Question

Prove that the angles of an equilateral triangle are ${60}^{\circ }$ each.

Answer 1

For an equilateral triangle, all sides are equal.

Assuming an equilateral $\Delta ABC,$

Then,

$AB=AC=BC$.

$\Rightarrow \angle A=\angle B=\angle C$

(Angles opp. to equal sides are equal)

For a triangle, by angle sum property,

$\angle A+\angle B+\angle C={180}^{\circ }$

Substituting

$\Rightarrow \angle A=\angle B=\angle C$

$\therefore \angle A+\angle A+\angle A={180}^{\circ }$

$\Rightarrow 3\angle A={180}^{\circ }$

$\Rightarrow \angle A={60}^{\circ }$

$\therefore \angle A=\angle B=\angle C={60}^{\circ }$

So, each angle of an equilateral triangle is $60º$.