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2009-12-17T12:46:00+01:00
W=tgα* cosβ
sinα = ⅔
z tożsamości trygonometrycznych: tgα = sinα/cosα
sin²α+ cos²α = 1
(⅔)² + cos²α = 1
4/9 + cos²α = 1
cos²α = 1 - 4/9
cos²α = 5/9
cosα = √5 /3

tgα = (2/3) / ( √5 /3) = 2√5 / 5

tgβ = ⅓
tgβ = sinβ/cosβ = 1/3
układamy układ równań:

sinβ/cosβ = 1/3
sin²β + cos²β = 1

sinβ =1/3 cosβ
(1/3 cosβ)² + cos²β =1
1/9 cos²β + cos²β =1
10/9 cos²β =1 /* (9/10)
cos²β =9/10
cosβ = 3√10 / 10

sinβ = (1/3) * (3√10 / 10) = √10 / 10

w = (2√5 / 5) * (3√10 / 10) = 3√50 / 25 = 15√2 / 25 = 3√2 /5