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Chapter 6: Problem 28

Find the inclination \(\theta\) (in radians and degrees) of the line. \(x-\sqrt{3} y+1=0\)

Short Answer

Expert verified
The inclination of the line \(x-\sqrt{3} y+1=0\) is \(\frac{\pi}{6}\) radians or 30 degrees.

Step by step solution

01

Convert the given equation into slope-intercept form

The given equation is \(x - \sqrt{3} y + 1 = 0\). Rearrange this to obtain the form \(y = mx + c\). So, \(\sqrt{3} y = x - 1 \). Divide the equation by \(\sqrt{3}\) to get \(y = \frac{1}{\sqrt{3}}x - \frac{1}{\sqrt{3}} \)
02

Calculate the Slope

From the slope-intercept form of the equation, \(m = \frac{1}{\sqrt{3}}\) is the slope of the line.
03

Inclination in Radians

Using the formula \(\theta = \text{atan}(m)\), substitute the value of \(m\) into the formula to get the inclination in radians: \(\theta = \text{atan}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}\) radians.
04

Inclination in Degrees

Convert the radian measurement to degrees by using the formula: degrees = radians * \(\frac{180}{\pi}\): \(\theta = \frac{\pi}{6} * \frac{180}{\pi} = 30\) degrees.

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