The normal form of the line $x+y+1=0$  is $x\cos \alpha +y\sin \alpha =p$ then  $\alpha =$

Reduce the equation $\mathrm{x}+\mathrm{y}+1=0$  in normal form.
Separate the constant term and then divide both sides with  $\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}=\sqrt{1^2+1^2}=\sqrt{2}$
  $\begin{array}{l}\frac{\mathrm{x}}{\sqrt{2}}+\frac{\mathrm{y}}{\sqrt{2}}=-\frac{1}{\sqrt{2}}\\ -\frac{\mathrm{x}}{\sqrt{2}}-\frac{\mathrm{y}}{\sqrt{2}}=\frac{1}{\sqrt{2}}\end{array}$
This can be rewrite as $x\cos \left(\frac{5\pi }{4}\right)+y\sin \left(\frac{5\pi }{4}\right)=\frac{1}{\sqrt{2}}$
  
Compare this equation with  $\mathrm{xcos\alpha }+\mathrm{ysin\alpha }=\mathrm{p}$
Hence,  $\mathrm{\alpha }=\frac{5\mathrm{\pi }}{4}=\boxed{3.93}$