Question

In the following figure, if $m($arc $DXE)={100}^{\circ }$ and $m($arc $AYC)={40}^{\circ }$, find $\angle DBE$ (in ${}^{\circ }$).

Answer 1

By inscribed angle theorem,
$m\angle AEB=\frac{1}{2}\times m\angle AYC$
$=\frac{1}{2}\times {40}^{\circ }={20}^{\circ }...\left(1\right)$
$m\angle EAD=\frac{1}{2}\times m\angle DXE$
$=\frac{1}{2}\times {100}^{\circ }$
$={50}^{\circ }....\left(2\right)$
$\angle DBE+\angle AEB=\angle EAD$
$\angle DBE+{20}^{\circ }={50}^{\circ }$
$\angle DBE={30}^{\circ }$