Solution:

(i)
∠ADB and ∠ACB are in the same segment.

∠ADB = ∠ACB = 50°

Now in
∆ADB,

∠DAB + X + ∠ADB = 180°

= 42^o + x + 50^o = 180^o

= 92^o + x = 180^o

x = 180^o – 92^o

x = 88^o

(ii) In the given figure we have

= 32^o + 45^o + x = 180^o

= 77^o + x = 180^o

x = 103^o

(iii) From the given number we have

∠BAD = ∠BCD

Because angles in the same segment

But ∠BAD = 20^o

∠BAD = 20^o

∠BCD = 20^o

∠CEA = 90^o

∠CED = 90^o

Now in triangle CED,

∠CED +
∠BCD + ∠CDE = 180^o

90^o + 20^o + x = 180^o

= 110^o + x = 180^o

x = 180^o – 110^o

x = 70^o

(iv) In
∆ABC

∠ABC + ∠ABC + ∠BAC = 180^o

(Because sum of a triangle)

69^o + 31^o + ∠BAC = 180^o

∠BAC = 180^o – 100^o

∠BAC = 80^o

Since ∠BAC and ∠BAD are in the same

Segment.

∠BAD = x^o = 80^o

(v) Given ∠CPB = 120^o , ∠ACP = 70^o

To find, x^o i,e., ∠PBD

Reflex ∠CPB = ∠BPO + ∠CPA

120⁰ = ∠BPD + ∠BPD

(BPD = CPA are vertically opposite ∠s)

2∠BPD = 120^o ∠PBD = 120⁰/2 = 60^o

Also ∠ACP and PBD are in the same segment

∠PBD + ∠ACP = 70⁰

Now, In ∆PBD

∠PBD + ∠PDB + ∠BPD = 180^o

(sum of all ∠s in a triangle)

70⁰ + x^o + 60⁰ = 180^o

x = 180^o – 130^o

x = 50^o

(vi) ∠DAB = ∠BCD

(Angles in the same segment of the circle)

∠DAB = 25⁰ (∠BCD = 25⁰ given)

In ∆DAP,

Ex, ∠CDA = ∠DAP + ∠DPA

x^o = ∠DAB + ∠DPA

x^o = 25^o + 35 ^o

x^o = 60^o