Mathematics

# In the given figure, $AB$ and $CD$ are straight lines through the centre $O$ of a circle. If $\angle AOC=80^o$ and $\angle CDE=40^o$, find $\angle ABC$

##### SOLUTION
We know that $\angle AOC$ and $\angle BOC$ from a linear pair

It can be written as

$\angle BOC =180^o -80^o$

By subtraction

$\angle BOC =100^o$

Using the angle sun property

$\angle ABC +\angle BOC +\angle DCE =180^o$

By substituting the values

$ABC +100^o +50^o +=180^o$

On further calculation

$\angle ABC=180^o-100^o -50^o$

By subtraction

$\angle ABC =180^o -150^o$

So we get

$\angle ABC =30^o$

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Subjective Medium Published on 09th 09, 2020
Questions 120418
Subjects 10
Chapters 88
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