• Matéria: Matemática
  • Autor: Washington2004
  • Perguntado 7 anos atrás

Calcule o valor de x em cada triângulo?

Anexos:

Respostas

respondido por: PauloRicardo86
5

a) \text{sen}~50^{\circ}=\dfrac{2\sqrt{3}}{x}


\dfrac{\sqrt{3}}{2}=\dfrac{2\sqrt{3}}{x}


\sqrt{3}\cdot x=2\cdot2\sqrt{3}


x=\dfrac{4\sqrt{3}}{\sqrt{3}}


x=4


b) \text{sen}~60^{\circ}=\dfrac{x}{2\sqrt{3}}


\dfrac{\sqrt{3}}{2}=\dfrac{x}{2\sqrt{3}}


2x=2\sqrt{3}\cdot\sqrt{3}


2x=6


x=3


c) \text{cos}~30^{\circ}=\dfrac{3\sqrt{3}}{x}


\dfrac{\sqrt{3}}{2}=\dfrac{3\sqrt{3}}{x}


\sqrt{3}\cdot x=2\cdot3\sqrt{3}


\sqrt{3}\cdot x=6\sqrt{3}


x=\dfrac{6\sqrt{3}}{\sqrt{3}}


x=6


d) \text{cos}~45^{\circ}=\dfrac{x}{2\sqrt{2}}


\dfrac{\sqrt{2}}{2}=\dfrac{x}{2\sqrt{2}}


2x=\sqrt{2}\cdot2\sqrt{2}


2x=4


x=2


e) \text{tg}~30^{\circ}=\dfrac{x}{2,5}


\dfrac{\sqrt{3}}{3}=\dfrac{x}{2,5}


3x=2,5\sqrt{3}


x=\dfrac{2,5\sqrt{3}}{3}


f) \text{tg}~45^{\circ}=\dfrac{3,8}{x}


1=\dfrac{3,8}{x}


x=3,8

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