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