MCQ
In the given figure, straight lines $AB$ and $CD$ intersect at $O$. If $\angle\text{AOC}+\angle\text{BOD}=130^\circ$ then $\angle\text{AOD}=?$
- ✓$115^\circ$
- B$125^\circ$
- C$110^\circ$
- D$65^\circ$
✓
Answer
Correct option: A.
$115^\circ$
We have,
$\angle\text{AOC}=\angle\text{BOD}$ [Vertically-Opposite Angles]
$\therefore\angle\text{AOC}+\angle\text{BOD}=130^\circ$
$\Rightarrow\angle\text{AOC}+\angle\text{AOC}=130^\circ [\therefore\angle\text{AOC}=\angle\text{BOD}]$
$\Rightarrow2\angle\text{AOC}=130^\circ$
$\Rightarrow\angle\text{AOC}=65^\circ$
Now,
$\angle\text{AOC}+\angle\text{AOD}=180^\circ$ [$\because$ COD is a straight line]
$\Rightarrow65^\circ+\angle\text{AOD}=180^\circ$
$\Rightarrow\angle\text{AOC}=115^\circ.$
$\angle\text{AOC}=\angle\text{BOD}$ [Vertically-Opposite Angles]
$\therefore\angle\text{AOC}+\angle\text{BOD}=130^\circ$
$\Rightarrow\angle\text{AOC}+\angle\text{AOC}=130^\circ [\therefore\angle\text{AOC}=\angle\text{BOD}]$
$\Rightarrow2\angle\text{AOC}=130^\circ$
$\Rightarrow\angle\text{AOC}=65^\circ$
Now,
$\angle\text{AOC}+\angle\text{AOD}=180^\circ$ [$\because$ COD is a straight line]
$\Rightarrow65^\circ+\angle\text{AOD}=180^\circ$
$\Rightarrow\angle\text{AOC}=115^\circ.$
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