Question

In ∆ABC, ∠A = 60°, ∠B = 80° and the bisectors of ∠B and ∠C meet at O. Find
(i) ∠C
(ii) ∠BOC.

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{all}\mathrm{three}\mathrm{angles}\mathrm{of}\mathrm{a}\mathrm{triangle}\mathrm{is}180°. \\ \mathrm{Hence},\mathrm{for}△\mathrm{ABC},\mathbf{}\mathrm{we}\mathrm{can}\mathrm{say}\mathrm{that}: \\ \angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180°(\mathrm{Sum}\mathrm{of}\mathrm{angles}\mathrm{of}△\mathrm{ABC}) \\ \Rightarrow 60°+80°+\angle \mathrm{C}=180° \\ \angle \mathrm{C}=180°-140° \\ \angle \mathrm{C}=40° \\ \mathrm{For}△\mathrm{OBC}: \\ \angle \mathrm{OBC}=\frac{\angle \mathrm{B}}{2}=\frac{80°}{2}(\mathrm{OB}\hspace{0.17em}\mathrm{bisects}\angle \mathrm{B}.) \\ \Rightarrow \angle \mathrm{OBC}=40° \\ \angle \mathrm{OCB}=\frac{\angle \mathrm{C}}{2}=\frac{40°}{2}(\mathrm{OC}\mathrm{bisects}\angle \mathrm{C}.) \\ \Rightarrow \angle \mathrm{OCB}=20° \\ \mathrm{If}\mathrm{we}\mathrm{apply}\mathrm{the}\mathrm{above}\mathrm{logic}\mathrm{to}\mathrm{this}\mathrm{triangle},\mathrm{we}\mathrm{can}\mathrm{say}\mathrm{that}: \\ \angle \mathrm{OCB}+\angle \mathrm{OBC}+\angle \mathrm{BOC}=180°(\mathrm{Sum}\mathrm{of}\mathrm{angles}\mathrm{of}△\mathrm{OBC}) \\ 20°+40°+\angle \mathrm{BOC}=180° \\ \angle \mathrm{BOC}=180°-60° \\ \angle \mathrm{BOC}=120° \end{gathered}$$