Mathematics

Two lines $$AB$$ and $$CD$$ intersect at a point $$O$$ such that $$\angle BOC+\angle AOD={280}^{o}$$, as shown in the figure. Find all four angles.


SOLUTION
From the figure we know that $$\angle AOD$$ and $$\angle BOC$$ are vertically opposite angles $$\angle AOD=\angle BOC$$

It is given that

$$\angle BOC+\angle AOC={ 280 }^{ o }$$

we know that $$\angle AOD=\angle BOC$$

so it can be written as

$$\angle AOD+\angle AOD={ 180 }^{ o }$$

$$2\angle AOD={ 280 }^{ o }$$

$$\angle AOD=\dfrac{180}{2}$$

$$\angle AOD=\angle BOC={ 140 }^{ o }$$

From the figure we know that $$\angle AOC$$ and $$\angle AOC$$ form a linear pair

So it can be written as

$$\angle AOC+\angle AOD={ 180 }^{ o }$$

substituting the values

$$\angle AOC+{ 140 }^{ o }={ 180 }^{ o }$$

$$\angle AOC={40}^{o}$$

From the figure we know that $$\angle AOC$$ and $$\angle BOD$$ are vertically opposite angles 

$$\angle AOC=\angle BOD={40}^{o}$$

Therefore $$\angle AOC={40}^{o}$$, $$\angle BOC={ 140 }^{ o }$$, $$\angle AOD={ 140 }^{ o }$$, $$\angle BOD={ 40 }^{ o }$$

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Subjective Medium Published on 09th 09, 2020
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