Question
In the given figure, straight lines AB and CD intersect at O. If $\angle\text{AOC}+\angle\text{BOD}=130^\circ$ then $\angle\text{AOD}=?$
- 65º
- 115º
- 110º
- 125º
✓
Answer
- 115º
Solution:
$\angle\text{AOC}+\angle\text{BOD}=1306^\circ$ (given)
But $\angle\text{AOC}=\angle\text{BOD}$ (Vartically Opposite angles)
$\Rightarrow2\angle\text{AOC}=130^\circ$
$\Rightarrow\angle\text{AOC}=65^\circ$
Since COD is a straight line,
$\angle\text{AOC}+\angle\text{AOD}=180^\circ$
$\Rightarrow65^\circ+\angle\text{AOD}=180^\circ$
$\Rightarrow\angle\text{AOD}=115^\circ$
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