Question

$O$ is the center of the circle. If $\angle OAB={20}^{\circ }$ and $\angle OCB={55}^{\circ }$, then what is the value of $\angle AOC$ ?

• A. ${40}^{\circ }$
• B. ${50}^{\circ }$
• C. ${60}^{\circ }$
• D. ${70}^{\circ }$

Answer 1

The correct option is D ${70}^{\circ }$

In $△OCB$,

$OC=OB$ ...[Radius of a circle]

$\Rightarrow$ $\angle OCB=\angle CBO={55}^o$ ...[Angles opposite of equal sides are equal].

Then, $\angle OCB+\angle CBO+\angle BOC={180}^o$ ...[Angle sum property]

$\Rightarrow$ ${55}^o+{55}^o+\angle BOC={180}^o$

$\Rightarrow$ ${110}^o+\angle BOC={180}^o$

$\Rightarrow$ $\angle BOC={180}^o-{110}^o$

$\Rightarrow$ $\angle BOC={70}^o$ ---- ( 1 ).

Now, in $△AOB$,

$\angle AOB+\angle OBA+\angle BAO={180}^o$ ...[Angle sum property]

$\Rightarrow$ $\angle AOP+\angle BOP+{20}^o+{20}^o={180}^o$ ...[Since, $\angle AOB=\angle AOP+\angle BOP$]

$\Rightarrow$ $\angle AOP+{70}^o+{40}^o={180}^0$ --- [From ( 1 )]

$\Rightarrow$ $\angle AOP={180}^o-{110}^o$

$\Rightarrow$ $\angle AOP={70}^o$.

$\therefore$ $\angle AOC=\angle AOP={70}^o$.

Hence, option $D$ is correct.