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

Answer: 70

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

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