Answer :
let ABC any equilateral triangle
let angle <A = x
since AB = BC
hence <A = <C = x
since BC = CA
hence <A = <B = x
sum of internal angle of any n sided polygon = (n-2)pi
<A + <B + <C = pi = 180 (here n = 3)
x + x + x = 180
3x = 180
x = 60
hence each angle of equilateral triangle is 60
let angle <A = x
since AB = BC
hence <A = <C = x
since BC = CA
hence <A = <B = x
sum of internal angle of any n sided polygon = (n-2)pi
<A + <B + <C = pi = 180 (here n = 3)
x + x + x = 180
3x = 180
x = 60
hence each angle of equilateral triangle is 60
here is the proof -
assume pqr any equilateral triangle ,assume angle <p =a pq = qr
as <p = <r = x ,as qr = rp
therefore <p = <q = x
sum of internal angle of any n sided polygon = (n-2)pi
<p + <q + <r = pi = 180 (here n = 3)
x + x + x = 180
3x = 180
x = 60
hence each angle of equilateral triangle is 60
assume pqr any equilateral triangle ,assume angle <p =a pq = qr
as <p = <r = x ,as qr = rp
therefore <p = <q = x
sum of internal angle of any n sided polygon = (n-2)pi
<p + <q + <r = pi = 180 (here n = 3)
x + x + x = 180
3x = 180
x = 60
hence each angle of equilateral triangle is 60