The angle between the pair of lines whose equation is $4{ x }^{ 2 }+10xy+m{ y }^{ 2 }+5x+10y=0$ is

The angle between 2 line represented by $ax^2+2hxy+by^2+2gx+2fy+c=0$

$\tan\theta=\Bigg|\dfrac{2\sqrt{h^2-ab}}{a+b}\Bigg|$

Given the equation of line i.e. $4x^2+10xy+my^2+5x+10y=0$

Compare above equation with standard equation

$a=4,h=5,b=m,g=\dfrac{5}{2},f=5,c=0$

Hence $\theta=\tan^{-1}\dfrac{2\sqrt{25-4m}}{m+4}$