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RangO!!
 2 years ago
Prove that: Δ=4.Rr cos a/2.cos b/2. cos y/2
RangO!!
 2 years ago
Prove that: Δ=4.Rr cos a/2.cos b/2. cos y/2

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RolyPoly
 2 years ago
Best ResponseYou've already chosen the best response.0What is Rr and what is \(\Delta\) here represents??

RangO!!
 2 years ago
Best ResponseYou've already chosen the best response.0Δ is the sign of delta !

RolyPoly
 2 years ago
Best ResponseYou've already chosen the best response.0Do you have a diagram for this question?

RangO!!
 2 years ago
Best ResponseYou've already chosen the best response.0the 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.

RolyPoly
 2 years ago
Best ResponseYou've already chosen the best response.0I 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....

RangO!!
 2 years ago
Best ResponseYou've already chosen the best response.0a, b , c are the sides _

RolyPoly
 2 years ago
Best ResponseYou've already chosen the best response.0No c in your question...

RolyPoly
 2 years ago
Best ResponseYou've already chosen the best response.0But I guess I'd better go, since I don't know how to solve your problem. I'm sorry!!

RolyPoly
 2 years ago
Best ResponseYou've already chosen the best response.0What is the radius of a triangle?

RolyPoly
 2 years ago
Best ResponseYou've already chosen the best response.0I mean radius is for circle right?

genius12
 2 years ago
Best ResponseYou've already chosen the best response.0Can you type that out in latex/equation editor?

AccessDenied
 2 years ago
Best ResponseYou've already chosen the best response.0There 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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