How do you use the law of cosines to show that for any triangle ABC, c^2<a^2+b^2 if C is acute and c^2>a^2+b^2 if C is obtuse?

1 Answer
Jan 6, 2017

Please see below.

Explanation:

According to cosine formula, in a triangle DeltaABC,

c^2=a^2+b^2-2abcosC

Now if /_C is acute, cosC>0, and hence -2abcosC<0

i.e. a^2+b^2-2abcosC<a^2+b^2

and hence c^2< a^2+b^2

and if /_C is obtuse, cosC<0, and hence -2abcosC>0

i.e. a^2+b^2-2abcosC>a^2+b^2

and hence c^2>a^2+b^2