Question

In cyclic quadrilateral ABCD, find the value of $\angle ADB$

Answer 1

ABCD is a cyclic quadrilateral.
$\therefore \angle BCD+\angle DAB={180}^{\circ }$(Sum of opp. angles of cyclic Quadrilatrial =${180}^{\circ })$
${110}^{\circ }+\angle DAB={180}^{\circ }\therefore \angle DAB={70}^{\circ }.$
$\angle DAE={180}^{\circ }-\angle DAB={180}^{\circ }-{70}^{\circ }={\angle }^{\circ }\to \left(1\right)$
$\angle ADB+\angle DAB+\angle DBA={180}^{\circ }\left(Anglyof\Delta ADB\right)$
$\angle ADB+{70}^{\circ }+{60}^{\circ }={180}^{\circ }$
$\therefore \angle ADB={180}^{\circ }-{130}^{\circ }={50}^{\circ }-\left(2\right)$