What is the angle between the asymptotes of the hyperbola $5x^2-2\sqrt 7 xy-y^2-2x+1=0$? 
What is the angle between the asymptotes of this hyperbola?
  $$5x^2-2\sqrt 7 xy-y^2-2x+1=0$$

I used $S+\lambda=0$ and used straight line condition to find combined equation to asymptotes. Then how to find angle between them?

Solve the quadratic equation for $y$:
$$y=-x\sqrt{7}-\sqrt{12x^2-2x+1},y=-x\sqrt{7}+\sqrt{12x^2-2x+1}$$
Let the assymptotes be $y=a_1x+b_1$ and $y=a_2x+b_2$. Then:
$$\lim_{x\to\infty} \frac{-x\sqrt{7}-\sqrt{12x^2-2x+1}}{a_1x+b_1}\stackrel{LR}{=}\lim_{x\to\infty} \frac{-\sqrt{7}-\frac{12x-1}{\sqrt{12x^2-2x+1}}}{a_1}= \lim_{x\to\infty} \frac{-\sqrt7-2\sqrt3}{a_1}=1\Rightarrow \\
a_1=-\sqrt7-2\sqrt3$$
Similarly, $a_2=-\sqrt7+2\sqrt3$. The angle between the lines with such slopes is:
$$\tan \phi=\left|\frac{a_1-a_2}{1+a_1a_2} \right|=\sqrt3 \Rightarrow \phi =60^\circ.$$
