Solution:

(a) Given,
∠AOC = 150° and AD = CD

We know that an angle subtends by an arc of a circle

at the center is twice the angle subtended by the same arc

at any point on the remaining part of the circle.

(i) ∠AOC = 2 × ∠ABC

∠ABC = ∠AOC/2 = 150^o/2 = 75^o

(ii) From the figure, ABCD is a cyclic quadrilateral

∠ABC + ∠ADC = 180^o

(Sum of opposite angels in a cyclic quadrilateral

Is 180^o)

75^o + ∠ADC = 180^o

∠ADC + 180^o – 75^o

∠ADC = 105^o

(b) (i) AC is the diameter of the circle

∠ABC = 90^o (Angle in a semi-circle)

(ii) ABCD is a cyclic quadrilateral

∠BAD + ∠BCD = 180^o

∠BAD + 75^o = 180^o

(∠BCD = 75^o)

∠BAD = 180^o -75^o = 105^o

But ∠EAF = ∠BAD

(Vertically opposite angles)

∠EAF = 105^o