MCQ
Two lines AB and CD intersect at O. If $\angle\text{AOC}+\angle\text{COB}+\angle\text{BOD}=270^\circ,$ then $\angle\text{AOC}=$
- A180º
- B70º
- C80º
- D90º
✓
Answer
- 90º
Solution:
Given that,
AB and CD intersect at O
$\angle\text{AOC}+\angle\text{COB}+\angle\text{BOD}=270^\circ\text{(i)}$
$\angle\text{COB}+\angle\text{BOD}=180^\circ$ (Linear pair) (ii)
Using (ii) in (i), we get
$\angle\text{AOC}+180^\circ=270^\circ$
$\text{AOC}=90^\circ.$
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