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Question

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


Solution

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 $$60º$$.

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Mathematics
RS Agarwal
Standard IX

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