найдите tgA если a)cosA=5/13 б)cosA=0,6 в)sinA=3/5 г)sinA=0,8. Created by 123451418. geometriya-ru

5 г)sinA=0,8...

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найдите tgA если a)cosA=5/13 б)cosA=0,6 в)sinA=3/5 г)sinA=0,8

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dnepr1

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tgA если:
a)cosA=5/13, sinA = √(1-cos²A) = √(1-(25/169)) = √(144/169) = 12/13.
  tgA = sinA/cosA = (12/13)*(13/5) = 12/5 = 2,4.
б)cosA=0,6, sinA = √(1-cos²A) = √(1-(36/100)) = √(64/100) = 8/10.
  tgA = sinA/cosA = (8/10)*(10/6) = 8/6 = 4/3 ≈ 1,333.
в)sinA=3/5, cosA =  √(1-sin²A) = √(1-(9/25)) = √(16/25) = 4/5.
  tgA = sinA/cosA = (3/5)*(5/4) = 3/4 = 0,75.
г)sinA=0,8, cosA =  √(1-sin²A) = √(1-(64/100)) = √(36/100) = 6/10.
  tgA = sinA/cosA = (8/10)*(10/6) = 8/6 = 4/3 ≈ 1,333.

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