Answer
$\arctan((15-g)/3)$ = 59.97
Work Step by Step
Acceleration varies linearly with μ, hence we can write $\ a=mμ+c $
putting points $\ (a1,0) $, $\ (0,μk2) $, $\ (-a1,μk3) $ in the avove equation, we'll get
$\ c=a1=3m/s^2 $
$\ m=-a1/μk2 =-15 m/s^2 $
hence, $\ a=3-15μ $
Now, from free body diagram of the block:
$\ N=mg+Fsin\theta $
$\ Fcos\theta - f =ma $
$\ f=Nμ $
$\ ma=Fcos\theta-μ(mg+Fsin\theta $
or $\ a=F/m cos\theta -μ(g-F/m sin\theta ) $
comparing with $\ a=3-15μ $,
we get $\ F/mcos\theta = 3 $
or $\ F/m = 3/cos\theta $
& $\ g+F/m sin\theta =15 $
putting value of $\ F/m , $
$\ g+3*sin\theta/cos\theta = 15 $
$\ tan\theta = (15-g)/3 $
$\ \theta=arctan((15-g)/3) $
$\ \theta=59.97^{\circ} $
