Prove that in a triangle, other than an equilateral triangle, angle opposite the longest side is greater than 2/3 of a right angle.

Solution:

Consider a triangle ABC with BC as the longest side.

Angle opposite to longest side is angle A

We have to prove that the angle opposite to the longest side is greater than 2/3 of a right angle.

In triangle ABC,

Since BC is the largest side

So, BC > AB

We know that the angle opposite to the longest side is greater.

∠A > ∠C -------------------- (1)

Similarly, BC > AC

So, ∠A > ∠B --------------- (2)

Adding (1) and (2),

∠A + ∠A > ∠C + ∠B

2∠A > ∠C + ∠B

Adding ∠A on both sides,

∠A + 2∠A > ∠C + ∠B + ∠A

Angle sum property states that the sum of all three interior angles of a triangle is always equal to 180 degrees.

So, ∠A + ∠B + ∠C = 180°

Now, 3∠A > 180°

∠A > 180°/3

∠A > 2/3 (90°)

Therefore, angle A is greater than 2/3 of a right angle.

✦ Try This: ABCD is a quadrilateral inscribed in a circle. Diagonals AC and BD are joined. If ∠CAD=60 degrees and ∠BDC=25 degrees. Find ∠BCD.

☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 7

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NCERT Exemplar Class 9 Maths Exercise 7.4 Problem 20

Prove that in a triangle, other than an equilateral triangle, angle opposite the longest side is greater than 2/3 of a right angle

Summary:

It is proven that in a triangle, other than an equilateral triangle, angle opposite the longest side is greater than 2/3 of a right angle

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