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
Questions 120418
Subjects 10
Chapters 88
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