Question

In the given figure, $O$ is the center of a circle in which $\angle OAB={20}^{\circ }$ and $\angle OCB={50}^{\circ }.$ Then, $\angle AOC=?$

(a) ${50}^{\circ }$

(b) ${70}^{\circ }$

(c) ${20}^{\circ }$

(d) ${60}^{\circ }$

Answer 1

The correct option is (d): ${60}^{\circ }$


We have:
$OA=OB$ (Radii of a circle)
$\Rightarrow \angle OBA=\angle OAB={20}^{\circ }$
In $△OAB,$ we have:
$\angle OAB+\angle OBA+\angle AOB={180}^{\circ }$ (Angle sum property of a triangle)
$\Rightarrow {20}^{\circ }+{20}^{\circ }+\angle AOB={180}^{\circ }$
$\Rightarrow \angle AOB=({180}^{\circ }-{40}^{\circ })={140}^{\circ }$
Again, we have:
$OB=OC$ (Radii of a circle)
$\Rightarrow \angle OBC=\angle OCB={50}^{\circ }$
In $△OCB,$ we have:
$\angle OCB+\angle OBC+\angle COB={180}^{\circ }$ (Angle sum property of a triangle)
$\Rightarrow 50°+50°+\angle COB={180}^{\circ }$
$\Rightarrow \angle COB=({180}^{\circ }-{100}^{\circ })={80}^{\circ }$
Since $\angle AOB={140}^{\circ },$ we have:
$\angle AOC+\angle COB={140}^{\circ }$
$\Rightarrow \angle AOC+{80}^{\circ }={140}^{\circ }$
$\Rightarrow \angle AOC=({140}^{\circ }-{80}^{\circ })={60}^{\circ }$