Wiedząc,że α jest kątem ostrym i sin α=√3/3 , oblicz tg α .
sin²α + cos²α = 1
cos²α = 1 - sin²α
cos α = √(1 - sin²α)
tg α = sin α / cos α
tg α = sin α / √(1 - sin²α) = ( √3/3 ) / √(1 - 3/9) = ( √3/3 ) / √(6/9) = ( √3/3 ) / √(2/3) = ( √3/3 ) / (√2/√3) = 1 / √2
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sin²α + cos²α = 1
cos²α = 1 - sin²α
cos α = √(1 - sin²α)
tg α = sin α / cos α
tg α = sin α / √(1 - sin²α) = ( √3/3 ) / √(1 - 3/9) = ( √3/3 ) / √(6/9) = ( √3/3 ) / √(2/3) = ( √3/3 ) / (√2/√3) = 1 / √2