MCQ
In Figure, if $\text{EC}\ ||\text{ AB}, \angle\text{ECD} = 70^\circ$ and $\angle\text{BDO} = 20^\circ,$ then $\angle\text{OBD}$ is: 
- A$60^\circ $
- B$70^\circ$
- C$20^\circ $
- ✓$50^\circ $
✓
Answer
Correct option: D.
$50^\circ $
$\text{EC}\ ||\text{ AB}$ and $CD$ is transverse to it.
Now $\angle\text{ECD} = \angle\text{AOD} = 70^\circ$ (Corresponding angles)
In $\angle\text{OBD}$
$\angle\text{OBD} + \angle\text{BOD} + \angle\text{ODB} = 180^\circ$
$\angle\text{BOD} = 180^\circ - \angle\text{AOD} = 180^\circ - 70^\circ = 110^\circ$
$\angle\text{ODB} = 20^\circ$ (Given)
So, $\angle\text{OBD} = 180^\circ - \angle\text{BOD} - \angle\text{ODB}$
$= 180^\circ - 110^\circ - 20^\circ$
$= 50^\circ$
Now $\angle\text{ECD} = \angle\text{AOD} = 70^\circ$ (Corresponding angles)
In $\angle\text{OBD}$
$\angle\text{OBD} + \angle\text{BOD} + \angle\text{ODB} = 180^\circ$
$\angle\text{BOD} = 180^\circ - \angle\text{AOD} = 180^\circ - 70^\circ = 110^\circ$
$\angle\text{ODB} = 20^\circ$ (Given)
So, $\angle\text{OBD} = 180^\circ - \angle\text{BOD} - \angle\text{ODB}$
$= 180^\circ - 110^\circ - 20^\circ$
$= 50^\circ$
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