Question
Prove that each angle of an equilateral triangle is 60°.
✓
Answer
Given to prove each angle of an equilateral triangle is 60°.
Let us consider an equilateral triangle ABC.
Such that AB = BC = CA
Now, AB = BC
$\angle\text{A}=\angle\text{C}\ ....(\text{i})$ [Opposite angles to equal sides are equal]
And $\text{BC}=\text{AC}$
$\angle\text{B}=\angle\text{A}\ ...(\text{ii)}$
From (i) and (ii), we get
$\angle\text{A}=\angle\text{B}=\angle\text{C}\ ...(\text{iii})$
We know that
Sum of angles in a triangle = 180
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180$
$\angle\text{A}+\angle\text{A}+\angle\text{A}=180$
$3\angle\text{A}=60$
$\angle\text{A}=60$
$\angle\text{A}=\angle\text{B}=\angle\text{C}=60$
Hence, each angle of an equilateral triangle is 60°.
Let us consider an equilateral triangle ABC.
Such that AB = BC = CA
Now, AB = BC
$\angle\text{A}=\angle\text{C}\ ....(\text{i})$ [Opposite angles to equal sides are equal]
And $\text{BC}=\text{AC}$
$\angle\text{B}=\angle\text{A}\ ...(\text{ii)}$
From (i) and (ii), we get
$\angle\text{A}=\angle\text{B}=\angle\text{C}\ ...(\text{iii})$
We know that
Sum of angles in a triangle = 180
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180$
$\angle\text{A}+\angle\text{A}+\angle\text{A}=180$
$3\angle\text{A}=60$
$\angle\text{A}=60$
$\angle\text{A}=\angle\text{B}=\angle\text{C}=60$
Hence, each angle of an equilateral triangle is 60°.
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