Question

If each angle of a triangle is less than the sum of the other two, show that the triangle is acute-angled.

Answer 1

Let ABC be the triangle.
Let $\angle A<\angle B+\angle C$
Then,
$$\begin{gathered} 2\angle A<\angle A+\angle B+\angle C\left[\text{Adding}\angle A\text{to both sides}\right] \\ \Rightarrow 2\angle A<180°\left[\because \angle A+\angle B+\angle C=180°\right] \\ \Rightarrow \angle A<\mathbf{90}\mathbf{°} \end{gathered}$$

Also, let $\angle B<\angle A+\angle C$
Then,
$$\begin{gathered} 2\angle B<\angle A+\angle B+\angle C\left[\text{Adding}\angle B\text{to both sides}\right] \\ \Rightarrow 2\angle B<180°\left[\because \angle A+\angle B+\angle C=180°\right] \\ \Rightarrow \angle B<\mathbf{90}\mathbf{°} \end{gathered}$$

And let $\angle C<\angle A+\angle B$
Then,
$$\begin{gathered} 2\angle C<\angle A+\angle B+\angle C\left[\text{Adding}\angle C\text{to both sides}\right] \\ \Rightarrow 2\angle C<180°\left[\because \angle A+\angle B+\angle C=180°\right] \\ \Rightarrow \angle C<\mathbf{90}\mathbf{°} \end{gathered}$$

Hence, each angle of the triangle is less than $90°$.
Therefore, the triangle is acute-angled.