Question

The internal bisectors of the angles B and C of a triangle ABC meet at O. Then, $\angle$BOC is equal to

• A. 90° + A
• B. 2A
• C. 90° + ½ A
• D. 180° – A

Answer 1

The correct option is C

90° + ½ A

$\angle$A+$\angle$B+$\angle$C = ${180}^{\circ }$

$\angle$B+$\angle$C = ${180}^{\circ }$ - $\angle$A

In $△$BOC, we have

$\frac{\angle B}{2}$ + $\frac{\angle C}{2}$ + $\angle$O = ${180}^{\circ }$

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