Question

In the given figure, $O$ is the center of the circle, $\angle OAB={30}^o$ and $\angle OCB={55}^o$. Find $\angle BOC$ and $\angle AOC$ respectively.

• A. ${50}^o,{30}^o$
• B. ${70}^o,{50}^o$
• C. ${120}^o,{50}^o$
• D. ${70}^o,{30}^o$

Answer 1

The correct option is B ${70}^o,{50}^o$

Givne $\angle OAB={30}^{};\angle OCB={55}^o$

In $△AOB,AO=OB$ (both are radius)

$\angle OBA=\angle BAO={30}^o$ (angles opposite to equal sides are equal)

In $△AOB$, $\angle AOB+\angle OBA+\angle BAO={180}^o$

$\angle AOB+{30}^o+{30}^o={180}^o$

$\angle AOB=180-60={120}^o$

Now In $△OCB$, $OC=OB$ (Radius)

$\angle OCB=\angle OBC={55}^o$ (Angles opposite to equal sides are equal)

In $△OCB$, $\angle COB+\angle OCB+\angle CBO={180}^o$

$\angle COB=180-55-55$

$=180-110$

$=\angle COB={70}^o=\angle BOC$

From $\angle AOB={120}^o$

$\angle AOC+\angle COB={120}^o$

$\angle AOC+{70}^o=120$

$\angle AOC=120-70$

$\angle AOC={50}^o$

$\therefore \angle AOC={50}^o,=\angle BOC={70}^o$