Question

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

Answer 1

SOLUTION:-

Given:

In a triangle, other than an equilateral ∆, angle opposite the longest side is greater than 2/3 right angle.

Proof:

In a right angled ∆:

∠B is the largest angle.

AC >BC

∠B > ∠A...............(1)

&

AC > AB

∠B > ∠C...............(2)

Add equation (1) & (2), we get:

∠B + ∠B > ∠A + ∠C

2∠B > ∠A +∠C

Add ∠B on the both sides,

2∠B +∠B > ∠A+ ∠B +∠C

=)3∠B > 180° [sum of three angles=180°]

=) ∠B>180°/3

=)∠B> 60°

=)∠B> 2/3 of 90°

So,

∠B> 2/3 × right angle. [2/3× 90=60°]

Proved.

Hope it helps ☺️