Question

These are numeric entry type questions.
In the given figure, $ABCDE$ is a pentagon, inscribed in a circle with centre ${ }^{\prime} O ^{\prime}, CD = DE$ and $AD = BD , \angle EAD =20^{\circ}$ and $\angle BOC =100^{\circ}$. Then, $\angle DAO +\angle BAO =$____

Answer 1

Since $AD = BD$ and $CD = DE , AE = BC$.
$\Rightarrow \angle BOC =\angle AOE =100^{\circ}$

$\Rightarrow \angle EAO =(\frac{180^{\circ}-100^{\circ}}{2})=40^{\circ} $
$\Rightarrow \angle DAO =40^{\circ}-20^{\circ}=20^{\circ} $
$\Rightarrow \angle AOD =180^{\circ}-2 \times 20^{\circ}=140^{\circ} $
$\Rightarrow \angle BOD =140^{\circ}$
$\therefore \angle BOA =360^{\circ}-2(140^{\circ})=80^{\circ} $
$\Rightarrow \angle BAO =\frac{180^{\circ}-80^{\circ}}{2}=50^{\circ} $
$\therefore \angle DAO +\angle BAO =20^{\circ}+50^{\circ}=70^{\circ}$