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Chapter 6: Problem 31

Find the inclination \(\theta\) (in radians and degrees) of the line. \(x+\sqrt{3} y+2=0\)

Short Answer

Expert verified
The inclination \(\theta\) (in radians and degrees) of the line \(x+\sqrt{3} y+2=0\) are \(\pi/6\) radians and 30 degrees respectively.

Step by step solution

01

Identify Variables from Equation

In the given equation, \(x+\sqrt{3} y+2=0\), the coefficient of \(x\) is \(a=1\), and the coefficient of \(y\) is \(b=\sqrt{3}\).
02

Calculate the Slope

The slope \(m\) of the line is the negative ratio of the coefficients of \(x\) and \(y\) in the given equation, which is \(-a/b\). So, \(m = -a/b = -1/\sqrt{3}\).
03

Calculate the Inclination in Radians

The inclination \(\theta\) of the line can be obtained by finding the arc-tangent of the slope. Therefore, \(\theta = atan(m)\), which results as \(\pi/6\) radians after substituting \(m = -1/\sqrt{3}\).
04

Convert the Inclination to Degrees

To convert the inclination from radians to degrees, multiply by 180 and divide by \(\pi\). Therefore, \(\theta = \(\pi/6 * 180/\pi = 30\) degrees.

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