Question

In the following figure, if $m($arc $DXE)={90}^{\circ }$ and $m($arc $AYC)={30}^{\circ }$, find $\angle DBE$

Answer 1

By inscribed angle theorem,
$m\angle AEB=\frac{1}{2}\times m\angle AYC$
$=\frac{1}{2}\times {30}^{\circ }={15}^{\circ }....\left(1\right)$
$m\angle EAD=\frac{1}{2}\times m\angle DXE$

$=\frac{1}{2}\times {90}^{\circ }={45}^{\circ }$ .... $\left(2\right)$

Now, for $△ABE,\angle DAE$ is an external angle. Hence

$m\angle ABE+m\angle AEB=m\angle DAE$

$\Rightarrow DBE+{15}^{\circ }={45}^{\circ }$

$\Rightarrow m\angle DBE={30}^{\circ }$