MCQ
Line sgements $AB$ and $CD$ intersect at $O$ such that $\text{AC}||\text{DB}.$ If $\angle\text{CAB} = 45^\circ$ and $\angle\text{CDB} = 55^\circ,$ then $\angle\text{BOD} =$
- ✓$80^\circ $
- B$90^\circ$
- C$100^\circ$
- D$135^\circ$
✓
Answer
Correct option: A.
$80^\circ $
$\text{AC}||\text{DB}$
And, AB is transverse to these parallel lines
So, $\angle\text{CAB} = \angle\text{ABD}$ (Alternate angles)
$\Rightarrow \angle\text{ABD} = 45^\circ$
Now In $\triangle\text{BOD}$
$\angle\text{BOD} + \angle\text{ODB} + \angle\text{DBA} = 180^\circ$
$\angle\text{DBA} = \angle\text{ABD} = 45^\circ, \ \angle\text{ODB} = 55^\circ$
So, $\angle\text{BOD} = 180^\circ - 45^\circ - 55^\circ$
$= 80^\circ$
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