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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.

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  • $\begingroup$ A good first step is to find out the value of $m$. Did you do that? $\endgroup$ – David K Feb 12 '16 at 19:03
  • $\begingroup$ yes I have done that but I cant get m to work out to prove its 150° $\endgroup$ – Darren Feb 12 '16 at 19:07
  • $\begingroup$ You made good edits to the question showing some thoughts on the problem. Your thoughts are mostly correct, except that you should expect the slope of the line to be negative rather than positive. But you're also (apparently) missing an important step, which is to get the equation of the line in the form $y = $ something, because it will be $y=mx+b$ and you can see immediately what $m$ really is. Then you can try to see how $m$ compares with $\sqrt3/3$. $\endgroup$ – David K Feb 12 '16 at 20:39
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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, $$\tan 150^{\circ}=\tan (180^{\circ}-30^{\circ})=-\tan 30^{\circ}=-\frac{1}{\sqrt 3}$$.

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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 $180^\circ+\theta$.

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  • $\begingroup$ I tired and is till cant figure it out could you please show me how to prove it. $\endgroup$ – Darren Feb 12 '16 at 19:23
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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 $$\theta=180°-30°=150°$$

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