Question
In the given figure, if $\angle\text{BAC}=60^\circ$ and $\angle\text{BCA}=20^\circ,$ find $\angle\text{ADC}.$
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Answer
Using angle sum property in $\triangle\text{ABC}, $$\angle\text{B}=180^\circ-(60^\circ+20^\circ)=100^\circ$
In cyclic quadrilateral ABCD, we have:$\angle\text{B}+\angle\text{C}=180^\circ$
$\angle\text{D}=180^\circ-100^\circ=80^\circ$
In cyclic quadrilateral ABCD, we have:$\angle\text{B}+\angle\text{C}=180^\circ$
$\angle\text{D}=180^\circ-100^\circ=80^\circ$
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