Question

In triangle ABC, angleA=60 degree, prove that BC2=AB2+AC2+AB.AC

Answer 1

Can you please double check the question? Is it "-AB.AC" instead of "+AB.AC" ?

Answer 2

use the cosine rule :

cos60 = (AB^2 + AC^2 - BC^2)/2ABAC

Answer 3

1/2 = (AB^2 + AC^2 - BC^2)/2*AB*AC
multiplify by 2 to both side, giving us
1 = (AB^2 + AC^2 - BC^2)/AB*AC

Answer 4

cross product out!
AB * AC = AB^2 + AC^2 - BC^2

Answer 5

and actually, Rolypoly is right, that should "-AB.AC" not +AB AC
i think there is a bit typo from u, @kumar69