Question

Prove that in a triangle, other than an equilateral triangle, angle opposite to the longest side is greater than $\frac{2}{3}$of a right angle.

Answer 1

Let AC be the longest side in the ∆ABC.

Now,

AC > AB

⇒ ∠B > ∠C .....(1) (In a triangle, greater side has greater angle opposite to it)

Also,

AC > BC

⇒ ∠B > ∠A .....(2) (In a triangle, greater side has greater angle opposite to it)

From (1) and (2), we have

∠B + ∠B > ∠A + ∠C

⇒ ∠B + ∠B + ∠B > ∠A + ∠B + ∠C

⇒ 3∠B > 180º (Using angle sum property of a triangle)

⇒ ∠B > $\frac{1}{3}$ × 180º

Or ∠B > $\frac{2}{3}$ × 90º

Thus, the angle opposite to the longest side is greater than $\frac{2}{3}$of a right angle.

Hence proved.