Question

In a $△ABC$, the sides $AB$ and $AC$ have been produced to $D$ and $E$. Bisectors of $\angle CBD$ and $\angle BCE$ meet at $O$. If $\angle A={64}^0$, then $\angle BOC$ is:

• A. ${52}^0$
• B. ${58}^0$
• C. ${26}^0$
• D. ${112}^0$

Answer 1

The correct option is B ${58}^0$

Given: $OB$ and $OC$ bisect $ext.\angle B$ and $ext.\angle C$, $\angle A={64}^{\circ }$.
Now, in $△OBC$,
we know, by angle sum property, the sum of angles $={180}^o$
$⟹$ $\angle OBC+\angle OCB+\angle BOC={180}^o$
$⟹$ $\frac{1}{2}(ext.\angle B+ext.\angle C)+\angle BOC={180}^o$ ...($OB$ and $OC$ bisect exterior angles)
$⟹$ $\frac{1}{2}({180}^o-\angle ABC+{180}^o-\angle ACB)+\angle BOC={180}^o$
$⟹$ $\frac{1}{2}({360}^o-(\angle ABC+\angle ACB\left)\right)+\angle BOC={180}^o$
$⟹$ $\frac{1}{2}({360}^o-({180}^o-\angle A\left)\right)+\angle BOC={180}^o$ (Angle sum property)
$⟹$ $\frac{1}{2}({180}^o+\angle A)+\angle BOC={180}^o$
$⟹$ $\angle BOC={180}^o-{90}^o-\frac{1}{2}\left({64}^o\right)$
$⟹$ $\angle BOC={58}^{\circ }$.

Therefore, option $B$ is correct.