Question

In the given figure, BO, CO are the angle bisectors of external angles of $\Delta$ABC. Then $\angle$BOC is __________.

• A. ${90}^o-\frac{1}{2}\angle$A
• B. ${90}^o+\frac{1}{2}\angle$ A
• C. ${180}^o-\frac{1}{2}\angle$A
• D. ${180}^o+\frac{1}{2}\angle$A

Answer 1

The correct option is A ${90}^o-\frac{1}{2}\angle$A

As $BO$ and $CO$ are the angle bisectors of external angles of$△ABC$, Then

$\angle 1=\angle 2\angle 4=\angle 3$

We know, $\angle A+\angle ABC+\angle ACB={180}^{\circ }\dots eqn\left(1\right)$

And $\angle ABC=180-2\angle 1\angle ACB=180-2\angle 4$

Putting it in the eqn (1), we get

$\angle A+180-2\angle 1+180-2\angle 4=180\Rightarrow \angle 1+\angle 4=90+\frac{1}{2}\angle A\dots eqn\left(2\right)$

Also we know from the figure, $\angle BOC+\angle 1+\angle 4={180}^{\circ }$

$\angle BOC=180-\angle 1-\angle 4$

From eqn (2)

$\angle BOC=180-90-\frac{1}{2}\angle A\Rightarrow \angle BOC={90}^{\circ }-\frac{1}{2}\angle A$