Question

In the figure at right, ABCD is a cyclic quadrilateral in which $\angle BCD={100}^o$ and $\angle ABD={50}^o$ find $\angle ADB$

Answer 1

We have;

$\angle BCD=100°;\angle ABD=50°$

We know that in a cyclic quadrilateral the sum of diagonal angles is $180°$

$\angle BCD+\angle DAB=180°$

$⟹\angle DAB=80°$

Now; in $\Delta ADB;$

$\angle ADB+\angle DAB+\angle ABD=180°$

$⟹\angle ADB+80°+50°=180°$

Hence;$\angle ADB=50°$