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