Show that the angles of an equilateral triang | vedclass

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(N/A) In $\Delta ABC$,we have:$AB = BC = CA$ $[\because \Delta ABC$ is an equilateral triangle$]$Since $AB = BC$,the angles opposite to these sides ar

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(N/A) In $\Delta ABC$,we have:$AB = BC = CA$ $[\because \Delta ABC$ is an equilateral triangle$]$Since $AB = BC$,the angles opposite to these sides are equal,so $\angle A = \angle C$ ......... $(1)$Similarly,since $AC = BC$,the angles opposite to these sides are equal,so $\angle A = \angle B$ ......... $(2)$From $(1)$ and $(2)$,we have:$\angle A = \angle B = \angle C$Let $\angle A = \angle B = \angle C = x$Since the sum of the angles in a triangle is $180^o$:$\angle A + \angle B + \angle C = 180^o$Substituting $x$ for each angle:$x + x + x = 180^o$$3x = 180^o$$x = \frac{180^o}{3} = 60^o$Therefore,$\angle A = \angle B = \angle C = 60^o$.Thus,each angle of an equilateral triangle is $60^o$.

Show that the angles of an equilateral triangle are $60^o$ each.

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