Prove that: Δ=4.Rr cos a/2.cos b/2. cos y/2

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- anonymous

Prove that: Δ=4.Rr cos a/2.cos b/2. cos y/2

- schrodinger

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- anonymous

help, pls

- anonymous

What is Rr and what is \(\Delta\) here represents??

- anonymous

radius and delta

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- anonymous

Δ is the sign of delta !

- anonymous

Do you have a diagram for this question?

- anonymous

the question written above is the exact question from the book :)
The whole question is this :
Prove for any Δ ABC : Δ=4.Rr cos a/2.cos b/2. cos y/2, where all the symbols have thier usual meaning.

- anonymous

I understand that a, b are the angles, but not y, and delta.
Δ in ΔABC means triangle
Δ itself have the meaning of change, or discriminant. But I don't know the meaning of delta in this case, i.e. Δ = 4 r cos....

- anonymous

a, b , c are the sides -_-

- anonymous

No c in your question...

- anonymous

But I guess I'd better go, since I don't know how to solve your problem. I'm sorry!!

- anonymous

What is the radius of a triangle?

- anonymous

I mean radius is for circle right?

- anonymous

Can you type that out in latex/equation editor?

- AccessDenied

There is a lot of ambiguity with the symbols used here, but I'll try to interpret what I understand to be the question?
\( \displaystyle \triangle = 4 R r \cos \frac{\alpha}{2} \cos \frac{\beta}{2} \cos \frac{\gamma}{2}\)
Triangle, which I assume is area of the triangle ABC, is equal to 4 times R (the circumradius) times r (the inradius) times the cosine of each angle (where the y is an attempted translation of gamma.
I've never seen this identity so I don't know for sure if it is correct, but I'll try to research it more.

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