What will be the angle between the lines $y-x-7$ $=0$ and $sqrt{3} y-x+6=0$ ?

We know that angle between the lines is

$an heta=left|frac{m_{1}-m_{2}}{1+m_{1} m_{2}} ight|$

Here, Equation of line is

y – x – 7 = 0

$Rightarrow m_{1}=1$

similarly,

$m_{2}=frac{1}{sqrt{3}}$

Now,

$an heta=left|frac{1-frac{1}{sqrt{3}}}{1+1 cdot frac{1}{sqrt{3}}} ight|$

$=left|frac{1-frac{1}{sqrt{3}}}{1+frac{1}{sqrt{3}}} ight|=left|frac{sqrt{3}-1}{sqrt{3}+1} ight|$

$=left|frac{sqrt{3}-1}{sqrt{3}+1} imes frac{(sqrt{3}-1)}{(sqrt{3}-1)} ight|$

$=left|frac{(sqrt{3}-1)^{2}}{sqrt{3}^{2}-1^{2}} ight|$

$=left|frac{sqrt{3}^{2}+1^{2}-2 sqrt{3}}{3-1} ight|$

$=left|frac{4-2 sqrt{3}}{2} ight|$

$an heta=(2-sqrt{3})$

$heta= an ^{-1}(2-sqrt{3})$