The acute angle between two lines whose D.R's are given by $l+m-n=0$ and $l^2+m^2–n^2$ = 0 is

The given relations between the direction ratios are $l+m-n=0$ and $l^2+m^2–n^2=0$

Suppose that $n=1$

$l+m=1,l^2+m^2=1$

Substitute $m=1-l$ in the above equation

$\begin{array}{rcl}{l}^2+{\left(1-l\right)}^2& =& 1\\ l^2+l^2-2l& =& 0\\ l^2-l& =& 0\\ l\left(l-1\right)& =& 0\end{array}$

Therefore, the direction ratios are 

Let $\theta$ be the acute angle between the lines then 

 $\begin{array}{rcl}\cos \theta & =& \frac{0+0+1}{\sqrt{2}\sqrt{2}}\\ & =& \frac{1}{2}\end{array}$

Therefore, the measure of the angle between the given two lines is $60°$