MCQ
In Fig. $AB || CO, \angle \text{OAB}=150^\circ$ and $\angle \text{OCO}=120^\circ.$ Then, $\angle \text{AOC}=$ 
- A$80^\circ$
- ✓$90^\circ$
- C$70^\circ$
- D$100^\circ$
✓
Answer
Correct option: B.
$90^\circ$
Construction: Draw a line $OE$ from the point $O$ parallel to $AB$ and $CD$
Since, $AB || OE$
$\therefore\angle \text{BAO}+\angle \text{AOE}=180^\circ$ [Angles on the same side of a transversal line are supplementary]
$\Rightarrow 150^\circ+\angle \text{AOE}=180^\circ$
$\Rightarrow \angle \text{AOE}=30^\circ$
Again, $CD || OE$
$\therefore \angle \text{DCO}+\angle \text{COE}=180^\circ$ [Angles on the same side of a transversal line are supplementary]
$\Rightarrow 120^\circ+\angle \text{COE}=180^\circ$
$\Rightarrow \angle \text{COE}=60^\circ$
Now, $\angle \text{AOC}=\angle \text{AOE}+\angle \text{COE}$
$=30^\circ+60^\circ$
$=90^\circ$
Hence, the correct answer is option $(b).$
Since, $AB || OE$
$\therefore\angle \text{BAO}+\angle \text{AOE}=180^\circ$ [Angles on the same side of a transversal line are supplementary]
$\Rightarrow 150^\circ+\angle \text{AOE}=180^\circ$
$\Rightarrow \angle \text{AOE}=30^\circ$
Again, $CD || OE$
$\therefore \angle \text{DCO}+\angle \text{COE}=180^\circ$ [Angles on the same side of a transversal line are supplementary]
$\Rightarrow 120^\circ+\angle \text{COE}=180^\circ$
$\Rightarrow \angle \text{COE}=60^\circ$
Now, $\angle \text{AOC}=\angle \text{AOE}+\angle \text{COE}$
$=30^\circ+60^\circ$
$=90^\circ$
Hence, the correct answer is option $(b).$
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