Question

If one angle of a triangle is less then the sum of the other two, show that the triangle is acute angled.

Answer 1

Consider $ABC$ as a triangle
According to the question it can be written as
$\angle A<\angle B+\angle C$
Add $\angle A$ to both the sides of the question
So we get
$\angle A+\angle A<\angle A+\angle B+\angle C$
We know that the sum of all the angles in a triangle is ${180}^{\circ }$.
So we can write it as
$\angle A+\angle B+\angle C={180}^{\circ }$
So we get
$2<A<{180}^{\circ }$
By division we get
$\angle A<180/2$
$\angle A<{90}^{\circ }$
In the same way we can also write
$\angle B<\angle A+\angle C$
Add $\angle B$ to both the sides of the equation
So we get
$\angle B+\angle B<\angle A+\angle B+\angle C$
We know that the sum of all the angles in a triangle is ${180}^{\circ }$.
So we can write it as
$\angle A+\angle B+\angle C={180}^{\circ }$
So we get
$2\angle B<{180}^{\circ }$
By division we get
$\angle B<180/2$
$\angle B<{90}^{\circ }$
So we know that
$\angle C<\angle A+\angle B$
Add $\angle C$ to both the sides of the equation
So we get
$\angle C+\angle C<\angle A+\angle B+\angle C$
We know that the sum of all the angles in a triangle is ${180}^{\circ }$.
So we can write it as
$\angle A+\angle B+\angle C={180}^{\circ }$
So we get
$2\angle C<{180}^{\circ }$
By division we get
$\angle C<180/2$
$\angle C<{90}^{\circ }$
Therefore, it is proved that the triangle $ABC$ is acute angled.