Q35 of 153 Page 148

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

AB || CD and BC is traversal.

So, ∠DCB = ∠ABC = 60^o

Now in triangle AEB, we have

∠ABE + ∠BAE + ∠AEB =180^o

∠AEB =180^o – 60^o – 50^o

= 70^o

In the given figure, ∠BAC = 40°, ∠ACB = 90° and ∠BED = 100°. Then ∠BDE = ?

In the given figure, BO and CO are the bisectors of ∠B and ∠C respectively. If ∠A = 50°, then ∠BOC = ?

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

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