MCQ
Line segments $AB$ and $CD$ intersect at $O$ such that $AC \| DB$. If $\angle\text{CAB}=45^\circ$ and $\angle\text{CDB}=55^\circ,$ then $\angle\text{BOD}=$
- A$100^\circ$
- ✓$80^\circ$
- C$90^\circ$
- D$135^\circ$
✓
Answer
Correct option: B.
$80^\circ$
$AC \| BD$
And, $AB$ is transverse to these parallal 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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