Question

In the figure given, $O$ is the center of the circle. If $\angle OBC={37}^o,$ the $\angle BAC$ is equal to:

• A. ${74}^o$
• B. ${106}^o$
• C. ${53}^o$
• D. ${37}^o$

Answer 1

The correct option is C ${53}^o$

Given, $\angle OBC={37}^o$.
Also, $OB=OC$.......(radius).
$\therefore \angle OCB=\angle OBC={37}^o$.

Then, by angle sum property, in $△BOC$,

$⟹$ $\angle BOC+\angle OBC+\angle OCB={180}^o$

$⟹$ $\angle BOC+{37}^o+{37}^0={180}^o$

$\therefore \angle BOC={180}^o-({37}^o+{37}^o)={106}^o$.
Now, we know, $\angle BAC=\frac{1}{2}\angle BOC=\frac{1}{2}\times {106}^o={53}^o$.

Therefore, option $C$ is correct.