MCQ
In the given figure, $AB \| CD$. If $\angle\text{BAO}=60^\circ$ and $\angle\text{OCD}=110^\circ$ then $\angle\text{AOC}=?$
- A$70^\circ$
- B$60^\circ$
- ✓$50^\circ$
- D$40^\circ$
✓
Answer
Correct option: C.
$50^\circ$
Let $\angle\text{AOC}=\text{x}^\circ$
Draw $YOZ \| CD \| AB.$
Now, $YO\| AB$ and $OA$ is the transversal.
$\Rightarrow\angle\text{YOA}=\angle\text{OAB}=60^\circ$ (alternate angles)
Again, $OZ \| CD$ and $OC$ is the transversal.
$\Rightarrow\angle\text{COZ}+\angle\text{OCD}=180^\circ$ (interior angles)
$\Rightarrow\angle\text{COZ}+110^\circ=180^\circ$
$\Rightarrow\angle\text{COZ}=70^\circ$
Now, $\angle\text{YOZ}=180^\circ$ (straight angle)
$\Rightarrow\angle\text{YOA}+\angle\text{AOC}+\angle\text{COZ}=180^\circ$
$\Rightarrow60^\circ+\text{x}+70^\circ=1806^\circ$
$\Rightarrow\text{x}=50^\circ$
$\Rightarrow\angle\text{AOC}=50^\circ$
Draw $YOZ \| CD \| AB.$
Now, $YO\| AB$ and $OA$ is the transversal.
$\Rightarrow\angle\text{YOA}=\angle\text{OAB}=60^\circ$ (alternate angles)
Again, $OZ \| CD$ and $OC$ is the transversal.
$\Rightarrow\angle\text{COZ}+\angle\text{OCD}=180^\circ$ (interior angles)
$\Rightarrow\angle\text{COZ}+110^\circ=180^\circ$
$\Rightarrow\angle\text{COZ}=70^\circ$
Now, $\angle\text{YOZ}=180^\circ$ (straight angle)
$\Rightarrow\angle\text{YOA}+\angle\text{AOC}+\angle\text{COZ}=180^\circ$
$\Rightarrow60^\circ+\text{x}+70^\circ=1806^\circ$
$\Rightarrow\text{x}=50^\circ$
$\Rightarrow\angle\text{AOC}=50^\circ$
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