Question

$A,B,C$ are three points on the circumference of a circle with centre $O$ such that $\angle OAC={53}^{\circ }$ and $\angle CBO={32}^{\circ }$, then $\angle AOB=$

• A. ${100}^{\circ }$
• B. ${120}^{\circ }$
• C. ${150}^{\circ }$
• D. ${170}^{\circ }$

Answer 1

The correct option is D ${170}^{\circ }$

Refer Figure:

Given that: $\angle OAC=53°$

$\angle CBO=32°$

Consider $△ACO$:

$AO=CO$ (radius of circle)

Thus, $\angle ACO=\angle OAC=53°$ (angles opposite to equal sides are equal)

$\angle AOC=180°-53°-53°=74°$ (Sum of three angles of a triangle is 180°)

Similarly in $△BCO$:

$BO=OC$

Hence, $\angle OBC=\angle OCB$

And $\angle COB=180°-32°-32°=116°$

$Reflex\angle AOB=\angle AOC+\angle COB=74°+116°=190°$

Thus $\angle AOB=360°-reflex\angle AOB$

$\angle AOB=360°-190°=170°$

Answer: $\angle AOB=170°$