The line $x+\sqrt{3} y-10=0$ makes an angle of $150$° with the positive sense of the $x$-axis. How can this be proven? I cant figure out how this is correct. I know that $\tan(a)=m$ of a line but I cant figure this out. Could someone show how to prove the line makes an angle of $150$° with the positive $x$-axis?
I know that tan a should probably = 30° so when you subtract it from 180 you get 150, so that should mean m=√3 divided by 3 but I cannot prove √3 divided by 3=m.    
 A: The equation of your line is also $y=-\frac{1}{\sqrt 3}x+\frac{10}{\sqrt 3}$ so you need just to verify that $\tan 150^{\circ}=-\frac{1}{\sqrt 3}$. In fact, $$\tan 150^{\circ}=\tan (180^{\circ}-30^{\circ})=-\tan 30^{\circ}=-\frac{1}{\sqrt 3}$$.
A: Hint 1: Observe that the slope of the line is negative.  Therefore, the angle is between $90^\circ$ and $180^\circ$.
Hint 2: You can use $\tan(\theta)=m$ where $\theta$ is the signed angle between the $x$-axis and the line (not necessarily with the positive direction).
Hint 3: Draw a picture and label the angle and you should see that the angle with the positive direction is $180^\circ+\theta$.
A: Use dot product of "direction vectors" of the line and of the x-axis. They are respectively $(1,\frac{-1}{\sqrt{3}})$ and $(1,0)$. The dot product $$(1,\frac{-1}{\sqrt{3}})(1,0)=1=||(1,0)||*||(1,\frac{-1}{\sqrt{3}})||cos{\theta}$$
$$1=cos{\theta}*1*(\frac {\sqrt{2}}{3})$$
$$cos{\theta}=\frac {\sqrt{3}}{2})$$
then $$\theta=\frac {\pi}{6}=30°$$
but $\theta$ is in negative direction. In positive direction $$\theta=180°-30°=150°$$
