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What is the inclination of the line joining $(3, 0)$ and $(2, \sqrt{3}) $

Answer: $\frac{2\pi}{3}$

$m = \tan(\alpha)$

So,

$$m = \frac{y_1 - y_2}{x_1 - x_2}$$

$$\Rightarrow m = \frac{0 - \sqrt{3}}{ 3 - 2} = -\sqrt{3}$$

$$\Rightarrow -\sqrt{3} = \tan(\alpha)$$

How to get angle from negative slope?

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  • $\begingroup$ take arctan and do something with pi. Did you graph the line? $\endgroup$
    – imranfat
    Sep 10, 2013 at 21:01
  • $\begingroup$ $\arctan$ works just fine with negative argument. $\endgroup$
    – Tunococ
    Sep 11, 2013 at 0:32

1 Answer 1

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There's no real problems here, we take \begin{align*} \theta &= \arctan(-\sqrt{3}) \\ &= -\arctan(\sqrt{3}) & \text{since } \arctan(x)=-\arctan(-x) \text{ for all } x \in \mathbb{R} \\ &= -\tfrac{\pi}{3}. \end{align*}

And, as we can see from the picture below, the angle should be negative:

Triangle

The answer of $2\pi/3=\pi-|\theta|$ is instead the angle marked $\varphi$; this is the angle of inclination.

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  • $\begingroup$ Thanks, i got it... Barely. This is what i dont like about math books. To solve this i needed common sense... which was nowhere in that chapter.. $\endgroup$ Sep 12, 2013 at 18:22
  • $\begingroup$ Have used this page for arctan table. Needed for next task. $\endgroup$ Sep 12, 2013 at 18:27

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