Question
Two lines $AB$ and $CD$ intersect at $O$. If $\angle\text{AOC}+\angle\text{COB}+\angle\text{BOD}=270^\circ,$ find the measure of $\angle\text{AOC},\angle\text{COB},\angle\text{BOD}$ and $\angle\text{DOA}.$
✓
Answer
Given: $\angle\text{AOC}+\angle\text{COB}+\angle\text{BOD}=270^\circ$
To find: $\angle\text{AOC},\angle\text{COB},\angle\text{BOD},\angle\text{DOA}$
Here, $\angle\text{AOC},\angle\text{COB},\angle\text{BOD}=270^\circ$[Complete angle]
$\Rightarrow 270 + AOD = 360 $
$\Rightarrow AOD = 360 - 270$
$ \Rightarrow AOD = 90$
Now, $AOD + BOD = 180$ [Linear pair]
$90 + BOD = 180 $
$\Rightarrow BOD = 180 - 90$
$ \Rightarrow BOD = 90 AOD = BOC = 90$ [Vertically opposite angles]
$BOD = AOC = 90$ [Vertically opposite angles]
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