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.

Re: trouble with Cosine Rule

Quote:

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?

Re: trouble with Cosine Rule

Quote:

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: $\displaystyle c^2=a^2+b^2-2ab\cos(C)$ you have different variables, $\displaystyle c~\&~C$ and $\displaystyle c\ne C$.

Re: trouble with Cosine Rule

Quote:

Originally Posted by

**Plato** It is easy: you are missing the point.

In that formula: $\displaystyle c^2=a^2+b^2-2ab\cos(C)$ you have different variables, $\displaystyle c~\&~C$ and $\displaystyle c\ne C$.

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