Question

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

• A. ${60}^{\circ }$
• B. ${66}^{\circ }$
• C. ${55}^{\circ }$
• D. ${80}^{\circ }$

Answer 1

Answer: (A), (C)

The correct options are
A ${60}^{\circ }$
C ${55}^{\circ }$

In $△ABC$,

$\angle ABC+\angle BAC+\angle ACB=180$

$60+70+\angle ACB=180$

$\angle ACB={50}^{\circ }$
$ext.\angle B=180-\angle ABC$

$2\angle OBC=180-70$ (BOB bisects $ext.\angle B$)

$\angle OBC={55}^{\circ }$
Similarly, $\angle OCB={65}^{\circ }$

Now, In $△OBC$,

$\angle OCB+\angle OBC+\angle BOC=180$

$65+55+\angle BOC=180$

$\angle BOC=180-120$

$\therefore \angle BOC={60}^{\circ }$