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)

$\Rightarrow {50}^{\circ }+{50}^{\circ }+\angle COB={180}^{\circ }$

$\Rightarrow \angle COB=({180}^{\circ }-{100}^{\circ })={80}^{\circ }$​

Since $\angle AOB={140}^{\circ },$ we have:

$\angle AOC+\angle COB={140}^{\circ }$​

$\Rightarrow \angle AOC+{80}^{\circ }={140}^{\circ }$​

$\Rightarrow \angle AOC=({180}^{\circ }-{80}^{\circ })={60}^{\circ }$