Question

O is the centre of the circle, find $\angle AOC,given\angle BAO={30}^{\circ }and\angle BCO={40}^{\circ }.$

• A. ${35}^{\circ }$
• B. ${140}^{\circ }$
• C. ${70}^{\circ }$
• D. Cannot be determined

Answer 1

Answer: (B)

${140}^{\circ }$

In $△ABO,AO=BO\Rightarrow \angle BAO=\angle ABO={30}^o$

In $△CBO,CO=BO,\Rightarrow \angle BCO=\angle CBO={40}^o$

$\Rightarrow \angle ABC=\angle ABO+\angle CBO=30+40={70}^o$

Angle subtended at the center is twice the angle subtended anywhere on the arc, so,

$\angle AOC=2\times \angle ABC={140}^o$

So option B is the right answer.