# Answer to Question #4350 in Abstract Algebra for sumaya

Question #4350

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

Expert's answer

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

& 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

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