Question

In the figure, If $\angle AOC={110}^{\circ },then\angle ABCin$ degrees is:

Answer 1

Given

$\angle AOC={110}^0$

$\Delta AOB$ and $\Delta AOC$ are similar

as $OA=OB=OC\left(radii\right)$

$\angle AOB+\angle BOC={110}^0$

$\Rightarrow 2\angle AOB={110}^0$

$\angle AOB={55}^0$

As in $\Delta AOB;OA=OB$

$\Rightarrow OAB=\angle OBA$ (properties of the triangle)

i,e. $2\angle OAB+\angle AOB={180}^0$

$2\angle OAB={125}^0$

$\angle OBA=\angle OAB=\frac{{125}^0}{2}$

Similarly from $\Delta OBC$

$\angle OBC=\angle OCB=\frac{{125}^0}{2}$

i.e, $\angle ABC=\frac{125}{2}+\frac{125}{2}={125}^0$