Question

Find the value of $\angle AEB$ in the given figure.

Answer 1

$T.P.T\angle AEB={60}^{\circ }$
CD =radius
$\Delta AOC,\Delta BOD$ are isosceles $\Delta$ Triangle.
$\Delta COD$ is an equilateral $\Delta$ Triangle
${180}^{\circ }-2x+{60}^{\circ }+180-2y=180$
2x+2y=240
x+y=120
In $\Delta AEB,x+y=\angle AEB={180}^{\circ }-(x+y)={180}^{\circ }-{120}^{\circ }={60}^{\circ }$