Question

In triangle $ABC,\overline{AB}=\overline{AC}$, angle $A={40}^{\circ }$. point $O$ is within the triangle with angle $OBC$ = angle $OCA$. The number of degrees in angle $BOC$ is :

• A. $110$
• B. $35$
• C. $140$
• D. $55$
• E. $70$

Answer 1

The correct option is A $110$

As AB=AC $\Rightarrow \angle B=\angle C$

$\therefore \angle B=\angle C=70$

Let $\angle OBC=\angle OCA=\alpha$

$\Rightarrow \angle OCB=70-\alpha$

Consider $△OBC$;

$\Rightarrow \angle BOC+\left(\alpha \right)+(70-\alpha )=180$

$\Rightarrow \angle BOC=110$