Question

the sum of two angles of a triangle is 116 and their difference is 24.find the measure of each angle of the triangle

Answer 1

$$\begin{gathered} \mathrm{Let}x,y\mathrm{and}z\mathrm{are}\mathrm{the}\mathrm{three}\mathrm{angles}\mathrm{of}\mathrm{a}\mathrm{triangle}\mathrm{then}, \\ \mathrm{According}\mathrm{to}\mathrm{question}: \\ x+y=116°............\left(1\right) \\ x-y=24°.............\left(2\right)\left[\mathrm{when}x>y\right] \\ \mathrm{and}y-x=24°.............\left(3\right)\left[\mathrm{when}y>x\right] \\ \underline{\mathbf{CASE}\mathbf{}\mathbf{1}\mathbf{.}}\mathrm{Adding}\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{get} \\ 2x=140° \\ \Rightarrow x=70° \\ \mathrm{Substituting}x=70°\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ 70°+y=116° \\ \Rightarrow y=116°-70°=46° \\ \mathrm{But},x+y+z=180°\left[\mathbf{sum}\mathbf{}\mathbf{of}\mathbf{}\mathbf{the}\mathbf{}\mathbf{angles}\mathbf{}\mathbf{of}\mathbf{}\mathbf{a}\mathbf{}\mathbf{triangle}\right] \\ 70°+46°+\mathit{}z=180° \\ \Rightarrow 116°+z=180° \\ \Rightarrow z=180°-116°=64° \\ \mathrm{Hence},x=70°,y=46°\mathrm{and}z=64° \end{gathered}$$


$$\begin{gathered} \underline{\mathbf{CASE}\mathbf{}\mathbf{2}\mathbf{.}}\mathrm{Adding}\left(1\right)\mathrm{and}\left(3\right)\mathrm{we}\mathrm{get}, \\ 2y=140° \\ \Rightarrow y=70° \\ \mathrm{Substituting}y=70°\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ x+70°=116° \\ \Rightarrow x=116°-70°=46° \\ \mathrm{But}x+y+z=180°\left[\mathbf{sum}\mathbf{}\mathbf{of}\mathbf{}\mathbf{the}\mathbf{}\mathbf{angles}\mathbf{}\mathbf{of}\mathbf{}\mathbf{a}\mathbf{}\mathbf{triangle}\right] \\ \Rightarrow 46°+70°+z=180° \\ \Rightarrow 116°+z=180° \\ \Rightarrow z=180°-116°=64° \\ \mathrm{Hence},x=46°,y=70°\mathrm{and}z=64° \end{gathered}$$