MCQ
In the given figure, $AB \| CD$. If $\angle\text{AOC}=30^\circ$ and $\angle\text{OAB}=100^\circ$ then $\angle\text{OCD}=?$
- ✓$130^\circ$
- B$150^\circ$
- C$80^\circ$
- D$100^\circ$
✓
Answer
Correct option: A.
$130^\circ$
Construction: Through $O$, draw $OE \| AB \| CD$
$\Rightarrow\angle\text{BAO}+\angle\text{EOA}=180^\circ$
$\Rightarrow100^\circ+\angle\text{EOA}=180^\circ$
$\Rightarrow\angle\text{EOA}=80^\circ$
So, $\angle\text{EOC}=\angle\text{EAO}-\angle\text{COA}=80^\circ-30^\circ=50^\circ$
Since $CD \| EO$
$\angle\text{OCD}+\angle\text{EOC}=1806^\circ$
$\Rightarrow\angle\text{OCD}+50^\circ=180^\circ$
$\Rightarrow\angle\text{OCD}=130^\circ$
$\Rightarrow\angle\text{BAO}+\angle\text{EOA}=180^\circ$
$\Rightarrow100^\circ+\angle\text{EOA}=180^\circ$
$\Rightarrow\angle\text{EOA}=80^\circ$
So, $\angle\text{EOC}=\angle\text{EAO}-\angle\text{COA}=80^\circ-30^\circ=50^\circ$
Since $CD \| EO$
$\angle\text{OCD}+\angle\text{EOC}=1806^\circ$
$\Rightarrow\angle\text{OCD}+50^\circ=180^\circ$
$\Rightarrow\angle\text{OCD}=130^\circ$
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