Question

In the figure, ABCD is a cyclic quadrilateral, If $\angle BCD={100}^{\circ }and\angle ABD={70}^{\circ },find\angle ADB$

Answer 1

$\angle BCD={100}^{\circ }and\angle ABD={70}^{\circ }$

$\because ABCD$ is a cyclic quadrilateral,

$\therefore \angle A+\angle C={180}^{\circ }$ (Sum of opposite angles)

$\Rightarrow \angle A+{100}^{\circ }={180}^{\circ }$

$\Rightarrow \angle A={180}^{\circ }-{100}^{\circ }={80}^{\circ }$

Now in $\Delta ABD$,

$\angle A+\angle ABD+\angle ADB={180}^{\circ }$

$\Rightarrow {80}^{\circ }+{70}^{\circ }+\angle ADB={180}^{\circ }$

$\Rightarrow {150}^{\circ }+\angle ADB={180}^{\circ }$

$\therefore \angle ADB={180}^{\circ }-{150}^{\circ }={30}^{\circ }$

Hence, $\angle ADB={30}^{\circ }$