# How do I find the inclination θ (in radians AND degrees) of the lines passing through the points? (-√3, -1), (0, -2)

Feb 17, 2018

$\setminus q \quad \setminus \quad \setminus \setminus \theta \setminus = \setminus \setminus {150}^{\circ} \setminus = \setminus \frac{5 \setminus \pi}{6} \setminus q \quad \text{is the inclination of the given line.}$

#### Explanation:

$\text{Recall the definition of the inclination of a line L:}$

$\text{If" \ \ m \ \ "is the slope of line" \ \ L \ \ "and" \ \ \theta \ \ "is the inclination of line" \ \ L, "then:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \tan \left(\setminus \theta\right) \setminus = \setminus m , \setminus q \quad \setminus q \quad \text{and} \setminus q \quad \setminus \quad \setminus \theta \setminus \in \left[0 , \setminus \pi\right) . \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus \setminus \left(1\right)$

$\text{So, to find the inclination of a line, we can start by finding its}$
$\text{slope, and then use eqn. (1).}$

$\text{The line we are given is the line through the two points:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \left(- \sqrt{3} , - 1\right) \setminus q \quad \setminus q \quad \text{and} \setminus q \quad \setminus q \quad \left(0 , - 2\right) .$

$\text{The slope of this line is:}$

$\setminus q \quad m \setminus = \setminus \frac{- 1 - \left(- 2\right)}{- \sqrt{3} - 0} \setminus = \setminus \frac{- 1 + 2}{- \sqrt{3}} \setminus = \setminus \frac{1}{- \sqrt{3}} \setminus = \setminus - \frac{1}{\sqrt{3}} .$

$\text{The slope of this line is:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad m \setminus = \setminus - \frac{1}{\sqrt{3}} .$

$\text{So, by eqn. (1), for the inclination" \ \ \theta \ \ "of this line:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus \tan \left(\setminus \theta\right) \setminus = \setminus - \frac{1}{\sqrt{3}} , \setminus q \quad \setminus q \quad \setminus \theta \setminus \in \left[0 , \setminus \pi\right) .$

$\text{As" \ \ tan(\theta) \ \ "is negative, and" \quad \theta \in [ 0, \pi ) , "we see that}$
$\setminus \setminus \theta \setminus \in \text{Quadrant II.}$

$\text{So, we have:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus \tan \left(\setminus \theta\right) \setminus = \setminus - \frac{1}{\sqrt{3}} , \setminus q \quad \setminus q \quad \setminus \theta \setminus \in \text{Quadrant II.} \setminus q \quad \setminus \quad \setminus \quad \setminus \setminus \left(2\right)$

$\text{Recognizing the connection between the value above, and}$
$\text{the 30-60-90 right triangle, and remembering the 30-60-90}$
$\text{right triangle, we recall:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus \quad \setminus \tan \left({30}^{\circ}\right) \setminus = \setminus \frac{1}{\sqrt{3}} .$

$\text{So, a Quadrant II angle which has tangent value the opposite}$
$\text{of the above, is the Quadrant II angle that has reference angle" \ 30^@, "that is to say, the angle} \setminus {180}^{\circ} - {30}^{\circ} :$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus \quad \setminus \quad \setminus \tan \left({180}^{\circ} - {30}^{\circ}\right) \setminus = \setminus - \frac{1}{\sqrt{3}} .$

$\setminus q \quad \setminus q \quad \therefore \setminus q \quad \setminus q \quad \setminus \quad \setminus \quad \setminus \setminus \tan \left({150}^{\circ}\right) \setminus = \setminus - \frac{1}{\sqrt{3}} .$

$\text{So:}$

$\setminus q \quad \setminus \quad \setminus \setminus \tan \left({150}^{\circ}\right) \setminus = \setminus - \frac{1}{\sqrt{3}} \setminus = \setminus m \setminus q \quad \text{and} \setminus q \quad \setminus \quad {150}^{\circ} \setminus \in \left[0 , \setminus \pi\right) .$

$\text{Thus:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus \theta \setminus = \setminus \setminus {150}^{\circ} \setminus \setminus \text{is the inclination of the given line.}$

$\text{In radians:} \setminus q \quad \setminus \quad \setminus {150}^{\circ} \setminus = \setminus 150 \setminus \cdot \setminus \frac{\pi}{180} \setminus = \setminus \frac{15}{18} \setminus \pi \setminus = \setminus \frac{5}{6} \setminus \pi \setminus = \setminus \frac{5 \setminus \pi}{6.}$

$\text{So, finally:}$

$\setminus q \quad \setminus \quad \setminus \setminus \theta \setminus = \setminus \setminus {150}^{\circ} \setminus = \setminus \frac{5 \setminus \pi}{6} \setminus q \quad \text{is the inclination of the given line.}$