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

• May 11th 2011, 11:43 AM
Zellator
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 $\displaystyle \Delta ABC$ are $\displaystyle AB = 6$ $\displaystyle BC=4$ and $\displaystyle CA=5$

$\displaystyle \cos <A =$
(The < means the angle)
But!
AB is not the Hypotenuse, since $\displaystyle 16+25=41$
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 (Bow)
• May 11th 2011, 11:51 AM
Plato
Quote:

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 $\displaystyle \Delta ABC$ are $\displaystyle AB = 6$ $\displaystyle BC=4$ and $\displaystyle CA=5$

$\displaystyle \cos <A =$
(The < means the angle)

You need to know the Law of Cosines.
$\displaystyle a^2=b^2+c^2-2bc\cos(\angle A)$
• May 11th 2011, 02:28 PM
Zellator
Yes! This makes sense!
Thanks so much Plato!