ed 213

Th dei seni applicato al triangolo ΑΒΡ

{x/SIN(β) = ΒΡ/SIN(α)

{ΒΡ/SIN(α) = ΑΒ/SIN(α + β)

SIN(α + β) = SIN(α)·COS(β) + SIN(β)·COS(α)

SIN(β) = COS(β) = √2/2

COS(α) = √(1 - SIN(α)^2)---> COS(α) = √(1 - 9/25)

COS(α) = 4/5

SIN(α + β) = 3/5·(√2/2) + √2/2·(4/5)---> SIN(α + β) = 7·√2/10

{x/(√2/2) = ΒΡ/(3/5)

{ΒΡ/(3/5) = 31·a/(7·√2/10)

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{ΒΡ = x/(√2/2)·(3/5)----> ΒΡ = 3·√2·x/5

{ΒΡ = 31·a/(7·√2/10)·(3/5)----> ΒΡ = 93·√2·a/7

quindi: 3·√2·x/5 = 93·√2·a/7---> x = 155·a/7 = AP

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Th dei seni applicato al triangolo ACP

y/SIN(α + β) = x/SIN(γ)

γ = pi - (2·α + β)

SIN(γ) = SIN(2·α + β)=

=SIN(2·α)·COS(β) + SIN(β)·COS(2·α)=

=2·SIN(α)·COS(α)·COS(β) + SIN(β)·(COS(α)^2 - SIN(α)^2)

SIN(γ) =2·(3/5)·(4/5)·(√2/2) + √2/2·((4/5)^2 - (3/5)^2)=

=12·√2/25 + √2/2·(7/25) = 31·√2/50

y = 155·a/7/(31·√2/50)·(7·√2/10)---> y = 25·a = ΑC

CΡ/SIN(α) = x/SIN(γ)

CΡ = 155·a/7/(31·√2/50)·(3/5)-----> CΡ = 75·√2·a/7

ΒC = CΡ + ΒΡ = 75·√2·a/7 + 93·√2·a/7----> ΒC = 24·√2·a

31a/sin 61,26° = AC/sin 45°

AC = 31a*√2/2 / 0,8768 = 25,00a

25/sin 45 = BC/sin 73,74

sin (2*36,87) = 0,8*0,6+0,6*0,8 = 0,9600

BC = 25*0,9600*2/2^0,5 = 33,941a = 24a√2

31a/ sin 98,13° = AP/sin 45°

AP = 15,5a*√2 /0,9899 = 22,143a = 155a/7