Q37 of 153 Page 148

In a ΔABC its is given that ∠A : ∠B : ∠C = 3 : 2 : 1 and CD ⊥ AC. Then ∠ECD = ?

As per question,

So,

∠A = 90^o

∠B = 60^o

∠C = 30^o

∠ACB + ∠ACD + ∠ECD = 180^o (sum of angles on straight line)

∠ECD = 180^o – 90^o – 30^o

= 60^o

In the given figure, AB || CD. If ∠EAB = 50° and ∠ECD = 60°, then ∠AEB = ?

In the given figure, ∠OAB = 75°, ∠OBA = 55° and ∠OCD = 100°. Then ∠ODC = ?

In the given figure, AB || CD. If ∠ABO = 45° and ∠COD = 100° then ∠CDO = ?

In the given figure, AB || DC, ∠BAD = 90°, ∠CBD = 28° and ∠BCE = 65°. Then ∠ABD = ?