In figure below, $A,B$ and $C$ are three points on a circle with centre $O$ such that $\angle BOC = 30^o$ and $\angle AOB = 60^o$. If $D$ is a point on the circle other than the arc $ABC$, find $\angle ADC$.
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Given:

$A,B$ and $C$ are three points on a circle with centre $O$ such that $\angle BOC = 30^o$ and $\angle AOB = 60^o$.

$D$ is a point on the circle other than the arc $ABC$.

To do:

We have to find $\angle ADC$.

Solution:

$\angle AOC = \angle AOB+\angle BOC$

$\angle AOC = 60^o+30^o$

$\angle AOC = 90^o$

We know that,

The angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the circle.

Therefore,

$\angle ADC = \frac{1}{2}\angle AOC$

$= \frac{1}{2}\times 90^o$

$= 45^o$

Hence, $\angle ADC =45^o$.

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Updated on: 10-Oct-2022

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