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Find the angle between the lines $2y - \sqrt{3}x - 5 = 0$ and $\sqrt{3}y-2x+6=0$.

I do know the formula for finding the angle $\alpha$ between two lines which is $$ \tan \alpha = \frac{m_2-m_1}{1+m_1m_2} $$ but I am trying to find the value of the fraction above and I am getting the value of $\frac{\sqrt{3}}{3}$ which isn't any value of any degree for $\tan \alpha$. Am I doing something wrong?

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    $\begingroup$ Where is the link to the images? $\endgroup$ – Taroccoesbrocco Aug 25 '18 at 12:50
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    $\begingroup$ i dont know i am new here how to do it ? $\endgroup$ – Gaurav Singh Aug 25 '18 at 12:51
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    $\begingroup$ Hi, welcome. I've reformatted the question to make it more readable. Click the "edit" link to see how the markdown and MathJax are used, so you can do the same thing in the future. $\endgroup$ – Matthew Leingang Aug 25 '18 at 12:56
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    $\begingroup$ thanks a lot phew.. $\endgroup$ – Gaurav Singh Aug 25 '18 at 12:56
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    $\begingroup$ What do you mean about $\frac{\sqrt{3}}{3}$ isn't the value for any degree for $\tan \theta$? Some angle has that as its tangent! :) But anyway, look carefully at the angles you know for trig functions and their tangents. $\endgroup$ – John Brevik Aug 25 '18 at 12:59
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$$m_1=\frac{\sqrt3}{2}$$ $$m_2=\frac{2}{\sqrt3}$$

Thus the angle between line 1 and line 2

$$\tan\alpha=\frac{\frac{2}{\sqrt3}-\frac{\sqrt3}{2}}{1+\frac{2}{\sqrt3}\frac{\sqrt3}{2}}$$ $$\tan\alpha=\frac{1}{4\sqrt3}$$

Thus the angle $\alpha$ is $\arctan(\frac{1}{4\sqrt3})\approx 8.213^\circ.$

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  • $\begingroup$ alright my calculaitons were wrong thanks for letting me know about the results . otherwise i woudnt ever know where i was wrong . $\endgroup$ – Gaurav Singh Aug 25 '18 at 14:10
  • $\begingroup$ Happy that you got your mistake $\endgroup$ – Deepesh Meena Aug 25 '18 at 14:11

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