Given that the terminal side of an <θ intersects the unit circle at the point
[tex]P(\frac{5}{6},\frac{-\sqrt[]{11}}{6})[/tex]
From the given point P:
[tex]\begin{gathered} x=\frac{5}{6} \\ y=\frac{-\sqrt[]{11}}{6} \\ \text{ s}ince,\text{ x is positive and y is negative, the angle lies in the 4th quadrant} \end{gathered}[/tex][tex]\begin{gathered} \tan \theta=\frac{\text{ opposite}}{\text{Adjacent}}=\frac{y}{x} \\ \tan \theta=\frac{\frac{-\sqrt[]{11}}{6}}{\frac{5}{6}}=\frac{-\sqrt[]{11}}{5} \\ \tan \theta=-0.6633 \\ \theta=326.44^0 \end{gathered}[/tex]
