MCQ
In the given figure, $\angle\text{OAB}=110^\circ$ and $\angle\text{BCD}=130^\circ$ then $\angle\text{ABC}$ is equal to:
- A$40^\circ$
- B$50^\circ$
- ✓$60^\circ$
- D$70^\circ$
✓
Answer
Correct option: C.
$60^\circ$
Through $B$ draw $\text{YBZ}\ \| OA \| CD$.
Now, $OA \| YB$ and $AB$ is the transversal.
$\Rightarrow\angle\text{OAB}+\angle\text{YBA}=180^\circ \ ($interior angles are supplementary$)$
$\Rightarrow\angle\text{YBA}=70^\circ$
Also, $CD \| BZ$ and $BC$ is the transversal.
$\Rightarrow\angle\text{DCB}+\angle\text{CBZ}=180^\circ\ ($interior angles are supplementary$)$
Now, $OA \| YB$ and $AB$ is the transversal.
$\Rightarrow\angle\text{OAB}+\angle\text{YBA}=180^\circ \ ($interior angles are supplementary$)$
$\Rightarrow\angle\text{YBA}=70^\circ$
Also, $CD \| BZ$ and $BC$ is the transversal.
$\Rightarrow\angle\text{DCB}+\angle\text{CBZ}=180^\circ\ ($interior angles are supplementary$)$
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