Question

Show that the angles of an equilateral triangle are ${60}^0$ each.

Answer 1

In $\Delta ABC,AB=AC=BC$.
$\therefore \angle A=\angle B=\angle C$, let this is $x^0$.
But sum of three angles is ${180}^0$.
$\angle A+\angle B+\angle C={180}^0$
$x+x+x={180}^0$
$3x={180}^0$
$\therefore x=\frac{180}{3}={60}^0$
$\therefore angleA={60}^0$
$\angle B={60}^0$
$\angle C={60}^0$