Trigonometria triangoli qualsiasi

Question

in un triangolo ABC si ha  Angolo ABC= beta=arccos4/5 angolo ACB= 2beta   Sapendo che perimetro è 15a determina misure dei lati

exProf · Accepted Answer

* β = arccos(4/5) ~= 37°* cos(β) = 4/5* sin(β) = 3/5* γ = 2*β* sin(γ) = sin(2*arccos(4/5)) = 24/25* α = π - (β + γ) = π - 3*β* sin(α) = sin(π - 3*β) = sin(3*β) = (3*cos^2(β) - sin^2(β))*sin(β)= (3*cos^2(β) - sin^2(β))*sin(β) == (3*(4/5)^2 - (3/5)^2)*3/5 = 117/125---------------* a/sin(α) = b/sin(β) = c/sin(γ) = 2*R ≡≡ a/(117/125) = b/(3/5) = c/(24/25) = 2*R* (a + b + c = 15) & (a/(117/125) = b/(3/5)) & (b/(3/5) = c/(24/25)) & (c/(24/25) = 2*R) ≡≡ (a = 45/8) & (b = 375/104) & (c = 75/13) & (R = 625/208)---------------http://www.wolframalpha.com/input?i=triangle+458+375104+7513+perimeterhttp://www.wolframalpha.com/input?i=triangle+458+375104+7513+circumcircle

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