Question

In the following figure, if $m(arcDXE={110}^{\circ }$ and $m\left(arcAYC\right)={50}^{\circ }$, find $\angle DBE$

Answer 1

By inscribed angle theorem,
$m\angle AEB=\frac{1}{2}\times \angle AYC$ (by inscribed angle theorem)
$=\frac{1}{2}\times {50}^{\circ }={25}^{\circ }.....\left(1\right)$
$m\angle EAD=\frac{1}{2}\times m\angle DXE$ (by inscribed angle theorem)
$=\frac{1}{2}\times {110}^{\circ }....\left(2\right)$
$={55}^{\circ }$
$\angle DBE+\angle AEB=\angle EAD$ (exterior angle theorem)
$\angle DBE+{25}^{\circ }={55}^{\circ }$
$\angle DBE={30}^{\circ }$