Question
In the given figure ABCD is a cyclic quadrilateral. If $\angle\text{BCD}=100^\circ$ and $\angle\text{ABD}=70^\circ,$ find $\angle\text{ADB}.$
✓
Answer
We have, $\angle\text{BCD}=100^\circ$ and $\angle\text{ABD}=70^\circ$$\therefore\angle\text{DAB}+\angle\text{BCD}=180^\circ$ [Opposite angles of cyclic quad.]
$\Rightarrow\angle\text{DAB}+100^\circ=180^\circ$
$\Rightarrow\angle\text{DAB}+180^\circ-100^\circ=80^\circ$
In $\triangle\text{DAB},$ by angle sum property$\angle\text{ADB}+\angle\text{DAB}+\angle\text{ABD}=180^\circ$
$\Rightarrow\angle\text{ADB}+80^\circ+70^\circ=180^\circ$
$\Rightarrow\angle\text{ADB}=180^\circ-80^\circ-70^\circ=30^\circ$
$\Rightarrow\angle\text{DAB}+100^\circ=180^\circ$
$\Rightarrow\angle\text{DAB}+180^\circ-100^\circ=80^\circ$
In $\triangle\text{DAB},$ by angle sum property$\angle\text{ADB}+\angle\text{DAB}+\angle\text{ABD}=180^\circ$
$\Rightarrow\angle\text{ADB}+80^\circ+70^\circ=180^\circ$
$\Rightarrow\angle\text{ADB}=180^\circ-80^\circ-70^\circ=30^\circ$
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