Odpowiedź:
h - wysokość trójkąta
sinα = 1/3
cosβ = 1/6
IBCI = 12
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cos²β = (1/6)² = 1/36
1 - sin²β = 1/36
sin²β = 1 - 1/36 = 35/36
sinβ = √(35/36) = √35/6
h/IBCI = sinβ = √35/6
h = IBCI * √35/6 = 12 * √35/6 =2√35
h/IACI = sinα = 1/3
h = IACI * 1/3
IACI = h : 1/3 = 2√35 : 1/3 = 2√35 * 3 = 6√35 ≈ 6 * 5,9160797.. ≈ 35,496478... ≈ 35,49648
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Odpowiedź:
h - wysokość trójkąta
sinα = 1/3
cosβ = 1/6
IBCI = 12
---------------------------------------------------
cosβ = 1/6
cos²β = (1/6)² = 1/36
1 - sin²β = 1/36
sin²β = 1 - 1/36 = 35/36
sinβ = √(35/36) = √35/6
h/IBCI = sinβ = √35/6
h = IBCI * √35/6 = 12 * √35/6 =2√35
h/IACI = sinα = 1/3
h = IACI * 1/3
IACI = h : 1/3 = 2√35 : 1/3 = 2√35 * 3 = 6√35 ≈ 6 * 5,9160797.. ≈ 35,496478... ≈ 35,49648