A straight line $L=0$ passes through  the point $\left(5,-4\right)$  is inclined at an angle of 60^o to the line $\sqrt{3}x+y-3=0$ ,then the equation of the line $L=0$ is

The equation of any line which makes an angle  $\alpha$ with the line $\mathrm{ax}+\mathrm{by}+\mathrm{c}=0$ and passing through the point $\left(x_1,y_1\right)$is $\mathrm{y}-{\mathrm{y}}_1=\tan \left(\mathrm{\theta }\pm \mathrm{\alpha }\right)\left(\mathrm{x}-{\mathrm{x}}_1\right)$ here  $\mathrm{tan\theta }=-\frac{\mathrm{a}}{\mathrm{b}}$

Here $\tan \theta =-\frac{\sqrt{3}}{1}\Rightarrow \theta =120°$  and  $\alpha =60°$

Equations of the required lines are $y+4=\tan \left(120°\pm 60°\right)\left(x-5\right)$

It implies that 

$\begin{array}{l}y+4=\tan \left(180°\right)\left(x-5\right)\\ y+4=0\end{array}$

or

$\begin{array}{c}y+4=\tan \left(60°\right)\left(x-5\right)\\ y+4=\sqrt{3}\left(x-5\right)\end{array}$

Therefore, the equations of the required lines are $\boxed{y+4=0,\sqrt{3}x-y-5\sqrt{3}-4=0}$