Mathematics

In the adjoining figure, $$DE$$ is a chord parallel to a diameter $$AC$$ of the circle with centre $$O$$. If $$\angle CBD=60^o$$, calculate $$\angle CDE$$.


SOLUTION
We know that the angles in the same segment of a circle are equal 

From the figure we know that $$\angle CAD $$ and $$\angle CBD$$ are in the segment $$CD$$

$$\angle CAD =\angle CBD =60^o$$

An angle in a semi-circle is a right angle

So we get

$$\angle ADC =90^o$$

Using the angle sum property

$$\angle ACD +\angle ADC+\angle CAD =180^o$$

By substituting the values

$$\angle ACD +90^o +60^o =180^o$$

On further calculation

$$\angle ACD =180^o -90^o -60^o$$

By subtraction

$$ACD =180^o -150^o$$

So we get

$$\angle ACD =30^o$$

We know that $$AC || DE$$ and $$CD$$ is a transversal 

From the figure we know that $$\angle ACD $$ and $$\angle CDE$$ are alternate angles 

So we get

$$\angle CDE =\angle ACD =30^o$$

Therefore, $$\angle CDE =30^o$$   
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