Question

In the triangle ABC, the bisectors of the exterior angles B and C meet at O. Given $\angle BAC={60}^{\circ }and\angle ABC={70}^{\circ }$. Find $\angle BOC$.

• A. ${60}^o$
• B. ${36}^o$
• C. ${54}^o$
• D. none of the above

Answer 1

The correct option is A ${60}^o$

In $△ABC$,
$\angle ABC+\angle BAC+\angle ACB=180$
$60+70+\angle ACB=180$
$\angle ACB={50}^{\circ }$
$ext.\angle B=180-\angle ABC$
$2\angle OBC=180-70$ (BO bisects $ext.\angle B$)
$\angle OBC={55}^{\circ }$
Similarly, $\angle OCB={65}^{\circ }$
Now, In $△OBC$,
$\angle OCB+\angle OBC+\angle BOC=180$
$65+55+\angle BOC=180$
$\angle BOC=180-120$
$\angle BOC={60}^{\circ }$