Question

In the triangle $ABC$, the bisectors of exterior angles $B$ and $C$ meet at $O$. Given that $\angle BAC={70}^{\circ }$ and $\angle ACB={50}^{\circ }$, find $\angle BOC$.

• A. ${55}^o$
• B. ${46}^o$
• C. ${36}^o$
• D. none of the above

Answer 1

The correct option is A ${55}^o$

In $△ABC$,
$\angle ABC+\angle ACB+\angle BAC={180}^o$ .....[By angle sum property]
$⟹$ $\angle ABC+{70}^o+{50}^o={180}^o$

$⟹$ $\angle ABC+{120}^o={180}^o$

$⟹$ $\angle ABC={180}^o-{120}^o$
$⟹$ $\angle ABC={60}^o$.
Also,
$\angle OCB=\frac{1}{2}({180}^o-\angle ACB)$ ......[Using property of linear pairs]
$⟹$ $\angle OCB=\frac{1}{2}({180}^o-{50}^o)$

$⟹$ $\angle OCB=\frac{1}{2}\left({130}^o\right)$
$⟹$ $\angle OCB={65}^o$.
Now,
$\angle OBC=\frac{1}{2}({180}^o-\angle ABC)$ ......[Using property of linear pairs]
$⟹$ $\angle OBC=\frac{1}{2}({180}^o-{60}^o)$

$⟹$ $\angle OBC=\frac{1}{2}\left({120}^o\right)$
$⟹$ $\angle OBC={60}^o$.
In $△OBC$,
$\angle OCB+\angle OBC+\angle BOC={180}^o$ .....[By angle sum property]
$⟹$ ${65}^o+{60}^o+\angle BOC={180}^o$
$⟹$ $\angle BOC={180}^o-{65}^o-{60}^o$
$⟹$ $\angle BOC={55}^o$.

Hence, option $A$ is correct.