Pré-Vestibular ⇒ (LINS - 1966) Trigonometria: Arco Duplo Tópico resolvido

[ALDRIN]

Tendo em vista as relações descritas na figura abaixo, calcular as distâncias [tex3]x\text{ e } y.[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]x\text{ e } y.[/tex3]

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  AF63.png (9.09 KiB) Exibido 3036 vezes

Resposta:

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[tex3]x=40\text{ m e } y=90 \text{ m}.[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]x=40\text{ m e } y=90 \text{ m}.[/tex3]

[adrianotavares]

Olá,Aldrin.

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  AF62.png (11.69 KiB) Exibido 3038 vezes

• [tex3]\left|\begin{array}{l}\triangle NCM:\text{ } \overline{MN}^2=\overline{MC}^2 +\overline{NC}^2\\ \triangle NPM:\text{ }\overline{MN}^2= 120^2 +(y-x)^2\\\triangle MAC:\text{ } \overline{MC}^2= 60^2 +x^2\\\triangle NBC:\text{ } \overline{NC}^2= 60^2 +y^2 \\
  \end{array}\right. \Longrightarrow xy=3600.[/tex3]

  Código: Selecionar tudoSHIFT+Click na eq = ZOOM

  [tex3]\left|\begin{array}{l}\triangle NCM:\text{ } \overline{MN}^2=\overline{MC}^2 +\overline{NC}^2\\ \triangle NPM:\text{ }\overline{MN}^2= 120^2 +(y-x)^2\\\triangle MAC:\text{ } \overline{MC}^2= 60^2 +x^2\\\triangle NBC:\text{ } \overline{NC}^2= 60^2 +y^2 \\
\end{array}\right. \Longrightarrow xy=3600.[/tex3]

• [tex3]\left|\begin{array}{l} \triangle MAB:\text{ } \text{tg} \alpha=\frac{x}{120}\\ \triangle NBA:\text{ } \text{tg}2 \alpha=\frac{y}{120} \\\text{tg} 2 \alpha = \frac{2\text{tg} \alpha}{1-\text{tg}^2 \alpha \end{array}\right. \Longrightarrow \left|\begin{array}{rl} \frac{\frac{2x}{120}}{1 - \frac{x^2}{120^2}}= \frac{y}{120}&\Longrightarrow \frac{2x}{120} \cdot \frac{120^2}{120^2-x^2}= \frac{y}{120}\\
  &\Longrightarrow 2\cdot 120^2\cdot x=(120^2-x^2)\cdot y\\
  &\Longrightarrow 2\cdot 120^2\cdot x^2=(120^2-x^2)\cdot xy\\
  &\Longrightarrow 2\cdot 120^2\cdot x^2=(120^2-x^2)\cdot 3600\\
  &\Longrightarrow 8x^2=120^2-x^2\\
  &\Longrightarrow 9x^2=120^2\\
  &\Longrightarrow x^2=\left(\frac{120}{3}\right)^2\\
  &\Longrightarrow x=40\text{ m}.\\
  &\Longrightarrow y=90\text{ m}.
  \end{array}\right.[/tex3]

  Código: Selecionar tudoSHIFT+Click na eq = ZOOM

  [tex3]\left|\begin{array}{l} \triangle MAB:\text{ } \text{tg} \alpha=\frac{x}{120}\\ \triangle NBA:\text{ } \text{tg}2 \alpha=\frac{y}{120} \\\text{tg} 2 \alpha = \frac{2\text{tg} \alpha}{1-\text{tg}^2 \alpha \end{array}\right. \Longrightarrow \left|\begin{array}{rl} \frac{\frac{2x}{120}}{1 - \frac{x^2}{120^2}}= \frac{y}{120}&\Longrightarrow \frac{2x}{120} \cdot \frac{120^2}{120^2-x^2}= \frac{y}{120}\\
&\Longrightarrow 2\cdot 120^2\cdot x=(120^2-x^2)\cdot y\\
&\Longrightarrow 2\cdot 120^2\cdot x^2=(120^2-x^2)\cdot xy\\
&\Longrightarrow 2\cdot 120^2\cdot x^2=(120^2-x^2)\cdot 3600\\
&\Longrightarrow 8x^2=120^2-x^2\\
&\Longrightarrow 9x^2=120^2\\
&\Longrightarrow x^2=\left(\frac{120}{3}\right)^2\\
&\Longrightarrow x=40\text{ m}.\\
&\Longrightarrow y=90\text{ m}.
\end{array}\right.[/tex3]