Question

 Calcule o ângulo entre as retas r : 4x − y + 4 = 0 e s : 4x − 3y + 8 = 0

Answer 1

A tangente trigonométrica do ângulo formado por duas retas é dada por;

$tg \alpha=\| \frac{m_s-m_r}{1+m_s.m_r} \|$

$m_s=\frac{-4}{-3}=\frac{4}{3} \\ \\ m_r=\frac{1}{4} \\ \\ tg \alpha= \|\frac{\frac{4}{3}-\frac{1}{4}}{1+\frac{4}{3}.\frac{1}{4}} \| \\ \\ tg \alpha=\|\frac{\frac{13}{12}}{\frac{4}{3}} \| \\ \\ tg \alpha=\frac{13}{12}.\frac{3}{4}=\frac{13}{9} \approx 1,44 \\ \\ \alpha \approx 55 ^o$

Answer 2

✅ Após resolver os cálculos, concluímos que o ângulo entre as referidas retas é:

         $\Large\displaystyle\text{ \begin{gathered}\boxed{\boxed{\:\:\:\bf \theta \cong 22,83^{\circ}\:\:\:}}\end{gathered} }$

Sejam as retas:

     $\Large\begin{cases} r: 4x - y + 4 = 0\\s: 4x - 3y + 8 = 0\end{cases}$

Quando estamos procurando a medida do ângulo entre as retas, na realidade estamos procurando o menor ângulo entre as retas e, para isso, devemos utilizar a seguinte fórmula:

$\Large\displaystyle\begin{gathered} \bf(I)\end{gathered}$      $\Large\displaystyle\text{ \begin{gathered} \theta = \arctan \bigg(\bigg|\frac{m_{r} - m_{s}}{1 + m_{r}\cdot m_{s}}\bigg|\bigg), \:\:\:\textrm{com}\:0 < \theta < 90^{\circ}\end{gathered} }$

Para resolver esta questão devemos:

• Calcular o coeficiente angular da reta "r".

               $\Large\displaystyle\begin{gathered} -y = -4x - 4\end{gathered}$    

                  $\Large\displaystyle\begin{gathered} y = 4x + 4\end{gathered}$

                 $\Large\displaystyle\begin{gathered} \therefore\:\:\: m_{r} = 4\end{gathered}$

• Calcular o coeficiente angular da reta s.

               $\Large\displaystyle\begin{gathered} -3y = -4x - 8\end{gathered}$

                   $\Large\displaystyle\begin{gathered} 3y = 4x + 8\end{gathered}$

                     $\Large\displaystyle\begin{gathered} y = \frac{4}{3}x + \frac{8}{3}\end{gathered}$

                     $\Large\displaystyle\begin{gathered} \therefore\:\:m_{s} = \frac{4}{3}\end{gathered}$

• Substituindo os valores dos coeficientes angulares na equação "I", temos:

                $\Large\displaystyle\text{ \begin{gathered} \theta = \arctan \bigg(\bigg|\frac{4 -\frac{4}{3}}{1 + 4\cdot\frac{4}{3}}\bigg|\bigg)\end{gathered} }$

                    $\Large\displaystyle\text{ \begin{gathered} = \arctan \bigg(\bigg|\frac{\frac{8}{3}}{1 + \frac{16}{3}}\bigg|\bigg)\end{gathered} }$

                    $\Large\displaystyle\text{ \begin{gathered} = \arctan \bigg(\bigg|\frac{\frac{8}{3}}{\frac{19}{3}}\bigg|\bigg)\end{gathered} }$

                    $\Large\displaystyle\begin{gathered} = \arctan \bigg(\bigg|\frac{8}{19}\bigg|\bigg)\end{gathered}$

                    $\Large\displaystyle\begin{gathered} = \arctan \bigg(\frac{8}{19}\bigg)\end{gathered}$

✅Portanto, o ângulo procurado é:

        $\Large\displaystyle\begin{gathered} \theta = \arctan (0,4210526316) \Longrightarrow \theta \cong 22,83^{\circ}\end{gathered}$

Saiba mais:

1. https://brainly.com.br/tarefa/52000661
2. https://brainly.com.br/tarefa/227145

Veja a solução gráfica da questão representada na figura: