Question

Question 6
Prove that a triangle must have atleast two acute angles.

Answer 1

Given:
$\Delta ABC$ is a triangle.
To prove:
$\Delta ABC$ must have two acute angles.
Proof:
Let us consider the following cases.
Case I :
When two angles are ${90}^{\circ }$
Suppose two angles,
$\angle B={90}^{\circ }and\angle C={90}^{\circ }$.


We know that, the sum of all three angles is ${180}^{\circ }$.
$\therefore \angle A+\angle B+\angle C={180}^{\circ }$ ...(i)
$\therefore \angle A+{90}^{\circ }+{90}^{\circ }={180}^{\circ }$
$\Rightarrow \angle A={180}^{\circ }-{180}^{\circ }=0^{\circ }$
So, no triangle is possible.

Case II :
When two angles are obtuse.
Suppose two angles $\angle B$ and $\angle C$ are more than ${90}^{\circ }$.
From equation (i),
$\angle A={180}^{\circ }-(\angle B+\angle C)={180}^{\circ }$ - [Angle greater than ${180}^{\circ }$]
[$\because \angle B+\angle C$ = more than ${90}^{\circ }$ + more than ${90}^{\circ }$ = more than ${180}^{\circ }$]
$\angle A$ = negative angle, which is not possible. So, no triangle is possible.

Case III :
When one angle is ${90}^{\circ }$ and other is obtuse.
Suppose $\angle B={90}^{\circ }and\angle C$ is obtuse.

From equation (i),
$\angle A+\angle B+\angle C={180}^{\circ }$
$\angle A={180}^{\circ }-({90}^{\circ }+\angle C)$
$={90}^{\circ }-\angle C$
= Negative angle [$\because \angle C$ is obtuse]
Hence, no triangle is possible.

Case IV :
When two angles are acute, then the sum of two angles is less than ${180}^{\circ }$, so that the third angle can be an acute or obtuse angle.