<s>[INST] find angle between two lines between $y-\sqrt{3}x-5=0$ and $\sqrt{3}y-x+6$ how to find to angle between these two lines $y-\sqrt{3}x-5=0$ and $\sqrt{3}y-x+6=0$
i tried so far like this
$y-\sqrt{3}x-5=0$
$y=\sqrt{3}x+5$ 
in the form of $y=mx+b$
got the value for $m_1=\sqrt{3}$
and for 
$\sqrt {3}y-x+6=0$
$y=\dfrac {x-6} {\sqrt {3}}$
$y=\dfrac {1} {\sqrt {3}}x-6$
$m_{2}=\dfrac {1} {\sqrt {3}}$
the formula to find angle between two lines
$\tan \theta =\dfrac {m_{2-}m_{1}} {1+m_{1}m_{2}}$
$\frac{\frac{1}{\sqrt{3}}-\sqrt{3}}{1+\sqrt{3}.\frac{1}{\sqrt{3}}}$
$\frac { - \frac { 2 } { \sqrt { 3 } } } { 1 + 1 }$
$\frac { - \frac { 2 } { \sqrt { 3 } } } { 2 }$
$- \frac { 2 } { 2 \sqrt { 3 } }$
$- \frac { 1 } { \sqrt { 3 } }$
is this right till now and how to find angle after wards . should i use tan inverse of $- \frac { 1 } { \sqrt { 3 } }$ or my algebra calculation is wrong . help me thank you
 [/INST] Recall that to find the angle we can refer to the lines parallel to the given and passing through the origin, that is


*

*$y=\sqrt{3}x$  

*$\sqrt{3}y=x$ 
then, using parametric form, the direction vectors are


*

*for $y=\sqrt{3}x$  $$v_1=(1,\sqrt 3)$$

*for $\sqrt{3}y=x$ $$v_2=(\sqrt 3,1)$$
finally we can compute the angle by dot product
$$\cos \theta = \frac{v_1\cdot v_2}{\|v_1\|\|v_2\|}=\frac{2\sqrt 3}{4}=\frac{\sqrt 3}{2}\implies \theta = 30°$$
As an alternative the angles between the two lines and the $x$ axis are


*

*for $y=\sqrt{3}x$  $$\tan \theta_1=\sqrt 3 \implies \theta_1=60°$$

*for $\sqrt{3}y=x$ $$\tan \theta_2=\frac1{\sqrt 3} \implies \theta_2=30°$$
$$\implies \theta =\theta_1-\theta_2= 30°$$
Then the acute angle between the two lines is equal to $30°=\frac{\pi}6$ and as a consequence the obtuse angle between the two lines is equal to $180°-30°=150°=\frac{5\pi}6$.
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