Cartesian product help?I'm interested how to solve the following problem:

[Chuckster]

I'm interested how to solve the following problem:
if we have a triangle, where a,b,c are sides of that triangle and we know that
(a+b+c)x(a+b-c)=3ab, we need to find the angle opposite to side c.

How to do this?

 

[HallsofIvy]

I have no clue what (a+b+c)x(a, b, -x)= 3ab means. Are a, b, and c numbers or vectors? What does "(a, b, -z)" mean?

 

[Chuckster]

I have no clue what (a+b+c)x(a, b, -x)= 3ab means. Are a, b, and c numbers or vectors? What does "(a, b, -z)" mean?

i messed up the post. check the edited version.

 

[micromass]

Hello Chuckster!

You still have not explained what (a+b+c)x(a+b-c)=3ab means. Are a,b,c just numbers and is x the regular multiplication? Also, what does the Cartesion product have to do with all of this?

When given three sider a,b,c, you can always find the angle opposite to c (called \gamma) by the cosine rule:

c^2=a^2+b^2-2ab\cos(\gamma).

I really have no clue what (a+b+c)x(a+b-c)=3ab has to do with this though...

 

[Chuckster]

Hello Chuckster!

You still have not explained what (a+b+c)x(a+b-c)=3ab means. Are a,b,c just numbers and is x the regular multiplication? Also, what does the Cartesion product have to do with all of this?

When given three sider a,b,c, you can always find the angle opposite to c (called \gamma) by the cosine rule:

c^2=a^2+b^2-2ab\cos(\gamma).

I really have no clue what (a+b+c)x(a+b-c)=3ab has to do with this though...

I know the law of cosines, but i can't find a way to use it here, because i don't know the numerical values of the pages, neither do i know the angles.

I just know i have the condition given that
(a+b+c)x(a+b-c)=3ab.

I'm guessing that x marks the Cartesian product of a+b+c and a+b-c and is equal to 3ab.
But, that doesn't sound logical or reasonable to me.

 

[micromass]

Did you try by expanding (a+b+c)(a+b-c) and then substitute

a^2+b^2-c^2=2ab\cos(\gamma)

Try if you can do something like that...

 

[Chuckster]

Did you try by expanding (a+b+c)(a+b-c) and then substitute

a^2+b^2-c^2=2ab\cos(\gamma)

Try if you can do something like that...

actually that was the key to solving this, but i found a version of this task where it said
(a+b+c)X(a+b-c), and i was in doubt what X means, and i thought it was a vector product.

however, it turned out to be just simple ol' multiplying.