(i) ∠AOC = 135^o

Therefore,

∠AOC + ∠BOC = 180^o (Linear pair)

135^o + ∠BOC = 180^o

∠BOC = 45^o

By degree measure theorem,

∠BOC = 2 ∠COB

45^o = 2x

x = 22 ^o

(ii) We have,

∠ABC = 40^o

∠ACB = 90^o (Angle in semi-circle)

In triangle ABC, by angle sum property

∠CAB + ∠ACB + ∠ABC = 180^o

∠CAB + 90^o + 40^o = 180^o

∠CAB = 50^o

Now,

∠COB = ∠CAB (Angle on same segment)

x = 50^o

(iii) We have,

∠AOC = 120^o

BY degree measure theorem,

∠AOC = 2 ∠APC

120^o= 2 ∠APC

∠APC = 60^o

Therefore,

∠APC + ∠ABC = 180^o (Opposite angles of cyclic quadrilateral)

60^o + ∠ABC = 180^o

∠ABC = 120^o

∠ABC + ∠DBC = 180^o (Linear pair)

120^o + x = 180^o

x = 60^o

(iv) We have,

∠CBD = 65^o

Therefore,

∠ABC + ∠CBD = 180^o (Linear pair)

∠ABC + 65^o = 180⁰

∠ABC = 115^o

Therefore,

Reflex ∠AOC = 2 ∠ABC (By degree measure theorem)

x = 2 * 115^o

= 230^o

(v) We have,

∠OAB = 35^o

Then,

∠OBA = ∠OAB = 35^o (Angle opposite to equal sides are equal)

In triangle AOB, by angle sum property

∠AOB + ∠OAB + ∠OBA = 180^o

∠AOB + 35^o + 35^o =180^o

∠AOB = 110^o

Therefore,

∠AOB + Reflex ∠AOB = 360^o (Complete angle)

110^o + Reflex ∠AOB = 360^o

Reflex ∠AOB = 250^o

By degree measure theorem,

Reflex ∠AOB = 2 ∠ACB

250^o= 2x

x = 125^o

(vi) We have,

∠AOB = 60^o

By degree measure theorem,

∠AOB = 2 ∠ACB

60^o = 2 ∠ACB

∠ACB = 30^o

x = 30^o

(vii) We have,

∠BAC = 50^o

∠DBC = 70^o

Therefore,

∠BDC = ∠BAC = 50^o (Angles on same segment)

In triangle BDC, by angle sum property

∠BDC + ∠BCD + ∠DBC = 180^o

50^o + x + 70^o = 180^o

120^o + x = 180^o

x = 60^o

(viii) We have,

∠DBO = 40^o

∠DBC = 90^o (Angle in semi-circle)

Therefore,

∠DBO + ∠OBC = 90^o

40^o + ∠OBC = 90^o

∠OBC = 50^o

By degree measure theorem,

∠AOC = 2 ∠OBC

x = 2 * 50^o

x = 100^o

(ix) In triangle DAB, by angle sum property

∠ADB + ∠DAB + ∠ABD = 180^o

32^o + ∠DAB + 50^o= 180^o

∠DAB = 98^o

Now,

∠DAB + ∠DCB = 180^o (Opposite angle of cyclic quadrilateral)

98^o + x = 180^o

x = 180^o – 98^o

= 82^o

(x) We have,

∠BAC = 35^o

∠BDC = ∠BAC = 35^o (Angle on same segment)

In triangle BCD, by angle sum property

∠BDC + ∠BCD + ∠DBC = 180^o

30^o + x + 65^o = 180^o

x = 80^o

(xi) We have,

∠ABD = 40^o

∠ACD = ∠ABD = 40^o (Angle on same segment)

In triangle PCD, by angle sum property

∠PCD + ∠CPD + ∠PDC = 180^o

40^o + 110^o + x = 180^o

x = 30^o

(xii) Given that,

∠BAC = 52^o

∠BDC = ∠BAC = 52^o (Angle on same segment)

Since, OD = OC

Then,

∠ODC = ∠OCD (Opposite angles to equal radii)

x = 52^o