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?

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Answer 1 · 2017-01-06T07:29:11

Please see below.

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#