Question

In the following figure, if $m\left(arcDXE\right)={120}^{\circ }$ and $m\left(arcAYC\right)={60}^{\circ }$, find $\angle DBE$.

Answer 1

By inscribed angle theorem,
$m\angle AEB=\frac{1}{2}\times \angle AYC$
$=\frac{1}{2}\times {60}^{\circ }={30}^{\circ }...\left(1\right)$
$m\angle EAD=\frac{1}{2}\times m\angle D\times E$
$=\frac{1}{2}\times {120}^{\circ }={60}^{\circ }....\left(2\right)$
$\angle DBE+\angle AEB=\angle EAD$
$\angle DBE+{30}^{\circ }=60$
$\angle DBE={30}^{\circ }$