Question

Find the value of $\angle AOB$ in the given figure.
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Answer 1

In $\Delta BOC$
$\angle BOC=\angle BCO=y[\because BC=OB]\therefore \angle OBC=180-2y\to \left(1\right)$
In $\Delta AOB.$
$\angle OAB=\angle OBA=z$ [OB =OA =radius]
$\angle AOB=180-2z\to \left(2\right)$
$\angle OBC=180+z=180-z\to \left(3\right)$[exterior angle property]
from (1)and (3)
$\Rightarrow 180-2y=180-z\Rightarrow z=2y$
COD is a straight angle
$\Rightarrow \angle AOD+\angle AOB+\angle BOC={180}^{\circ }$
$x=180-2z+y={180}^{\circ }$
x-2z+y=0
x-2(2y)+y=0
x-4y+y=0
x-3y=0
x=3y
If $x={60}^{\circ }$
$y=\frac{x}{3}={20}^{\circ }z=2y={40}^{\circ }\angle AOB={180}^{\circ }-2z={180}^{\circ }-{80}^{\circ }={100}^{\circ }$