(i) x = 100

Given AB||CD, ∠ABE = 35° and ∠EDC = 65°

Draw a line PEQ||AB or CD

∠1 = ∠ABE = 35°[AB||PQ and alternate angle] _______________ (i)

∠2 = ∠EDC = 65°[CD||PQ and alternate angle] _______________ (ii)

From equations (i) and (ii)

∠1 + ∠2 = 100°

⇒ x = 100°

(ii) x=280

Given AB||CD, ∠ABE = 35° and ∠EDC = 65°

Draw a line POQ||AB or CD

∠1 = ∠ABO = 55°[AB||PQ and alternate angle] _______________ (i)

∠2 = ∠CDO = 25°[CD||PQ and alternate angle] _______________ (ii)

From equations (i) and (ii)

∠1 + ∠2 = 80°

Now,

∠BOD + ∠DOB = 360°

⇒ 80° + x° = 360°

⇒ x = 280°

(iii) x=120

Given AB||CD, ∠BAE = 116° and ∠DCE = 124°

Draw a line EF||AB or CD

∠BAE + ∠PAE = 180° [Because PAB is a straight line]

⇒ 116° + ∠3 = 180°

⇒ ∠3 = 180° - 116°

⇒ ∠3 = 64°

Therefore,

∠1 = ∠3 = 64° [Alternate angles] ____________________ (i)

Similarly, ∠4 = 180° - 124°

∠4 = 56°

Therefore,

∠2 = ∠4 = 56° [Alternate angles] ____________________ (ii)

From equations (i) and (ii)

⇒ ∠1 + ∠2 = 64° + 56°

⇒ x = 120°