Sketch an angle $\theta$ in standard position such that $\theta$ has the least possible positive measure, and the point $(-1, \sqrt{3})$ is on the terminal side of $\theta$. Then find the values of the six trigonometric functions for the angle. applicable. Do not use a calculator.
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Step 1 ：Sketch an angle \(\theta\) in standard position such that \(\theta\) has the least possible positive measure, and the point \((-1, \sqrt{3})\) is on the terminal side of \(\theta\).

Step 2 ：The point \((-1, \sqrt{3})\) lies in the second quadrant. In standard position, an angle's terminal side is determined by rotating counter-clockwise from the positive x-axis. The least possible positive measure for an angle with its terminal side in the second quadrant is \(120^\circ\) or \(\frac{2\pi}{3}\) radians.

Step 3 ：To find the six trigonometric functions for the angle, we can use the definitions of the functions in terms of the coordinates of the point on the terminal side of the angle. The radius (r) can be found using the Pythagorean theorem, \(r = \sqrt{x^2 + y^2} = 2\).

Step 4 ：Using the coordinates of the point \((-1, \sqrt{3})\) and the radius \(r = 2\), we can calculate the six trigonometric functions as follows: \(\sin(\theta) = \frac{\sqrt{3}}{2}\), \(\cos(\theta) = -\frac{1}{2}\), \(\tan(\theta) = -\sqrt{3}\), \(\csc(\theta) = \frac{2}{\sqrt{3}}\), \(\sec(\theta) = -2\), \(\cot(\theta) = -\frac{\sqrt{3}}{3}\).

Step 5 ：\(\boxed{\text{Final Answer: The angle } \theta \text{ in standard position that has the least possible positive measure and the point } (-1, \sqrt{3}) \text{ on its terminal side is } 120^\circ \text{ or } \frac{2\pi}{3} \text{ radians. The six trigonometric functions for this angle are: } \sin(\theta) = \frac{\sqrt{3}}{2}, \cos(\theta) = -\frac{1}{2}, \tan(\theta) = -\sqrt{3}, \csc(\theta) = \frac{2}{\sqrt{3}}, \sec(\theta) = -2, \cot(\theta) = -\frac{\sqrt{3}}{3}}\)