# In Fig. 9.43, if EC||AB, ∠ECD =70° and ∠BDO =20° , then ∠OBD isA. 20°B. 50°C. 60°D. 70°

Given,

EC AB

ECD = 70o

BDO = 20o

Since,

EC AB

And, OC cuts them so

ECD = 1 (Alternate angle)

1 = 70o

1 + 3 = 180o (Linear pair)

3 = 110o

In

BOD + OBD + BDO = 180o

3 + ODB + 20o = 180o

ODB = 50o

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