✅ Funções circulares:
Se o ângulo "α" é igual :
[tex]\large\displaystyle\text{$\begin{gathered}\alpha = 30^o \end{gathered}$}[/tex]
E, sabendo que:
[tex]\large\displaystyle\text{$\begin{gathered}\overline{BD} = tg\ \alpha \end{gathered}$}[/tex]
[tex]\large\displaystyle\text{$\begin{gathered}\overline{OC} = cos\ \alpha \end{gathered}$}[/tex]
Então, temos:
[tex]\large\displaystyle\text{$\begin{gathered}\frac{\overline{BD}}{\overline{OC}} = \frac{tg\ \alpha}{cos\ \alpha} \end{gathered}$}[/tex]
[tex]\large\displaystyle\text{$\begin{gathered}= \frac{\frac{sen\ \alpha}{cos\ \alpha} }{cos\ \alpha} \end{gathered}$}[/tex]
[tex]\large\displaystyle\text{$\begin{gathered}= \frac{sen\ \alpha}{cos\ \alpha} . \frac{1}{cos\ \alpha} \end{gathered}$}[/tex]
[tex]\large\displaystyle\text{$\begin{gathered}= \frac{sen\ \alpha}{cos^{2}\alpha } \end{gathered}$}[/tex]
Portanto:
[tex]\large\displaystyle\text{$\begin{gathered}\frac{\overline{BD}}{\overline{OC}} = \frac{sen\ \alpha}{cos^{2}\alpha} \end{gathered}$}[/tex]
Se:
[tex]\large\displaystyle\text{$\begin{gathered}\alpha = 30^o \end{gathered}$}[/tex]
Então:
[tex]\large\displaystyle\text{$\begin{gathered}sen\ 30^o = \frac{1}{2} \end{gathered}$}[/tex]
[tex]\large\displaystyle\text{$\begin{gathered}cos\ 30^o = \frac{\sqrt{3} }{2} \end{gathered}$}[/tex]
Então:
[tex]\large\displaystyle\text{$\begin{gathered}\frac{\overline{BD}}{\overline{OC}} = \frac{\frac{1}{2} }{(\frac{\sqrt{3} }{2})^{2}} \end{gathered}$}[/tex]
[tex]\large\displaystyle\text{$\begin{gathered}= \frac{\frac{1}{2} }{\frac{(\sqrt{3} )^{2} }{2^{2} } } \end{gathered}$}[/tex]
[tex]\large\displaystyle\text{$\begin{gathered}= \frac{\frac{1}{2} }{\frac{3}{4} } \end{gathered}$}[/tex]
[tex]\large\displaystyle\text{$\begin{gathered}= \frac{1}{2} . \frac{4}{3} \end{gathered}$}[/tex]
[tex]\large\displaystyle\text{$\begin{gathered}= \frac{4}{6} \end{gathered}$}[/tex]
[tex]\large\displaystyle\text{$\begin{gathered}= \frac{2}{3} \end{gathered}$}[/tex]
✅ Portanto, o resultado da questão é:
[tex]\large\displaystyle\text{$\begin{gathered}\frac{\overline{BD}}{\overline{OC}} = \frac{2}{3} \end{gathered}$}[/tex]
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