Sketch the least positive angle $\theta$ and find the values of the six trigonometric functions of $\theta$ if the terminal side of an angle $\theta$ in standard position is defined by $-\sqrt{2} x+y=0, x \leq 0$.

Step 1 ：The given equation of the line is \(-\sqrt{2} x+y=0\). This line passes through the origin and has a slope of \(\sqrt{2}\). Since \(x \leq 0\), the terminal side of the angle is in the second quadrant.

Step 2 ：The least positive angle \(\theta\) is the angle between the positive x-axis and the terminal side of the angle. In the second quadrant, the angle \(\theta\) is \(180^\circ - \alpha\), where \(\alpha\) is the angle between the negative x-axis and the terminal side of the angle.

Step 3 ：The slope of the line is \(\tan(\alpha) = \sqrt{2}\). So, \(\alpha = \tan^{-1}(\sqrt{2})\).

Step 4 ：The six trigonometric functions of \(\theta\) are \(\sin(\theta)\), \(\cos(\theta)\), \(\tan(\theta)\), \(\csc(\theta)\), \(\sec(\theta)\), and \(\cot(\theta)\).

Step 5 ：We can find these values using the identities \(\sin(\theta) = \sin(180^\circ - \alpha)\), \(\cos(\theta) = -\cos(\alpha)\), \(\tan(\theta) = -\tan(\alpha)\), \(\csc(\theta) = \csc(180^\circ - \alpha)\), \(\sec(\theta) = -\sec(\alpha)\), and \(\cot(\theta) = -\cot(\alpha)\).

Step 6 ：\(\alpha = 0.955\) radians, \(\theta = 2.186\) radians, \(\sin(\theta) = 0.816\), \(\cos(\theta) = -0.577\), \(\tan(\theta) = -1.414\), \(\csc(\theta) = 1.225\), \(\sec(\theta) = -1.732\), and \(\cot(\theta) = -0.707\).

Step 7 ：\(\boxed{\text{Final Answer: The least positive angle } \theta \text{ is } 2.186 \text{ radians. The values of the six trigonometric functions of } \theta \text{ are } \sin(\theta) = 0.816, \cos(\theta) = -0.577, \tan(\theta) = -1.414, \csc(\theta) = 1.225, \sec(\theta) = -1.732, \text{ and } \cot(\theta) = -0.707.}\)