Question

in a triangle ABC the lengths of the sides, in cm, are AB=c BC=a AC=b the angle ACB=120 show that c^2=a^2+b^2+ab

Answer 1

sine theorem:
& AB^2=BC^2+AC^2-2*BC*AC*sin(ACB)
sin(120)=-1/2
so:
c^2=a^2+b^2-2*a*b*(-1/2)=a^2+b^2+ab