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Chapter 6: Problem 39

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

Short Answer

Expert verified
The inclination of the given line is \(-\frac{\pi}{6}\) radians or \(-30^{\circ}\).

Step by step solution

01

Express the line equation in slope-intercept form

The line equation can be rewritten as \(y = -\frac{1}{\sqrt{3}}x - \frac{2}{\sqrt{3}}\). Therefore, the gradient \(m\) of this line is \(-\frac{1}{\sqrt{3}}\).
02

Find the inclination in radians

Using the formula \(tan(\theta) = m\), solving for \(\theta\) gives \(\theta = atan(-\frac{1}{\sqrt{3}})\). Therefore, \(\theta = -\frac{\pi}{6}\)
03

Express the inclination in degrees

The inclination in degrees can be found by converting the radians value \(-\frac{\pi}{6}\) to degrees using the formula \(\theta_{deg} = -\frac{\pi}{6} \times \frac{180}{\pi} = -30^{\circ}\).

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Most popular questions from this chapter

Find the distance between the point and the line. Point \((-2,6)\) Line \(y=-x+5\)

In Exercises \(129-132,\) determine whether the statement is true or false. Justify your answer. If \(\left(r_{1}, \theta_{1}\right)\) and \(\left(r_{2}, \theta_{2}\right)\) represent the same point in the polar coordinate system, then \(\left|r_{1}\right|=\left|r_{2}\right|\).

Determine whether the statement is true or false. Justify your answer. If the vertex and focus of a parabola are on a horizontal line, then the directrix of the parabola is vertical.

(a) Show that the distance between the points \(\left(r_{1}, \theta_{1}\right)\) and \(\left(r_{2}, \theta_{2}\right)\) is \(\sqrt{r_{1}^{2}+r_{2}^{2}-2 r_{1} r_{2} \cos \left(\theta_{1}-\theta_{2}\right)}\) (b) Simplify the Distance Formula for \(\theta_{1}=\theta_{2} .\) Is the simplification what you expected? Explain. (c) Simplify the Distance Formula for \(\theta_{1}-\theta_{2}=90^{\circ}\) Is the simplification what you expected? Explain.

Find the distance between the parallel lines. (Graph can't copy) $$\begin{aligned} &x+y=1\\\ &x+y=5 \end{aligned}$$
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