# Math Help - Cos and Sin in a triangle that has no right angle

1. ## Cos and Sin in a triangle that has no right angle

Hey MHF!
Here something I found.
I'm still learning geometry again, so this may seem easy for you guys.

The sides of $\Delta ABC$ are $AB = 6$ $BC=4$ and $CA=5$

$\cos
(The < means the angle)
But!
AB is not the Hypotenuse, since $16+25=41$
So, I figure this is not a Triangle Rectangle, but something else.
No right angle on this one, right?

I'm still kind of amateur in Geometry, so go easy! hahaha
Thanks

2. Originally Posted by Zellator
Hey MHF!
Here something I found.
I'm still learning geometry again, so this may seem easy for you guys.

The sides of $\Delta ABC$ are $AB = 6$ $BC=4$ and $CA=5$

$\cos
(The < means the angle)
You need to know the Law of Cosines.
$a^2=b^2+c^2-2bc\cos(\angle A)$

3. Yes! This makes sense!
Thanks so much Plato!