Question

In the figure, in ∆ABC, the angle bisectors of ∠B and ∠C meet at a point O. Find the measure of ∠BOC.
Figure

Answer 1

$$\begin{gathered} \mathrm{In}∆ABC,\mathrm{using}\mathrm{angle}\mathrm{sum}\mathrm{property},\mathrm{we}\mathrm{have}: \\ \angle CAB+\angle ABC+\angle BCA=180° \\ \Rightarrow 70°+\angle ABC+\angle BCA=180° \\ \therefore \angle ABC+\angle BCA=110°-----\left(1\right) \\ \mathrm{Now}, \\ BO\mathrm{is}\mathrm{the}\mathrm{angle}\mathrm{bisector}\mathrm{of}\angle ABC. \\ \Rightarrow \angle OBC=\frac{1}{2}\angle ABC \\ \mathrm{and} \\ CO\mathrm{is}\mathrm{the}\mathrm{angle}\mathrm{bisector}\mathrm{of}\angle BCA. \\ \Rightarrow \angle OCB=\frac{1}{2}\angle BCA \\ \therefore \angle OBC+\angle OCB=\frac{1}{2}\left(\angle ABC+\angle BCA\right)-----\left(2\right) \\ \mathrm{In}∆OBC,\mathrm{using}\mathrm{angle}\mathrm{sum}\mathrm{property},\mathrm{we}\text{get}: \\ \angle BOC+\angle OBC+\angle OCB=180° \\ \mathrm{Using}\left(2\right)\mathrm{we}\text{get}: \\ \angle BOC+\frac{1}{2}\left(\angle ABC+\angle BCA\right)=180° \\ \mathrm{Now},\mathrm{using}\left(1\right)\mathrm{we}\text{get}: \\ \angle BOC+\frac{1}{2}\left(110°\right)=180° \\ \Rightarrow \angle BOC+55°=180° \\ \Rightarrow \angle BOC=180°-55°=125° \\ \mathrm{Therefore},\angle BOC=125° \end{gathered}$$