Question

Suppose a triangle has sides a, b, and c, and let be the angle opposite the side of length b. If cos > 0, what must be true?

A. a^2 + c^2 < b^2

B. a^2 + b^2 = c^2

C. a^2 + c^2 > b^2

D. a^2 + b^2 > c^2

**i don't get this one :( @terenzreignz :)

Answer 1

Well, remember that
\[\large \cos(\theta) = \frac{a^2+c^2-b^2}{2ac}\]

Answer 2

mhmm :)

Answer 3

Well then, since a, b, and c are all positive, then 2ac is positive, so it all falls on
a^2 + c^2 - b^2

Answer 4

yes :)

so would that mean a^2 + c^2 > b^2 ??

Answer 5

is that right? :)