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 $\angle DCE$

SOLUTION
From the figure we know that $\angle CED =90^o$

Consider $\triangle CED$

Using the angle sun property

$\angle CED +\angle EDC +\angle DCE =180^o$

By substituting the values

$90^o +40^o +\angle DEC =180^o$

On further calculation

$\angle DCE=180^o-90^o -40^o$

By subtraction

$\angle DCE =180^o -130^o$

so we get

$\angle DCE =50^o$

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Subjective Medium Published on 09th 09, 2020
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