Question

ABC is a triangle in which ∠B = ∠C and ray AX bisects the exterior angle DAC. If ∠DAX = 70°, find ∠ACB.

Answer 1

$$\begin{gathered} \mathrm{Here}, \\ \angle \mathrm{CAX}=\angle \mathrm{DAX}(\because \mathrm{AX}\mathrm{bisects}\angle \mathrm{CAD}) \\ \Rightarrow \angle \mathrm{CAX}=70° \\ \angle \mathrm{CAX}+\angle \mathrm{DAX}+\angle \mathrm{CAB}=180° \\ 70°+70°+\angle \mathrm{CAB}=180° \\ \angle \mathrm{CAB}=180°-140° \\ \angle \mathrm{CAB}=40° \\ \angle \mathrm{ACB}+\angle \mathrm{CBA}+\angle \mathrm{CAB}=180°(\mathrm{Sum}\mathrm{of}\mathrm{the}\mathrm{angles}\mathrm{of}△\mathrm{ABC}) \\ \angle \mathrm{ACB}+\angle \mathrm{ACB}+40°=180°(\because \angle \mathrm{C}=\angle \mathrm{B}) \\ 2\angle \mathrm{ACB}=180°-40° \\ \angle \mathrm{ACB}=\frac{140°}{2} \\ \mathbf{\Rightarrow }\mathbf{\angle }\mathbf{ACB}\mathbf{=}\mathbf{70}\mathbf{°} \end{gathered}$$