Q. 64.0( 10 Votes )

# Prove that a tria

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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