• Matéria: Matemática
  • Autor: isaque4667
  • Perguntado 7 anos atrás

ajuda nessa ajudas

Anexos:

Respostas

respondido por: gryffindor05
1

(a)

cos(30°) =  \dfrac{16}{x} =  >  \dfrac{ \sqrt{3} }{2}  = \dfrac{16}{x} =  >  \sqrt{3} x = 32  \\  =  > x = \dfrac{32}{ \sqrt{3} } =  > x = \dfrac{32}{ \sqrt{3} } \cdot\dfrac{ \sqrt{3} }{ \sqrt{3} } =  > x = \dfrac{32 \sqrt{3} }{3}

(b)

sen(y) =  \dfrac{13}{26}  =  > y =  arcsen \begin{pmatrix}\dfrac{13}{26} \end{pmatrix} = > y = arcsen\begin{pmatrix}\dfrac{1}{2} \end{pmatrix} = > y = 30°

(c)

sen(60°) =  \dfrac{w}{18}  =  > \dfrac{ \sqrt{3} }{2} = \dfrac{w}{18} =  > 2w = 18 \sqrt{3} \\  =  > w = \dfrac{18 \sqrt{3} }{2} =  > w = 9 \sqrt{3}

(d)

cos(45°) =  \dfrac{20}{z}  =  >  \dfrac{ \sqrt{2} }{2}  = \dfrac{20}{z} =  >  \sqrt{2} z = 40 \\  =  > z = \dfrac{40}{ \sqrt{2} }  =  > z = \dfrac{40}{ \sqrt{2} } \cdot\dfrac{ \sqrt{2} }{ \sqrt{2} } =  > z = \dfrac{40 \sqrt{2} }{2} \\  =   >  z = 20 \sqrt{2}


isaque4667: valeu
gryffindor05: arrumei a letra b
isaque4667: fecho e nozes
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