(i) 50°

Given, ∠BAE = 110° and ∠ACD = 120°

∠ACB + ∠ACD = 180° [Because BD is a straight line]

⇒ ∠ACB + 120° = 180°

⇒ ∠ACB = 60°_______________ (i)

In triangle ABC,

∠BAE = ∠ABC + ∠ACB

⇒ 110° = x + 60°

⇒ x = 50°

(ii) 120°

In triangle ABC,

∠A + ∠B + ∠C = 180° [Sum of angles of triangle ABC]

⇒ 30° + 40° + ∠C = 180°

⇒ ∠C = 110°

∠BCA + ∠DCA = 180° [Because BD is a straight line]

⇒ 110° + ∠DCA = 180°

⇒ ∠DCA = 70°_________________ (i)

In triangle ECD,

∠AED = ∠ECD + ∠EDC

⇒ x = 70°+ 50°

⇒ x = 120°

(iii) 55°

Explanation:

∠BAC = ∠EAF = 60°[Opposite angles]

In triangle ABC,

∠ABC + ∠BAC = ∠ACD

⇒ X°+ 60°= 115°

⇒ X°= 55°

(iv) 75°

Given AB||CD

Therefore,

∠BAD = ∠EDC = 60°[Alternate angles]

In triangle CED,

∠C + ∠D + ∠E = 180°[Sum of angles of triangle]

⇒ 45° + 60° + x = 180°[∠EDC = 60°]

⇒ x = 75°

(v) 30°

Explanation:

In triangle ABC,

∠BAC + ∠BCA + ∠ABC = 180°[Sum of angles of triangle]

⇒ 40° + 90° + ∠ABC = 180°

⇒ ∠ABC = 50°________________ (i)

In triangle BDE,

∠BDE + ∠BED + ∠EBD = 180°[Sum of angles of triangle]

⇒ x° + 100° + 50° = 180°[∠EBD = ∠ABC = 50°]

⇒ x° = 30°

(vi) x=30

Explanation:

In triangle ABE,

∠BAE + ∠BEA + ∠ABE = 180°[Sum of angles of triangle]

⇒ 75° + ∠BEA + 65° = 180°

⇒ ∠BEA = 40°

∠BEA = ∠CED = 40°[Opposite angles]

In triangle CDE,

∠CDE + ∠CED + ∠ECD = 180°[Sum of angles of triangle]

⇒ x° + 40° + 110° = 180°

⇒ x° = 30°