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In Fig. ABCD is a quadrilateral inscribed in a circle with centre O. CD is produced to E such that ∠AED = 95° and ∠OBA = 30°. Find ∠OAC.

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∠ADE = 95°(Given) 

Since, 

OA = OB, so 

∠OAB = ∠OBA 

∠OAB = 30° 

∠ADE + ∠ADC = 180° 

(Linear pair) 

95° + ∠ADC = 180° 

∠ADC = 85° 

We know that, 

∠ADC = 2∠ADC 

∠ADC = 2 x 85° 

∠ADC = 170° 

Since, 

AO = OC 

(Radii of circle) 

∠OAC = ∠OCA 

(Sides opposite to equal angle) ... (i) 

In triangle OAC, 

∠OAC + ∠OCA + ∠AOC = 180° 

∠OAC + ∠OAC + 170° = 180° 

[From (i)] 

2∠OAC = 10° 

∠OAC = 5° 

Thus, 

∠OAC = 5°

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