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

$\text{ Reduce the equation }x+y+1=0\text{ in normal form. }$

$\text{ Separate the constant term and then divide both sides with }\sqrt{a^2+b^2}=\sqrt{1^2+1^2}=\sqrt{2}$

$\begin{array}{l}\frac{x}{\sqrt{2}}+\frac{y}{\sqrt{2}}=-\frac{1}{\sqrt{2}}\\ -\frac{x}{\sqrt{2}}-\frac{y}{\sqrt{2}}=\frac{1}{\sqrt{2}}\end{array}$

This can be rewrite as 

$x\cos \left(\frac{5\pi }{4}\right)+y\sin \left(\frac{5\pi }{4}\right)=\frac{1}{\sqrt{2}}$

$\text{ Compare this equation with }x\cos \alpha +y\sin \alpha =p$

$\text{ Hence, }\alpha =\frac{5\pi }{4}=3.93$