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2010-03-23T19:38:18+01:00
Zadanie 2.50
z tw. cosinusów:
(19d)² = a² + b² - 2*ab*cos120° = a² + b² + 2*ab*cos60°
(7d)² = a² + b² - 2*ab*cos60°

(a² + b² + 2*ab*cos60°)/19² = (a² + b² - 2*ab*cos60)/7²
(a² + b² + ab)/361 = (a² + b² - ab)/49
312a² + 312b² - 410ab = 0
156a² + 156b² - 205ab = 0
Δ = 42025b² - 97344b² < 0, nie ma takich a i b

zadanie 2.51
W równoległoboku przekątne przecinają się w połowie, dlatego wysokości to odpowiednio 2l i 2k

z tw. cosinusów:
d² = a² + b² - 2ab*cosα
D² = a² + b² - 2ab*cos(180° - α) = a² + b² + 2ab*cosα

sinα = 2k/b => b = 2k/sinα
sinα = 2l/a => a = 2l/sinα

d² = (2l/sinα)² + (2k/sinα)² - 2(2l/sinα)(2k/sinα)*cosα = 4(k² + l² - 2kl*cosα)/sin²α
D² = (2l/sinα)² + (2k/sinα)² + 2(2l/sinα)(2k/sinα)*cosα = 4(k² + l² + 2kl*cosα)/sin²α

d = 2√(k² + l² - 2kl*cosα)/sinα
D = 2√(k² + l² + 2kl*cosα)/sinα

zadanie 2.52
z tw. cosinusów:
q² = a² + b² - 2ab*cosα
p² = a² + b² - 2ab*cos(180° - α) = a² + b² + 2ab*cosα

4ab*cosα = p² - q²
ab = (p² - q²)/4cosα
P = ab*sinα = (p² - q²)sinα/4cosα = (p² - q²)*tgα/4

zadanie 2.53
z tw. cosinusów:
|BC|² = |AB|² + |AC|² - 2|AB|*|AC|*cosα
|CD|² = |AD|² + |AC|² - 2|AD|*|AC|*cosβ

19 = 9 + 25 - 2*5*3*cosα => cosα = 15/2*5*3 = 1/2
21 = 16 + 25 - 2*4*5*cosβ => cosβ = 20/2*4*5 = 1/2

0 < α, β < π (z tw. odwrotnego do tw. Pitagorasa)
cosα = cosβ = 1/2 => α = β = π/3
α + β = 2π/3

z tw. cosinusów:
|BD|² = |CD|² + |BC|² - 2*|BC|*|CD|*cos(α + β)
|BD|² = 19 + 21 - 2*√399*cos(2π/3) = 40 - 2*√399*cos(2π/3) = 40 + 2√399*cos(π/3) = 40 + √399

zadanie 2.54
suma kątów przy jednym ramieniu to 180°
|<DAB| = α
|<CDA| = 180° - α
|<ABC| = β
|<BCD| = 180° - β

z tw. cosinusów:
|DB|² = |AD|² + |AB|² - 2*|AB|*|AD|*cosα
|AC|² = |AD|² + |DC|² - 2*|AD|*|DC|*cos(180° - α) = |AD|² + |DC|² + 2*|AD|*|DC|*cosα
|AC|² = |BC|² + |AB|² - 2*|BC|*|AB|*cosβ
|DB|² = |BC|² + |CD|² - 2*|BC|*|CD|*cos(180° - β) = |BC|² + |CD|² + 2*|BC|*|CD|*cosβ

cosα = (|AD|² + |AB|² - |DB|²)/2*|AB|*|AD|
cosα = (|AC|² - |AD|² - |DC|²)/2*|AD|*|DC|
cosβ = (|BC|² + |AB|² - |AC|²)/2*|BC|*|AB|
cosβ = (|DB|² - |BC|² - |CD|²)/2*|BC|*|CD|

(|AD|² + |AB|² - |DB|²)/2*|AB|*|AD| = (|AC|² - |AD|² - |DC|²)/2*|AD|*|DC|
(|BC|² + |AB|² - |AC|²)/2*|BC|*|AB| = (|DB|² - |BC|² - |CD|²)/2*|BC|*|CD|

(|AD|² + |AB|² - |DB|²)/|AB| = (|AC|² - |AD|² - |DC|²)/|DC|
(|BC|² + |AB|² - |AC|²)/|AB| = (|DB|² - |BC|² - |CD|²)/|CD|

(9 + 56,25 - |DB|²)/3 = (|AC|² - 9 - 14,0625)/3,75
(27,5625 + 56,25 - |AC|²)/3 = (|DB|² - 27,5625 - 14,0625)/3,75

(65,25 - |DB|²)/3 = (|AC|² - 23,0625)/3,75
(83,8125 - |AC|²)/3 = (|DB|² - 41,625)/3,75

21,75 - |DB|²/3 = |AC|²/3,75 - 6,15
27,9375 - |AC|²/3 = |DB|²/3,75 - 11,1

|AC|² = 15,6*3,75 - 1,25*|BD|² = 58,5 - 1,25*|BD|²
|AC|² = 16,8275*3 - 0,8|DB|² = 50,4825 - 0,8|DB|²

0,45|BD|² = 8,0175
|BD|² = 801,75/45 = 1,6035/9
|BD| = √1,605/3 ≈ 1,3/3
|AC|² = 58,5 - 1,25*1,6035/9 = 58,277292
|AC| ≈ 7,63

zadanie 2.55
suma kątów przy jednym ramieniu to 180°
|<DAB| = α
|<CDA| = 180° - α
|<ABC| = β
|<BCD| = 180° - β

z tw. cosinusów:
|DB|² = |AD|² + |AB|² - 2*|AB|*|AD|*cosα
|AC|² = |AD|² + |DC|² - 2*|AD|*|DC|*cos(180° - α) = |AD|² + |DC|² + 2*|AD|*|DC|*cosα
|AC|² = |BC|² + |AB|² - 2*|BC|*|AB|*cosβ
|DB|² = |BC|² + |CD|² - 2*|BC|*|CD|*cos(180° - β) = |BC|² + |CD|² + 2*|BC|*|CD|*cosβ

|AD|² + |AB|² - 2*|AB|*|AD|*cosα = |BC|² + |CD|² + 2*|BC|*|CD|*cosβ => cosβ = (|AD|² + |AB|² - 2*|AB|*|AD|*cosα - |BC|² - |CD|²)/2*|DC|*|AD|
|AD|² + |DC|² + 2*|AD|*|DC|*cosα = |BC|² + |AB|² - 2*|BC|*|AB|*(|AD|² + |AB|² - 2*|AB|*|AD|*cosα - |BC|² - |CD|²)/2*|DC|*|AD|

|AD|³*|DC| + |DC|³*|AD| + 2*|AD|²*|DC|²*cosα = |BC|²*|DC|*|AD| + |AB|²*|DC|*|AD| - |BC|*|AB|*|AD|² - |BC|*|AB|³ + 2*|AB|²*|BC|*|AD|*cosα + |BC|³ *|AB| + |CD|²*|BC|*|AB|

|AD|³*|DC| + |DC|³*|AD| + 2*|AD|²*|DC|²*cosα = |BC|²*|DC|*|AD| + |AB|²*|DC|*|AD| - |BC|*|AB|*|AD|² - |BC|*|AB|³ + 2*|AB|²*|BC|*|AD|*cosα + |BC|³ *|AB| + |CD|²*|BC|*|AB|

cosα = (|AD|³*|DC| + |DC|³*|AD| - |BC|²*|DC|*|AD| - |AB|²*|DC|*|AD| + |BC|*|AB|*|AD|² + |BC|*|AB|³ - |BC|³ *|AB| - |CD|²*|BC|*|AB|)/(2*|AD|²*|DC| + 2*|AB|²*|BC|*|AD|)
cosβ = (|AD|² + |AB|² - 2*|AB|*|AD|*cosα - |BC|² - |CD|²)/2*|DC|*|AD|

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