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BRU2705
@BRU2705
December 2019
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Seja a um ângulo agudo. Se sen a= 2 sobre 3, calcule cos a e tg a.
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lucasblack
Sen *2 a+ cos *2 a =1
(2/3)*2+cos*2 a = 1
4/9+cos*2 a=1
cos*2 a= 1-4/9
cos a= √( 5/9)
cos a= √5/3
tg a= sen a / cos a
tg a= (2/3)/(√5/3)
tg a = 6/3√5
tg a = 6√5/(3√5•√5)
tg a= 6√5/15
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Lista de comentários
(2/3)*2+cos*2 a = 1
4/9+cos*2 a=1
cos*2 a= 1-4/9
cos a= √( 5/9)
cos a= √5/3
tg a= sen a / cos a
tg a= (2/3)/(√5/3)
tg a = 6/3√5
tg a = 6√5/(3√5•√5)
tg a= 6√5/15