Answer
$ F_t = \frac{\mu_k mg}{cos\theta + \mu_k sin \theta}$
Work Step by Step
We first find the x and y components of each of the forces:
$ F_n = mg - F_tsin\theta$ and $F_tcos\theta- \mu_k F_n= 0 $
Using substitution, we find:
$F_tcos\theta- \mu_k F_n= 0 $
$F_tcos\theta- \mu_k (mg - F_tsin\theta)= 0 $
$ F_t = \frac{\mu_k mg}{cos\theta + \mu_k sin \theta}$
