If each angle of a triangle is less than the sum of the other two,

Solution:

Let ABC be the triangle.

Let $\angle A<\angle B+\angle C$

Then,

$2 \angle A<\angle A+\angle B+\angle C \quad$ [Adding $\angle A$ to both sides]

$\Rightarrow 2 \angle A<180^{\circ} \quad\left[\because \angle A+\angle B+\angle C=180^{\circ}\right]$

$\Rightarrow \angle A<\mathbf{9 0}^{\circ}$

Also, let $\angle B<\angle A+\angle C$

Then,

$2 \angle B<\angle A+\angle B+\angle C \quad$ [Adding $\angle B$ to both sides]

$\Rightarrow 2 \angle B<180^{\circ} \quad\left[\because \angle A+\angle B+\angle C=180^{\circ}\right]$

$\Rightarrow \angle B<\mathbf{9 0}^{\circ}$

And let $\angle C<\angle A+\angle B$

Then,

$2 \angle C<\angle A+\angle B+\angle C \quad$ [Adding $\angle C$ to both sides]

$\Rightarrow 2 \angle C<180^{\circ} \quad\left[\because \angle A+\angle B+\angle C=180^{\circ}\right]$

$\Rightarrow \angle C<\mathbf{9 0}^{\circ}$

Hence, each angle of the triangle is less than $90^{\circ}$.

Therefore, the triangle is acute-angled.