Question

One angle of a triangle is greater than the sum of the other two. What can you say about the measure of this angle? What type of a triangle is this?

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{three}\mathrm{angles}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{triangle}\mathrm{be}\angle \mathrm{a},\angle \mathrm{b}\mathrm{and}\angle \mathrm{c}. \\ \mathrm{We}\mathrm{know}:\angle \mathrm{a}>\angle \mathrm{b}+\angle \mathrm{c}.....\left(\mathrm{i}\right)\left(\mathrm{Given}\right) \\ \mathrm{We}\mathrm{also}\mathrm{know}\mathrm{that}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{all}\mathrm{the}\mathrm{angles}\mathrm{of}\mathrm{a}\mathrm{triangle}\mathrm{is}\mathrm{equal}\mathrm{to}180°. \\ \therefore \angle \mathrm{a}+\angle \mathrm{b}+\angle \mathrm{c}=180° \\ \Rightarrow \angle \mathrm{b}+\angle \mathrm{c}=180°-\angle \mathrm{a} \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}\angle \mathrm{b}+\angle \mathrm{c}\mathrm{from}\mathrm{equation}\left(\mathrm{i}\right): \\ \angle \mathrm{a}>180°-\angle \mathrm{a} \\ \Rightarrow 2\angle \mathrm{a}>180° \\ \Rightarrow \angle \mathrm{a}>90° \\ \mathrm{Thus},\mathrm{the}\mathrm{angle}\mathrm{is}\mathrm{more}\mathrm{than}90°. \\ \mathrm{Hence},\mathrm{we}\mathrm{can}\mathrm{conclude}\mathrm{by}\mathrm{saying}\mathrm{that}\mathrm{the}\mathrm{given}\mathrm{triangle}\mathrm{is}\mathrm{an}\mathrm{obtuse}\mathrm{triangle}. \end{gathered}$$