Question
Show that the angles of an equilateral triangle are $60^\circ $ each.
✓
Answer
Let $ABC$ is an equilateral triangle. We know that all the sides of an equilateral triangle are equal.
$\therefore AB = BC = CA ....(1)$
To prove :- $\angle A = \angle B = \angle C = 60^\circ $
Proof :-
In $\Delta ABC$ we have:-
AB = AC [from (1)]
$\Rightarrow \angle C = \angle B ....(2)$
[$\because$ Angles opposite to equal sides of a triangle are equal]
Again from $(1),$
$BC = AC$
$\Rightarrow \angle A = \angle B .....(3)$
[$\because$ Angles opposite to equal sides of a triangle are equal] .
From $(2) \& (3) ;$
$\Rightarrow \angle A = \angle B=\angle C ....(4)$
Now,
$\angle A + \angle B + \angle C = 180 ^ { \circ }$ [$\because$ Angle sum property of a triangle]
$\Rightarrow \angle A + \angle A + \angle A = 180 ^ { \circ }$ [From$ (4) ]$
$\Rightarrow 3 \angle A = 180 ^ { \circ }$|
$\Rightarrow \angle A = \frac { 180 ^ { \circ } } { 3 } = 60 ^ { \circ }$
$\therefore \angle A = \angle B = \angle C = 60 ^ { \circ }$[ from $(4) ]. $
Hence, each angle of an equilateral triangle is equal to $60^\circ .$
$\therefore AB = BC = CA ....(1)$
To prove :- $\angle A = \angle B = \angle C = 60^\circ $
Proof :-
In $\Delta ABC$ we have:-
AB = AC [from (1)]
$\Rightarrow \angle C = \angle B ....(2)$
[$\because$ Angles opposite to equal sides of a triangle are equal]
Again from $(1),$
$BC = AC$
$\Rightarrow \angle A = \angle B .....(3)$
[$\because$ Angles opposite to equal sides of a triangle are equal] .
From $(2) \& (3) ;$
$\Rightarrow \angle A = \angle B=\angle C ....(4)$
Now,
$\angle A + \angle B + \angle C = 180 ^ { \circ }$ [$\because$ Angle sum property of a triangle]
$\Rightarrow \angle A + \angle A + \angle A = 180 ^ { \circ }$ [From$ (4) ]$
$\Rightarrow 3 \angle A = 180 ^ { \circ }$|
$\Rightarrow \angle A = \frac { 180 ^ { \circ } } { 3 } = 60 ^ { \circ }$
$\therefore \angle A = \angle B = \angle C = 60 ^ { \circ }$[ from $(4) ]. $
Hence, each angle of an equilateral triangle is equal to $60^\circ .$
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