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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

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