Question

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$\huge\sf\red{Question}$

$\sf\bold\green{ Chapter-:Traingle}$

If AC > AB > BC for ΔABC, then ∠ABC is always greater than

A)30° B)45° C)90° D)60° ​

$\sf\bold\pink{Explanation \:is\: must }$​

Answer 1

$\huge\tt\red\diamond{\underline{\underline{Answer}}}$

$60\degree$

Explanation:-

Since the triangle is not an equilateral triangle and we know the angles are always dependent on their opposite sides.

As in △ABC AC is the longest side.$\angle{ABC}$ will always be greater than 60°

I hope itz correct :D

@ᎮѕуcнσAεѕтнεтíc

Answer 2

$\large\underline{\sf{Solution-}}$

Given that,

↝ In ΔABC, AC > AB > BC

$\bf\implies \: \triangle \: ABC \: is \: not \: equilateral.$

$\bf\implies \: \angle \: ABC \: \ne \: 60 \degree$

Again, it is given that,

$\rm :\longmapsto\:AC > AB$

$\bf\implies \:\:AC^{2} > AB^{2} - - - (1)$

and

$\rm :\longmapsto\:AC > AB > BC$

$\rm \implies\:\:AC > BC$

$\bf \implies\:\:AC ^{2} > BC^{2} - - - (2)$

On adding equation (1) and equation (2), we get

$\bf\implies \:2AC^{2} = AB^{2} + BC^{2}$

$\bf\implies \: AB^{2} + BC^{2} \ne \: AC ^{2}$

$\bf\implies \: \triangle \: ABC \: is \: not \: right \: angle.$

Now,

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

So,

$\rm :\longmapsto\:AC > AB$

$\bf\implies \: \angle \: B \: > \: \angle \: C$

Also,

$\rm :\longmapsto\: AB > BC$

$\bf\implies \: \angle \: C \: > \: \angle \: A$

From these, we concluded that,

$\bf\implies \: \angle \: B \: > \: \angle \: C \: > \: \angle \: A$

As, We know that, sum of all interior angles of a triangle is supplementary.

So,

$\bf\implies \:\angle \: B \: > \: 60 \degree$