Question

The internal bisectors of ∠B and ∠C of ∆ABC meet at O. If B + C = 100°, then ∠BOC = __________.

Answer 1

In ∆ABC, ∠B + ∠C = 100°.

Also, OB and OC are the bisectors of ∠B and ∠C, respectively.

$\therefore \angle \mathrm{OBC}=\frac{\angle \mathrm{B}}{2}.....\left(1\right)$

Similarly, $\angle \mathrm{OCB}=\frac{\angle \mathrm{C}}{2}.....\left(2\right)$

In ∆BOC,

∠OBC + ∠OCB + ∠BOC = 180º (Angle sum property of triangle)

$$\begin{gathered} \Rightarrow \frac{\angle \mathrm{B}}{2}+\frac{\angle \mathrm{C}}{2}+\angle \mathrm{BOC}=180°\left[\mathrm{Using}\left(1\right)\mathrm{and}\left(2\right)\right] \\ \Rightarrow \frac{\angle \mathrm{B}+\angle \mathrm{C}}{2}+\angle \mathrm{BOC}=180° \\ \Rightarrow \frac{100°}{2}+\angle \mathrm{BOC}=180°\left(\mathrm{Given}\right) \\ \Rightarrow \angle \mathrm{BOC}=180°-50°=130° \end{gathered}$$


The internal bisectors of ∠B and ∠C of ∆ABC meet at O. If B + C = 100°, then ∠BOC = ____130º____.