Matematica esercizio 181,182,184

angolo C = 180-(100+42) = 38°

15/sin 38° = 18/sin B

sin B = 18*sin 38°/15 = 6/5*0,6157 = 0,7388

angolo B = 47,63°

angolo AMB = 180-100 = 80°

angolo A = 42+(180-(80+47,63) = 94,37°

15/sin 38 = BC/sin 94,37°

BC = 15*sin 94,37°/sin 38° = 15*0,9971/0,6157 = 24,3 

CM = BC/2 = 12,15

sin MAB = sin (180-(80+47,63)) = 0,792 

angolo C = angolo A = 65°

angolo BDC = 180-(65+55) = 60°

angolo ADB = angolo DBC = 55°

angolo ABD = 180-(55+65) = 60°

angolo BDC = angolo ABD = 60°

BD/sin 65 = AB/sin 55

AB = 10*sin 55/ sin 65 = 9,04  

BD/sin 65° = AD/sin 60°

AD = 5*√3/sin 65° = 9,56 

perimetro ABCD = 2(9,04+9,56) = 2*18,6 = 37,2

d*tan35 = (d-1000)*tan 50°

0,700d = 1,192d-1192

1192 = 0,492d

d = 1192/0,492 = 2.423 m

h = d*tan 35° = 2423*0,7 = 1.696 m 

lunghezza l = √6^2+15^2-2*6*15*0,3 = √261-54 = √207 cm 

Ex.181

TH seni

ΒD/SIN(α) = ΑD/SIN(β) = ΑΒ/SIN(δ)

12/SIN(60°) = ΑD/SIN(45°) = ΑΒ/SIN(75°)

ΑD = 12/SIN(60°)·SIN(45°)-------> ΑD = 4·√6

ΑΒ = 12/SIN(60°)·SIN(75°)-------> ΑΒ = 2·√6 + 6·√2

perimetro=2(AD+AB)=2·(4·√6 + 2·√6 + 6·√2) = 12·√6 + 12·√2

perimetro=12·√2·(√3 + 1)

Ex.182

COS(β°) = 2/3·√2----> β = 19.47°

SIN(β) = √(1 - (2/3·√2)^2)------> SIN(β) = 1/3

SIN(α) = SIN(pi/6)----> SIN(α) = 1/2

Th seni

ΜΝ/SIN(α) = LΜ/SIN(β) = ΝL/SIN(α + β)

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

SIN(α + β) = 1/2·(2/3·√2) + 1/3·COS(α)

SIN(α + β) = 1/2·(2/3·√2) + 1/3·(√3/2)

SIN(α + β) = √3/6 + √2/3

SIN(α + β) = (√3 + 2·√2)/6

ΜΝ = 60/(1/3)·(1/2)----> ΜΝ = 90 cm

ΝL = 60/(1/3)·((√3 + 2·√2)/6)----> ΝL = 30·(√3 + 2·√2) cm

Th Carnot

ΑΑ' = x

30^2 = 20^2 + x^2 - 2·20·x·COS(60°)

900 = x^2 - 20·x + 400----> x^2 - 20·x - 500 = 0

Risolvo ed ottengo:

x = 10 - 10·√6 ∨ x = 10·√6 + 10

quindi: x = 34.495 m

Ancora Carnot:

(10·√6 + 10)^2 = 20^2 + 30^2 - 2·20·30·COS(β)

200·√6 + 700 = 1300 - 1200·COS(β)

COS(β°) = (1300 - (200·√6 + 700))/1200

COS(β°) = 1/2 - √6/6

β° = 84.736°