✅ Após resolver os cálculos, concluímos que o coeficiente angular da reta cuja inclinação é 45° é:
[tex]\Large\displaystyle\text{$\begin{gathered}\boxed{\boxed{\:\:\:\bf m_{r} = -\sqrt{3}\:\:\:}}\end{gathered}$}[/tex]
Seja o ângulo de inclinação:
[tex]\Large\displaystyle\text{$\begin{gathered} \theta = 120^{\circ}\end{gathered}$}[/tex]
Sabemos que o coeficiente angular da uma reta no plano cartesiano é numericamente igual à tangente do referido ângulo, isto é:
[tex]\Large\displaystyle\text{$\begin{gathered} \bf I\end{gathered}$}[/tex] [tex]\Large\displaystyle\text{$\begin{gathered} m_{r} = \tan\theta\end{gathered}$}[/tex]
Substituindo o valor do ângulo na equação "I", temos:
[tex]\Large\displaystyle\text{$\begin{gathered} m_{r} = \tan 120^{\circ}\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} = \frac{\sin 120^{\circ}}{\cos 120^{\circ}}\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} = \frac{\dfrac{\sqrt{3}}{2}}{-\dfrac{1}{2}}\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} = \frac{\sqrt{3}}{\!\diagup\!\!\!\!2}\cdot\bigg(-\frac{\!\diagup\!\!\!\!2}{1}\bigg)\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} = \frac{-\sqrt{3}}{1}\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} = -\sqrt{3}\end{gathered}$}[/tex]
✅ Portanto, o coeficiente angular é:
[tex]\Large\displaystyle\text{$\begin{gathered} m_{r} = -\sqrt{3}\end{gathered}$}[/tex]
[tex]\LARGE\displaystyle\text{$\begin{gathered} \underline{\boxed{\boldsymbol{\:\:\:Bons \:estudos!!\:\:\:Boa\: sorte!!\:\:\:}}}\end{gathered}$}[/tex]
Saiba mais:
