Question

$ABCD$ is a cyclic quadrilateral. Find $\angle ADB.$

Answer 1

we know that in a cyclic quadrilateral opposite angles are supplementary

$\begin{array}{c}\therefore \angle DAB+\angle BCD={180}^{\circ }\\ \Rightarrow \angle DAB+{120}^{\circ }={180}^{\circ }\\ \Rightarrow \angle DAB={60}^{\circ }\\ In\Delta DAB\\ \angle DAB+\angle ABD+\angle ADB={180}^{\circ }\\ {60}^{\circ }+{50}^{\circ }+\angle ADB={180}^{\circ }\\ \therefore \angle ADB={180}^{\circ }-{110}^{\circ }\\ Hence,\angle ADB={70}^{\circ }.\end{array}$