Question

One of the exterior angles of a triangle is 80°, and the interior opposite angles are equal to each other. What is the measure of each of these two angles?

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{us}\mathrm{assume}\mathrm{that}\mathrm{A}\mathrm{and}\mathrm{B}\mathrm{are}\mathrm{the}\mathrm{two}\mathrm{interior}\mathrm{opposite}\mathrm{angles}. \\ \mathrm{We}\mathrm{know}\mathrm{that}\angle \mathrm{A}\mathrm{is}\mathrm{equal}\mathrm{to}\angle \mathrm{B}. \\ \mathrm{We}\mathrm{also}\mathrm{know}\mathrm{that}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{interior}\mathrm{opposite}\mathrm{angles}\mathrm{is}\mathrm{equal}\mathrm{to}\mathrm{the}\mathrm{exterior}\mathrm{angle}. \\ \mathrm{Hence},\mathrm{we}\mathrm{can}\mathrm{say}\mathrm{that}: \\ \angle \mathrm{A}+\angle \mathrm{B}=80° \\ \mathrm{Or}, \\ \angle \mathrm{A}+\angle \mathrm{A}=80°(\because \angle \mathrm{A}=\angle \mathrm{B}) \\ 2\angle \mathrm{A}=80° \\ \angle \mathrm{A}=\frac{80°}{2}=40° \\ \mathbf{\angle }\mathbf{A}\mathbf{=}\mathbf{\angle }\mathbf{B}\mathbf{=}\mathbf{40}\mathbf{°} \\ \mathrm{Thus},\mathrm{each}\mathrm{of}\mathrm{the}\mathrm{required}\mathrm{angles}\mathrm{is}\mathrm{of}40°. \end{gathered}$$