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.

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Solution

## 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.

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