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

Hint: Use the angle Sum Property of Triangle which states that Sum of three interior angles of a triangle is equal to ${180^0}$.

Complete step-by-step answer:

According to the question, it is given that,
Each angle of a triangle is less than the sum of the other two angle in a triangle
Now consider First case,
$
  A < B + C{\text{ }}......\left( 1 \right) \\
  B < C + A{\text{ }}......\left( 2 \right) \\
  C < A + B{\text{ }}......\left( 3 \right) \\
$
Now, by angle sum property of triangle, we can write
\[
  A + B + C = {180^0} \\
  B + C = {180^0} - A \\
\]
By putting the above value in equation (1), we get
\[
  A < {180^0} - A \\
  2A < {180^0} \\
  A < {90^0} \\
\]
Similarly, for Second case,
\[
  A + B + C = {180^0} \\
  C + A = {180^0} - B \\
\]
By putting the above value in equation (2), we get
\[
  B < C + A \\
  B < {180^0} - B \\
  2B < {180^0} \\
  B < {90^0} \\
\]
Similarly, for third case,
\[
  A + B + C = {180^0} \\
  A + B = {180^0} - C \\
\]
By putting the above value in equation (3), we get
\[
  C < A + B \\
  C < {180^0} - C \\
  2C < {180^0} \\
  C < {90^0} \\
\]
Hence Proved, so we found out that all angles are acute then it is proven that this triangle is acute.

Note: In this question, observe the question carefully and apply the property directly. The exterior angle of a triangle is equal to the sum of its opposite interior angles. Triangle is the smallest polygon which has three sides and three interior angles.