA = 180° – (50° + 30°) = 100° 

Also, DPB being a straight line, 

∠APB = 180° – ∠DPA 

= 180° – 100° = 80°.

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