Cos and Sin in a triangle that has no right angle

**Zellator** · May 11th 2011, 11:43 AM

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 $CA=5$

$\cos <A =$
(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

**Plato** · May 11th 2011, 11:51 AM

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 $CA=5$

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

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

**Zellator** · May 11th 2011, 02:28 PM

Yes! This makes sense!
Thanks so much Plato!

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