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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 $-\sqrt{10} x+y=0, x \leq 0$
Which of the graphs shows $-\sqrt{10} x+y=0$, where $x \leq 0$ ?

Step 1 ：Given the equation $-\sqrt{10} x+y=0$, where $x \leq 0$, we can rewrite it as $y=\sqrt{10}x$. This is a line passing through the origin with a positive slope. Since $x \leq 0$, the line is in the second quadrant.

Step 2 ：The slope of the line is $\sqrt{10}$, which is the tangent of the angle $\theta$. So, we can use the arctangent function to find the angle. The least positive angle $\theta$ is approximately $1.26$ radians.

Step 3 ：The six trigonometric functions of $\theta$ can be found using the right triangle formed by the line, the x-axis, and the line perpendicular to the x-axis passing through the point where the line intersects the unit circle. The x-coordinate of this point is the cosine of the angle, the y-coordinate is the sine of the angle. The other trigonometric functions can be found using these two values.

Step 4 ：\(\sin(\theta) \approx 0.95\), \(\cos(\theta) \approx 0.30\), \(\tan(\theta) \approx 3.16\), \(\csc(\theta) \approx 1.05\), \(\sec(\theta) \approx 3.32\), \(\cot(\theta) \approx 0.32\)

Step 5 ：\(\boxed{\text{The least positive angle } \theta \text{ is approximately } 1.26 \text{ radians. The six trigonometric functions of } \theta \text{ are approximately as follows:}}\)

Step 6 ：\(\boxed{\sin(\theta) \approx 0.95, \cos(\theta) \approx 0.30, \tan(\theta) \approx 3.16, \csc(\theta) \approx 1.05, \sec(\theta) \approx 3.32, \cot(\theta) \approx 0.32}\)