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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