Question

$O$ is the circumcentre of $△ABC$. If $\angle BOC={120}^{\circ }$, then $\angle BAC=$

• A. ${60}^{\circ }$
• B. ${90}^{\circ }$
• C. ${80}^{\circ }$
• D. ${40}^{\circ }$

Answer 1

The correct option is A

${60}^{\circ }$

Draw the circum-circle and join $AO$ and produce it to $D$.

For $△ABO,\angle BOD$ is the exterior angle.

So, $\angle BOD=\angle OBA+\angle BAO$

$\Rightarrow \angle BOD=2\angle BAO$ ... (i) [$\angle OBA$ and $\angle BAO$ are equal as $△AOB$ is an isosceles triangle, $OA$ and $OB$ bieng circum-radius]

For $△ACO,\angle COD$ is the exterior angle.

So, $\angle COD=\angle OCA+\angle CAO$

$\Rightarrow \angle COD=2\angle CAO$ ... (ii) [$\angle OCA$ and $\angle CAO$ are equal as $△AOC$ is an isosceles triangle, $OA$ and $OC$ bieng circum-radius]

Add equation (i) and (ii):

$\angle BOD+\angle COD=2(\angle BAO+\angle CAO)$

$\Rightarrow \angle BOC=2\angle BAC$

$\Rightarrow \angle BAC=\frac{\angle BOC}{2}$

$\Rightarrow \angle BAC=\frac{120}{2}={60}^{\circ }$