Question

In the given figure, points A, B, and C lie on the circumference of the circle centered at O. if $\angle$OAB measures ${50}^o$ and $\angle$BCO measures ${60}^o$, calculate $\angle AOC$.

• A. ${110}^o$
• B. ${120}^o$
• C. ${130}^o$
• D. ${140}^o$

Answer 1

The correct option is D ${140}^o$

As per given figure, $OA=OB=OC$ ....(radius of circle )

In $△AOB$,

$OA=OB$ and $\angle OAB={50}^0$ ...(Given )

Then, $\angle OAB=OBA={50}^0$

So, $\angle AOB=180-50-50={80}^0$

In $△BOC$,

$OC=OB$ and $\angle OCB={60}^0$ ....(Given )

Then $\angle OBC=OCB={60}^0$

So, $\angle BOC=180-60-60={60}^0$

Then, $\angle AOC=\angle AOB+\angle BOC=80+60={140}^0$