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