In the figure, if $\angle BAC = 60^o$ and $\angle BCA = 20^o$, find $\angle ADC$.
"
Given:
$\angle BAC = 60^o$ and $\angle BCA = 20^o$.
To do:
We have to find $\angle ADC$.
Solution:
In $\triangle ABC$,
$\angle BAC + \angle ABC + \angle ACB = 180^o$ (Sum of the angles in a triangle is $180^o$)
$60^o + \angle ABC + 20^o = 180^o$
$\angle ABC + 80^o = 180^o$
Therefore,
$\angle ABC = 180^o -80^o= 100^o$
$ABCD$ is a cyclic quadrilateral.
This implies,
$\angle ABC + \angle ADC = 180^o$
$100^o + \angle ADC = 180^o$
$\angle ADC = 180^o- 100^o$
$= 80^o$
Hence, $\angle ADC = 80^o$.
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