# 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?

Jan 6, 2017

#### Explanation:

According to cosine formula, in a triangle $\Delta A B C$,

${c}^{2} = {a}^{2} + {b}^{2} - 2 a b \cos C$

Now if $\angle C$ is acute, $\cos C > 0$, and hence $- 2 a b \cos C < 0$

i.e. ${a}^{2} + {b}^{2} - 2 a b \cos C < {a}^{2} + {b}^{2}$

and hence ${c}^{2} < {a}^{2} + {b}^{2}$

and if $\angle C$ is obtuse, $\cos C < 0$, and hence $- 2 a b \cos C > 0$

i.e. ${a}^{2} + {b}^{2} - 2 a b \cos C > {a}^{2} + {b}^{2}$

and hence ${c}^{2} > {a}^{2} + {b}^{2}$