91Ó°ÊÓ

Quadrants in a plane Cartesian coordinate system are generally numbered from 1 to 4, typically, being the central angles measured in standard position (i.e., counterclockwise from the positive x-axis, negative angles go clockwise). So:\[\begin{align*}1^{\text{st}} \text{Quadrant} & : 0 < θ < \frac{\pi}{2} \2^{\text{nd}} \text{Quadrant} &: \frac{\pi}{2} < θ < \pi \3^{\text{rd}} \text{Quadrant} &: \pi < θ < \frac{3\pi}{2} \4^{\text{th}} \text{Quadrant} &: \frac{3\pi}{2} < θ < 2\pi \\end{align*}\]We can rewrite this for negative angles:\[\begin{align*}4^{\text{th}} \text{Quadrant} & : 0 > θ > -\frac{\pi}{2} \3^{\text{rd}} \text{Quadrant} &: -\frac{\pi}{2} > θ > -\pi \2^{\text{nd}} \text{Quadrant} &: -\pi > θ > -\frac{3\pi}{2} \1^{\text{st}} \text{Quadrant} &: -\frac{3\pi}{2} > θ > -2\pi \\end{align*}\]

As specified, -\(\frac{\pi}{6}\) which falls in the range 0 > θ > -\(\frac{\pi}{2}\), belongs to the 4th Quadrant.

Since -\(\frac{11\pi}{9}\) it's greater than -2\(\pi\) and less than -\(\pi\), we will find the positive coterminal angle by adding 2\(\pi\) until the value is between 0 and -2\(\pi\). The positive coterminal angle is -\(\frac{11\pi}{9}\) + 2\(\pi\) = \(\frac{7\pi}{9}\), which belongs to the 1st Quadrant.