To determine the inclination of a line given its slope, we first need to understand the relationship between the slope and the angle of inclination. The slope [tex]\( m \)[/tex] of a line is given by the tangent of the angle [tex]\( \theta \)[/tex] that the line makes with the positive direction of the x-axis. This relationship can be expressed as:
[tex]\[ m = \tan(\theta) \][/tex]
Given that the slope [tex]\( m \)[/tex] is [tex]\( \frac{-1}{\sqrt{3}} \)[/tex], we need to find the angle [tex]\( \theta \)[/tex] such that:
[tex]\[ \tan(\theta) = \frac{-1}{\sqrt{3}} \][/tex]
The tangent of the angle [tex]\( \theta \)[/tex] yields a negative value, indicating that [tex]\( \theta \)[/tex] lies in either the second or the fourth quadrant, as the tangent function is negative in these quadrants.
To find the exact angle, we can use the arctangent function:
[tex]\[ \theta = \arctan\left(\frac{-1}{\sqrt{3}}\right) \][/tex]
This computation gives us the angle in radians. Based on typical trigonometric values, the reference angle where [tex]\(\tan\left(\frac{-1}{\sqrt{3}}\right)\)[/tex] is:
[tex]\[ \theta = -\frac{\pi}{6} \][/tex]
Since [tex]\(-\frac{\pi}{6}\)[/tex] is in the fourth quadrant, we need to convert this to positive radians by adding [tex]\(\pi\)[/tex] (to shift it to the second quadrant, where the line's inclination would be):
[tex]\[ \theta = \pi - \left(-\frac{\pi}{6}\right) = \pi + \frac{\pi}{6} = \frac{6\pi}{6} + \frac{\pi}{6} = \frac{7\pi}{6} \][/tex]
Therefore, the actual angle in the second quadrant remaining between [tex]\(0\)[/tex] and [tex]\(2\pi\)[/tex] would be:
[tex]\[ \theta = \frac{5\pi}{6} \][/tex]
As a final check, converting [tex]\(\frac{5\pi}{6}\)[/tex] radians to degrees:
[tex]\[ \frac{5\pi}{6} \times \frac{180^\circ}{\pi} = 5 \times 30^\circ = 150^\circ \][/tex]
This confirms that the inclination angle is indeed [tex]\( \frac{5\pi}{6} \)[/tex] radians, which is the correct answer.
Thus, the correct choice is B) [tex]\( \frac{5 \pi}{6} \)[/tex].
