In Fig. 16.201, <

∠ADE = 95^o (Given)

Since,

OA = OB, so

∠OAB = ∠OBA

∠OAB = 30^o

∠ADE + ∠ADC = 180^o (Linear pair)

95^o + ∠ADC = 180^o

∠ADC = 85^o

We know that,

∠ADC = 2 ∠ADC

∠ADC = 2 * 85^o

∠ADC = 170^o

Since,

AO = OC (Radii of circle)

∠OAC = ∠OCA (Sides opposite to equal angle) (i)

In triangle OAC,

∠OAC + ∠OCA + ∠AOC = 180^o

∠OAC + ∠OAC + 170^o = 180^o [From (i)]

2 ∠OAC = 10^o

∠OAC = 5^o

Thus,

∠OAC = 5^o