In this problem
we have that
sin(theta) is positive and cos(theta) is negative
That means
the angle theta lies on the II quadrant
Remember that
[tex]\cot (\theta)=\frac{\cos(\theta)}{\sin(\theta)}[/tex]Find out the value of cos(theta)
[tex]\sin ^2(\theta)+\cos ^2(\theta)=1[/tex]substitute the given value
[tex](\frac{\sqrt[]{48}}{8})^2+\cos ^2(\theta)=1[/tex][tex]\cos ^2(\theta)=1-\frac{48}{64}[/tex][tex]\begin{gathered} \cos ^2(\theta)=\frac{16}{64} \\ \cos ^{}(\theta)=-\frac{4}{8} \end{gathered}[/tex]Find out the value of cot(theta)
substitute given values
[tex]\cot (\theta)=-\frac{4}{\sqrt[\square]{48}}[/tex]simplify
[tex]\cot (\theta)=-\frac{4}{\sqrt[\square]{48}}\cdot\frac{\sqrt[]{48}}{\sqrt[]{48}}=-\frac{4\sqrt[]{48}}{48}=-\frac{\sqrt[]{48}}{12}=-\frac{4\sqrt[]{3}}{12}=-\frac{\sqrt[]{3}}{3}[/tex]Find out the angle theta
using a calculator
angle in II quadrant