Question

In a triangle, an exterior angle at a vertex is 95° and its one of the interior opposite angles is 55°. Find all the angles of the triangle.

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{interior}\mathrm{opposite}\mathrm{angles}\mathrm{is}\mathrm{equal}\mathrm{to}\mathrm{the}\mathrm{exterior}\mathrm{angle}. \\ \mathrm{Hence},\mathrm{for}\mathrm{the}\mathrm{given}\mathrm{triangle},\mathrm{we}\mathrm{can}\mathrm{say}\mathrm{that}: \\ \angle \mathrm{ABC}+\angle \mathrm{BAC}=\angle \mathrm{BCO} \\ \Rightarrow 55°+\angle \mathrm{BAC}=95° \\ \mathrm{Or}, \\ \angle \mathrm{BAC}=95°-55° \\ =\mathbf{\angle }\mathbf{BAC}\mathbf{=}\mathbf{40}\mathbf{°} \\ \mathrm{We}\mathrm{also}\mathrm{know}\mathrm{that}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{all}\mathrm{angles}\mathrm{of}\mathrm{a}\mathrm{triangle}\mathrm{is}180°. \\ \mathrm{Hence},\mathrm{for}\mathrm{the}\mathrm{given}△\mathrm{ABC},\mathrm{we}\mathrm{can}\mathrm{say}\mathrm{that}: \\ \angle \mathrm{ABC}+\angle \mathrm{BAC}+\angle \mathrm{BCA}=180° \\ \Rightarrow 55°+40°+\angle \mathrm{BCA}=180° \\ \mathrm{Or}, \\ \angle \mathrm{BCA}=180°-95° \\ =\mathbf{\angle }\mathbf{BCA}\mathbf{=}\mathbf{85}\mathbf{°} \end{gathered}$$