Prove that a triangle must have at least two acute angles.

(N/A) Let the angles of a triangle be $\angle A, \angle B,$ and $\angle C$. According to the angle sum property of a triangle,$\angle A + \angle B + \angle C = 180^{\circ}$.
Case $1$: If the triangle is an acute-angled triangle,all three angles are less than $90^{\circ}$. Thus,it has three acute angles.
Case $2$: If the triangle is a right-angled triangle,one angle is $90^{\circ}$. Let $\angle A = 90^{\circ}$. Then $\angle B + \angle C = 180^{\circ} - 90^{\circ} = 90^{\circ}$. Since the sum of two positive angles is $90^{\circ}$,both $\angle B$ and $\angle C$ must be less than $90^{\circ}$,meaning they are both acute.
Case $3$: If the triangle is an obtuse-angled triangle,one angle is greater than $90^{\circ}$. Let $\angle A > 90^{\circ}$. Then $\angle B + \angle C = 180^{\circ} - \angle A$. Since $\angle A > 90^{\circ}$,$\angle B + \angle C < 90^{\circ}$. This implies that both $\angle B$ and $\angle C$ must be less than $90^{\circ}$,meaning they are both acute.
Conclusion: In all possible cases,a triangle must have at least two acute angles.