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