# Math Help - trouble with Cosine Rule

1. ## trouble with Cosine Rule

Hi all,

x² = a² + b² - c²

these figures fit the equation

a=10, b=5, c=2 giving x=11

121 = 100 + 25 - 4

then it occured to me that the original equation looked very like the Cosine Rule

c² = a² + b² - 2ab Cos(c)

so that

2ab Cos(c) = a² + b² - c²

so combining this with my original equation I would have thought

x² = 2ab Cos(c) which it doesn't ¿ 2 x 10 x 5 x Cos(2) = 99.94......

what did I miss or misunderstand?

Many thanks,

Pro

2. ## Re: trouble with Cosine Rule

hi procyon

it shouldn't be cos(c) but in there should be an angle opposite to the side c in the triangle with sides a,b and c.

3. ## Re: trouble with Cosine Rule

Originally Posted by anonimnystefy
hi procyon

it shouldn't be cos(c) but in there should be an angle opposite to the side c in the triangle with sides a,b and c.
Wow, thanks for the quick reply

Not sure that I understand you though

I'll put the equations together

a² + b² - c² = x²
a² + b² - c² = 2ab.Cos(c)

Surely 2ab.Cos(c) must equal x² for any given a, b & c (10, 5 & 2 in this case)

hmmm...
just had a thought, a triangle can't have sides 10, 5 & 2. Is this perhaps where it falls down?

4. ## Re: trouble with Cosine Rule

Originally Posted by procyon
Wow, thanks for the quick reply
Not sure that I understand you though
I'll put the equations together
a² + b² - c² = x²
a² + b² - c² = 2ab.Cos(c)
Surely 2ab.Cos(c) must equal x² for any given a, b & c (10, 5 & 2 in this case)
It is easy: you are missing the point.
In that formula: $c^2=a^2+b^2-2ab\cos(C)$ you have different variables, $c~\&~C$ and $c\ne C$.

5. ## Re: trouble with Cosine Rule

Originally Posted by Plato
It is easy: you are missing the point.
In that formula: $c^2=a^2+b^2-2ab\cos(C)$ you have different variables, $c~\&~C$ and $c\ne C$.
Hi Plato,

Many thanks for clearing that up for me. That's the trouble when you suddenly remember a formula through the mists of time

I had thought 2ab.Cos(c) not 2ab.Cos(ø)

thanks again