Question

In $∆$ABC (fig.2.20), the angle bisectors of $\angle$B and $\angle$C meet at O. if $\angle$A = 70$°$, find $\angle$BOC.

Answer 1

$$\begin{gathered} \mathrm{In}△\mathrm{ABC},\mathrm{we}\mathrm{have}: \\ \angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180° \\ \Rightarrow 70°+\angle \mathrm{B}+\angle \mathrm{C}=180°\left(\mathrm{Given}\right) \\ \Rightarrow \angle \mathrm{B}+\angle \mathrm{C}=110° \\ \Rightarrow \frac{1}{2}\left(\angle \mathrm{B}+\angle \mathrm{C}\right)=\frac{110}{2}=55°.....\left(\mathrm{i}\right) \\ \mathrm{In}△\mathrm{OBC},\mathrm{we}\mathrm{have}: \\ \angle \mathrm{BOC}+\angle \mathrm{OBC}+\angle \mathrm{OCB}=180° \\ \Rightarrow \angle \mathrm{BOC}+\frac{1}{2}\left(\angle \mathrm{B}+\angle \mathrm{C}\right)=180°(\because \angle \mathrm{B}\mathrm{and}\angle \mathrm{C}\mathrm{are}\mathrm{bisected}) \\ \Rightarrow \angle \mathrm{BOC}+55°=180°\left[\mathrm{From}\left(\mathrm{i}\right)\right] \\ \Rightarrow \angle \mathrm{BOC}=125° \end{gathered}$$