Question

In the figure, the bisector of B and C meet at O. Then $\angle$ BOC is

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

Answer 1

The correct option is A

${90}^{\circ }+\frac{1}{2}\angle A$

From the given figure,
$\angle$BOC = 180${}^{\circ }-(\angle$2 + $\angle$4)

= 180${}^{\circ }$ - ($\frac{\angle B}{2}$ + $\frac{\angle C}{2}$ )

= 180${}^{\circ }$ - ($\frac{{180}^{\circ }-\angle A}{2}$)
[$\because$ 180${}^{\circ }$ - $\angle$ A = $\angle$ B + $\angle$ C]

= 180${}^{\circ }$ - 90${}^{\circ }$ + $\frac{\angle A}{2}$

= 90${}^{\circ }$ + $\frac{\angle A}{2}$