Question

Suppose a triangle has sides a, b, and c, and that a^2 + b^2 < c^2. Let (theta) be the measure of the angle opposite the side of length c. Which of the following must be true? Check all that apply.

Please explain. Thank you!

Answer 1

hint:
so we know we don't have a right triangle ...
by law of cosines we have:
hmmm we are given that
\[0=a^2+b^2-c^2-2ab \cos(\theta) <c^2-c^2-2abcos(\theta)=-2abcos(\theta) \\ 0 < -2 ab \cos(\theta) \\ \text{ multiplying the negative 1 on both sides gives } 0>2 ab \cos(\theta)\]

Answer 2

oops I gave one answer :p
and kinda gave the other