Question

12. Prove that in a triangle, other than an equilateral triangle, angle
opposite the longest side is greater than ; of a night angle.​

Answer 1

Step-by-step explanation:

Given -- ΔABC other than equilateral triangle.

So let me just establish the relation between the sides.

Let  AB = AD = DB

ABD is an equilateral triangle since AB = AD = DB

∠ABD = ∠ADB = ∠BAD = 60°

Longest side= BC

Angle opposite to longest side = ∠BAC

Here it is clear that ∠BAC = ∠BAD + ∠CAD

So ∠BAC > ∠BAD

∠BAC > 60°

Hence proved!

Hope This Helps You!

Answer 2

Given -- ΔABC other than equilateral triangle.

The construction is already given in the picture.So let me just establish the relation between the sides.

Let  AB = AD = DB

ABD is an equilateral triangle since AB = AD = DB

∠ABD = ∠ADB = ∠BAD = 60°

Longest side= BC

Angle opposite to longest side = ∠BAC

Here it is clear that ∠BAC = ∠BAD + ∠CAD

So ∠BAC > ∠BAD

∠BAC > 60°

Hence proved!

Hope This Helps You!