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


Let us draw a ΔABC as below:


We have to prove that a triangle must have at least two acute angles.


Let us assume a case where two angles are right angles, i.e., 90° each.


Let these angles be B and C


B = 90°, and C = 90°


We know that the sum of the three angles of a triangle is equal to 180°.


A + B + C = 180°


A + 90° + 90° = 180° (Since, B = 90°, and C = 90°)


A = 0°, which is not possible because then, no triangle would exist.


Thus, two angles of a triangle cannot be 90° each.


Let us assume another case where two angles of the triangle are obtuse angle, i.e., each of the angles is more than 90°.


Let the obtuse angles be B and C


B + C>180° because each of them is more than 90°.


We know that the sum of the three angles of a triangle is equal to 180°.


A + B + C = 180°


A = 180° - (B + C)


A = a negative value, which is not possible.


Thus, no such triangle is possible which has its two angles greater than 90°.


Again, let us assume a case where one angle is 90° and another angle is an obtuse angle, i.e., greater than 90°.


Let’s say B = 90°, and C is obtuse, i.e., C > 90°.


We know that the sum of the three angles of a triangle is equal to 180°.


A + B + C = 180°


A + 90° + C = 180°


A = 180° - 90° - C


A = 90° - C


Since, C > 90°, the value of A becomes negative.


Thus, no such triangle exists such that one angle is 90° and other angle is an obtuse angle.


Let us assume the case when two angles are acute, i.e., both the angles are less than 90°.


Let these angles be B and C.


The sum of these two angles is less than 180°.


B + C<180°


We know that the sum of the three angles of a triangle is equal to 180°.


A + B + C = 180°


A = 180° - (B + C)


A = a positive value, since B + C<180°


A = an acute angle


Thus, it is proved that a triangle should have atleast two acute angles.


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